A NOTE ON

DIGITAL SIGNAL
PROCESSING
TH
[6 SEM ETC -ETT603]
PREPARED BY

SRI PARAMANANDA GOUDA

[Lecturer (PTGF), E & TC]
(Email-ID  tulu257@gmail.com)

U.C.P. ENGINEERING SCHOOL

BERHAMPUR- 760010
Under Govt. of Odisha, Skill Development and Technical Education Dept
Affiliated to A.I.C.T.E under S.C.T.E. & V.T
Phone / Fax: - 0680 2291826 (O) 0680 2291648 (R)
E-Mail:-principalucpes@gmail.com, Website:-www.ucpesbam.in
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <INTRODUCTION TO DSP> [Page 1.1]
[CHAPTER-1]
---------------------------------------- Introduction to DSP ---------------------------------------
 DIGITAL: -
 The word “Digital” means that the
processing is done with a digital
computer or special purpose digital
hardware. It is discrete in nature.
 SIGNAL: -
 Signal is defined as any physical
quantity that varies with time,
space, or any other in dependent
variable or variables.
 More precisely Signal is a function
of set of independent variables.
 The signal itself carries some kind of information for observation.
 The term "Signal" includes, among
others, audio, video, speech, image,
communication, geophysical, sonar,
radar, medical and musical signals.
 PROCESSING: -
 Changing the basic nature of signal to
obtain the desired shaping of the input
signal is called processing
 Signal Processing is concerned with the
representation, transformation and
manipulation of signals and the information they contains.
 Signal Processing is two types depending upon the type of signals to be processed : -
 Analog Signal Processing.
 Digital Signal Processing.
ANALOG SIGNAL PROCESSING (ASP)
 In analog signal processing, continuous-amplitude with continuous-time signals is processed.
DIGITAL SIGNAL PROCESSING(DSP): -
 Digital signal processing is concerned with the representation of signals by sequence of numbers or
symbols and the processing of these sequences.
 The purpose of such processing may be to estimate characteristic parameters or transform a signal into
a form which is in some sense of more desirable.
 In other word DSP is a mathematical manipulation of discrete-time signals to get more desirable
properties of the signals, such as less noise or distortion.
 SIGNAL PROCESSING SYSTEMS: -
 A System may also be defined as a physical device that performs an operation on a signal.
 For example, a filter used to reduce the noise and interference from a corrupting signal to a desired
information bearing signal is called a System.
 In this case the filter performs some operations on the signal, which has the effect of reducing
(filtering) the noise and interference from the desired information bearing signal.
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <INTRODUCTION TO DSP> [Page 1.2]
 In general, the system is characterized by the type of operation that it performs on the signal.
 For example, if the operation on the signal is linear, then the system is called Linear.
 If the operation is nonlinear, then the system is said to be Nonlinear.
 Signal Processing Systems are of two types depending on the type f signal to be processed.
 Continuous-time Systems
 Discrete-time Systems.
Continuous-time Systems
 Continuous time systems are systems in which
both input & output are continuous time signals.
 H(s) is the transfer function of a discrete time
system. Figure shows the block diagram for
Continuous-time Systems.
Discrete-time Systems
 Discrete time systems are the systems for which both input and output are discrete time signals.
 H (z) is the transfer function of a Discrete time system. Figure shows its block diagram.
 ELEMENTS OF DIGITAL SIGNAL PROCESSING SYSTEM: -

 The input signal is applied to the anti aliasing filter. The low pass filter removes the high frequency
noise and to band-limit the signals.
 The sample and hold provides the discrete time signal to A/D converter.
 The ADC converts the along signal into its equivalent digital signal.
 The Digital Signal Processor may be a large programmable digital computer, programmed to perform
the desired operations on the input signals.
 The output of DSP is converted to analog signal by DAC.
 The high frequency components in DAC output are released by the reconstruction filter.
 ADVANTAGES OF DSP OVER ASP : -
1. Digital signal processing operations can be changed by changing the program in digital programmable
system. This means that these are flexible system.
2. There is a better control of accuracy in digital systems compared to analog systems.
3. Digital signals are easily stored on magnetic media (tape or disk) without loss of signal quality.
4. Sophisticated signal processing algorithms can be implemented by DSP method.
5. Digital circuits are less sensitive to tolerance of component values.
6. Digital systems are independent of temperature, ageing and other external parameters.
7. Digital circuits can be reproduced easily in large quantities at comparatively lower cost.
8. Cost of processing per signal in DSP is reduced by time sharing of a processors among multiple signals
9. As adaptive filters are used in DSP which can be easily adjusted in digital implementation.
10. Digital systems can be cascaded without any loading effect.
11. As digital Codes are often used in the transmission of information. So it becomes more data secure.
12. Digital signals typically use less bandwidth. So we can send more information in the same space
13. Data Transmission is at a higher rate.
14. There is minimal electromagnetic interference in digital technology.
15. It enables multi-directional transmission simultaneously etc.
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <INTRODUCTION TO DSP> [Page 1.3]
 LIMITATIONS OF DSP: -
1. System Complexity: - System complexity increased in the digital
processing of analog signals because of the devices such as A/D and
D/A converters and their associated filters are used here.
2. Bandwidth limited by Sampling Rate: - Processing of signals
beyond higher frequencies (beyond GHz) and below lower
frequencies (a few Hz) involves limitations
3. Power Consumption: - Processing of signals involves more power consumption.
4. Information is lost because we only take samples of the signal at intervals.
 APPLICATIONS OF DSP: -
 As a matter of fact, there are various application areas of DSP due to the availability of high resolution
spectral analysis. Some of these areas are can be listed as under: -
1. Speech Processing
2. Image Processing
3. Radar Signal Processing
4. Sonar Signal Processing
5. Digital Communication
6. Spectral Analysis
7. Instrumentation and control
 Few other application of DSP are: -
1. Transmission lines.
2. Advanced Optical Fiber Communication
3. Analysis of sound & Vibration signals
4. Implementation of speech recognition algorithms.
5. VLSI Technology
6. Telecommunication networks
7. Microprocessor Systems
8. Satellite Communications
9. Telephony Transmission
10. Astronomy
11. Industrial noise control
12. Biomedical Engineering etc
 CLASSIFICATION OF SIGNALS: -
 There are various methods of classifying signals based on different features.
 Based on type of Independent Variables : -
 Continuous Time Signal & Discrete Time Signal
 Depending upon the number of independent variable : -
 1-Dimentional Signal, 2-Dimentional Signal & M-Dimensional Signal
 Depending upon the certainty by which the signal can be uniquely described : -
 Deterministic Signal & Random Signal
 Based on repetition nature: -
 Periodic Signal & Non-Periodic Signal
 Based on reflection: -
 Even Signal (Symmetric) & Odd Signal (Asymmetric/Anti-symmetric)
Continuous Time Signal: -
 It is also referred as Analog Signal i.e. the signal is
represented continuously in time.
 In simple words, a signal x (t) is said to be a continuous time
signal if it is defined for all time.
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <INTRODUCTION TO DSP> [Page 1.4]
Discrete Time Signal: -
 Signals are represented as sequence at discrete time intervals. Thus it has discrete values only.
1-Dimentional Signal (1-D)
 It is a function of single independent variables. Ex -
Speech Signal, Music Signal are function of time.
2-Dimentional Signal (2-D)
 It is a function of two independent variables. Ex – Image
Signal, each frame of a B/W video signal etc.
M-Dimensional Signal (M-D)
 It is function of m independent variable in time. Ex – Colour Signal (3-D) composed of RGB.
Deterministic Signal
 A signal that can be uniquely determined by a well defined process such as a mathematical expression
or rule or table is called a deterministic signal.
 For examples A Sinusoidal Signal can be expressed as v (t) = Vm Sin ωt, for t ≥0
 A Square Signal can be expressed as x (t) = A for 0 < t < T/2 and x (t) = - A for T/2 < t < T
 A Exponential function can be expressed as x (n) = ean for n ≥ 0
Random Signal
 A signal that is generated in a random fashion and cannot be predicted ahead of
time is called random signal. Ex: - Speech Signal, ECG Signal, EEG Signal etc.
Periodic Signal
 The periodic signal means, the signal which repeats every finite interval of time or a continuous time
signal x(t) is a function that satisfies the condition x(t+T) = x(t), for all „t‟. Ex-Sine, Square wave etc

Non-Periodic Signal
 Any signal x (t) for which there is no value of T to satisfy the
relation given in equation x (t) = x (t +T) is known as non-
periodic signal. i.e. x (t+T) ≠ x (t), for all „t‟.
Even Signal (Symmetric)
 A Signal x (t) or x (n) is referred to as even signal, if it is identical to its time reversal counter part i.e.
with its about origin. x (-t) = x (t) for all „t‟. Even Signals are Symmetric w.r.t vertical axis.

`
Odd Signal (Asymmetric/Anti-symmetric )
 A Signal is said to be odd if x (- t) = - x (t) for all „t‟

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <INTRODUCTION TO DSP> [Page 1.5]
Properties of EVEN and ODD Signals
 The Sum of two Even signals are Even signal.
 The Sum of two Odd signals are Odd signal.
 The Sum of an Even signal and an Odd signal is neither Even nor Odd Signal.
 The Product of two Even signal is Even signal.
 The Product of two Odd signal is Even signal.
 The Product of an Even signal and an Odd signal is an Odd Signal.
 PROOFS
 Let f1(x)=8x2 f1(-x) = 8(-x)2 = 8x2  Even Siganl ; f2(x)=6x4 f2(-x) = 6(-x)4 = 6x4  Even Signal
 f3(x)=7x3 f3(-x) = 7(-x)3 = -7x3  Odd Signal ; f4(x)=2x5 f4(-x) = 2(-x)5 = -2x5  Odd Signal
 Now for f(x) = f1(x) + f2(x) = 8x2 + 6x4  f(-x) = 8(-x)2 + 6(-x)4 = 8x2 + 6x4 = f(x)  EVEN Signal
 For, f(x) = f3(x) + f4(x) = 7x3 + 2x5  f(-x) = 7(-x)3 + 2(-x)5= - (7x3 + 2x5) = - f(x)  ODD Signal
 For, f(x) = f1(x) + f3(x) = 8x2 + 7x3  f(-x) = 8(-x)2 + 7(-x)3 = 8x2 - 7x3 ≠ f(x)  Neither Even nor Odd
 For, f(x) = f1(x) x f2(x) = 8x2 x 6x4 = 48x6  f(-x) = 48(-x)6 = 48x6 = f(x)  EVEN Signal
 For, f(x) = f3(x) x f4(x) = 7x3 x 2x5 = 14x8  f(-x) = 14(-x)8 = 14x8 = f(x)  EVEN Signal
 For, f(x) = f1(x) x f3(x) = 8x2 x 7x3 = 56x5  f(-x) = 56(-x)5 = 56x5 = f(x)  ODD Signal
Check the following Signals are Even or Odd or Neither Even nor Odd.
 x(t) = e-at  Neither Even nor Odd  y(t) = x(2t)2  Even
 x(t) = A Sin t  Odd  y(t) = t x(t)  Neither Even nor Odd
-a|t|
 x(t) = e  Even  y(t) = t Sin t  Even
t/2
 x(t) = e  Neither Even nor Odd  y(t) = At x(t)  Neither Even nor Odd
 x(t) = (Sint)/8  Odd  y(t) = x(t) Cos200πt  Neither Even nor Odd
 x(t) = Sin(t/8)  Odd  x(t) = 2et  Neither Even nor Odd
 y(t) = x(2t)  Neither Even nor Odd  x(t) = (Sint)/t  Neither Even nor Odd
Even & Odd Components of any Signal  Xe (n) = ½ [X (n) + X (-n)] Xo (n) = ½ [X (n) - X (-n)]

 MULTICHANNEL AND MULTIDIMENSIONAL SIGNALS

Multi Channel Signals
 Signals which are generated by multiple sources or multiple sensors are called Multichannel signals.
 The resultant signal is the vector sum of signals from all channels.
 These signals are represented in vector form. A 3- channel signal is expressed as : -
 For Ex: In electrocardiography, 3-lead and 12-lead electrocardiograms (ECG) are often used in
practice, which result in 3-channel and 12-channel signals.
Multidimensional Signals
 If the signal is a function of a single independent variable, the signal is called a one-dimensional signal.
 On the other hand, if the signal is a function of multi (many) independent variables, then it is called as
multidimensional signal.
 For Ex: Speech signal is a 1-Dimensional signal because amplitude of signal depends upon single
independent variable, namely, time.
 A black & white picture signal is an example of 2-Dimensional signal because brightness of the signal
at each point is a function of two spatial independent variable, namely, x and y. Variables x and y are
width and height of the picture element.
 A colored picture signal is an example of 3-Dimensional signal because
brightness of the signal at each point is a function of three independent
variables, namely, x, y & time (t). A 3-Dimensional It can be represented as

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <INTRODUCTION TO DSP> [Page 1.6]
 CONTINUOUS-TIME Vs DISCRETE-TIME SIGNALS
Continuous-Time Signals:
 A signal that varies continuously with time is known as continuous time.
 Continuous-time signals or analog signals are defined for every value of
time and they take on values in the continuous interval (a, b), where a
can be - ∞ and b can be ∞.
 Mathematically, these signals can be described by functions of a continuous variable.
 A signal of continuous in amplitude and time is known as a continuous time signal or an analog signal.
 This (a signal) will have some value at every instant of time.
 The electrical signals derived in proportion with the physical quantities such as temperature, pressure,
sound etc. are continuous signals.
 The other examples of continuous signals are sine wave, cosine wave, triangular wave etc.
 It can be represented as x (t) where time „t‟ represents the independent variable.
 The speech waveform and the signals x (t) = cos πt, x2{t) = 𝑒 −𝑡 , etc. are examples of it.
Discrete Signal or Discrete-Time Signal:
 A signal that has values only at discrete instant of time is known as
Discrete-Time Signal. Discrete-time signals are defined only at
certain specific values of time.
 These time instants need not be equidistant, but in practice they are
usually taken at equally spaced intervals for computational
convenience and mathematical tractability or manipulate.
 These signals are continuous in amplitude and discrete in time.
 They are denoted by x (nT) where n = 0, 1, 2, 3… and T is the time interval between two values.
 Continuous-Valued Versus Discrete-Valued Signals
 The values of a continuous time or discrete time signal can be continuous or discrete.
 If a signal takes on all possible values on finite/infinite range, it is said to be continuous valued signal.
 Whereas, if signal takes on values from a finite set of possible values, it is said to be discrete valued.
 These values are equidistant and hence can be expressed
as an integer multiple of the distance between two
successive values. A discrete time signal having a set of
discrete values is called a digital signal.
 The figure shows a digital signal that takes on one of
four possible values. In order for a signal to be processed
digitally, it must be discrete in time and its values must
be discrete (i.e. it must be a digital signal).
 If the signal to be processed is in analog form, it is
converted to a digital signal by sampling the analog signal at discrete instants in time, obtaining a
discrete time signal, and then by quantizing its values to a set of discrete values.
 The process of converting continuous-valued signal into a discrete-valued signal, called Quantization.
 The Concept Of Frequency In Continuous-Time and Discrete-Time Signals
 The concept of frequency is familiar to students in engineering & sciences.
 For Ex, the design of a radio receiver, a high-fidelity system, or a spectral filter for color photography.
 From physics we know that frequency is closely related to a specific type of periodic motion called
harmonic oscillation, which is described by sinusoidal functions.
 The concept of frequency is directly related to concept of time.
 Actually, it has the dimension of inverse time. Thus we should
expect that the nature of time (continuous or discrete) would
affect the nature of the frequency accordingly.
Continuous-Time Sinusoidal Signals: -
 A simple harmonic oscillation is mathematically described by the following continuous-time sinusoidal
signal: xa (t) = A Cos (Ωt + θ), -∞ < t < ∞. The subscript „a‟ used with x (t) denotes an analog signal.

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <INTRODUCTION TO DSP> [Page 1.7]
 This signal is completely characterized by three parameters:
 A is the amplitude of the sinusoidal signal, Ω is the frequency in radian per sec and θ is the phase in
radian. Instead of Ω, we often use the frequency F in cycles per sec (Hz). Where Ω = 2πF=2π/TP.
 So the above equation can be written as: xa (t) = A Cos (2πFt + θ), -∞ < t < ∞.
 The analog sinusoidal signal is characterized by the following properties : -
 For every fixed value of the frequency F, xa (t) is periodic. Indeed, it can easily be shown by
using elementary trigonometry, that xa (t + Tp) = xa (t), where Tp = 1/F.
 Continuous time sinusoidal signals with different frequencies are themselves distinct.
 Increasing the frequency F results in an increase in the rate of oscillation of the signal, in the
sense that more periods are included in a given time interval.
 NOTE: -
 The relation we have described for sinusoidal signals carry over to the class of complex exponential
signals xa (t) = A e j(Ωt+θ) . This can be easily be seen by expressing these signals in terms of sinusoids
using the Euler‟s Identity e ±jɸ = Cos ɸ ± j Sin ɸ.
𝐀 𝐀 -
 Thus xa (t) = A Cos (Ωt+θ) can be expressed as xa (t) = 𝟐 e j(Ωt+θ) + 𝟐 e j(Ωt+θ)
Discrete-Time Sinusoidal Signals: -
 A discrete-time sinusoidal signal may be expressed as
x (n) = A Cos (ωn + θ) , -∞ < n < ∞.
 Where n is an integer variable, called the sample
number, A is the amplitude of the sinusoidal signal, ω
is the frequency in radians per sample and θ is the
phase in radian.
 Instead of ω, we often use the frequency variable f in
cycles per sample defined by ω = 2πf.
 So the above equation can be written as: x (n) = A Cos (2πfn + θ), -∞ < n < ∞.
 In contrast to continuous time sinusoids, discrete time sinusoids are characterized by following points:-
 A discrete time sinusoid is periodic only if its frequency f is a rational number.
 Discrete time sinusoids whose frequencies are separated by an integer multiple of 2n are identical.
 The highest rate of oscillation in a discrete-time sinusoid is attained when ω = π (or ω = - π) or,
equivalently, f = 1/2 (or f = - 1/2).
HARMONICALLY RELATED COMPLEX EXPONENTIALS: -
 Sinusoidal signals and complex exponentials play a major role in the analysis of signals and systems.
 In some cases we deal with sets of harmonically related complex exponentials (or sinusoids).
 These are sets of periodic complex exponentials with fundamental frequencies that are multiple of a
single positive frequency. Although we confine our discussion to complex exponentials, the same
properties clearly hold for sinusoidal signals.
 We consider harmonically related complex exponentials in both continuous time and discrete time.
Continuous-Time Exponentials: -
 The basic signals for continuous-time, harmonically related exponentials are
Sk (t) = ejk Ω 0 t = ejk 2πF 0 t ---- (1) Where k = 0, ± 1, ± 2, ± 3 ……
 We note that for each value of k, Sk (t) is periodic with fundamental period 1/ (kF0) = Tp/k or
fundamental frequency kF0.
 Since a signal that is periodic with period Tp/k is also periodic with period k (Tp/k) = Tp for any
positive integer k, we see that all of the Sk (t) have a common period of Tp.
 Furthermore F0 is allowed to take any value and all members of the set and distinct, in the sense that if
k1 ≠ k2, then Sk1 (t) ≠ Sk2 (t). From the above basic signal
equation (1) we can construct a linear combination of
harmonically related complex exponential of the form.
 Where ck = 0, ± 1, ± 2, ± 3 ……are arbitrary complex
constants. The signal xa (t) is periodic with fundamental period Tp = 1/F0.
 The above representation of xa(t) is called Fourier Series expnasion for xa (t). The complex valued
constants ck are the Forier Series coefficients and the signals sk (t) is called kth harmonic of xa (t).
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <INTRODUCTION TO DSP> [Page 1.8]
Discrete-Time Exponentials: -
 Since a discrete-time complex exponential is periodic if its relative frequency is a rational number, we
choose f0 = 1/N and we define the sets of harmonically related complex exponentials by
Sk (n) = 𝐞𝐣𝟐𝛑𝐤𝐟𝟎 𝐭 , Where k = 0, ± 1, ± 2, ± 3 ……

 In contrast to the continuous time case we note that

 This means that, there are only N distinct periodic complex exponentials in the set.
 Furthermore all members of the set have a common periodic of N samples.
 Clearly we can choose any consecutive N complex exponentials, say from k = n0 to k = n0 + N – 1 to
form a harmonically related set with fundamental frequency f0 = 1/N.
 Most often, for convenience we choose the set that corresponding to n0 = 0, that is the set
Sk (n) = ej2πkn /N , , Where k = 0, 1, 2, 3 …… N – 1
 As in the case of continuous-time signals, it is obvious that the
linear combination results in a periodic signal with
fundamental period N.
 We shall see later, this is the Fourier series representation for a periodic discrete-time sequence with
Fourier coefficients {CK}. The sequence SK(n) is called the kth harmonic of x (n).
 Along with Euler‟s Identity, Euler also described a way to represent a complex exponential signal in
terms of its real and imaginary parts through Euler’s Relation:  Cos (ωt) = (e jwt + e –jwt) /2
 ANALOG-TO-DIGITAL AND DIGITAL-TO-ANALOG CONVERSION
 Most signals of practical interest, such as speech, biological signals, seismic signals, radar signals,
sonar signals, and various communications signals
such as audio and video signals, are analog.
 To process analog signals by digital means, it is
first necessary to convert them into digital form,
that is, to convert them to a sequence of numbers
having finite precision.
 This procedure is called analog-to-digital (A/D)
conversion, and the corresponding devices are called A/D converters (ADCs).
 Conceptually, we view A/D conversion as a three step process. This process is illustrated in Fig. above.
SAMPLING.
 This is the conversion of a continuous time signal in to a discrete-time signal obtained by taking
“samples‟" of the continuous-time signal at discrete-time instants.
 Thus, if xa(t) is the input to the sampler, the output is xa(nT) ≡ x(n), where T is called sampling interval
QUANTIZATION.
 This is the conversion of a discrete-time continuous-valued
signal into a discrete-time, discrete-valued (digital) signal.
 The value of each signal sample is represented by a value
selected from a finite set of possible values.
 The difference between the un-quantized sample x (n) &
quantized output xq (n) is called quantization error.
CODING.
 In the coding process, each discrete value xq{n) is represented
by a b-bit binary sequence.
 Although we model the A/D converter as a sampler followed by a quantizer and coder, in practice the
A/D conversion is performed by a single device that takes xa(t) and produces a binary -coded number.
 The operations of sampling and quantization can be performed in either order but in practice, sampling
is always performed before quantization.
 In many cases of practical interest (e.g., speech processing) it is desirable to convert the processed
digital signals into analog form.
 Obviously, we cannot listen to the sequence of samples representing a speech signal or see the numbers
corresponding to a TV signal.)

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <INTRODUCTION TO DSP> [Page 1.9]
 The process of converting a digital signal into an analog signal is know n as digital-to-analog (D/A)
conversion. All D/A converters “connect the dots‟" in a digital signal by performing some kind of
interpolation, whose accuracy depends on the quality of the D/A conversion process.
 Below figure illustrates a simple form of D/A conversion, called a zero-order holder a staircase
approximation. Other approximations are possible, such as linearly connecting a pair of successive
samples (linear interpolation), fitting a quadratic through three successive samples (quadratic
interpolation), and so on. Sampling and quantization are treated in this section.
 In particular, we demonstrate that sampling do does not result in a loss of information, nor does it
introduce distortion in the signal if the signal bandwidth is finite.
 In principle, the analog signal can be reconstructed from the samples, provided that the sampling rate is
sufficiently high to avoid the problem commonly called aliasing.
 On the other hand, quantization is a noninvertible or irreversible process that results in signal
distortion. We shall show that the amount of distortion is dependent on the accuracy, as measured by
the number of bits, in the A /D conversion process.
 The factors affecting the choice of the desired accuracy of the A /D converter are cost and sampling
rate. In general, the cost increases with an increase in accuracy and or sampling rate.
SAMPLING OF ANALOG SIGNALS
 Among many ways to sample an analog signal, periodic or uniform sampling is the type of sampling
used most often in practice. This is described by the relation x (n) = xa (nT) -∞<n<∞
 Where x(n) is the discrete-time signal obtained by taking samples of the analog signal xa(t) every T Sec
 This procedure is illustrated in Fig. below.
 The time interval T between successive samples is called the Sampling Period or Sample Interval.
 The reciprocal 1/T = FS is called the Sampling rate (samples per second) or Sampling frequency (Hz)
 Periodic sampling establishes a relationship between the time variables t and n of continuous-time and
discrete-time signals, respectively.
 Indeed, these variables are linearly related through the sampling period T or, equivalently, through the
sampling rate FS = 1/T as t = nT = n / Fs.
 As a consequence of there exists a relationship between the frequency variable F (or Ω) for analog
signals and the frequency variable f (or ω) for discrete-time signals.
 To establish this relationship, consider an analog
sinusoidal signal of the form xa(t)=ACos(2πFt + θ),
when sampled periodically at a rate Fs = 1/ T
samples per second. Yields
𝟐𝛑𝐧𝐅
Xa(nT) ≡ x(n) = A Cos (2πFnT+θ)=A Cos[ 𝐅 + θ]
𝐬
 By comparing the above equation with the general
expression of discrete – time Sinusoidal signal i.e. x
(n) = A Cos (2πfn + θ), we note that the frequency
variables F and f are linearly related as f = F/Fs.
 Where Fs = Sampling frequency, F = Frequency of analog signal and f = frequency of digital signal or,
equivalently, ω = ΩT.
 The relation in above justifies the name relative
or normalized frequency, which is sometimes
used to describe frequency variable.
 We can use to determine the frequency F in Hz
only if the sampling frequency Fs is known.
 For e.g. a sinusoidal with F = 4 kHz, samples at
every T = 0.1m sec, means Ts = 1/ T= 104 Hz 
ω = 2πF/Fs = (2π x 4 x 103) / 104 = 4π/5 rad.
 From these relations we observe that the
fundamental difference between continuous-time
and discrete-time signals is in their range of
values of the frequency variables F and f, or Ω and ω.
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <INTRODUCTION TO DSP> [Page 1.10]
 Periodic sampling of a continuous-time signal implies a mapping of the infinite frequency range for the
variable F (or Ω) into a finite frequency range for the variable F (or Ω).
 Since the highest frequency in a discrete -time signal is ω = π or f = ½ , it follows that, with a sampling
rate Fs, the corresponding highest values of F and Ω are Fmax = Fs/2 = 1/2T and Ωmax = πFs = π/2
 Thus, sampling introduces an ambiguity, since the highest frequency in a continuous-time signal that
can be uniquely distinguished when such signal is sampled at a rate Fs =1/T is Fmax=Fs/ 2, or Ωmax=πFs.
THE SAMPLING THEOREM
 Let us suppose that any analog signal can be represented as a sum
of sinusoids of different amplitudes, frequencies & phases that is,
Where N denotes the number of frequency components.
 All signals, such as speech and video themselves to such a representation over any short time segment.
 The amplitudes, frequencies, & phases usually change slowly with time from onetime segment to
another. However, suppose that the frequencies do not exceed some known frequency, say Fmax.
 For example, Fmax = 3000 Hz for the class of speech signals and Fmax = 5 MHz for television signals.
 Since the maximum frequency may vary slightly from different realizations among signals of any given
class (e.g., it may vary slightly from speaker to speaker), we may wish to ensure that Fmax does not
exceed some predetermined value by passing the analog signal through a filter that severely attenuates
frequency components above Fmax.
 Thus we are certain that no signal in the class contains frequency components (having significant
amplitude or power) above Fmax. In practice, such filtering is commonly used prior to sampling.
 From our knowledge of Fmax, we can select the appropriate sampling rate.
 We know that the highest frequency in an analog signal that can be unambiguously reconstructed when
the signal is sampled at a rate Fs = 1/T is Fs/2.
 Any frequency above Fs/2 or below - Fs/2 results in samples that are identical with a corresponding
frequency in the range Fs/2 ≤ F ≤ Fs/2.
 To avoid the ambiguities resulting from aliasing, we must select the sampling rate to be sufficiently
high. That is, we must select Fs/ 2 to be greater than Fmax.
 Thus to avoid the problem of aliasing, Fs is selected so that Fs > 2Fmax. Where Fmax is the largest
frequency component in the analog signal.
 With the sampling rate selected in this manner, any frequency component, say |Fi| < Fmax, in the analog
signal is mapped in to a discrete time sinusoid with frequency - ½ ≤ fi = Fi/Fs ≤ ½ or -π ≤ ωi = 2πfi ≤ π.
 Since, | f | = 1/2 or | ω | = π is the highest (unique) frequency in a discrete- time signal, the choice
of sampling rate according to above equation avoids the problem of aliasing.
 In other words, the condition Fs > 2Fmax ensures that all the sinusoidal components in the analog signal
are mapped in to corresponding discrete time frequency components with frequencies in the
fundamental interval.
 Thus all the frequency components of the analog signal are represented in sampled form without
ambiguity, and hence the analog signal can be reconstructed without distortion from the sample values
using an appropriate interpolation (digital-to -analog conversion) method. The „appropriate” or ideal
interpolation formula is specified by the Sampling Theorem.
 SAMPLING THEOREM DEFINATION.
 If the highest frequency contained in an analog signal xa (t) is Fmax = B and the signal is sampled at a
rate Fs > 2Fmax = 2B, then Xa (t) can be exactly recovered from its sample values using the interpolation
𝐒𝐢𝐧 𝟐𝛑 𝐁𝐭
function. g(t) = 𝟐𝛑 𝐁𝐭 . Thus Xa (t) may be expressed as 
Where xa(n/Fs) = xa(nT ) = x(n) are the samples of xa(t).
 When the sampling of xa (t) is performed at the minimum
sampling rate Fs = 2B, the reconstruction formula becomes 
 The sampling rate FN = 2B = 2 Fmax is called the Nyquist rate.
 As can be observed from either above equations, the reconstruction of
xa(t) from the sequence x(n) is a complicated process, involving a
weighted sum of interpolation function g(t) and its time shifted versions
g(t-nT) for - ∞ < n < ∞, where weighting factors are the samples x(n).

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <INTRODUCTION TO DSP> [Page 1.11]
 Because of the complexity and the infinite number of samples required in above equations, these
reconstruction formulas are primarily of theoretical in nature.
 Example: - Consider a Signal xa(t) = 3 Cos 50πt + 10 Sin 300 πt – Cos 100 πt, Find Nyquist rate.
Solution:  The frequencies present in the signal above are F1 = 25 Hz. F2 = 150 Hz. F3 = 50 Hz
Thus Fmax = 150 Hz and Fs > 2 Fmax = 300 Hz  The Nvquist rate is FN = 2Fmax. Hence FN = 300 Hz.
QUANTIZATION OF CONTINUOUS - AMPLITUDE SIGNALS
 A digital signal is a sequence of numbers (samples) in which each number is represented by a finite
number of digits (finite precision).
 The process of converting a discrete - time continuous - amplitude signal in to a digital signal by
expressing each sample value as a finite (instead of infinite) number of digits, is called Quantization.
 The error introduced in representing the continuous valued signal by a finite set of discrete value levels
is called quantization error or quantization noise.
 We denote the quantizer operation on the samples x (n) as Q[x(n)] and let Xq(n) denote the sequence of
quantized samples at the output of the quantizer. Hence Xq(n) = Q[x(n)].
 Then the quantization error is a sequence eq(n) defined as the difference between the quantized value
and the actual sample value. Thus eq(n) = xq(n) - x(n)
 Let us consider the discrete-time signal Obtained by sampling the analog
exponential signal xa(t)= 0.9t , t ≥ 0 with a sampling frequency fs= 1Hz.
 Table shows the values of the first 10 samples of x(n), reveals that the
description of the sample value x(n) requires n significant digits. It is obvious that this signal cannot
be processed by using a calculator or a digital computer since only the first few samples can be stored
and manipulated. For example, most calculators process numbers with only eight significant digits.

 However, let us assume that we want to use only one significant digit. To eliminate the excess digits,
we can either simply discard them (truncation) or discard them bv rounding the resulting number.
 The resulting quantized signals xq(n) are shown in Table. We discuss only quantization by rounding,
although it is just as easy to treat truncation. The rounding process is graphically illustrated in Fig b.
 The values allowed in the digital signal are called the quantization levels, whereas the distance Δ
between two successive quantization levels is called the quantization step size or resolution.
 The rounding quantizer assigns each sample of x(n) to the nearest quantization level.
 In contrast, a quantizer that performs truncation would have assigns each sample of x(n) to the
quantization level below it.
 The quantization error eq(n) in rounding is limited to the range of that is
 If xmin and xmax represent the minimum and maximum value of x(n ) and
L is the number of quantization levels, then
 We define the dynamic range of the signal as xmix and xmin.
 Here, we have xmix= 1, xmin = 0, & L=11, which leads to Δ =0.1.
 Note if dynamic range is fixed, increasing the number of quantization
levels, L results in a decrease of the quantization step size.
 So quantization error decreases & accuracy of quantizer increases.
 In practice we can reduce quantization error to an insignificant
amount by choosing a sufficient number of quantization levels.
 Theoretically, quantization of analog signals always results in a loss
of information. This is a result of ambiguity by quantization.
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <INTRODUCTION TO DSP> [Page 1.12]
 Indeed, quantization is an irreversible or noninvertible process (i.e. a many-to-one mapping) since
all samples in a distance A /2 about a certain quantization level are assigned the same value.
 This ambiguity makes the exact quantitative analysis of quantization extremely difficult.
CODING OF QUANTIZED SAMPLES
 The coding process in an A/D converter assigns a unique binary number to each quantization level.
 If we have L levels we need at least L different binary numbers. With a word length of b bits we can
create 2b different binary numbers. Hence we have 2b > L. or equivalently, b > log2L.
 Thus the number of bits required in the coder is the smallest integer greater than or equal to log2 L.
 In our example it can easily be seen that we need a coder with b = 4 bits.
 Commercially available A/D converters may be obtained
with finite precision of b=16 or less.
 Generally, for the high sampling speed and fine
quantization, the devices used become more expensive.
DIGITAL-TO-ANALOG CONVERSION
 To convert a digital signal into an analog signal we can use
a digital-to-analog converter.
 As stated previously, the task of a D/A converter is to interpolate between samples
 The sampling theorem specifies the optimum interpolation for a band limited signal.
 However, this type of interpolation is too complicated and hence impractical, as indicated previously.
 From a practical view point, the simplest D/A converter is the zero order hold which simply holds
constant the value of one sample until the next one is received.
 Additional improvement can be obtained by using linear interpolation as shown in fig to connect
successive samples with straight-line segments.
 The zero-order hold and linear interpolator are analyzed. Better interpolation can be achieved by using
more sophisticated higher order interpolation techniques.
 In general, suboptimum interpolation techniques result in passing frequencies above folding frequency.
 Such frequency components are undesirable and are usually removed by passing the output of the
interpolator through a proper analog filter, which is called a Post Filter or Smoothing Filter.
 Thus D/A conversion usually involve a suboptimum interpolator followed by a post filter.
ANALYSIS OF DIGITAL SIGNAL SYSTEMS VS DISCRETE-TIME SIGNAL SYSTEMS
 We have seen that a digital signal is defined as a function of an integer in dependent variable and its
values are taken from a finite set of possible values.
 The usefulness of such signals is a consequence of the possibilities offered by digital computers.
 Computers operate on numbers, which are represented by a string of 0's and l's.
 The length of this string (word length) is fixed and finite and usually is 8, 12, 16 or 32 bits.
 The effects of finite word length in computations cause complications in the analysis of digital signal
processing systems. To avoid these complications, we neglect the quantized nature of digital signals
and systems in much o four analysis and consider them as discrete-time signals and systems.
-------------- ALL THE BEST -------------------- ALL THE BEST ----------------
 Define Analog, Discrete and Digital Signals with diagram?
 Draw the block diagram of Simple DSP System.
 Write the Advantages of DSP over ASP.
 Write the different Applications of DSP.
 What is Continuous and Discrete Time Signal with Fig?
 What are 1, 2 & Multi dimensional signals with example?
 What do you mean by deterministic and random signal?
 What is periodic and non-periodic signal with fig?
 What is Symmetric and Asymmetric Signal with fig?
 Write the properties of Even and Odd Signals.
 What do you mean by multichannel and multi dimensional signals with examples?
 What is Sampling & Quantization and where we use it.
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.1]

[CHAPTER-2]
---------------------- DISCRETE TIME SIGNALS & SYSTEMS ----------------------
 SIGNAL: -
 Discrete-time Signals are defined for discrete
values of independent variables (time).
 Discrete-time signal is not defined at instants
between two successive samples.
 Discrete time signals are represented in 2 ways
 S (n), N1 ≤ n ≤ N2, where N1 and N2 are the
first and last sample point respectively in a
given discrete time signal. It represents non-
uniformly spaced samples and is shown in
below fig1.
 Whereas S (nTs), N1 ≤ n ≤ N2, represents
uniformly spaced samples & is shown in below fig2.

 REPRESENTATION OF DISCRETE-TIME SIGNALS: -

 Discrete time signals can be represented in the following ways : -
 Graphical Representation
 Functional Representation
 Tabular Representation
 Sequence Representation
GRAPHICAL REPRESENTATION
 Discrete-time signals can be represented by a graph
when the signal is defined for every integer value of
n for -∞ < n < ∞. This is illustrated in figure.
FUNCTIONAL REPRESENTATION
 Discrete-time signals can be represented functionally as given below
2, for n = 1, 3, −2
S (n) = 4, for n = 2, 0
0, elsewhere
TABULAR REPRESENTATION
 Discrete-time signal can also be represented by a table as,
n …. -3 -2 -1 0 1 2 3 ……
S(n) …. 0 2 0 4 2 4 2 ….
SEQUENCE REPRESENTATION
 An infinite duration (-∞ ≤ n ≤ ∞) signal with the time as origin (n = 0) & indicated by the symbol ↑ as
S (n) = {……, 0, 2, 0, 4, 2, 4, 2,….}

 A Finite duration signal can be represented as S (n) = {0, 2, 1, 4, 3, 1, 2}

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.2]
 SOME ELEMENTARY DISCRETE-TIME SIGNALS: -
 There are some basic signals which play an important role in the study of discrete-time signals and
systems. These signals are given below : -
 Unit-Sample (Impulse) Sequence, δ (n)
 Unit-Step Sequence, u (n)
 Unit-Ramp Sequence, r (n)
 Exponential Sequence, e (n)
 Sinusoidal Sequence, s (n)
UNIT-SAMPLE (IMPULSE) SEQUENCE [Delta] : -
 A Discrete Time Unit-Sample (Impulse) Sequence is denoted as δ (n).
 Its amplitude is 1 at n = 0 and for all other values of n; its
𝟏, 𝐧 = 𝟎
amplitude is zero i.e. δ (n) = .
𝟎, 𝐧 ≠ 𝟎
 In Sequence form it represented as δ (n) = {…0, 0, 𝟏, 0, 0, …}
 A continuous time delta function is denoted by δ (t).
 The graphical representation of delta function for Discrete Time
and Continuous Time signal is as shown in side figure:
UNIT-STEP SEQUENCE
 Unit step signal means the signal has unit amplitude in positive axis & zero amplitude in negative axis.
 There are two types of unit step signal as follows: Discrete & Continuous time unit step signal.
 A discrete time unit step signal is denoted by u (n).
 Its value is unity (1) for all positive values of n. that means
its value is one for n = 0. While for other values of n, its
𝟏, 𝐧 ≥ 𝟎
value is zero. i.e. u (n) =
𝟎, 𝐧 < 0
 In the form of sequence it can written as, u (n) = {…0, 0,
0, 0, 𝟏 , 1, 1, 1, 1, .}. The graphical representations of unit step functions are shown above.
UNIT-RAMP SEQUENCE
 A discrete time unit ramp signal is denoted by r (n).
 Its value increases linearly with sample number n.
𝐧, 𝐧 ≥ 𝟎
 Mathematically it is defined as, r (n) =
𝟎, 𝐧 < 0
 From above equation, it is clear that the value of
signal at a particular interval is equal to the number of
interval at that instant. r(n) = { 𝟎 , 1, 2, 3, 4, ……..}
 For Ex; for first interval signal has amplitude 1, for second it has amplitude 2, for third it is 3, & so on.
 The graphical representation of Unit Ramp function for Discrete & Continuous Time signal is in fig.
EXPONENTIAL SEQUENCE
 A discrete time exponential signal is expressed as, x (n) = an.
 Here „a‟ is some real constant. If „a‟ is complex number then x (n)
is written as, x (n) = e (n) = r ebn for all n & both r & b are real.
 We can define two cases depending on the parameter b is positive
or negative as:  (i) Growing exponential for b > 0 (ii) Decaying exponential for b < 0
 Now depending upon value of a, we have four different cases as under:
(i) Case 1: a > 1 (ii) Case 2: 0 < a < 1 (iii) Case 1: a < -1 (iv) Case 1: -1 < a < 0

(i) Decaying Exponential (ii) Rising Exponential (iii) Double Sided Growing & (iv) Double Sided Decaying Exponential Signal
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.3]
SINUSOIDAL SEQUENCE
 There are two types of sinusoidal sequences;
that are called as the sine and cosine sequence.
 A Sine Sequence is defines as s (n) = A Sin ω0n,
for all values of n.
 Cosine Sequence is defines as s (n) = A Cos ω0 n, for all values of n.
SINC FUNCTION: -
𝐒𝐢𝐧 (𝛑𝐱)
 The Sinc Function is mathematically expressed as Sinc (x) = 𝛑𝐱
for x = 0
 Sin (x) = 1 at x = 0; Sin x = 0 at x = ±1, ±2, ±3 …
SIGNUM FUNCTION: -
𝟏, 𝐭 > 0
 The CT Signum Function is mathematically defined as, sgn (t) =
−𝟏, 𝐭 < 0
 It is an Odd and Anti-symmetric Function.
 A discrete time Signum function can be obtained by
sampling the CT Signum function.
 It is train of samples of values +1 for positive n and 1 for
negative n as shown in figure.
RECTANGULAR FUNCTION: -
 A rectangular pulses of Unit Amplitude and Unit duration is
𝟏 𝟏
𝟏, − 𝟐 ≤ 𝐭 ≤ 𝟐
represented as rect (t) =
𝟎, 𝐎𝐭𝐡𝐞𝐫𝐰𝐢𝐬𝐞
 A rectangular pulses of A Amplitude and T duration is
𝐓 𝐓
𝐭 𝐀, − ≤ 𝐭 ≤
represented as rect 𝐓 = 𝟐 𝟐. It is centered about the y-axis i.e. about t = 0.
𝟎, 𝐎𝐭𝐡𝐞𝐫𝐰𝐢𝐬𝐞
 Relationship between Step, Ramp & Delta Signals:-
d
Relation between Unit Step & Unit Ramp Function: - r (t) = u (t) and u t dt = r (t)
dt
d
Relation between Unit Step & Delta Function: - u (t) = δ (t) and δ t dt = u (t)
dt
d2
Relation between Unit Ramp & Delta Function : - r (t) = δ t dt and r (t) = δ (t)
dt 2

 CLASSIFICATION OF DISCRETE TIME SIGNAL.

 There are various methods of classifying signals based on different features. Which are as follows: -
 Analog and Digital Signals
 Continuous Time Signal and Discrete Time Signals
 Deterministic Signal and Random Signals
 Multi Channel and Multi Dimensional Signals
 Energy and Power Signals
 Periodic Signal and Non-Periodic or Aperiodic Signals
 Even Signal (Symmetric) and Odd Signal (Asymmetric/Anti-symmetric)
 Causal and Non-Causal Signals
 Analog & Digital Signals, Continuous & Discrete Time Signals and Deterministic Signal & Random
Signals, Multi Channel & Multi Dimensional Signals etc, we have discussed in 1st chapter.
Energy Signal And Power Signals
 For a discrete-time signal x(n) the Energy E is defined as 
 Average Power of discrete-time signal x (n) is defined as 

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.4]
 The Energy Signal is one which gas Finite Energy and Zero Average Power.
 So, x(t) is an Energy Signal if : 0 < E < ∞ and P = 0, where E=Energy & P=Power of the signal x(t).
 The Power Signal is one which has Finite Average Power and Infinite Energy.
 Hence, Hence, x (t) is an Power Signal if: 0 < P < ∞ and E = ∞.
 If signal does not satisfy any of the above condition, then is neither an Energy Signal nor Power Signal.
 Some Important Formulas: -
1
∞
if a < 1
∞
 1 + a + a + a + a + a + …….+ a =
2 3 4 5
n=0 a
n
= 1−a
1 {Infinite Series}
if a > 1
a−1
1−a N +1
if a < 1
 1 + a + a2 + a3 + a4 + a5 + …….+ aN = N
n=0 a
n
= 1−a
{Finite Series}
a N +1 −1
if a > 1
a−1
∞ ∞ ∞
 1+e+e +e +…….+e =
2 3
n=0 e
n
= ∞; ♣ e =∞ ♣ e-∞ = 1/∞ = 0
 Nn=−N 1n = 2N + 1 ♣ ∞ n
n=−∞ 1 = ∞ ♣ limn→∞ a = a ; ejθ = 1
e jω t − e −(jω t ) e jω t − e −(jω t )
 Sin(ωt) = ♣ Cos(ωt) = ♣ e jωt = Cos(ωt) + jSin(ωt)
2j 2
 2jSinθ = ejθ − e−jθ ♣ 2Cosθ = ejθ + e ♣ ejθ = Cosθ + jSinθ  ejθ = 1 −jθ

 SOME EXAMPLES: - Determine the Value of Power and Energy of the following Signals. Find
whether the signals are Power, Energy or neither Power nor Energy Signals.
1 n π
(i) x (n) = u (n). (iii) x (n) = Sin n
3 4
π π
j n+
x (n) = e 2 4
(ii) (iv) x (n) = e2n u (n)
 SOLUTION ( i ) n
Given that x (n) = (1/3) u (n).
 We know that the Energy of the Signal, E = ∞ 𝐧=−∞ | 𝐱 𝐧 |
𝟐
[Putting the value of x(n)]
2 2 2
∞ 1 n 0 1 n ∞ 1 n
E= n=−∞ u n = n=−∞ u n + n=0 u n (limit divided)
3 3 3
2 2
0 1 n ∞ 1 n 1, n ≥ 0
= n=−∞ 0 + n=0 1 As, Unit Step Function u n =
3 3 0, n < 0
2
∞ 1 n ∞ 1 n 𝟏 𝟗 ∞ n 1
=0+ n=0 = n=0 = 𝟏 = 𝟖  E = 9/8 (Finite) As, n=0 a = 1−a
3 9 𝟏−
𝟗
𝟏 𝐍
 Now the Power of the Signal, 𝐏 = 𝐥𝐢𝐦 𝟐𝐍+𝟏 𝐧=−𝐍 | 𝐱 𝐧 |𝟐
𝐍→∞
2 2
1 0 1 n N 1 n
 P = limN→∞ 2N+1 n=−N u n + n=0 u n
3 3
2 2
1 0 1 n N 1 n 1, n ≥ 0
= limN→∞ 2N+1 n=−N 0 + n=0 1 As, u n =
3 3 0, n < 0
N +1
1 N 1 n 1 N 1 n 1 1−1 9
= limN→∞ 2N+1 0 + n=0 = limN→∞ 2N+1 n=0 = limN→∞ 2N+1 x
9 9 1−1 9
∞+1
1 1−1 9 1 1 N n 1−a N +1
= 2 x ∞+1 X =∞ =0 P=0 As, =0 As, n=0 a =
1−1 9 ∞ 1−a
 Here the Energy is Finite and Power is Zero. Therefore the given signal is an Energy Signal.
𝛑 𝛑
 SOLUTION ( ii ) Given that x (n) = 𝐞𝐣 𝟐𝐧+𝟒
 We Know that Energy of the Signal, E = ∞ 𝐧=−∞ | 𝐱 𝐧 |
𝟐
[Putting the value of x(n)]
π π 2 πn π 2 π
ej n+ j j
E= ∞
n=−∞ 2 4 = ∞
n=−∞ e 2 xe 4 [As, ejθ = 1. Let ejθ = ej 2 and putting ejθ ]
(1/2) jθ 2 2
E= ∞ n=−∞ e
jθn
x e = ∞ n
n=−∞ 1 x 1
1/2
= ∞ n=−∞ 1
2n
x1
∞ 2n ∞ 2 n ∞ n ∞ n
= n=−∞ 1 = n=−∞ 1 = n=−∞ 1 = ∞  E = ∞ As, n=−∞ 1 = ∞
𝟏 𝐍 𝟐
 Now the Power of the Signal, P = 𝐥𝐢𝐦𝐍→∞ 𝟐𝐍+𝟏 𝐧=−𝐍 | 𝐱 𝐧 |
𝛑 𝛑 𝟐
1 1
𝐞𝐣
N 𝐧+ N n
 P = limN→∞ 2N+1 n=−N
𝟐 𝟒 = limN→∞ 2N+1 n=−N 1 [Solved Above]

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.5]
1
= limN→∞ 2N+1 x 2N + 1 = limN→∞ 1 = 1  P = 1 (Finite) As, Nn=−N 1n = 2N + 1
 Here the Energy is Infinite and Power is Finite. Therefore the given signal is an Power Signal
𝛑
 SOLUTION ( iii ) Given that x (n) = Sin 𝟒 𝐧
∞
 We Know that the Energy of the Signal, E = 𝐧=−∞ | 𝐱 𝐧 |𝟐 [Putting the value of x(n)]
π π
π 2 π 1−cos 2 n 1−cos n
∞ ∞
E= n=−∞ sin n = n=−∞ Sin2 n = ∞
n = −∞
4
= ∞
n = −∞
2
=∞
4 4 2 2
𝟏 1 π 𝟐
𝐍 N
 Now the Power of the Signal, 𝐏 = 𝐥𝐢𝐦 𝟐𝐍+𝟏 𝐧=−𝐍 | 𝐱 𝐧 |𝟐 = lim 2N+1 n=−N Sin n
𝐍→∞ N→∞ 4
𝛑
1 𝟏−𝐜𝐨𝐬 𝐧 𝟏−𝐜𝐨𝐬𝟐𝛉 π
N
 P = lim 2N+1 n=−N
𝟐
As, Sin2 θ = also cos =0
N→∞ 𝟐 𝟐 2
1 𝟏 N 𝟏 1 N 𝟏 1
= lim n=−N (1 − 0) = 𝟐 lim 2N+1 n=−N 𝟏 = 𝟐 𝐱 lim 2N+1 2N + 1
N→∞ 2N+1 𝟐 N→∞ N→∞
𝟏 𝟏 𝟏 N n
= 𝟐 𝐱 lim 1 = 𝟐 𝐱 1 = 𝟐 As, n=−N 1 = 2N + 1
N→∞
 Here the Energy is Infinite and Power is Finite. Therefore the given signal is an Power Signal
 SOLUTION ( iv ) Given that x (n) = 𝐞𝟐𝐧 u (n)
 We Know that the Energy of the Signal, E = ∞ 𝐧=−∞ | 𝐱 𝐧 |
𝟐
[Putting the value of x(n)]
∞ 2n 2 0 2n 2 ∞ 2n 2
 E = n=−∞ e u(n) = n=−∞ e u(n) + n=0 e u(n)
1, n ≥ 0
= 0n=−∞ e2n 0 2 + ∞ n=0 e
2n
12=0+ ∞ n=0 e
2n 2
= ∞ n=0 e
4n
As, u n =
0, n < 0
= e4x0 + e4x1 + e4x2 + … + ∞ = e0 + e4 + e8 + … + ∞ = 1 + e4 + e8 + … + e∞ = ∞
𝟏
 Now The Power of the Signal, 𝐏 = 𝐥𝐢𝐦 𝟐𝐍+𝟏 𝐍𝐧=−𝐍 | 𝐱 𝐧 |𝟐
𝐍→∞
1 N 1 N
 P = limN→∞ 2N+1 n=−N e2n u(n) 2
= limN→∞ 2N+1 n=0 e
4n
[Solved Above]
1 1 1 e 4(N +1) −1 1
= limN→∞ 2N+1 N
n=0 e
4n
= limN→∞ 2N+1 N
n=0 e4 n
= limN→∞ 2N+1 = limN→∞ 2N+1 e∞ = ∞
e 4 −1
 Here both Energy & Power are infinite. Thus the given signal is neither Power nor an Energy Signal.
Periodic Signal And Aperiodic Signals
 A Signal is periodic with period N if and only if x (n + N) = x (n) for all n
 The Smallest value of N for which the above equation holds is known as fundamental period.
 If the above equation does not satisfy even for one value of n then the discrete time signal is Aperiodic.
 We have already observed that the sinusoidal signal of the form x (n) = A Sin 2π f0n is periodic when
f0 is a rational number, that is if f0 can be expressed as f0 = k/N, where k & N are integers.
 The energy of a periodic signal x(n) over a signal period say over the interval 0 ≤ n ≤ N - 1, is finite if
x (n) takes on finite values over the period.
 However, the energy of the periodic signal for -∞ ≤ n ≤ ∞ is infinite.
 On the other hand, the average power of the periodic signal is finite and is equal to the average power
over a single period.
 Thus if x (n) is a periodic signal with fundamental period N and takes on finite values, its power is
𝟏
given by P = 𝐍 𝐍−𝟏 𝟐
𝐧=𝟎 𝐱(𝐧) .  Consequently, Periodic signals are Power Signals.
 Let x(n) = A sin 𝜔0 𝑛 + 𝜃 ….. (i) Then it is periodic if and only if x(n + N) = x(n)
 x(n + N) = A sin ω0 n + N + θ = A sin ω0 n + ω0 N + θ = A sin ω0 n + θ + ω0 N …… (ii)
m 𝟐𝛑𝐦
Equation (i) & (ii) equal if and only if 𝛚𝟎 𝐍 = 2πm  ω0 = 2π N and N = 𝛚 (Minimum, m = 1)
𝟎
 Thus, the discrete time signal is periodic if fundamental frequency „ω0’ is a rational multiple of 2π,
else it is Aperiodic.
 SOME EXAMPLES: - Determine whether each of the following signals are periodic or not.
𝟑 𝟏
2𝜋 𝜋 3𝜋
(i) x (n) = 𝑒 𝑗 6𝜋𝑛 (ii) x (n) = 𝒆𝒋𝟓 𝒏+
𝟐 (iii) x (n) = Cos 𝑛 (iv) x(n) = Cos 𝑛 + Cos 𝑛
3 3 4
 SOLUTIONS (i)
2πm 2πm m
 For the given signal x (n) = 𝒆𝒋𝟔𝝅𝒏 ≡ 𝒆𝒋ω 0 𝒏  ω0 = 6π  N = N= =
ω0 6π 3
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.6]
3
 But the minimum value of „m‟ for which „N‟ is integer is 3. So, N = 3 = 1.
 As fundamental frequency is multiple of π. Therefore, the signal is Periodic. Fundamental Period, N=1.
𝟑 𝟏
3
 SOLUTION (ii):- Given signal x (n) = 𝒆𝒋𝟓 𝒏+
𝟐  ω0 = 5 which is not a multiple of π. So Aperiodic.
𝟐𝝅 𝟐𝝅
 SOLUTION (iii):- For the given signal x (n) = Cos 𝒏 ≡ Cos ω0 𝒏  ω0 =
𝟑 𝟑
2πm 2πm 2πm x 3
N= = = = 3m
ω0 𝟐𝛑/𝟑 2π
 To make N is an integer, the minimum value of m = 1, so N = 3. Therefore, the signal is Periodic.
𝜋 3𝜋
 SOLUTION (iv):- For x(n) = Cos 3 𝑛 + Cos 4 𝑛
𝛑 2πm
 The fundamental period of the signal Cos 𝐧 is N1 = = 6m = 6 (for m = 1)
𝟑 π/3
2πm 8m N 6 3
 Similarly N2 = 3𝜋/4 = = 8 (for m = 3)  N 1 = 8 = 4  N = 4N1 = 3N2 = 24  N = 24 Periodic
3 2
𝑛𝜋
 Also ♣ x(n) = sin( 4 )  N=8 ; Periodic ♣ x(n) = 𝑒 𝑗 2𝑛  N = nπ ; Aperiodic
𝜋
♣ x(n)= Cos 4 𝑛 + cos 2𝑛  N1= 8, N2 = π  N1/N2 = 8/π; Aperiodic,
♣ x(n) = e2n ≡ 𝑒 ω 0 𝑛  ω0 = 2  N = 2πm/2 = πm  Aperiodic
♣ x(n) = Sinπt ≡ Sin ω0t  ω0 = π  N = 2πm/π = 2m  N=2 for m = 1 Periodic
♣ x(n) = Sin3n ≡ Sin ω0n  ω0 = 3  N = 2πm/3  Aperiodic
♣ x(n) = Sin(π + 0.2n) ≡ Sin (ω0n + θ)  ω0 = 0.2  N = 2πm/0.2 = 10πm  Aperiodic
𝐣𝛑
𝐧 π 2πm 8πm
♣ x(n) = 𝒆 𝟒 ≡ 𝑒 jω 0 𝑛  ω0 = 4  N = = = 8m  N = 8 for m = 1  Periodic
π/4 π
𝟔𝛑𝐧 6π 2πm 14πm 7m
♣ x(n) = Sin ≡ Sin ω0n  ω0 =  N = 6π/7 = =  N = 7 for m = 3  Periodic
𝟕 7 6π 3
𝐧𝛑 π 2πm
♣ x(n) = Sin ≡ Sin ω0n  ω0 = 8  N = = 16m  Periodic
𝟖 π/8
𝟐𝛑𝐧 𝟐𝛑𝐧 𝟐𝛑𝐧 2πm
♣ x(n) = Cos + Cos Fundamental period of the signal Cos is N1 = 2π/5 = 5m = 5
𝟓 𝟕 𝟓
2πm N 5
(For m = 1); Similarly N2 = 2𝜋/7 = 7m = 7 (m = 1)  N 1 = 7 N =7N1=5N2= 35  N = 35 Periodic
2
♣ x(t) = 2Cost + Cos (t/3)  ω1 = 1  N1 = 2πm/1 = 2πm/1  Aperiodic;
N1 2πm
Similarly, ω2 = 1/3  N2 = (2πm)/(1/3) = 6πm  Aperiodic; Now, N = = 6πm = 1/3  Periodic
N2
♣ x(t) = 2Cost + 7Cost  ω1 = 1  N1 = 2πm/1 = 2πm  Aperiodic;
N 2πm
But ω2 = 1  N2 = 2πm / 1 = 2πm  APeriodic;  Now, N = N 1 = 2πm = 1  Periodic
2
♣ x(t) = 2Cos (1.5πt) + Sin (3.5πt)  ω1 = 1.5π  N1 = 2πm/1.5π = 2/1.5 = 4m/3
 N1 = 4 for m = 3  Periodic; Similarly ω2 = 3.5π  N2 = 2πm/3.5π = 4m/7
N 4
 N2 = 4 for m = 7  Periodic  Now, N = N 1 = 4 = 1  Periodic
2
𝐣𝐧𝛑 𝐣𝐧𝛑
nπ nπ nπ nπ
♣ x(n) = Re 𝒆 𝟏𝟐 + Im 𝒆 𝟏𝟖  x(n) = Re Cos + jSin + Im Cos + jSin
12 12 18 18
𝐧𝛑 𝐧𝛑 π 2πm
 x(n) = 𝐂𝐨𝐬 + 𝐣𝐒𝐢𝐧  ω1 = 12  N1 = 𝜋/12 = 24m  Periodic;
𝟏𝟐 𝟏𝟖
π 2πm N1 24m
Similarly, ω2 = 18  N2 = 𝜋/18 = 36m  Periodic; Now, N = = 36m = 2/3 (6)  Periodic
N2
 The sum of two periodic signals x1(n) and x2(n) with period N1 and N2 may be periodic or may not be
periodic depending on the relationship between N1 and N2.
 For the sum to be periodic, the ratio of time periods (N1/N2) must be a rational number or ratio of two
integers. Otherwise the sum is not periodic.
Symmetric (Even) Signal And Anti-symmetric (Odd) Signals
 A real valued signal x (n) is called Symmetric (EVEN) x (-n) = x (n) for all n. [ Ex: - Cos ωn ]
 On the other hand a signal x (n) is called Anti-symmetric (ODD) if x (-n) = - x (n). [ Ex: - Sin ωn ]
 A signal x (n) can be expressed as sum of Even & Odd components. i. e. X (n) = Xe (n) + X0 (n)
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.7]
Where Xe (n) is Even component and X0 (n) is Odd component.
 Now in above equ. replacing n by –n, we get X (-n) = Xe (-n) + X0 (-n)  X (-n) = Xe (n) - X0 (n)
[As x (-n)= - x (n), For Odd and x(-n) = x (n), For Even Signals]
 Now adding above two equations, we get 2Xe(n) = X (n) + X (-n), So the expression for Even Signal is

Xe (n) = ½ [X (n) + X (-n)] Similarly by Subtracting for Odd Component, we get

Properties of EVEN and ODD Signals Xo (n) = ½ [X (n) - X (-n)]

 The Sum of two Even signals are Even signal  The Product of two Odd signal is Even signal
 The Sum of two Odd signals are Odd signal  The Product of an Even signal & Odd signal.
 Product of two Even signal is Even signal [E+E=E, E.E=E, 0+0=0, 0.0=0, E+0=NENO, E.0=0]
 SOME EXAMPLES
 x(n) = cos 𝜔𝑛 x(-n) = cos 𝜔(−𝑛) = cos(−𝜔𝑛) = cos 𝜔𝑛 = x(n) EVEN [As, (cos −𝜃) = cos 𝜃]
 x(n) = sin 𝜔𝑛  x(-n) = sin 𝜔(−𝑛) = sin(−𝜔𝑛) = - x(n)  ODD <x(t) = e-t  No Even or Odd>
sin 𝑛 sin −𝑛 − sin 𝑛 sin 𝑛 sin n
 x(n) = 𝑛  x(-n) = −𝑛 = −𝑛 = 𝑛 = x(n) EVEN is also known as Sin c
n
𝟐𝐧𝟐 𝟑𝐧 2(−n)2 3(−n) 2n 2 −3n
 x(n) = 𝐒𝐢𝐧 x𝐂𝐨𝐬  x (-n) = Sin x Cos = Sin x Cos ODD
𝟑𝐧 𝟐𝐧𝟐 3(−n) 2(−n)2 −3n 2n 2
 Find Even & Odd Components of the signal x(t) = 1+ t +3t + 5t + 9t  x(-t) = 1- t +3t - 5t3 + 9t4
2 3 4 2

 xe (t) = ½ [X (n) + X (-n)] = 1 +3t2 + 9t4 AND xo (t) = ½ [X (n) - X (-n)] = t + 5t3
 Find Even & Odd Components of the signal x (t) = Cos t + Sin t + Cos t . Sin t  x (-t) = Cos (-t) +
Sin(-t) + Cos(-t) . Sin(-t) = Cos t - Sin t – Cos t . Sin t  xe (t) = Cos t & xo (t) = Sin t + Cos t . Sin t
 We have already discussed some more problems in 1st Chapter.
Causal and Non-Causal Signals: -
 A Signal is said to be Causal if its value
is Zero for n < 0. Otherwise the signal is
Non-Causal. For example X1(n) = an u
(n) & X2 (n) = {𝟏, 2, -3, -1, 2} are
examples of Causal Signals. [u(t), r(t)]
 Whereas X3 (n) = an u (- n + 1) and X4 (n) = {-2, 𝟒, 2, -1, 3} are examples of Non-Causal Signals
 QNS:-Represent Graphically (i) x1(n) = {1,2, 𝟎 ,-1,1} (ii) x1(n) = {0,0,-1,2, 𝟑} (iii) x1(n)={𝟎,1,-1,1,-1}
 Sketch (i) x(n) = 2-n for -2 ≤ n ≤ 2 [≡ {4,2,1, 𝟎 ,0.5,0.25}] (ii) x(n) = 2n2 for -2 ≤ n ≤ 2 [≡ {8,2, 𝟎,2,8}]
 x(n) = {2,5,1,𝟐, 2,0,1,7}, Find y1(k) = 3k=−3 x n and y2(k) = 3k=−3 x n . {Answer  13 & 0}
 SIMPLE MANIPULATION OF DISCRETE TIME SIGNAL: -
 Signal processing is a group of basic operations applied to an input signal resulting in another signal as
the output. The mathematical transformation from one signal to another is represented as y (n)=T[x(n)]
 The basic set of operations are
1) Shifting < Delays or Advances > 4) Scalar Multiplication
2) Time Scaling 5) Signal Multiplier
3) Time Reversal < Folding> 6) Signal Addition
SHIFTING
 The Shift operation takes the input sequence and shift the values by an integer increment of the
independent variable. The shifting may Delays or Advances the sequence in time.
 Mathematically this can be represented as y (n) = x (n – k), Where x (n) is input and y (n) is output.
 If k is Positive the Shifting Delays the sequence. If k is Negative the shifting Advances the sequence.

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.8]

Ex1: x(n) = {2, 1, 𝟐,1, 3, 5}, So, x(n-2) = {𝟐,1, 2, 1, 3, 5} [Delay] & x(n+2) = {2, 1, 2, 1, 𝟑, 5} [Advance]
Ex2: x (n) = {2, 1, 5, 9, 𝟖, 6, 1, 2, 3}  x (n+1) = {2, 1, 5, 9, 8, 𝟔, 1, 2, 3}
x (n-4) = {𝟐, 1, 5, 9, 8, 6, 1, 2, 3}, x (n+2) = {2, 1, 5, 9, 8, 6, 𝟏, 2, 3}
 Find the Values of the Following questions
 ∞ n=0 δ n e
2n
= e2n n=0 = 1  ∞ n=−∞ δ n − 2 sin2n = sin2n n=2 = 𝐒𝐢𝐧 𝟒
∞
 n=0 δ n + 1 e −2n
=0  ∞ n=−∞ δ n + 1 x(n) = x(n) n=−1 = x (-1)
 n=−∞ δ n − 2 Cos 2n + δ n − 1 Sin 2n = Cos 4 + Sin 2
5

SCALING : -
 This is accomplished by replacing „n’ by „λn’ in the sequence x (n). Then y (n) = x (λn)

 From the above, figure (a) is a discrete time signal x (n). If it is scaled by 2 means that y (n) = x (2n).
 It means that what we were getting at x (2n) that will get at y (n). That is y (1) = x (2), y (2) = x (4) etc.
 Similarly If it is scaled by ½ means that y (n) = x (½ n). i.e. y (2) = x (1), y (4) = x (2), y (6) = x (3) etc.
 The process of reducing sampling rate is often referred as Down Sampling or Decimation.
 Whereas the process of increasing sampling rate is referred as Up Sampling.
 Example If x(n) = {1, 2, 3, 4, 5, 4, 3, 2, 1}  x(2n) = {1, 3, 5, 3, 1} <Down Sampling>
-4 -3 -2 -1 1 2 3 4 -2 -1 1 2

 x (n/2) = {1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 4, 0, 3, 0, 2, 0, 1} < Up Sampling >

-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8

TIME REVERSAL < FOLDING> : -

 Time reversal of a signal x (n) can be obtained by folding the signal about n = 0 & is denoted as x (- n).

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.9]
 x (- n) is obtained by folding the sequence about “n = 0” or origin.
 Here for n (- n - k), If k = +ve  Advance & if k = -ve  Delay
 Example: x(n) = {𝟐, 1, 2, 1}  x(-n) = {1, 2, 1, 𝟐}
Thus x (- n - 2) < Means k = +ve, Adv Signal> = {1, 2, 1, 2, 0, 𝟎}
And x (- n + 2) <Means k = -ve, Delayed = {1, 𝟐, 1, 2}
SCALAR MULTIPLICATION : -
 A Scalar multiplier is shown in figure.
 Here the signal x (n) is multiplied by a scale factor a.
 For example if x (n) = {1, 2, 1, -1} and a = 2 then y (n) = a x(n) = {2, 4, 2, -2}
SIGNAL MULTIPLIER : -
 Figure shows the multiplication of two signal sequences to form another sequence.
 For ex, if x1 (n) = { -1, 2, -3, -2} and x2 (n) = { 1, -1, -2, 1},
Then, x1 (n) . x2 (n) = {-1.1, 2.-1, -3.-2, -2.1} = {-1, -2, 6, -2}
SIGNAL ADDITION : -
 Two signal sequences can be added to form another sequence as shown in figure.
 For example, if x1(n) = { 1, 3, 2, 5} & x2 (n) = { 1, 2, 4, 3},
Then, x1(n) + x2(n) = {1+1, 3+2, 2+4, 5+3} = {2,5,6,8}.
 QUESTION: - Prove That δ (n) = u(n) – u(n-1)
 Solution: - u(n) = {… , 0, 0, 0, 𝟏, 1, 1, 1, …..}, u (n-1) = {… , 0, 0, 𝟎, 1, 1, 1, 1, …..},
Thus, u(n) – u(n-1) = {… , 0, 0, 0, 1, 0, 0, 0, ….. } = δ (n) (Proved)
 QUESTION: - Prove That u(n) = 𝐤=−∞ 𝛅(𝐤) 𝐍

 Solution: - We know that u(-15)= 0, u(-8)=0, u(-2)=0, u(0)=1, u(3)=1, u(7)=1, u(12)=1 and so on.
Now from the expression, u(-8) = δ (-∞) + …. + δ (-10) + δ (-9) + δ (-8) = 0+…+0+0+0 = 0
Similarly, u (0) = δ (-∞) + …. + δ (-2) + δ (-1) + δ (0) = 0+…+0+0+1 = 1
u (7) = δ (-∞) + …. + δ (0) + δ (1) +…..+ δ (7) = 0+…+1+0+…..+0 = 1, Hence proved
 EX: - A DTS is as shown in fig1. Sketch (i) x(n-3) (ii) x(3-n) (iii) x(2n) (iv) x(n) u(3-n) (v) x[(n-1)2]

 EX: - A Continuous Time Signal x (t) is as shown in fig1. Sketch (i) x(t/2)
 EX: - A Continuous Time Signal x (t) is as shown in fig2. Sketch (i) x(3t)

1  &2 
1, for − 1 ≤ t ≤ 1
 EX: - For a Continuous Time Signal defined as x(t) = . Sketch the following
0, Otherwise
(i) x(2t) (ii) x(-2+t) (iii) x(t+1) (iv) x(t+1) u(t) (v) x(t) . δ (t)
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.10]

 EX: - A DTS is given by x (n) = {1, 𝟏, 1, 1, 2} Sketch the following signals: - (i) x(n-2) (ii) x(n+1)
(iii) x (3-n) (iv) x(n).u(n-1) (v) x(n-1) δ (n-1) (vi) Even samples of x(n) (vii) Odd samples of x(n)

 EX: - Sketch a DTS x(n) = 2-n for -2 ≤ n ≤ 2 and obtain y1(n) = 2 x(n) + δ (n) & y2(n) = x(n) . u (2-n)

 EX: - A Discrete Time Signal is as shown in fig. Sketch (i) x(n-3) (ii) x(3-n) (iii) x(2n) (iv) x(n) u(3-n)

 EX: - A CTS is as shown in fig. Sketch (i) x(t-2) (ii) x(1-t) (iii) [x(t) + x(1-t)]. u(t-1)

 EX: - A DTS is given by x (n) = {3, 2, 1, 𝟎, 1, 2, 3}.Then Sketch y (n) = [x (n) + x (n-1)] – x (n+1)

 EX: - Sketch the Even and Odd parts of the signal x(n) = u(n).

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.11]
 EX: - Sketch the even and odd parts of the signal x(n) = an u(n).

NOTE: -
 In time shifting operation equation [y (n) = x (n – k)], if k=1,
we get unit delay element that delays the signal by one sample.
 If the input signal is x (n) the output is x (n -1).
 In fact the sample x (n-1) is stored in memory at time n-1 & is
recalled from memory at time n to form y (n) = x (n-1)
 Thus this basic building block requires memory.
 Similarly If k = - 1, we get unit advance element. The diagram Shows unit delay and advance element.
 DISCRETE TIME SYSTEM.
 A discrete-time system is a device or algorithm that
operates on a discrete-time input signal x (n),
according to some well defined rule, to produce
another discrete-time signal y (n) called output signal.
 The relation between input signal x (n) and the output signal y(n) is Representation for the output
signal y (n ) = 0.25 y (n - 1) + 0.5 x (n) + 0.5 x (n - 1) is shown below.

 CLASSIFICATION OF DISCRETE TIME SYSTEM.

 In the analysis as well as in the design of systems, it is desirable to classify the systems according to
the general properties that they satisfy.
 Discrete-time systems are classified according to their general properties and characteristics. They are:-
(1) Static and Dynamic Systems. (4) Linear and non-linear Systems.*
(2) Time-variant & Time-invariant Systems.* (5) FIR and IIR Systems.
(3) Causal and non-causal Systems. (6) Stable and Unstable Systems.
STATIC AND DYNAMIC SYSTEMS.
 A discrete-time system is called Static or Memory less if it‟s output at any instant n depends at most
on the input samples at the same time, but not on the past or future samples of the input.
 In any other case, the system is said to be Dynamic or to have memory.
 If the output of a system at time n is completely determined by the input samples in the interval from
(n - N) to n (N ≥ 0), the system is said to have memory of duration N.
 If N = 0, the system is Static. If 0 < N < ∞, the system is said to have Finite Memory.
 Where as if N = ∞, system is said to have Infinite Memory.
 The systems described by the following input output equations are Static or memory less: -
(1) y (n) = a x (n) (2) y (n) = n x (n) + bx3 (n)
 Note that there is no need to store any of the past inputs or outputs to compute the present output.
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.12]
 However, the systems described by the following input output relations are Dynamic systems:-
(1) y (n) = x (n) + 3 x (n - 1) (3) y (n) = 𝑛𝑘=0 𝑥 (𝑛 − 𝑘)
(2) y (n) = x (n) + x (n – 1) (4) y (n) = ∞𝑘=0 𝑥 (𝑛 − 𝑘)
 The system described by equations (1), (2) & (3) has finite memory whereas the system described by
equation (4) has infinite memory. These are example of dynamic systems as they need memory.
 We observe that static or memory less systems are described in general by I/O equations of the form
y(n) = Ґ [x (n) , n] and do not include delay elements (memory).
 EXAMPLES: -
(1) y(n) = x(2n), Here y(0) = x(0), but y(1) = x(2) which is the future input Thus is a Dynamic System.
(2) y(n) = x2(n), Here y(0) = x2(0), y(1) = x2(1), y(2) = x2(2) and so on, Thus it is a Static System.
(3) y(n) = x2(n) + x(n)  Static; (4) y(n) = x(n) + 1/x(n-1) (5) x(n/2), x(n2) etc  Dynamic System.
TIME-VARIANT AND TIME-INVARIANT SYSTEMS.
 We can subdivide the general class of sytem into 2 categories, time-variant & time-invariant system.
 A system is called time-invariant if its input-output characteristics do not change with time.
 To test if any given system is time-invariant, first apply an arbitray sequence x (n) and find y (n).
 Thus we can write y (n) = Ґ [x (n)].
 Now delay the same input signal by k units of time to get x (n - k), and again apply to the same system.
 If the characteristics of the system do not change with time resulting output y (n – k), then the system is
said to be time-invariant or shift invariant system.
Ґ Ґ
 Hence x (n) → y (n) and x (n - k) → y (n - k) for every input signal x (n) and every time shift k.
 In general we can write the expression for out put as  y (n, k) = Ґ [x (n - k)]
 Now if this output y (n, k) = y (n – k), for all possible values of k, the system is Time Invariant.
 On the other hand, if the output y (n, k) ≠ y (n – k) for every values of k, the system is Time Variant.
 Examples of time Invariant Fig (a) & some time Variant systems Fig. (b) to (d) is shown in figure.

 EXPLATION
(A) This system is described by the input-output equations y (n) = Ґ [x (n)] = x (n) – x (n – 1)
 Now if the input x(n) is delyed by k units in time and applied to the system, it is clear from the block
diagram that the output will be y (n , k) = x (n – k) – x (n – k – 1)
 On the other hand, if we delay y (n) by k units in time, we obtain y (n - k) = x (n – k) – x (n – k – 1)
 Since y (n, k) = y (n – k), for all possible values of k, the system is Time Invariant System.
(B) This system is described by the input-output equations y (n) = Ґ [x (n)] = n x (n)
 The response of this system to x (n –k) is y (n, k) = n x (n – k) ----- (1)
 Now if we delay y (n) by k units in time for the given expression i.e. y (n) = n x (n), we obtain
y (n – k) = (n – k) x (n – k) = [{n x (n – k)} – {k x (n – k)}] ------------ (2)
 Since y (n, k) ≠ y (n – k), for all possible values of k, the system is Time Variant System.
(C) This system is described by the input-output equations y (n) = Ґ [x (n)] = x (- n)
 The response of this system to x (n –k) is y (n, k) = x (- n - k) ------------- (1)
 Now if we delay y (n) by k units in time for the given expression i.e. y (n) = x (- n), we obtain
y (n – k) = x {-(n – k)} = x (- n + k) ------------- (2)
 Since y (n, k) ≠ y (n – k), for all possible values of k, the system is Time Variant System
(D) This system is described by the input-output equations y (n) = Ґ [x (n)] = x (n) Cos ω0n
 The response of this system to x (n –k) is y (n, k) = x (n - k) Cos ω0n ------------- (1)
 Now if we delay y (n) by k units in time for the given expression i.e. y (n) = x (- n), we obtain
y (n – k) = x (n -k) Cos ω0 (n – k) ------------- (2)
 Since y (n, k) ≠ y (n – k), for all possible values of k, the system is Time Variant System
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.13]
 EXAMPLES: - *** y(n, k ) = F[x(n-k)] ; y (n-k) = n  n-k ***
y(n, k) = sin x(n − k)
1) y(n) = sin x(n)   Same → Time Invariant
y(n − k) = sin x(n − k)
y(n, k) = nx(n − k)
2) y(n) = nx(n)   Not Equal → Time Variant
y n − k = (n − k)x(n − k)
y(n, k) = x n − k Cos200πn
3) y(n)=x(n) Cos200πn   Not Same → Time Variant
y(n − k) = x n − k Cos 200π(n − k)
y n, k = x n − k + x(n − 1 − k)
4) y(n) = x(n) + x(n-1)   Same → Time Invariant
y n − k = x n − k + x(n − k − 1)
y n, k = nx 2 n − k
5) y(n) = nx2(n)   Not Equal → Time Variant
y n − k = n − k x2 n − k
n
y n, k = x −k
2
6) y(n) = x(n/2)  n−k
 Not Equal → Time Variant
y n−k = x 2
y(t,k) = sin [x(t+2−k)]
7) y(t) = sin [x(t+2)]  y(t−k) = sin [x(t−k+2)] Same  Time invarient
y(n,k) = nx (n−k)+4x(n−1−k)–2x(n+1−k)
8) y(n)=nx(n)+4x(n-1)–2x(n+1) Time Varient
y(n−k) = (n−k)x(n−k)+4x(n−k−1)–2x(n−k+1)
y(n,k) = anx (n−k)
9) y(n) = anx(n) y(n−k) = a(n−k)x(n−k) Not Same  Time Varient
y(n,k) = ax (n−1−k) + bx (n−2−k)
10) y(n) = ax(n-1) + bx(n-2)  y(n−k) = ax (n−k−1) + bx (n−k−2) Same  Time Invarient
y(n,k) = n x(n−k) 2
11) y(n) = n x(n) 2
 y(n−k) = (n−k) x(n−k) 2 Not Same  Time Varient
y(n,k) = a x(n−k) 2 + bx (n−k)
12) y(n) = a x(n) 2 + bx(n)  Same  Time Invarient
y(n−k) = a x(n−k) 2 + bx (n−k)
y(n,k) = 7x(n−k) + 5n
13) y(n) = 7x(n) + 5n  y(n−k) = 7x(n−k) + 5(n−k) Not Same  Time Varient
14) y(n) = 3y 2 (n-1) - nx(n) + 4x(n-1) - 2x(n+1)
y(n,k) = 3y 2 (n−1−k) − nx (n−k) + 4x(n−1−k) − 2x(n+1−k)
 y(n−k) = 3y 2 (n−k−1) − n−k x(n−k) + 4x(n−k−1) − 2x(n−k+1) Not Same  Time Varient
y(n,k) = 12x(n−1−k) + 11x(n−2−k)
15) y(n) = 12x(n-1) + 11x(n-2)  y(n−k) = 12x(n−k−1) + 11x(n−k−2)  Time Invarient
y n,k = T x n−k =nx n−k
16) y(n) = nx(n)   Time Varient
y(n−k) = (n−k) x n−k
y n,k = T x n−k =e x n −k
17) y(n) = ex(n)   Time Invarient
y(n−k) = e x n −k
y n,k = T x n−k =x 2n−k
18) y(n) = x(2n) y n−k = x 2 n−k = x 2n−2k  Time Varient

19) y(n) = m n
k=0 a k x n − k − k=1 b k y n − k [Here k is there so take a new term m]
y n,m = T x n−m = m n
k =0 a(k)x n−k−m − k =1 b(k)y n−k−m
 Time invarient
y(n−m) = m n
k =0 a(k)x n−m−k − k =1 b k y n−m−k {𝑝𝑢𝑡 𝑛→𝑛−𝑚 }
y n,k = T x n−k =x n−k +nx n+1−k
20) y(n) = x(n)+ nx(n+1)   Time varient
y n−k = x n−k +(n−k) x n+1−k
CAUSAL AND NON-CAUSAL SYSTEMS.
 A system is said to be causal if the output of the system at any time n depends only on prsent and past
inputs, but does not depend on future inputs. This can be represented mathematically as
y (n) = F [x (n), x (n – 1), x (n - 2), …… ]
 If systems depend not only on present & past but also future inputs then said to be a non-causal system.
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.14]
 EXAMPLES:- Determine if the system described by the following equations are Causal or Non-causal
𝟏
(i) y(n) = x(n) + 𝒙 (𝒏−𝟏) (ii) y (n) = x (n2)
𝟏
 SOLUTION (i) Given that y(n) = x(n) + 𝒙 (𝒏−𝟏)
𝟏 𝟏 𝟏
 So, For n = - 1, y (- 1) = x(-1) + 𝒙 (−𝟐) ; For n = 0, y (0) = x(0) + 𝒙 (−𝟏) ; For n = 1, y (1) = x(1) + 𝒙 (𝟎)
 For all the values of n the output depends on present and past inputs. Therefore, the system is Causal.
(i) Given y(n) = x (n2); So, For n = - 1, y(-1) = x(1) ; For n = 0, y(0) = x(1); For n = 1, y(1) = x(1),
 For negative values of n, the system depends on future inputs. So, the system is Non-Causal.
 y (n) = Ax(n)+B ; ax(n)+bx(n-1) ; x(t) cos(t+1) ; x(n) +x(n-1) ; 0.8x(n-1)  Causal Systems
 y(n) = x(2n) ; x(n)+3x(n+4) ; x(-n) & y(n-2) = x(n) are Non-Causal Systems.
EX: (1) y(t) = x(-t2)  Causal (2) y(n) = x(n-1)  Causal (3) y(t) = x(t+1)  Non-causal
(4) y(n) = x(n) + x(n+1)  Non-causal (5) y(t) = 0.2x(t) – x(t-1)  Causal
LINEAR AND NON-LINEAR SYSTEMS
 A system that satisfies the superposition principle is said
to be a linear system.
 Superposition principle states that the response of the
system to a weighted of a signal be equal to the
corresponding weighted sum of the outputs of system to
each of the individual input signals.
 A system is said to be Linear if and only if
T [a1 x1(n) + a2 x2(n)] = a1T[x1(n)] + a2T[x2(n)]
for any arbitrary constant a1 & a2
 A relaxed system that does not satisfy the Superposition
principle is called Non-Linear System.
 EXAMPLES:- Determine if the system described by the following equations are Linear or Non-Linear
𝟏
(i) y (n) = nx(n) (ii) y(n) = x(n2) (iii) y(n) = x(n) + 𝒙 (𝒏−𝟏) (iv) y(n) = x2(n)
 SOLUTION
(i) Given that y(n) = n x(n). The outputs due to the signals x1(n) & x2(n) are [Here T=n, Scalar Multiplication]
y1 (n) = T[x1(n)] = n x1(n) …….(1.1) & y2 (n) = T[x2(n)] = n x2(n) ………….(1.2)
The weighted sum of outputs is y3(n) = a1T[x1(n)] + a2T[x2(n)] = a1 nx1(n) + a2 nx2(n) ……… (1.3)
The output due to weighted sum of inputs is, y4(n) = T[a1x1(n) + a2x2(n)] = na1x1(n)+ na2x2(n) …(1.4)
 The equation 1.3 and 1.4 are equal. Superposition principle is satisfied. The system is LINEAR
(ii) Given that y(n) = x(n2)  The outputs due to the signals x1(n) and x2(n) are [Here T=n2, Time Square]
y1 (n) = T[x1(n)] = x1 (n2) ……… (2.1) & y2 (n) = T[ x2(n) ] = x2(n2) ………(2.2)
The weighted sum of outputs is y3(n) = a1y1(n) + a2y2(n) = a1x1 (n2) + a2x2 (n2) ………(2.3)
The output due to weighted sum of inputs is y4 (n) = T[a1x1(n) + a2x2(n)] = T[a1x1(n)] + T [a2x2(n)]
= a1T [x1(n)] + a2T [x2(n)] = a1x1(n2) + a2x2(n2)] ----(2.4) < As Linear Satisfy Additive and scaling Property >
 Equation 2.3 & 2.4 are equal, Superposition principle is satisfied. So, system in Linear.
𝟏
(iii) Given y(n) = x(n) + 𝒙 (𝒏−𝟏) . For two input sequence x1 (n) and x2 (n) the corresponding outputs are
𝟏 𝟏
y1 (n) = T [x1 (n)] = x1(n) + 𝒙 ------- (3.1) & y2 (n) = T [x2 (n)] = x2(n) + 𝒙 --------- (3.2)
𝟏 (𝒏−𝟏) 𝟐 (𝒏−𝟏)
 The output due to weighted sum of inputs is y4 (n) = T [a1 x1(n) + a2 x2(n)]
𝟏
Y4 (n) =a1 x1(n) + a2 x2(n) + 𝒂 ------------ (3.3) [Here T= p(n) + 1/p(n-1)]
𝟏 𝒙𝟏 𝒏−𝟏 +𝒂𝟐 𝒙𝟐 (𝒏−𝟏)
 On the other hand, the linear combination of the two outputs is
𝒂𝟏 𝒂𝟐
Y3 (n) =a1 y1(n) + a2 y2(n) = a1 x1(n) + 𝒙 (𝒏−𝟏) + a2 x2(n) + 𝒙 -------------------- (3.4)
𝟏 𝟐 (𝒏−𝟏)
 Equations 3.3 & 3.4 are not equal, superposition principle is not satisfied, So system is Non-Linear.
(iv) Given y(n) = x2(n). For two input sequence x1 (n) and x2 (n) the corresponding outputs are
y1(n) = T [x1(n)] = x12(n) ---- (4.1) & y2(n) = T [x2(n)] = x22(n) ----- (4.2) [Here T= Square the Signal]

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.15]
Thus, linear combination of the two outputs will be y3(n)= a1y1(n) + a2y2(n) = a1 x12 (n) + a2 x22 (n) ---(4.3)
Now the response of the system to the linear combination of inputs will be, y4(n) = T [a1x1(n) + a2x2(n)]
= [a1x1(n) + a2x2(n)]2 = a12x12(n) + a22x22(n) + 2 a1a2x1(n)x2(n) ……. (4.4)
 Equations 4.3 & 4.4 are not equal, superposition principle is not satisfied, So system is Non-Linear.
EXAMPLES
 y(t) = ax(t) + b; y1(t) = T[x1(t)] = ax1(t) + b1; y2(t) = ax2(t) + b
y3(t) = y1(t) + y2(t) = a1x1(t)+ b1 + a2x2(t)+ b2 ;  y4(t) = f [x1(t) + x2(t)] = a[x1(t) + x2(t)] + b  NL
 y(t) = Sin[x(t+2)]; y1(t) = Sin[x1(t+2)], y2(t) = Sin[x2(t+2)]
y3(t) = Sin[x1(t+2)] + Sin[x2(t+2)] ; y4(t) = Sin[x1(t+2) + x2(t+2)]  NL
Also, y3(t) = a1Sin[x1(t+2)] + a2Sin[x2(t+2)]; y4(t) = Sin[a1x1(t+2) + a2x2(t+2)]  NL
 y(t) = x(n). x(n-2); y1(t) = x1(n). x1(n-2) , y2(t) = x2(n). x2(n-2)
y3(t) = a1x1(n). x1(n-2) + a2x2(n). x2(n-2) , y4(t) = [a1x1(n) + a2x2(n)] [a1x1(n-2) + a2x2(n-2)]  NL
 y(t)=ex(n);y1(t)=ex 1 (n) ,y2(t)=ex 2 (n) ; y3(t)=a1ex 1 (n) +a2ex 2 (n) ;y4(t)=ea 1 x 1 n +a 2 x 2 (n)=ea 1 x 1 n .ea 2 x 2 (n)NL
 y(t) = 5sinx(t); y1(t) = 5sinx1(t), y2(t) = 5sinx2(t) ;
y3(t) = a15sinx1(t) + a25sinx2(t) ; y4(t) = 5sin[a1x1(t) + a2x2(t)]  NL
 y(t) = 7x(t) + 5; y1(t) = 7x1(t) + 5,y2(t) = 7x2(t) + 5
y3(t) = 7x1(t) + 5 + 7x2(t) + 5 = 7(x1(t) + x2(t)) + 10; y4(t) = 7[x1(t) + x2(t)]+5  NL
 y(t) = n[x(n)]2 ; y1(t) = n[x1(n)]2, y2(t) = n[x2(n)]2 ; y3(t) = n[{x1(n)}2 + {x2(n)}2] <F[x1(n)] + F[x2(n)]>
y4(t) = F[x1(n) + x2(n)] = n[x1(n) + x2(n)]2 = n[[{x1(n)}2 + {x2(n)}2 + 2 x1(n) .x2(n)]  NL
FIR AND IIR SYSTEMS
 Linear time –invariant systems can be classified according to the type of impulse response.
 If the impulse response sequence is of finite duration, the system is called a Finite Impulse Response
[FIR] System. On the other hand, an Infinite Impulse Response [IIR] System has an impulse
response that is of infinite duration.
 An example of a FIR systems is described by
−𝟏, 𝒏 = 𝟏, 𝟐
𝟏, 𝒏 = −𝟏, 𝟑
h (n) =  An example of an IIR system is described by h (n) = n u (n).
𝟐, 𝒏 = 𝟎, 𝟒
𝟎, 𝑶𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆
STABLE AND UNSTABLE SYSTEMS.
 An LTI system is stable if it produces a bounded output sequence for every bounded input (BIBO).
 If for some bounded input sequence x(n), output is unbounded (infinite), the system is called Unstable.
 Let x (n) be a bounded input sequence, h (n) be the impulse response of the system and y (n) be the
output sequence. Taking the magnitude of the output, we have | y (n) | = ∞ 𝒌= − ∞ 𝒉 𝒌 𝒙 (𝒏 − 𝒌)
 We know that the magnitude of the sum term is less than or equal to the sum term of the magnitude,
hence, | y (n) | ≤ ∞𝒌= − ∞ 𝒉 𝒌 𝒙 (𝒏 − 𝒌)
 Let the bounded values of the input is equal to M, then the above equation can be written as
| y (n) | ≤ M ≤ ∞ 𝒌= − ∞ h (k)
 The above condition will be satisfied when ∞ 𝒌= − ∞ 𝒉 𝒌 ≤ ∞ ∞
𝒌= − ∞ 𝐡 𝐤 ≤ M ≤ ∞
 So the necessary and sufficient condition for stability is 
 EXAMPLES:- Test the Stability of the System whose impulse response is h(n) = (½)n u(n)
 SOLUTION: - The necessary and sufficient condition for stability is ∞ 𝒏= − ∞ 𝒉 𝒏 ≤ ∞
∞
 Given that, h(n) = (½)n u(n)  ∞ 𝑛= − ∞ ℎ 𝑛 = 0
𝑛= − ∞ (½) 𝑛
𝑢(𝑛) + 𝑛= 0 (½) 𝑛
𝑢(𝑛)
∞ 𝑛 2 3 1
𝑛= 0 (½) = 1+ ½ + ½ + ½ + …..∞ = 1− 1/2 = 2 < ∞ 1
As, 1+ a + a2+ a3+ …..∞ = 1− 𝑎
So, the given system is stable.
 EXAMPLES: - Test if the following systems are stable or not.
a) y(n) = x(n)u(n); If the input x(n) is bounded, i.e. │x(n)│ ≤ M then │y(n)│= │x(n)││u(n)│ ≤ M
That is │y(n)│ ≤ M. hence the system is Stable.
b) y(n) = sign[x(n)]; If the input x(n) is bounded, i.e. │x(n)│ ≤ M then │y(n)│=│sign[x(n)]│is also
bounded. Hence the system is Stable.

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c) y(n) = x(2n) ; If │x(n)│ ≤ M then│y(n)│= │x(2n)│ is also bounded. Hence the system is Stable.
d) y(n) = x(n) + nx(n+1); If the input│x(n) │≤ M then output │y(n)│= │x(n) + nx(n+1)│; The output
increases with increasing n. Hence the bounded input produce an unbounded output. Hence Unstable.
𝐧+𝐧𝟎 n+n 0 n+n 0
e) y(n)= 𝐤=𝐧−𝐧 𝟎
𝐱(𝐤); If│x(n)│≤ M then,│y(n)│= k=n−n 0
│x(k)│≤ k=n−n 0
M ≤(2n0+1)M Stable.
f) y(n) = ex(n); If │x(n) │≤ M then, │y(n)│= │ ex(n)│≤ e│[x(n)]│ ≤ eM. Hence the system is Stable.
g) y(n) = ax(n) + b; If │x(n)│≤ M then,│y(n)│= │ ax(n) + b│≤ a│x(n)│+│b│≤ a│M│+│b│Stable.
h) y(n) = x(-n); If │x(n) │≤ M then, │y(n)│=│x(-n)│ ≤ M; Hence the system is Stable.
i) y(n) = x(n) cos(ω0n); If│x(n)│≤ M then,│y(n)│=│x(n) cos(ω0n)│ ≤│x(n)││ cos(ω0n)│
If │ cos(ω0n)│ ≤ M y(n) is bounded and the system is Stable.
j) y(t) = sin[x(t+2)] = ∞ t=−∞ sin[x(t + 2)]; For every bounded value of x(t+2), it‟s sin function value is
always bounded means finite. So, the given system is Stable.
 For each response listed below, determine whether corresponding system is (i) Causal (ii) Stable.
a) h(n) = 𝟐𝐧 u(-n); h(n) ≠ 0 for n < 0. So the system is non-causal.
∞ ∞ n ∞ n ∞ n
For stability, n=−∞ │h(n)│< ∞ = n=−∞ │2 u(−n)│= n=−∞ 2 = n=0(1/2)
1
= 1 + ½ + 1/22 … ∞ = = 2 < ∞ ; So, the system is Stable.
1−1/2
b) h(n) = sin (nπ)/2; h(n) ≠ 0 for n < 0. So the system is non-causal.
For stability, ∞ ∞
n=−∞ │h(n)│< ∞ = n=−∞ │sin (nπ)/2│;
For odd values of n, │sin (nπ)/2│= 1; ∞ n=−∞ │h(n)│= ∞ ; The system is Unstable.
c) h(n) = δ(n) + sin nπ; For all integer values of n, sin nπ = 0. The system is Causal as h(n) = 0 for n < 0.
For the system to be stable it has to satisfy the conditioin, ∞
n=−∞ │h(n)│< ∞
∞
In this case, n=−∞ │h(n)│ =│h(-∞)│+ …+│h(-1)│+│h(0)│+│h(1)│+…+│h(∞)│
= 0+…+ 0 + 1 + 0 + …+0 ; Therefore, the system is Stable.
𝟐𝐧
d) h(n) = 𝐞 u(n-1); The system is causal as h(n) = 0 for n < 0. {u(n-1) = 0 for n < 1}
∞ ∞ ∞
n=−∞ │h(n)│< ∞ i.e.,
2n
For stability, n=−∞ │h(n)│ = n=−∞ │e u(n − 1)│
= ∞ 2 4 6
n=−∞ │e │= e + e + e +…∞ = ∞; Therefore, the system is Unstable.
2n

 Test the causality and stability of the following systems,

(i) y(n) = x(n) – x(-n-1) + x(n-1); The output depends on future values of input. Therefore non causal.
The impulse response is y(n) = δ(n+1) - δ(-n-1) + δ(n-1)
The impulse response is absolutely summable. Hence the system is Stable.
(ii) y(n) = sin[x(n)]; Causal ; if │x(n)│< M, then│y(n)│=│sin(x(n)│< M  The system is BIBO Stable.
 INTER CONNECTION OF DISCRETE -TIME SYSTEM.
 Discrete-time systems can be interconnected to form larger systems.
 There are two basic ways in which systems can be interconnected: - Series (Cascade) or Parallel.
These interconnections are illustrated in figure. Note that the two interconnected systems are different.
 In Cascade Interconnection, output of first system is y1 (n) = Ґ1 [x (n)]
And output of the first system is y (n) = Ґ2 [y1 (n)] = Ґ2 { Ґ1 [x (n)] }
 We observe that systems Ґ1 and Ґ2 can be combined into a single
overall system such as ҐC ≡ Ґ2 Ґ1
 we can express the output of the combined system as y (n) = ҐC [x (n)]
 In general, the order in which the operations Ґ1 and Ґ2 are performed is
important Ґ2 Ґ1 ≠ Ґ1 Ґ2 for an arbitray systems.
 However, if the systems Ґ1 and Ґ2 are linear and time invariant system,
then (a) ҐC is time invariant system and (b) Ґ2 Ґ1 = Ґ1 Ґ2 i.e, the order in
which the system processes is not important.
 In the Parallel Interconnection, the output of the systems Ґ1 is y1(n)
and the output of the systems Ґ2 is y2(n).
 Hence output of the parallel interconnection is y3 (n) = y1 (n) + y2 (n) =
Ґ1[x (n)] + Ґ2[x (n)] = (Ґ1 + Ґ2) [x (n)] = ҐP [x (n)] , Where ҐP = Ґ1 + Ґ2
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
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 In general, we can use parallel & cascade interconnection of system to construct large complex system.
 Conversely, we can take a larger system and break it down into smaller subsystems for purpose of
analysis and implementation.
 DISCRETE TIME LINEAR TIME-INVARIANT (LTI) SYSTEM.
 We had classified systems in accordance with a number of characteristic properties or categories,
namely: linearity, causality, stability, and time-invariance. Having done so.
 We now turn our attention to analysis of the important class of Linear, Time-Invariant (LTI) systems.
 In particular, we shall demonstrate that such systems are characterized in the time domain simply by
their response to a unit sample sequence. We shall also demonstrate that any arbitrary input signal can
be decomposed and represented as a weighted sum of unit sample sequences.
 As a consequence of the linearity and time-invariance properties of the system, the response of the
system to any arbitrary input signal can be expressed in terms of unit sample response of the system.
 The general form of the expression that relates the unit sample response of the system and the arbitrary
input signal to the output signal, called the convolution sum or convolution formula, is also derived.
 Thus we are able to determine the output of any LTI system to any arbitrary input signal.
 RESOLUTION OF A DISCRETE-TIME SIGNAL INTO IMPULSES: -
 Suppose we have an arbitrary signal x (n) that we wish to
resolve in to a sum of unit sample sequences.
 To utilize the notation established in the preceding section,
we select elementary signals xk (n) to be xk (n) = δ(n - k).
Where k represents the delay of the unit sample sequence.
 To handle an arbitrary signal x (n) that may have non zero
values over an infinite duration, the set of unit impulses
must also be infinite, to cover infinite number of delays.
 Now we will multiply two sequences x (n) and δ (n - k).
 Since δ (n - k) is zero everywhere except at n = k, where
its value is 1, the result of this multiplication is another
sequence that is zero everywhere except at (n - k), where
its value is x (k), as shown in Fig.
 Thus x (n) δ (n - k) = x (k) δ (n - k) is a sequence that is
zero everywhere except at n = k, where its value is x (k).
 If we will repeat multiplication of x (n) with δ (n-m),
where m is another delay (m ≠ k), the result will be a sequence that is zero everywhere except at n = m,
where its value is x (m). Hence x(n) δ(n - m) = x(m) δ (n — m)
 In other words, each multiplication of the signal x (n) by a unit impulse at some delay k, [i.e., δ (n-k)],
in results out the single value x (k) of the signal x (n) at the delay where the unit impulse is nonzero.
 Consequently , if we repeat this multiplication over all possible delays, - ∞ < k < ∞, and sum all the
product sequences, the result will be a sequence equal to the sequence x (n), that is,
∞
x (n) = 𝒌= − ∞ 𝐱 𝐤 𝛅 (𝐧 − 𝐤)

 We emphasize that the right-hand side of above equation is the summation of an infinite number of unit
sample sequences where the unit sample sequence δ (n - k) has an amplitude value of x (k).
 Thus it gives the resolution of or decomposition of any arbitrary signal x (n) into a weighted (scaled)
sum of shifted unit sample sequences.
 EXAMPLES:- Consisder a finite-duration sequence as x (n) = {2, 𝟒, 0, 3}. Resolve the sequences x(n)
into a sum of weighted impulse sequences.
 SOLUTION: - Since the sequece x (n) is nonzero for the time instants n = -1, 0, 2, we need three
impulses at delays k = -1, 0, 2. So by follwing the above equation, we find that
x (n) = 2δ (n + 1) + 4δ (n) + 3δ (n - 2) {0. δ (n - 1) = 0 is missing here}
 EXAMPLES:- Represent the sequence x (n) = {4, 2, -1, 𝟏, 3, 2, 1, 5} as sum of shifted unit impulse.
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
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 SOLUTION: - Since the sequece x (n) is nonzero for the time instants n = -3, -2, -1, 0, 1, 2, 3, 4. So by
follwing the equation we find that x (n) = ∞ 𝒌= − ∞ 𝒙 𝒌 𝜹 (𝒏 − 𝒌)
= x(-3)δ(n+3) + x(-2)δ(n+2) + x(-1)δ(n+1) + x(0)δ(n) + x(1)δ(n-1) + x(2)δ(n-2) + x(3)δ(n-3) + x(4)δ(n-4)
= 4δ (n+3) + 2δ (n+2) – δ (n+1) + δ (n) + 3δ (n-1) + 2δ (n-2) + δ (n-3) + 5δ (n-4)
 Response of LTI Systems to Arbitrary Inputs: The CONVOLUTION Sum
 A discrete-time system performs an operation on input signal based on predefined criteria to produce a
modified output signal. The input signal x (n) is the system excitation and y (n) is system response.
 If the input to the system is a unit impulse i.e. x (n) = δ (n) then the output of the system is known as
impulse response denoted by h (n) where h (n) = T [δ (n)].
 We know that any arbitrary sequence x (n) can be represented as a weighted sum of discrete impulses.
 Now the system response is given by y(n) = T [x(n)] = T [ ∞ 𝒌= − ∞ 𝒙 𝒌 𝜹 (𝒏 − 𝒌)]
∞
 For a linear system it can be reduced to y (n) = 𝒌= − ∞ 𝒙 𝒌 𝑻 [𝜹 (𝒏 − 𝒌)]
 The response to the shifted impulse sequence can be denoted by h(n, k) is defined as h(n,k)=T[δ(n-k)]
 For a time invariant system h(n, k) = h(n – k)
 Substituting this in the above equation, we obtain y (n) = ∞ 𝒌= − ∞ 𝒙 𝒌 𝒉 (𝒏 − 𝒌)
 For a linear time-invariant system if the input sequence x (n) and impulse response h(n) is given, we
can find the output y (n) by using the equation y (n) = ∞ 𝒌= − ∞ 𝒙 𝒌 𝒉 (𝒏 − 𝒌) .
 Which is known as Convolution Sum and can be represented as y (n) = x (n) * h (n), Where * denotes
the convolution operation. This is an extremely powerful result that allows us to compute the system
output for any input signal excitation.
 The convolution sum of two sequences can be found by using following steps.
Step-1 Choose an initial value of n, the starting time for evaluating the output sequence y (n). If x (n)
starts at n = n1 and h (n) starts at n = n2 then n = n1 + n2 is a good choice.
Step-2 Express both sequences in terms of the index k.
Step-3 Fold h (k) about k = 0 to obtain h (- k) and shift by n to the right if n is positive and left if n is
negative to obtain h (n – k).
Step-4 Multiply the two sequences x(k) & h(n-k) elements by element and sum the products to get y(n)
Step-5 Increment the index n, shift the sequence h (n-k) to right by one sample and do Step - 4.
Step-6 Repeat Step – 5 until the sum of products is Zero for all remaining values of n.
 PROPERTIES OF CONVOLUTION
(1) Commutative law : - x (n) * h (n) = h (n) * x (n)
(2) Associative Law : - [x (n) * h1(n)] * h2 (n) = x (n) *[ h1(n) * h2 (n)] = [x (n) * h1(n)] * h2 (n)
(3) Distributive Law : - x (n) * [h1(n) + h2 (n)] = [x (n) * h1(n) + x (n) * h2 ]
 EXAMPLES:- (A) Find the convolution sum of two sequences
X (n) = n+2 for 0 ≤ n ≤ 3 and h (n) = an u (n) for all n.
 We know that y(n) = ∞ 3
k=−∞ x k . h(n − k) = k=0 x k . h(n − k) (As, h(n) = 0 for n < 0)
= 3k=0 k + 2 . an−k u(n − k) = 2an u(n) + 3an-1 u(n-1) + 4an-2 u(n-2) + 5an-3 u(n-3)
(B) Find y(n) if x(n) = h(n) = {𝟏, 2, -1} [Total no. of samples M = 3, N = 3 M+N-1 = 3+3-1 = 5]
[No. of Samples in LHS & RHS are xl = hl = 0 yl = xl + hl = 0+0 = 0 & xr = yr =2 yr = xr + hr = 2+2=4]
 Now x(k) ={𝟏, 2, -1}, h(k) ={𝟏, 2, -1}  h(-k) ={-1, 2, 𝟏}
 h (1-k) ={-1, 𝟐, 1} h (2-k) ={-𝟏, 2, 1}
h (3-k) ={𝟎, -1, 2, 1} h (4-k) ={𝟎, 0, -1, 2, 1}
So, y(n) = {1.1, 2.1+1.2, -1.1+2.2+1.-1, 0.1+-1.2+2.-1+1.0, 0.1+0.2+-1.-1+2.0} = {𝟏, 4, 2, -4, 1}
 EXAMPLES:- Find the convolution sum of two sequences x (n) = {𝟑, 2, 1, 2} and h (n)= {1, 𝟐, 1, 2}
 SOLUTION: - < METHOD -1 > From the problem x (n) starts at n1 = 0 and h (n) starts at n2 = -1.
 Thus the starting value of n = n1 + n2 = 0 + (-1) = -1
 For n = -1 , y (-1) = ∞𝒌= − ∞ 𝒙 𝒌 𝒉 (−𝟏 − 𝒌), from Figure , we get y (-1) = 3.1 = 3
 Similarly, for n = 0, y (0) = ∞𝒌= − ∞ 𝒙 𝒌 𝒉 (−𝒌) = 3.2 + 2.1 = 8
∞
 For n = 1, y (1) = 𝒌= − ∞ 𝒙 𝒌 𝒉 (𝟏 − 𝒌) = 3.1 + 2.2 + 1.1 = 8
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
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 For n = 2, y (2) = ∞ 𝒌= − ∞ 𝒙 𝒌 𝒉 (𝟐 − 𝒌)
= 3.2 + 2.1 + 1.2 + 2.1 = 12
 For n = 3, y (3) = ∞ 𝒌= − ∞ 𝒙 𝒌 𝒉 (𝟑 − 𝒌)
= 3.0 + 2.2 + 1.1+ 2.2 = 9
 For n = 4, y (4) = ∞ 𝒌= − ∞ 𝒙 𝒌 𝒉 (𝟒 − 𝒌)
= 1.2 + 2.1 = 4
 For n = 5, y (5) = ∞ 𝒌= − ∞ 𝒙 𝒌 𝒉 (𝟓 − 𝒌)
= 2.2 = 4
 For n = 6, y (6) = ∞ 𝒌= − ∞ 𝒙 𝒌 𝒉 (𝟔 − 𝒌)
= 0.0 = 0
Hence y (n) = {3, 𝟖, 8, 12, 9, 4, 4}
 To check the correctness of the result, sum
all the samples in x (n) and multiply with
the sum of all the samples in h (n).
 This value must be equal to sum of all the
samples in y (n).
 In the given problem ∞ 𝒌= − ∞ 𝒙 𝒏 = 8,
∞ ∞
𝒌= − ∞ 𝒉 𝒏 = 6, 𝐤= − ∞ 𝒚 𝒏 = 48.
 As ∞ 𝒌= − ∞ 𝒙 𝒏 .
∞
𝒌= − ∞ 𝒉 𝒏 =
∞
𝒌= − ∞ 𝒚 𝒏 is proved. Thus result is correct. [8 x 6 = 48]
 SOLUTION: - < METHOD -2 >
 From the problem tabulate the sequence
x(n) and shifted version of h (n) as shown
in table below.
 Starting value of n = n1 + n2 = 0 + (-1) = -1
 The Staring value of n = -1
 For n = -1, y (-1) = 3.1 = 3
For n = 0, y (0) = 3.2 + 2.1 = 8
For n = 1, y (1) = 3.1 + 2.2 + 1.1 = 8 ; For
n = 2, y (2) = 3.2 + 2.1 + 1.2 + 2.1 = 12 ;
For n = 3, y (3) = 3.0 + 2.2 + 1.1 + 2.2 = 9 ; For n = 4, y (4) = 1.2 + 2.1 = 4 ;
For n = 5, y (5) = 2.2 = 4; For n = 6, y (6) = 0.0 = 0 ;  Hence y (n) = {3, 𝟖, 8, 12, 9, 4, 4}
 SOLUTION: - < METHOD -3 > Given that x (n) = {3, 2, 1, 2} and h (n) = {1, 𝟐, 1, 2}
 Write down the sequence x (n) and h (n) as shown in figure.
 Multiply each and every sample in h (n) with samples of x (n) & tabulate values.
 Divide the elements in table by drawing diagonal lines as shown in fig.
 Starting from the left sum all the elements in each strip and write down in the
same order.  3, 6 + 2, 3 + 4 + 1, 6 + 2 +2 + 2 , 4 + 1 + 4, 2 + 2, 4
 3, 8, 8, 12, 9, 4, 4
 The starting value of n = -1, mark symbol ↑ at the time of origin (n = 0).
 Hence y (n) = {3, 8, 8, 12, 9, 4, 4}
 EXAMPLES:- Find the convolution sum of two signals
𝟏, 𝒏 = −𝟐, 𝟎, 𝟏
h (n) = δ (n) - δ (n - 1) + δ (n - 2) - δ (n - 3) and x (n) = 𝟐, 𝒏 = −𝟏
, 𝑶𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆
 SOLUTION:-
 Given that x (n) = {1, 2, 𝟏, 1} and h (n) = { 𝟏, -1, 1, -1}
 The starting value of n = n1 + n2 = - 2 + 0 = - 2
 Now applying method - 3, we get
 y (n) = {1, - 1 + 2, 1 - 2 + 1, -1 + 2 - 1 + 1, -2 + 1 – 1, -1 + 1, -1}
= {1, 1, 0, 1, -2, 0, -1}  Hence y (n) = {1, 1, 𝟎, 1, -2, 0, -1}

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.20]
 By Applying General Convolution Method it can be Calculated as: -
 From Fig , n=-2 , y(-2)= ∞ k= − ∞ x k h (−2 − k) = 1.1 = 1
Similarly For n = -1 , y (-1) = ∞ k= − ∞ x k h (−1 − k)
= - 1.1 + 1.2 = -1 + 2 = 1
 For n = 0, y (0) = ∞ k= − ∞ x k h (−k)
= 1.1 + (-1.2) + 1.1 = 1 – 2 + 1 = 0
 For n = 1, y (1) = ∞ 𝒌= − ∞ 𝒙 𝒌 𝒉 (𝟏 − 𝒌) = -1.1 + 1.2 +
(-1.1) + 1.1 = - 1 + 2 -1 +1 = 1
 For n = 2, y (2) = ∞ 𝒌= − ∞ 𝒙 𝒌 𝒉 (𝟐 − 𝒌) = -1.2 + 1.1 +
(-1.1) = -2 +1 -1 = - 2
 For n=3, y(3)= ∞ 𝒌= − ∞ 𝒙 𝒌 𝒉(𝟑 − 𝒌) = -1.1 + 1.1 = 0
 For n = 4, y (4) = ∞ 𝒌= − ∞ 𝒙 𝒌 𝒉 (𝟒 − 𝒌) = -1.1 = -1

 y (n) = {1, 1, 𝟎, 1, -2, 0, -1}

 INTERCONNECTION OF LTI SYSTEM :-
 Parallel Connection of LTI Systems: -
 Consider two LTI systems with impulse responses h1 (n)
& h2 (n) connected in Parallel as shown in fig.
 From the figure the output of the system 1 is
y1(n) = x (n) * h1 (n) and The output of system 2
is y2 (n) = x (n) * h2 (n)
 Thus the output of the Total System is given by
y (n) = y1 (n) + y2 (n)
= [x (n) * h1 (n)] + [x (n) * h2 (n)]
= ∞ ∞
𝑘= − ∞ 𝑥 𝑘 ℎ1 (𝑛 − 𝑘) + 𝑘= − ∞ 𝑥 𝑘 ℎ2 (𝑛 − 𝑘)
∞
= 𝑘= − ∞ 𝑥 𝑘 [ ℎ1 (𝑛 − 𝑘)] + ℎ2 (𝑛 − 𝑘)]
= ∞ 𝑘= − ∞ 𝑥 𝑘 ℎ (𝑛 − 𝑘)  y (n) = x(n) * h (n) , Where h (n) = h1 (n) + h2 (n)
 Thus if the two systems are connected in parallel then overall impulse response is equal to Sum of two
impulse responses.
 Cascade Connection of LTI Systems: -
 Consider two LTI systems with impulse responses h1 (n) & h2 (n) connected in cascade as shown in fig.

 From the fig the output of system 1 is y1 (n) = x (n) * h1 (n) = ∞𝑘= − ∞ 𝑥 𝑘 ℎ1 (𝑛 − 𝑘) and
∞ ∞
The output of system 2 is y (n) = y1 (n) * h2 (n) = 𝑘= − ∞ 𝒌= − ∞ 𝒙 𝒌 𝒉𝟏 𝒏 − 𝒌 . ℎ2 (𝑛 − 𝑘) and
 By putting the expression for y1 (n), we can find out the expression for y (n) as x(n) * h (n).
 Where h (n) = ∞ 𝑘= − ∞ ℎ1 𝑛 ℎ2 (𝑛 − 𝑘) = h1 (n) * h2 (n)
 Thus the impulse response of two LTI systems connected cascade is the Convolution Product of the
individual impulse responses.
 Example: Obtain convolution sum of two discrete-time signals given:x(n) = 𝐞−𝐧 , h(n) = 3𝐧𝟐 for all n.
𝟐

 Solution: - x(k) = e−k , h(n-k) = 3(n − k) 2  y(n) = ∞𝐤=−∞ 𝐱 𝐤 . 𝐡(𝐧 − 𝐤) = ∞k=−∞ e−k . 3(n − k) 2
2 2

∞ −k 2 ∞ −k 2 ∞ 2 −k 2 ∞ −k 2
=3 k=−∞ e n 2 + k 2 − 2nk = 3n 2 k=−∞ e +3 k=−∞ k e - 6n3 k=−∞ ke
 Example: - Find convolution sum for x(n) = 𝟐 u(n), h(n) = 𝟏/𝟑 u(n)
𝐧 𝐧
[As y(n) = x(n)*h(n)]
n +1 −1 n +1 −1
∞ k n−k n ∞ k k n ∞ k n 6 n 6
y(n) = k=0 2 1/3 = 1/3 k=0 2 3 = 1/3 k=0 6 = 1/3 6−1
= 1/3 5
 Example: - Find convolution sum of the given signals below.
 x(n) = u(n-1), h(n) = 3nu(-n-1), ANS: 0.5(3)n for n < 0; 0.5 for n ≥ 0.
5[1− −1/2 n +1
 x(n) = u(n), h(n) = 5(1/2)nu(n), ANS: 1+1/2

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.21]
 FIND THE CONVOLUTION SUM OF THE BELOW SIGNAL.
Q1. x(n) = u(n), h(n) = 𝟐𝐧 u(n)  x(k) = u(k), h(n-k) = 2n−k u(n-k)  y(n) = ∞k=−∞ x k . h(n − k)
∞ n−k 0 n−k ∞ n−k
= k=−∞ u(k). 2 u(n − k) = k=−∞ u(k). 2 u(n − k) + k=0 u(k). 2 u(n − k)
= ∞ k=0 2
n−k
= ∞ n −k
k=0 2 . 2 = 2n ∞ −1 k
k=0(2 ) = 2
n −∞ k 1
k=0(1/2) = ½ + (1/2) +(1/2) +…
2

1− 1/2 n +1 1− 1/2 n +1
= 2[1- (1/2) n+1] = 2n . 2[1 - (1/2) n+1] = 2n+1 [1- (1/2) n+1] [By Formula, = 2−1 ]
1−1/2
2
= 2n+1 - (1/2) n+1. 2n+1 = 2n+1 - (1) n+1 = 2n+1 – 1 (ANS)
∞ ∞ 1 n−k
Q2.x(n)= 2nu(n), h(n)= (1/3)nu(n)  y(n) = k=−∞ x k . h(n − k) = k=−∞ 2 ku k . u(n − k)
3
0 1 n−k N 1 n−k N 1 n−k
= k=−∞ 2 k u(k) . u(n − k) + k=0 2 k u(k) . u(n − k) = k=0 2
k
. 3
3 3
N k 1 n 1 −k 1 n N k 1 −k 1 n N k k 1 n N k
= k=0 2 . = k=0 2 . = k=0 2 . 3 = k=0 6
3 3 3 3 3 3
1 n 6 n +1 −1 6 n +1 −1 𝟏 𝐧 𝟔 𝐧+𝟏 −𝟏
= (6 + 61 +62 +…+6n+1 ) = = = (ANS)
3 6−1 5 𝟑 𝟓
∞ ∞ −1 n−k
Q3.x(n) = u(n), h(n) = 5 (-1/2)nu(n)  y(n) = k=−∞ x k . h(n − k) = k=0 u n .5 u(n − k)
2
N −1 n −1 −k −1 n N k −1 n 1− −2 n +1
= k=0 5 2 2
= 5 2 k=0 −2 =5 2
− 1−(−2)
−1 n 1− −2 n +1 𝟓 −𝟏 𝐧 𝐧+𝟏
= − =𝟑 𝟏 − −𝟐 (ANS)
2 3 𝟐
 Systems with Finite-Duration and Infinite-Duration Impulse Response: -
 Up to this point we have characterized a LTI system in terms of its Impulse Response h (n).
 It is also subdivided the class of linear time-invariant systems into two types, those that have a Finite-
duration Impulse Response (FIR) and those that have an Infinite-duration Impulse Response (IIR).
 Thus an FIR system has an impulse response that is zero outside of some finite time interval.
 Without loss of generality, we focus our attention on causal FIR systems, so that h(n)= 0, n<0 & n>M.
 The convolution formula for such a system reduces to y (n) = 𝑴−𝟏 𝒌= 𝟎 𝒉 𝒌 𝒙 (𝒏 − 𝒌).
 A useful interpretation of this expression is obtained by observing that the output at any time n is
simply a weighted linear combination of the input signal samples x (n), x (n - 1)…… x (n - M + 1).
 In other words, the system simply weights, by the values of the impulse response h (k), k = 0, 1… M-1,
the most recent M signal samples and sums the resulting M products.
 In effect, the system acts as a window that views only the most recent M input signal samples in
forming the output. It neglects or simply “forgets” all prior input samples [i.e., x(n - M), x (n - M -1),..]
 Thus we say that an FIR system has a finite memory of length-M samples.
 In contrast, an IIR linear time-invariant system has an infinite-duration impulse response.
 Its output, based on the convolution formula, is y (n) = ∞ 𝐤= 𝟎 𝐡 𝐤 𝐱 (𝐧 − 𝐤).
 Where causality has been assumed, although this assumption is not necessary.
 Now the system output is a weighted [by the impulse response h(k)] linear combination of the input
signal samples x (n), x (n - 1), x (n - 2 )........
 Since this weighted sum involves the present and all the past input samples, we say that the system has
an infinite memory. We investigate characteristics of FIR & IIR systems in more detail in later chapter.
 Discrete-Time Systems Described By Difference Equations: -
 Up to this point we have treated linear and time-invariant systems that are characterized by their unit
sample response h (n). In turn h (n) allow s us to determine the output y (n) of the system for any given
input sequence x (n) by means of the convolution summation. y (n) = ∞ 𝐤= 𝟎 𝐡 𝐤 𝐱 (𝐧 − 𝐤).
 In general, then, we have shown that any linear time-invariant system is characterized by the input-
output relationship.
 Moreover, the convolution summation formula suggests a means for the realization of the system.
 In the case of FIR systems, such a realization involves additions, multiplications, and a finite number
of memory locations. Consequently, an FIR system is readily implemented directly, as implied by the
convolution summation.

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.22]
 If the system is IIR, however, its practical implementation as implied by convolution is clearly
impossible, since it requires an infinite number of memory locations, multiplications, and additions.
 A question that naturally arises, then, is whether or not it is possible to realize IIR systems other than in
the form suggested by the convolution summation.
 Fortunately, the answer is yes, there is a practical and computationally efficient means for
implementing a family of IIR systems, as will be demonstrated in this section.
 Within the general class of IIR systems, this family of discrete-time systems is more conveniently
described by difference equations.
 This family or subclass of IIR systems is very useful in a variety of practical applications, including the
implementation of digital filters, and the modeling of physical phenomena and physical systems.
 Recursive and Non-recursive Discrete-Time Systems: -
 As indicated above, the convolution summation formula expresses the output of linear time-invariant
system explicitly and only in terms of the input signal.
 However, this need not be the case, as is shown here.
 There are many systems where it is either necessary or
desirable to express the output of the system not only in terms
of the present and past values of the input, but also in terms of
the already available past output values.
 The following problem illustrates this point. Suppose that we wish to compute the cumulative average
𝟏 𝐧
of a signal x(n) in the interval 0 < k < n, defined as y (n) = 𝐱 (𝐤) , n =0, 1, 2, 3……
𝐧+𝟏 𝐤=𝟎
 As implied by above equation, the computation of y (n) requires the storage of all the input samples
x(k) for 0 < k < n. Since n is increasing, our memory requirements grow linearly with time.
 However, that y (n) can be computed more efficiently by utilizing the previous output value y (n - 1).
 Indeed, by a simple algebraic rearrangement of above, we obtain
(n + 1) y (n) = 𝐧𝐤=𝟎 𝐱 (𝐤) + x (n), n =0, 1, 2, 3……
𝐧 𝟏
= n y(n - 1) + x (n) And hence, y (n) = 𝐧+𝟏 𝐲 (𝐧 − 𝟏) + 𝐧+𝟏 𝐱 (𝐧)
 Now, the cumulative average y (n) can be computed recursively by multiplying previous output value
y(n - 1) by n / (n + 1), multiplying the present input x(n) by 1/(n + 1), and adding the two products.
 Thus the computation of y(n) by means of above equation requires two multiplications, one addition,
and one memory location, as illustrated in Fig. This is an example of a recursive system.
 Thus the output of a causal and practically realizable recursive system can be expressed in general as
y(n) = F[y(n-1), y(n-2),……..,y(n-N), x(n), x(n-1),………, x(n-M)]
 This is a recursive equation specifying a prouder for computing the system output in terms of previous
values of the output and present & past inputs.
 Similarly, if y (n) depends only on the present and past inputs, then is called as Non-recursive system.
 The non-recursive system can be expressed as y(n) = F[x(n), x(n-1),………, x(n-M)]
 Solution of Linear Constant-Coefficient Difference Equations
 Given a linear constant-coefficient difference equation as the input-output relationship describing a
linear time-invariant system, our objective is to determine an expression for y(n).
 The method that is developed is termed as the direct method.
 An alternative method based on the z-transform is described in ch-3. Basically, the goal is to determine
the output y(n), n > 0, of the system given a specific input x(n), n> 0, and a set of initial conditions.
 The direct solution method assumes that the total solution is sum of two parts: y(n) = yh(n) + yp(n).
 The part yh(n) is known as the Homogeneous or complementary solution,
 Whereas yp(n) is called the Particular solution.
 To find the homogeneous solution yh(n) of a difference equation we begin the problem of solving the
linear constant-coefficient difference equation given by 𝐍𝐤=𝟎 𝐚𝐤 𝐲(𝐧 − 𝐤) = 0.
 Then the most general solution to the homogeneous difference equation is given by
 yh(n) = C1λ1n + C2λ2n + C3λ3n +….+ CNλNn , where C1,C2.........CN are Weighting Coefficients.
 These coefficients are determined from the initial conditions specified for the system.
 Since the input x(n) = 0, so above equation can be used to obtain the zero-input response of the system.
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.23]
 EXAMPLE: - Determine the zero-input response of the system described by the homogeneous
second-order difference equation y(n) - 3y(n - 1) - 4y(n - 2) = 0 ------------ (1)
 Solution First we determine the solution to the homogeneous equation. Here x (n) = 0.
 We assume the solution to be the exponential yh(n) = λn
 Upon substitution of this solution into above expression, we obtain the characteristic equation
 λn - 3 λn-1- 4 λn-2 = 0  λn-2 (λ2- 3 λ- 4) = 0  λn-2 (λ2- 4 λ+ λ - 4) = 0  λn-2(λ- 4) (λ+1) = 0
 Thus λ−4=0→ λ=4
λ+1→ λ= −1
and the general form of the solution to the homogeneous equation is
yh(n) = C1 λ1 + C2 λ2n = C1 (-1)n + C2 (4)n --------------- (2)
n

 From the difference equation (1) ,

y (0) = 3 y(-1) + 4y(-2) & y (1) = 3y(0) + 4y(-1) = 3[3 y(-1) + 4y(-2)] +4y(-1) = 13 y(-1) +12y(-2)
 On the other hand, from the equation (2) we obtain y(0) = C1+ C2 and y(1) = - C1+ 4C2
 By equation above two sets of relation, we have C1+ C2 = 3 y(-1) + 4y(-2)
- C1+ 4C2 = 13 y(-1) +12y(-2)
 The solution of these two equations is C1= -1/5 y(-1) + 4/5 y(-2) & C2= 16/5 y(-1) + 16/5 y(-2)
 Therefore, the zero-input response of the system is given by
yh(n) = [-1/5 y(-1) + 4/5 y(-2)] (-1)n + [16/5 y(-1) + 16/5 y(-2)] (4)n
 If y(-2) = 0 and y(-1) = 5, then C1= -1 and C2 = 16. Hence yh(n) = (-1)n+1 + (4)n+2 n ≥ 0
 The Impulse Response of a Linear Time-Invariant Recursive System: -
 The impulse response of a linear time-invariant system was previously defined as the response of the
system to a unit sample excitation [i.e., x (n) = δ (n)].
 In the case of a recursive system, h (n) is simply equal to the zero -state response of the system when
the input x (n) = δ (n) and the system is initially relaxed.
 For example, in the simple first-order recursive system given above, the zero-state response is
yzs (n) = 𝐧𝐤=𝟎 𝐚𝐤 𝐱(𝐧 − 𝐤)
When, x (n) = δ (n) is substituted into, we obtain yzs (n) = 𝐧𝐤=𝟎 𝐚𝐤 𝛅(𝐧 − 𝐤) = an, n ≥ 0
 Hence the impulse response of the first order recursive system is, h(n) = an u(n)
 In general case of an arbitrary, linear time-invariant recursive system, the zero-state response expressed
in terms of the convolution summation is yzs (n) = 𝐧𝐤=𝟎 𝐡(𝐤)𝐱(𝐧 − 𝐤), n ≥ 0
 When the input is an impulse [i.e., x(n)=δ(n)], it is reduces to yzs (n) = h (n)
 Now, let us consider the problem of determining the impulse response h (n) given a linear constant-
coefficient difference equation description of the system.
 In the preceding subsection, we have established the fact that the total response of the system to any
excitation function consists of the sum of two solutions of the difference equation: the solution to the
homogeneous equation plus the particular solution to the excitation function.
 In case where excitation is an impulse, particular solution is 0, since x (n) = 0 for n > 0. i.e. yp (n) = 0
 Consequently, the response of the system to an impulse consists only of the solution to homogeneous
equation, with the {Ck} parameters evaluated to satisfy the initial conditions dictated by the impulse.
 Following example shows the procedure for obtaining h (n) given the difference equation for a system.
 EXAMPLE: - Find the natural response of the system described by difference equation :
y (n) + 2y (n - 1) + y (n - 2) = x (n) + x (n - 1) with initial condition y(-1) = y(-2) = 1
 SOLUTION: Given y (n) + 2y (n - 1) + y (n - 2) = x (n) + x (n - 1) ………………(i)
 The homogeneous equation can be obtained by equating the input terms to zero. That is
y (n) + 2y (n - 1) + y (n - 2) = 0
 The homogeneous solution is of the formyh(n) = λn ……………(ii)
n n-1 n-2
 Substituting equation (ii) in equation (i) we get λ + 2 λ + λ = 0
 λn-2 [λ2 + 2λ + 1] = 0  λ2 + 2λ + 1 = 0  (λ + 1)2 = 0  λ1 = -1, λ2 = -1 …(iii)
 The roots are repeated, therefore the general form of homogeneous solution is
yh(n) = C1 λ1n + C2 λ2n = C1(-1)n + C2n(-1)n ……………(iv)
 From equation (iv) , y(0) = C1, y(1) = - C1 - C2 ……………(v)
 From the homogeneous equation, y(0) + 2y(-1) + y(-2) = 0 ; Given y (-1) = 1; y (-2) = 1
Therefore y(0) + 2 + 1 = 0  y (0) = -3 ……………(vi a)
 Similarly y(1) + 2y(0) + y(-1) = 0  y(1) + 2(-3) + 1 = 0  y (1) = 5 ……………(vi b)
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <DISCRETE TIME SIGNALS & SYSTEMS > [Page 2.24]
 Comparing equation (v) and (vi), we get C1= y(0) = - 3, -C1 – C2 = y(1) = 5  C2 = - 2
 Putting values of C1 and C2 in equation (iv), we get
 The Natural response, Yn(n) = -3(-1)n – 2n(-1)n n ≥ 0 = -3 (-1)n u(n) – 2n (-1)n u(n)
 PRACTICE PROBLEM: - Find natural response of the system described by difference equation :
y (n) - 4y (n - 1) + 4y (n - 2) = x (n) - x (n - 1) when initial conditions are y (-1) = y (-2) = 1
 y(n) - 4y(n-1) + 4y(n-2) = 0 …(i)  λn - 4 λn-1+ 4 λn-2 = 0 …(ii)
λn-2 [λ2 - 4 λ+ 4] = 0  λ2 - 2 λ- 2 λ + 4 = 0  (λ- 2) (λ- 2) = 0 …(iii)  λ1 = 2, λ2 = 2
yh (n) = C1λ1𝑛 + C2λ𝑛2 = C1(2) n + C2(2) n ..… (iv)
 Putting value of n in equ (iv) n = 0, 1.  y(0) = C1 ; y(1) = 2C1 + 2C2
 From homogeneous equ is y(0) - 4y(-1) + 4y(-2) = y(0) - 4+4  y(0) = 0 = C1
y(1) - 0+4 = 0  y(1) = -4 2C1 + 2C2 = -4 ; Comparing this two, 2C2 = -4 C2 = -2
 Putting the value of C1 and C2 in equn (iv), we get, 0.(2) + (-2)n (2) = -2n (2)n 4(n) n ≥ 0 (ANS)
n

 EXAMPLE: - Determine the impulse response h(n) for the system described by the second-order
difference equation y (n) = 0.6y(n-1) – 0.08 y(n-2) + x(n)
 SOLUTION: Given that : y (n) = 0.6y(n-1) – 0.08 y(n-2) + x(n)
 We know the total response y (n) = yh(n) + yp(n) ……………… (i)
 For impulse x(n) = δ(n), the particular solution yp(n) = 0  y(n) = yh(n)
 Homogeneous solution can be found by substituting x(n) = 0  y(n)-0.6y(n-1)+0.08y(n-2) = 0 ……(ii)
 Let the solution yh(n) = λn …………(iii).
 Substituting equ(iii) in equ (ii),  λn - 0.6 λn-1 + 0.08 λn-2 = 0  λn-2 [λ2 - 0.6 λ + 0.08] = 0
 λ2 - 0.6 λ + 0.08. Thus the roots of the characteristic equation are λ1 = 0.4, λ2 = 0.2
 The general form of the solution of the homogeneous equation is given by
yh(n) = C1 λ1n + C2n λ2n = C1 (0.4)n + C2(0.2)n ………………(iv)
 From equn (iv), y(0) = C1 + C2 & y(1) = 0.4 C1 + 0.2 C2 ……………(v)
 From the difference equation we have y(0) = 0.6y(-1) – 0.08 y(-2) + x(0) = 1
y(-1) = y(-2) = 0, x(0)=δ(0)=1  y(1)= 0.6y(0) – 0.08 y(-1) + x(1)=0.6(1) – 0.08(0) + 0 =0.6…(v)
 Comparing equation (iv) & (v), C1 + C2 = 1, 0.4 C1 + 0.2 C2 = 0.6 After Solving we get, C1=2, C2=-1
 Substituting the values in equation (iv) yields y(n) = 2 (0.4)n u(n) - (0.2)n u(n)
 EXAMPLE: - Determine the impulse response h(n) for the system described by difference equation
y(n) + y(n-1) – 2y(n-2) = x(n-1) + 2x(n-2)
 SOLUTION: Given that: y(n) + y(n-1) – 2y(n-2) = x(n-1) + 2x(n-2)
 The homogeneous solution includes an impulse term. Total response is given by y(n) = yh(n) + yp(n).
 For input x(n) = δ(n), the particular solution yp(n) = 0  y(n) = yh(n)
 Homogeneous solution can be found by input terms = zero, that is y(n) + y(n-1) – 2y(n-2) = 0...(i)
 Let the solution yh(n) = λn. Substituting this solution in equation (i) we obtain the characteristic
equation λn + λn-1 - 2λn-2 = 0  λn-2 [λ2 + λ - 2] = 0  λ2 + λ – 2 = 0
 Therefore, the roots are 1, -2 and the general form of the solution to the homogeneous equation is
yh(n) = c1 (1)n + c2n (-2)n + Aδ(n) ………………(ii)
 From the difference equation y (0) + y(-1) – 2y(-2) = x(-1) + 2x(-2), y(0) = 0 <for n=0>
y(1) + y(0) – 2y(-1) = x(0) + 2x(-1), y(1) = 1  y(0) = 0, y(1) = 1, y(2) = 1 ………………(iii)
 Putting n=0, n=1 & n=2 in equ (ii), we get y(0) = C1 + C2 + A, y(1) = C1 – 2C2, y(2) = C1 – 4C2….(iv)
 From which C1 = 1, C2 = 0, A = -1.
 Substituting these values in equation (ii)  yield y(n) = u(n) - δ(n) = u(n-1)
 PRACTICE PROBLEM: - Determine the impulse response h(n) for the system described by the
n
second-order difference equation y(n) - 4y(n-1) + 4y(n-2) = x(n-1) [ANS: 2 (2)n]
 PRACTICE PROBLEM: - Determine the impulse response of a discrete-time LTI system whose
difference equation y(n) = y(n-1) + 0.5y(n-2) + x(n) + x(n-1) [ANS: 1.37(1.37)n - 0.37(0.37)n]
 PRACTICE PROBLEM: - Determine solution for the system described by the difference equation
y(n) = 5/6 y(n-1) -1/6 y(n-2) + x(n) for x(n) = 2nu(n) [A -(1/2)n u(n) + 2/5 (1/3)n u(n)+8/5 2nu(n)]
-------------- ALL THE BEST -------------------- ALL THE BEST ----------------
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <Z-TRANSFORM & ITS APPLICATION> [Page 3.1]

[CHAPTER-3]
---------------THE
Z-TRANSFORM & ITS APPLICATION TO LTI SYSTEM------------
 INTRODUCTION: -
 Transform techniques are an important
tool in the analysis of signals and
Linear Time-Invariant (LTI) systems.
 In this chapter we introduce the Z -
Transform, develop its properties, and
demonstrate its importance in the
analysis and characterization of linear
time-invariant systems.
 The Z - transform plays the same role
in the analysis of discrete-time signals and LTI systems as the Laplace transform does in the analysis of
continuous-time signals and LTI systems. [ Convolution (CTS) = Multiplication (Z-Transform) ]
 For example, we shall see that in the Z - domain (complex Z-plane) the convolution of two time-
domain signals is equivalent to multiplication of their corresponding z-transforms.
 This property greatly simplifies the analysis of the response of an LTI system to various signals.
 In addition, the Z -Transform provides us with a means of characterizing an LTI system, and its
response to various signals, by its pole-zero locations.
 THE DIRECT z-TRANSFORM: -
 In this section we introduce the Z-transform of a discrete-time signal, investigate its convergence
properties, and briefly discuss the Inverse Z-transform.
X(z) ≡ ∞𝒏=−∞ 𝒙(𝒏)𝒛
−𝒏
 The Z-Transform of a discrete-time signal x(n) is defined as the power series
 Where z is a complex variable. In polar form z can expressed as z = rejω. Where r = radius of the circle.
 So the above equation becomes X (z) = ∞ 𝒏=−∞ 𝒙(𝒏)(𝐫 𝐞 )
𝐣𝛚 −𝒏
= ∞ 𝒏=−∞{𝒙(𝒏)(𝐫
−𝒏
)}𝐞− 𝐣𝛚𝒏 .
 Hence from the above expression, it is clear that if r =1, then the z – transform evaluated on the unit
circle produces the discrete-time Fourier transform of the signal x (n).
 The relation is sometimes called the Direct Z - Transform because it transforms the time-domain
signal x (n) in to its complex-plane representation X (z).
 The inverse procedure [i.e., obtaining x (n) from X (z)] is called the Inverse Z – Transform.
 For convenience, the Z - transform of a signal x (n) is denoted by X (z) ≡ Z{x(n)} ; x(n) = Z-1{ X(z) }
 Whereas the relationship between x (n) and X (z) is indicated by x (n) Z  X(z)
 Region of Convergence (ROC): -
 The Region of Convergence (ROC) of X (z) is the set of all values of z in the z-plane for which the
magnitude of X (z) is finite. Thus any time we cite a z - transform we should also indicate its ROC.
 Since the z - transform is an infinite power series; it exists only for those values of z for which this
series converges. We illustrate these concepts by some simple examples.
 Properties of Region of Convergence (ROC) for Z-Transform: -
1) The ROC is a ring or disk in the z-plane centered at the origin. The ROC cannot contain any pole.
2) If x(n) is a causal sequence then ROC is the entire z-plane except at z = 0;
3) If x (n) is a non-causal sequence then the ROC is the entire z-plane except at z = ∞.
4) If x (n) is a finite duration, two sided sequence the ROC is entire z-plane except at z = 0 and z = ∞.
5) If x (n) is an infinite duration, two sided sequence the ROC will consists of a ring in the z-plane,
bounded on the interior and exterior by a pole, not containing any poles.
6) The ROC of a LTI stable system contains the unit circle. The ROC must be a connected region.
 TWO-SIDED Z-TRANSFORM: -
 The Z-Transform of a discrete-time signal x(n) may be expressed as
∞ −𝒏
Z {x (n)} = X (z) = 𝒏=−∞ 𝒙(𝒏)𝐳 -------------------- (3. 1)
 Where z is a complex variable. This expression is generally known as Two-sided z-Transform
 A signal that has finite duration on both the left and right hand sides is known as two sided sequences.
 For such type of sequence the ROC is entire z-plane except at z = 0 and z = ∞.
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <Z-TRANSFORM & ITS APPLICATION> [Page 3.2]
 EXAMPLE: -Find the z - transform and ROC of the signal x(n) = { 2, -1, 3, 2, 𝟏, 0, 2, 3, -1}
∞ −𝑛
ANS: - Given x (n) = { 2, -1, 3, 2, 𝟏, 0, 2, 3, -1}, we know that X (z) = 𝑛=−∞ 𝑥(𝑛)z
Thus X(z) = 2z4 – z3 + 3z2 + 2z + 1 + 2z-2 + 3z-3 - 3z-4. X(z) converges for all value except z = 0 & ∞.
 ONE-SIDED Z-TRANSFORM < RIGHT HAND SEQUENCE >: -
 If discrete time signal x(n) is causal signal i.e. x (n) = 0 for n < 0, then the z-transform is One-Sided z-
Transform (Right Hand Sequence) and is expressed as Z{x(n)}= X (z)= ∞ 𝒏 = 𝟎 𝒙(𝒏)𝐳
−𝒏
----- (3.2)
 For such type of sequence the ROC is entire z-plane except at z = 0.
 EXAMPLE: -Find the z – transform and ROC of the signal x(n) = { 𝟏, 0, 3, -1, 2}
ANS: - Expanding the equation (3.2), we get X+ (z) = x(0) + x(1) Z-1 + x(2) Z-2 + x(3) Z-3 + ………
Thus x (Z) = 1 + 3z-2 - z-3 + 2z-4. The X+ (z) converges for all value of z except z = 0.
 ONE-SIDED Z-TRANSFORM < LEFT HAND SEQUENCE >: -
 If a discrete time signal x(n) is non-causal signal i.e. x (n) = 0 for n > 0, then z-transform is also called
as One-sided z-Transform (Left Handed) & is expressed as Z{x(n)}=X(z)= 𝟎𝒏 =−∞ 𝒙(𝒏)𝐳 −𝒏 --- (3.3)
 EXAMPLE: -Find the z – transform and ROC of the signal x(n) = { -3, -2, -1, 0, 𝟏 }
ANS: - Expanding the equation (3.3), we get X - (z) = x(0) + x(-1) Z1 + x(-2) Z2 + x(-3) Z3 + ………
Thus x - (Z) = 1 - z2 - 2z3 - 3z4. The X - (z) converges for all value of z except z = ∞.
 EXAMPLE: - Determine the z - transforms of the following Finite Duration signals.
1) x1 (n) = {𝟏, 2, 5, 7, 0, 1} 4) x4 (n) = {2, 4, 𝟓, 7, 0, 1} 7) x7 (n) = δ (n + k), k > 0
2) x2 (n) = {1, 2, 𝟓, 7, 0, 1} 5) x5 (n) = δ (n) 8) x8 (n) = {𝟏, 4, 3, 2, 3, 4}
3) x3 (n) = {3, 𝟒, 2, 5, 7, 0,1} 6) x6 (n) = δ (n - k), k > 0 9) x9 (n) = {2, 3, 𝟏, 9, 9}

10) x10 (n) = {2n, for 0 ≤ n ≤ 5 and 0, otherwise} i.e x10 (n) = {𝟎, 2, 4, 6, 8, 10}
 SOLUTION: - From definition of Z - Transform, we have
1) X1(z) = 1 + 2z-1 + 5z-2 + 7z-3 + z-5, ROC: Entire z-plane except z = 0
2) X2(z) = z2 + 2z + 5 + 7z-1 + z-3, ROC: Entire z-plane except z = 0 and z = ∞
2 -1 -2 -3 -5
3) X3(z) = 3z + 4 + 2z + 5z + 7z + z ROC: Entire z-plane except z = 0 and z = ∞
4) X4(z) = 2z2 + 4z + 5 + 7z-1 + z-3, ROC: Entire z-plane except z = 0 and z = ∞
5) X5(z) = l [i.e. δ(n)  z 1], ROC: Entire z-plane
-k -k
6) X6(z) = z [i.e., δ(n - k)  z  z ], k > 0, ROC: Entire z-plane except z = 0
7) X7(z) = z k [i.e., δ(n + k)  z  z k], k > 0, ROC: Entire z-plane except z = ∞
-1 -2 -3 -4 -5
8) X8(z) = 1 + 4z + 3z + 2z + 3z +4z , ROC: Entire z-plane except z = 0
9) X9(z) = 2z2 + 3z + 1 + 9z-1 + 9z-2, ROC: Entire z-plane except z = 0 and z = ∞
-1 -2 -3 -4 -5
10) X10(z) = 2z + 4z + 6z + 8z +10z ROC: Entire z-plane except z = 0
Finite Duration Signals with their ROC Infinite Duration Signals with their ROC

 Z- TRANSFORM AND ROC OF INFINITE DURATION SEQUENCES: -

 EXAMPLE-1:- Determine Z– Transform of the signal x(n) = (½)n u(n)
 SOLUTION:- Given that x(n) = (½)n u(n)  X (z) = Z [x(n)]
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <Z-TRANSFORM & ITS APPLICATION> [Page 3.3]
∞ −𝑛
 Thus X (z) = 𝑛=−∞ 𝑥(𝑛) 𝑧 = ∞ 𝑛=−∞ [ ½
n
u n ] 𝑧 −𝑛
1 n
∞
= 𝑛=0 ½ 𝑧 = ∞
n −𝑛
𝑛=0 ½ 𝑧 −1 n
= ∞
𝑛=0 2z [As u (n) = 0, n<0]
𝟏 𝟐𝒛 1
 X(z) = 𝟏 = [As 1 + A + A2 + A3 + … = 1−𝐴 if |A| < 1]
𝟏− 𝟐𝒛−𝟏
𝟐𝐳
𝟏
 ROC: <𝟏 ½<Z
𝟐𝐳
 EX2: - Find z - Transform of x(n) = 𝒂𝒏 u(n) = 𝒂𝒏 , n ≥ 0 & 𝟎, n < 0}
 SOLUTION: - Given that x(n) = a n u(n)  x (z) = Z [x(n)] [Causal]
 Thus X (z) = ∞𝑛=−∞ 𝑥(𝑛) 𝑧
−𝑛
= ∞ n
𝑛=−∞ [a u n ] 𝑧
−𝑛
[As, u(n)=0, n<0]
∞ 1 𝒛
= 𝑛=0 a 𝑧 = 𝒏=𝟎 𝐚 𝒛
∞ n −𝑛 −𝟏 𝐧
= −1
=
1− a 𝑧 𝒛− 𝐚
 ROC: a 𝑧 −1 < 1  a/z < 1  a < z  ROC is exterior of a circle
𝒛
having radius |a|. x(n) = 𝜶𝒏 u(n)  z  X(z) = ROC: |z| > |α| ;
𝒛−𝜶
 The ROC is the exterior of a circle having radius |α|.
 Fig shows a graph of the signal x (n) and its
corresponding ROC.
 Here, in general, α need not be real.
 If we set α = 1, we obtain the z-transform of the unit
𝟏 𝒛
step signal x (n) = u(n)  z  X(z) = 𝟏−𝐳−𝟏 = 𝒛−𝟏 ROC: |z| > 1
 EXAMPLE-3: - Find the z-Transform and the ROC of the signal x(n) = - an u (- n - 1) [Anti Causal ]
 SOLUTION: - Given that x(n) = - an u (- n - 1) X(z) = Z [x(n)]
 Thus X (z) = ∞n=−∞ x(n) z
−n
= ∞ n
n=−∞ [−a u −n − 1 ] z
−n
[u(- n -1) = {…, 1, 1, 1, 𝟎, 0, 0, 0, …}]
= n=−∞ [−a u −n − 1 ] z + n=0[−a u −n − 1 ] z
−1 n −n ∞ n −n
[As, u (- n - 1) = 0, n ≥ 0; u(- n-1) = 1, n ≤ -1]
= 𝑛=−∞ [−a ] 𝑧 = − n=−∞ a z
−1 n −𝑛 −1 −1 −n

[Let n = -m Thus n  - ∞  m  ∞ and n  - 1  m  1 and taking

the limit from lower to higher i.e. from 1 to ∞]
∞ z m
 X (z) = − 𝐦=𝟏 𝐚−𝟏 𝐳 𝐦
=− ∞
m=1 a
z z 2 z 3 z 4
 =− + + + +⋯
a a a a
z z z 2 z 3 z 4 z 1 z a −z 𝐳
=− 1+ + + + +⋯ =- = -a x a−z = a−z = ROC: z < a
a a a a a a 1−z a 𝐳−𝐚
 The ROC is now the interior of a circle having radius b as shown in fig.
 EX-4:-Find z-Transform and the ROC of the signal x(n) = an u(n) + bn u(-n-1)
 SOLUTION: - Given that x(n) = an u(n) + bn u(-n - 1) [Two Sided Signal ]
 Thus X(z) = ∞ n −𝑛
𝑛=0 a 𝑧 + −1 n −𝑛
𝑛=−∞ b 𝑧 = ∞ n=0 az
−1 n
+ ∞ −1 n
n=1 b z
−1
 The First power series converses if az < 1 or z > a .
 The Second power series converses if b−1 z < 1 or z < b .
 In determining the convergence of X (z), we consider two different cases.
 CASE-1 |b| < |a |: In this case the above two ROC don’t overlap, as shown in
Fig. (a). Consequently, we cannot find values of z for which both power series
converge simultaneously. Clearly, in this case, X (z) does not exist.
CASE-2 |b| > |a |: In this case there is a ring in the z - plane where both power
series converge simultaneously, as shown in Fig.(b).
 That is for an infinite duration two sided signal the ROC is a ring in the z-plane.
𝐳 𝐳
 Now X (z) = 𝐳−𝐚 - 𝐳−𝐛 ROC: a < z < b .
PRACTICE PROBLEMS: - Find the Z-Transform of the following Signals:
1 1
 X(n) = (½)n u( n) Answer : - X(Z) = 1 ; ROC: z >
1− z −1 2
2
−1 1
 X(n) = (1/3) u(- n - 1)
n
Answer: - X(Z) = 1 ; ROC: z <
1− z −1 3
3

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <Z-TRANSFORM & ITS APPLICATION> [Page 3.4]
1 1 1 1
 X(n) = (1/2)n u( n) + (1/3)n u(- n - 1) Ans: - X(Z) = 1 - 1 ; ROC: < z <
1− z −1 1− z −1 2 3
2 3
 PROPERTIES OF THE Z-TRANSFORM : -
 The z-transform is a very powerful tool for the study of discrete-time signals and systems.
 The power of this transform is a consequence of some very important properties that the transform
possesses. In this section we examine some of these properties.
 In the treatment that follows, it should be remembered that when we combine several z-transforms, the
ROC of the overall transform is, at least, the intersection of the ROC of the individual transforms.
𝑧 𝑧 𝑧 𝑧
LINEARITY:- If x1(n) x1(z) ; x2(n) x2(z) ; x3(n) x3(z)  x3(n) x3(z) = a1x1(z) + a2x2(z)
Where z{x3(n)} = a1 x1(n) + a2 x2(n) for any constant a1 and a2.
 PROOF: - x3(n) = ∞ 𝑛=−∞ x3 n z
−n
= ∞ 𝑛=−∞ a1 x1 n + a 2 x2 n z
−n
= ∞ 𝑛=−∞ a1 x1 n z
−n

+ ∞𝑛=−∞ a 2 x2 (n)z
−n
= a1 ∞ −n ∞
𝑛=−∞ x1 n z +a 2 𝑛=−∞ x2 (n)z
−n
= a1 x1(z) + a2 x2(z) [Proved]
 EXAMPLE: - Find the z-Transform of the signal x (n) = [5(3) - 4(2) ] u(n).
n n

 SOLUTION:- We have X(z) = ∞ 𝐧=−∞ 𝐱(𝐧)𝐳

−𝐧
= ∞ n
n=−∞ 5(3) − 4 2
n
u n z −n
= ∞ n
n=0 5(3) − 4 2
n
z −n = ∞ n −n
n=0 5(3) z - ∞n=0 4 2 z =5 ∞
n −n −1 n
n=0(3z ) - 4 n=0 2z
∞ −1 n
−1
 The First power series converges when 3z < 1, i.e. for z > 3.
−1
 The Second power series converges when 2z < 1, i.e. for z > 2. [As, ROC : ROC1 ∩ ROC2]
5 4 5z 4z
 Hence x (z) converges for z > 3. Now X(z) = 1−3z −1 - 1−2z −1 ROC: z > 3  X(z) = z−3 - z−2
 Using Property: - We can determine the signal by putting x1(n) = 3nu(n) & x2(n) = 2nu(n),
 Then x (n) can be written as x(n) = 5x1(n) - 4x2(n).
 According to linearity property, the z-transform of the signal x(n) is given by X(Z) = 5x1(Z) - 4x2(Z).
z z
 But we know that Z {an u(n)}= z−a  Z{ x1(n)}= Z{3n u(n)}= z−3 ; ROC : |z| > 3
z
 Similarly Z{ x2(n)}= Z{2nu(n)}= z−2 ; ROC : |z| > 2
5z 4z
 Now the overall z-transform is given by x(Z) = - . ROC: ROC1 ∩ ROC2 i.e. |z| > 3
z−3 z−2
1 1
TIME REVERSAL: - If x(n)  z  X (z); ROC: r1 < |z| < r2  x(-n)  z  X(z-1) ; < |z| < 𝑟
𝑟2 1
 PROOF: Z{x(-n)} = ∞ 𝑛=−∞ 𝑥 −𝑛 𝑧
−𝑛
= ∞ 𝑛=−∞ 𝑥 𝑙 𝑧 =
𝑙 ∞
𝑛=−∞ 𝑥 𝑙 𝑧
−1 −𝑙
= X (z-1) [Put = l = -n]
SCALING IN THE Z-DOMAIN:- If x(n)  z  X(z); ROC: r1 < |z| < r2  anx (n)  z  X(a-1z)
 The ROC: |a| r1 < |z| < |a| r2 for any constant a, real or complex.
 PROOF: - Z{ anx (n)} = ∞ n
𝒏=−∞ { a x (n)} 𝑧
−𝑛
= ∞ 𝒏=−∞ x n a 𝑧
n −𝑛

= 𝒏=−∞ x n (a𝑧 ) = 𝒏=−∞ x n (𝑎 z) = X(𝑎 z)  Z{ anx (n)} = X(𝒂−𝟏 𝐳) (Proved)

∞ −1 𝑛 ∞ −1 −𝑛 −1

 Since the ROC of X(z) is r1 < |z| < r2  r1 < |𝑎−1 z | < r2  ROC of X(𝑎−1 z) is |a| r1 < |z| < |a| r2
TIME SHIFTING:- If x(n)  z  X(z) then, x(n - m)  z  z – m X (z)
 The ROC of z - m X (z) is the same as that of X (z) except for z = 0 if m > 0 and z = ∞ if m < 0.
 PROOF: - We have X(z) = ∞ 𝐧=−∞ 𝐱(𝐧)𝐳
−𝐧
 z {x(n-m)} = ∞ n=−∞ x(n − m)z
−n

= ∞ n=−∞ x(n − m)z

−n+ m−m
= ∞ n=−∞ x(n − m)z
−(n− m)−m
= ∞ n=−∞ x n − m z
− n−m
. z −m
= z −m ∞ n=−∞ x n − m z
− n−m
= 𝒛−𝒎 ∞ 𝒍=−∞ 𝒙 𝒍 𝒛
−𝒍
= z – m X (z) [Proved]
[Where n – m = l Thus n  - ∞  l  - ∞ and n  ∞  l  ∞] Thus z {x(n-m)}= Z {x(n-m)} = z – m X (z)]
 The properties of linearity and time shifting are the key features that make the z-transform extremely
useful for the analysis of discrete-time LTI systems. [Similarly z {x (n + m)}= Z {x(n+m)} = z m X (z)]
 EX: -If x1(n) = {1 , 4, 3, 2, 3, 4} & x2(n) = {1, 2, 5 , 7, 0, 1}Find the z-transform of x1(n-2) & x2(n+3).
 SOLUTION:- Applying Shifting property, we can write if Z{x(n)} = X(z)  Z{x(n - m)} = z – m X (z)
 But Z{x1(n)} = X1(z) = 𝟓𝒏=𝟎 𝒙 𝒏 𝒛−𝒏 = 1 + 4z-1 + 3z-2 + 2z-3 + 3z-4+ 4z-5,
 Thus Z{x1(n-2)} = z-2 {1 + 4z-1 + 3z-2 + 2z-3 + 3z-4+4z-5}= z-2 + 4z-3 + 3z-4 + 2z-5 + 3z-6+4z-7
 Similarly Z{x2(n + 3)} = z3 { z2 + 2z + 5 + 7z-1 + z-3} = z5 + 2z4 + 5 z3 + 7z2 + 1}
 EX: -Find the Z-Transform of x(n) = 2n u(n-2) by using properties of z-Transform.
1 z z −2 z −2 z z −1
 SOLUTION:- We know Z{u(n)} = =  By Time Shifting Z{u (n-2)}= or =
1−z −1 z−1 1−z −1 z−1 z−1
n (2−1 z)−2 4z −2 (2−1 z)−1 2 z −1 𝟒 𝐳 −𝟏
 Now applying Scaling property Z{2 u (n-2)}= = 1−2z −1 or = (z−2)/2 =
1−(2−1 z)−1 (2−1 z)−1 𝐳−𝟐

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <Z-TRANSFORM & ITS APPLICATION> [Page 3.5]
𝟏 𝐧
 EXAMPLE:- Find the z-Transform of the sequence x(n) = 𝐮(𝐧 − 𝟏)
𝟑
1 n 1 n z
 SOLUTION: - We know the z-Transform sequence u(n − 1) = z u(n) = 1
3 3 z−
3
1 n z 𝟏 𝟑
 Using time Shifting property Z[x(n-1)] = z −1 X(z) z u(n − 1) = z −1 1 = 𝟏 = 𝟑𝐳−𝟏
3 z− 𝐳−
3 𝟑
 EXAMPLE: - Find the Z-Transform for the sequence x(n) = u(-n)
1 𝟏
 SOLUTION: -We know Z{u(n)}= 1−z −1 ; ROC:|z| >1 By reversal property Z{u(-n)}= 𝟏−𝐳; ROC:|z|<1
Sin n+1 ω
 EXAMPLE: - Find the Z-Transform for the sequence x(n) = r n u(n); 0<r<1
Sin ω
n Sin n+1 ω n Sin n+1 ω
 SOLUTION: - X(z) = ∞
𝐧=−∞ 𝐱(𝐧)𝐳 −𝐧 = ∞
n=−∞ r u(n)z = ∞ n=0 r
−n
z −n
Sin ω Sin ω
∞ rn ∞ −1 jω n e
jω
∞ −1 −jω n e
−jω
= n=0 2j Sin ω ej(n+1)ω − e−j(n+1)ω z −n = n=0 rz e - n=0 rz e
2j Sin ω 2j Sin ω
∞ 1
 Both series converge if the ROC is rz −1 e−jω < 1, or z > r [As n=0 a =n
]
1−a
1 e jω e −jω 1 1 𝐳𝟐
X(z) = 2j Sin ω − 1−rz −1 e −jω = = 1−2rz −1 = 𝐳𝟐 −𝟐𝐫 𝐂𝐨𝐬 𝛚 𝐳+𝐫 𝟐
1−rz −1 e jω 1−rz −1 e jω 1−rz −1 e −jω Cos ω +r 2 z −2

 Find its Z – Transform and ROC of Cos 𝐰𝟎 𝐧 u(n).

 SOLUTION: - z{x(n)} = ∞ 𝑛=−∞ x(n)z
−n
= z{Cos w0 n u(n)} = ∞
𝑛=−∞ Cos w0 n u(n)z
−n
jw n
e 0 +e −jw 0 n
∞
= 0
𝑛=−∞ Cos w0 n u(n)z −n + 𝑛=0 Cos w0 n u(n)z −n = ∞𝑛=0 2
z −n
1 ∞ 1
= x 𝑛=0 e
jw 0 n
+ e−jw 0 n z −n = 2 x ∞ 𝑛=0 e
jw 0 n −n
z + ∞
𝑛=0 e
−jw 0 n −n
z
2
1 ∞ jw 0 −1 n ∞ −jw 0 −1 n 1 1 1
=2x 𝑛=0(e z ) + 𝑛=0(e z ) =2x 1−e jw 0 z −1
+ 1−e −jw 0 z −1
1 1−e −jw 0 z −1 +1−e jw 0 z −1 1 2−e −jw 0 z −1 −e jw 0 z −1
=2x =2x
1−e jw 0 z −1 1−e −jw 0 z −1 1−e −jw 0 z −1 −e jw 0 z −1 +e jw 0 z −1 e −jw 0 z −1
1 2−z −1 (e −jw 0 +e jw 0 ) 1 2−2z −1 Cos w 0 1 2(1−z −1 Cos w 0 )
=2x =2x =2x
1−z −1 e −jw 0 +e jw 0 +z −2 1−2z −1 Cos w0 +z −2 1−2z −1 Cos w 0 +z −2
𝟏−𝐂𝐨𝐬 𝐰𝟎 𝐳 −𝟏
= 𝟏−𝟐𝐂𝐨𝐬 𝐰 −𝟏 +𝐳 −𝟐
(Ans) ; ROC: Z > ejw 0  Z > 1, Z > e−jw 0  Z > 1
𝟎𝐳
 Find its Z – Transform and ROC of 𝐚𝐧 Cos 𝐰𝟎 𝐧 u(n)
1−Cos w 0 z −1 1−Cos w 0 (a −1 z)−1
 z{Cosw0 n u(n)}= 1−2Cos w  z{an Cos w0 n u(n)} = 1−2Cos w [By scaling property]
0 z −1 +z −2 0 az
−1 +a 2 z −2

 Find its Z – Transform and ROC of Sin 𝐰𝟎 𝐧 u(n).

 SOLUTION: - z{x(n)} = ∞ 𝑛=−∞ x(n)z
−n
 z{Sin w0 n u(n)} = ∞
𝑛=−∞ Sin w0 n u(n)z
−n

e jw 0 n
−e −jw 0 n
= 0
𝑛=−∞ Sin w0 n u(n)z
−n
+ ∞ 𝑛=0 Sin w0 n u(n)z
−n
= ∞ 𝑛=0 z −n
2j
−1 ∞ 1
= 2j 𝑛=0 ejw 0 n − e−jw 0 n z −n = 2j ∞ 𝑛=0 e z − ∞
jw 0 n −n
𝑛=0 e
−jw 0 n −n
z
1 ∞ 1 1 1 1 (1−e −jw 0 z −1 )−(1−e jw 0 z −1 )
= 𝑛=0(e z ) − ∞
jw 0 n −1 −n
𝑛 =0(e
−jw 0 n −1 −n
z ) = jw − −jw =
2j 2j 1−e 0 z −1 1−e 0z −1 2j 1−e jw 0 z −1 1−e −jw 0 z −1
1 1−e −jw −1 jw
0 z −1+e 0 z −1 1 jw
(e 0 −e −jw 0 )z −1
= 2j = 2j
1−e −jw 0 z −1 −e jw 0 z −1 +e jw 0 z −1 e −jw 0 z −1 1−z −1 e −jw 0 +e jw 0 + e jw 0 e −jw 0 z −2
1 2j Sin w 0 n z −1 𝐒𝐢𝐧 𝐰𝟎 𝐧 𝐳 −𝟏
= 2j = 𝟏−𝟐𝐂𝐨𝐬 𝐰 ROC: Z > ejw 0 = Z > 1 ; Z > e−jw 0 = Z>1 (ANS)
1−2Cos w 0 z −1 +z −2 𝟎 𝐳 −𝟏 +𝐳 −𝟐
 Find its Z – Transform and ROC of 𝐚𝐧 Sin 𝐰𝟎 𝐧 u(n).
Sin w z −1
 SOLUTION: - z{Sin w0 n u(n)} = 1−2Cos w 0z −1 +z −2  z{an Sin w0 n u(n)} (Using Scaling Property)
0
Sin w 0 (a −1 z)−1 𝐒𝐢𝐧 𝐰𝟎 𝐚𝐳 −𝟏
= 1−2Cos w −1 −1 −1 −2
= (ANS)
0 (a z) +(a z) 𝟏−𝟐𝐂𝐨𝐬 𝐰𝟎 𝐚𝐳 −𝟏 +𝐚𝟐 𝐳 −𝟐

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <Z-TRANSFORM & ITS APPLICATION> [Page 3.6]
𝑧 𝑧 d
DIFFERENTIATION: If x(n) x(z) then n x(n) -z dz x(z)
 PROOF: z{x(n)} = ∞ 𝑛=−∞ x(n)z
−n
(Taking differentiation both side)
d d d d −n
 dz z{x(n)} = dz 𝑛=−∞ x(n)z  dz z{x(n)} = ∞
∞ −n
𝑛=−∞ x(n) dz z = ∞ 𝑛 =−∞ x n −n z
−n−1

= z −1 ∞ 𝑛=−∞ x n −n z
−n
(multiplying –z both side)
d 1 𝐝
 - z dz z{x(n)} = - z x z ∞𝑛=−∞ x n z
−n
−n = ∞ 𝑛=−∞ n x n z
−n
- z𝐝𝐳 (z) = z{n x(n)} [Proved]
 EXAMPLE: - Find the z-transform of the sequence x (n) = 𝐧 𝐚𝐧 𝐮(𝐧).
1
 SOLUTION: - We know z an u(n) = 1−az −1 z > a.
d d 1 az −1
 By differentiation property, z n x(n) = -zdz X(z)  z n an u(n) = -z dz = : z > a
1−az −1 1−az −1 2
az −1 az
 Thus, z n an u(n) = = ; ROC: z > a
1−az −1 2 z −a 2
𝟏 𝐧 𝟏 −𝐧
 EXAMPLE: - Find the z-transform of the signal x (n) = − 𝟓 u (n) + 5 u (-n-1) also find ROC.
𝟐
1 n 1 −n
 SOLUTION:- We have X(z) = ∞
n=−∞ x(n)z
−n
= ∞
n=−∞ −5 u(n) + 5 u(−n − 1) z −n
2
∞ 1 n ∞ 1 n ∞
= n=0 −5 z −n + 5 −1 n −n
n=−∞ 2 z = n=0 −5 z −n + 5 n=1 2−1 z n

1 −1 1
 First series converges for − 5 z < 1, i.e. z > 5. Second series converges for z/2 < 1, i.e. z > 2.
1 1 5 1
 Therefore, the ROC is 5 < z < 2 . Thus, X(z) = 1+0.2z −1 - 1−2z −1 ROC: 5 < z < 2
𝟏 𝟏
 EXAMPLE: - Find the z-transform of the signal x(n) = 𝟐 𝛅(𝐧) + 𝛅(𝐧 − 𝟏) - 𝟑 𝛅(𝐧 − 𝟐) & find ROC.
𝟏 𝟏
 SOLUTION: - Referring the table we find X(z) = 𝟐 + 𝐳 −𝟏- 𝟑 𝐳 −𝟐 ; ROC: All values of z except z = 0.
 EXAMPLE: - Find the z-transform of the signal x(n) = u(n-2) also find ROC.
z
 SOLUTION: - We know z[u(n)] = z−1. Using Time shifting property we have Z[x(n-m)] = z −m X(z).
𝐳 𝐳 −𝟏
 Similarly Z[u(n-2)] = 𝐳 −𝟐 𝐳−𝟏 = 𝐳−𝟏 ROC: z > 1
𝟏 𝐧
 EX: - Find z-transform of x(n) = 𝐧 + 𝟎. 𝟓 𝐮(𝐧) & ROC
𝟑
1 n
 SOLUTION: - Given that x(n) = n + 0.5 u(n)
3
1 n 1 n
= n u(n) + 0.5 u(n)
3 3
𝟏 𝐧 𝟏
 From the table we find 𝒁 𝐮(𝐧) = 𝟏 .
𝟑 𝟏− 𝐳 −𝟏
𝟑
𝐝
 Using Differential property, Z[nx(n)] = −𝐳 𝐝𝐳 𝐗(𝐳)
1 −1
1 n d 1 z
 =𝑍 n u(n) = −z dz 1 = 3
1 −1 2
3 1− z −1 1− z
3 3
1 −1
z 0.5 1
3
 X(z) = 1 −1 2
+ 1 ; ROC: z > 3
1− z 1− z −1
3 3

 PRACTICE PROBLEM:- Find the z-transform of the following sequence and find ROC in each case.
1 (0.3)4
(i) x(n) = (0.4)n u(n) + (0.3)n u(n-4) ANS: 1−0.4z −1 + 1−0.3z −1 ROC: z > 0.4
−z −2 [1+az −1 ]
(ii) x(n) = n2 an u(n) ANS: ROC: z >a
(1−az −1 )3
 EX: - Find inverse z-transform of X(z) = 𝐥𝐨𝐠 𝟏 − 𝟎. 𝟓𝐳 −𝟏
; z > 0.5 using Differential Property.
d 0.5z −2
 SOLUTION: - Given X(z) = log 1 − 0.5z −1 . Differentiating on both sides we get dz X(z) = 1−0.5z −1
d −0.5z −1
 Multiplying both sides by –z obtain −z dz X(z) = 1−0.5z −1
1 z −1
As 0.5 n u n = and Z 0.5 n−1
u n−1 =
1 − 0.5z −1 1 − 0.5z −1

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <Z-TRANSFORM & ITS APPLICATION> [Page 3.7]
d −0.5z −1 n−1
 From the differentiation property, we get Z[nx(n)] = −z dz X(z) = 1−0.5z −1 = −0.5Z 0.5 u n−1
𝟎.𝟓 𝐧
 Therefore, n x(n) = −0.5 0.5 n−1 u n − 1  x(n) = − 𝐧 𝐮(𝐧 − 𝟏)
 PRACTICE PROBLEM: - Find Inverse z-Transform of
1 1 a 1 1
(A) X(z) = log 1−az −1 ; z > a ; ANS: n an u(n-1) & (B) X(z) = log a−z −1 ; z > ; ANS: a-n u(n-1)
a n
1 d d 1 d 1 d 1
 SOLUTION (A): - x(z) = log 1−az −1  dz x(z) = dz log 1−az −1  dz x(z) = 1
dz 1−az −1
1−a z −1
d d d d
1−az −1 x x 1− (1−az −1 .1) 0− (−az −1 ) a z −1 −1 a z −2
= (1 − az −1 ) = dz dz
= dz
= dz
=
(1−az −1 )2 1−az −1 1−az −1 1−az −1
d −a z −2+1 a 𝐳 −𝟏 a z −1 𝐚𝐧
 -zdz x(z) = = 𝟏−𝐚𝐳−𝟏  n x(n) = 1−az −1 = a . an-1 u(n-1)  x(n) = u(n-1) (ANS)
1−az −1 𝐧
𝟐𝐧 𝐟𝐨𝐫 𝐧 < 0
𝟏 𝐧
 EXAMPLE: - Find Z-transform of x(n) = 𝐟𝐨𝐫 𝐧 = 𝟎, 𝟐, 𝟒
𝟐
𝟏 𝐧
𝐟𝐨𝐫 𝐧 = 𝟏, 𝟑, 𝟓
𝟑
1 n 1 n
 SOLUTION : - x(z) = −∞ n=−∞ x(n)z
−n
= −1n=−∞ 2 z
n −n
+ ∞ n=0 2 z −n [Even] + ∞n=0 3 z −n [Odd]
 For First part of above equation, Let n = -m, if n = ∞, m = -∞, then n = 1, m = -1. Putting these values,
z z z m z 1 𝒛
∞
1
𝑚 =∞ 2−m z m = −1 m
𝑚 =1(2 z) =2 1+ + +⋯ =2 z = 𝟏−𝒛𝟐
2 2 1− 𝟐
2
∞ 1 2p 1 2p
 For the Second part, p=0 2 z −2p as even so, 2
p
1 2 −2 1 −2 p 1 𝐳𝟐
= ∞ p=0 2 z = ∞
p=0 4 z = 1 −2  (Multiplying by z 2 ) = 𝟐 𝟏
1− z 𝐳 − 𝟒
4
1 2q+1 1 2q 1 1 1 1 −1 1 2q −2q
 For the Third part, ∞ q=0 3 z −(2q+1) = ∞ q=0 3 z −2q −1
z = z ∞
q=0 z
3 3 3
q z −1 𝐳
z −1 ∞ 1 2 −2 z −1 ∞ 1 −2 q z −1 1
= 3 q=0 3 z = 3 q=0 9 z = 3 1 −2 (Multiplying z ) = z 1−1 3z 2 = 𝟐 𝟏𝟑
2 2
1− z 9 𝐳 − 𝟗
9
𝒛 𝐳 𝟐 𝐳
 Combining all part, we get x(z) = 𝟏−𝒛𝟐 + 𝟐 𝟏 + 𝟐 𝟏𝟑 (ans)
𝟐 𝐳 − 𝟒 𝐳 − 𝟗
CONVOLUTION OF TWO SEQUENCES: -
 If x1(n)  z  X1(z) & x2(n)  z  X2(z)  x(n) = x1(n) * x2(n)  z  X(z) = X1(z) X2(z)
 The ROC of X (z) is, at least, the intersection of that for X1(z) and X2(z).
 PROOF: The convolution of x1(n) & x2(n) is x(n) = x1(n) * x2(n) = ∞ 𝑘=−∞ 𝑥1 𝑘 𝑥2 (𝑛 − 𝑘)
∞
 The z-transform of x(n) is X(z) = ∞ 𝒏=−∞ 𝒙 𝒏 𝒛 −𝒏
= ∞
𝑛=−∞ 𝒌=−∞ 𝒙𝟏 𝒌 𝒙𝟐 (𝒏 − 𝒌) 𝑧
−𝑛

 Upon interchanging the order of the summations and applying the time-shifting property, we obtain,
X(z) = ∞ 𝑘=−∞ 𝑥1 𝑘
∞
𝑛=−∞ 𝑥2 𝑛 − 𝑘 𝑧
−𝑛
= ∞ 𝑘=−∞ 𝑥1 𝑘 . 𝒛
−𝒌
X2(z) [As Z{x(n-k)}=Z-k x(Z)]
= X2(z). ∞ 𝑛=−∞ 𝑥1 𝑘 𝑧
−𝑘
= X2(z). Z{x1(k)} = X2(z). X1(z)  Z{x(n) * h(n)} = X(z). H(z)
1, 0 ≤ 𝑛 ≤ 5
 EXAMPLE: - Find the convolution sum x(n) of the signals x1(n) = {1, -2, 1} & x2(n)=
0, 𝑂𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
 SOLUTION: - By applying Z-transform to x1(n), we have X1(Z) = 1 - 2z-1 + z-2
 Similarly Z{x2(n)}= X2(Z) = 1 + z-1 + z-2 + z-3 + z-4 + z-5
 According to convolution property of z-transform, X(Z) = X1(z) . X2(z) = 1- z-1 – z-6 + z-7
 Hence x (n) = {1, -1, 0, 0, 0, 0, -1, 1}
 The same result can also be obtained by using the convolution sum formula in time domain approach.
CORRELATION OF TWO SEQUENCES:-
 If x1(n)  z  X1(z) & x2(n)  z  X2(z)
∞
 Then 𝑟𝑥 1 𝑥 2 (𝑙) = 𝑛=−∞ 𝑥1 𝑛 𝑥2 (𝑛 − 𝑙)  z  𝑅𝑥 1 𝑥 2 (𝑧) = X1(z) X2(z-1)
 PROOF: We recall that 𝑟𝑥 1 𝑥 2 (𝑙) = x1(𝑙) * x2(-𝑙). Using the convolution and time-reversal properties,
we easily obtain, 𝑅𝑥 1 𝑥 2 (𝑧) = Z{x1(𝑙)} Z{x2(-𝑙)} = X1(z) X2(z-1)
 The ROC of 𝑅𝑥 1 𝑥 2 (𝑧) is at least the intersection of that for X1(z) and X2(z-1).

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <Z-TRANSFORM & ITS APPLICATION> [Page 3.8]
 As in the case of convolution, the cross-correlation of two signals is more easily done via polynomial
multiplication and then inverse transforming the result.
 EXAMPLE: - Determine the autocorrelation sequence of the signal x(n) = an u(n), -1 < a 1.
 SOLUTION: -Since the autocorrelation sequence of a signal is its correlation with itself, gives
𝑹𝒙𝒙 (𝒛) = 𝒁{𝒓𝒙𝒙 𝒍 } = X(z) X(z-1)
1
 We have X(z) = 1−𝑎𝑧 −1 ROC: z > a (Causal signal)
1 1
 But by using property we obtain X(z-1) = 1−𝑎𝑧 ROC: z > (Anti-causal signal)
a
1 1 𝟏 1
 Thus 𝑅𝑥𝑥 (𝑧) = 1−𝑎𝑧 −1 x 1−𝑎𝑧 = 𝟏−𝒂 ROC: a < z <
𝒛+𝒛−𝟏 +𝒂𝟐 a
 Since the ROC of 𝑅𝑥𝑥 (𝑧) is a ring, 𝑟𝑥𝑥 𝑙 is a two-sided signal, even if x(n) is causal.
 To obtain 𝑟𝑥𝑥 𝑙 , we observe that the z-transform of the sequence with b = 1/a is simply (1-a2) 𝑅𝑥𝑥 (𝑧).
𝟏
 Hence it follows that 𝒓𝒙𝒙 𝒍 = 𝟏−𝒂𝟐 𝒂 𝒍 , -∞ < 𝑙 < ∞
INITIAL VALUE THEOREM: -If x(n) is causal [i.e. x(n) = 0 for n < 0], then x(0) = 𝐥𝐢𝐦𝐳 ∞ 𝐗(𝐳)
 PROOF: Since x(n) is causal, We have, X(z) = ∞ n=0 x(n)z
−n
= x(0) + x(1) z −1 + x(2) z −2 + …
 Obviously, as z  ∞, z −n 0 since n > 0.
FINAL VALUE THEOREM: - If x(z) = z{x(n)} and the poles of x(z) are all inside the unit circle
then the final value of the sequence x(∞) can be determined by using the expression,
x(∞) = 𝐥𝐢𝐦𝐧 ∞ 𝐗(𝐧) = 𝐥𝐢𝐦 𝐳 𝟏 [ 𝟏 − 𝐳 −𝟏 𝐱(𝐳)]
PARSEVAL S RELATION: -If x1(n) and x2(n) are complex-valued sequences, then
𝟏 𝐳
∞
𝐧=−∞ 𝐱 𝟏 𝐧 𝐱𝟐∗ (𝐧) = 𝟐𝛑𝐣 𝐜
𝐗 𝟏 (𝐯) 𝐗 𝟐 𝐯 𝐯 −𝟏 𝐝𝐯.
Provided that 𝑟1𝑙 𝑟2𝑙 < 1 < 𝑟1𝑢 𝑟2𝑢 , where 𝑟1𝑙 < z < 𝑟1𝑢 & 𝑟2𝑙 < z < 𝑟2𝑢 are the ROC of X1(z) and X2(z).
MULTIPLICATION OF COMPLEX SEQUENCES:- If X1(n) z X1(z) & X2(n)  z  X2(z)
𝟏 𝒛
X(n) = x1(n) x2(n)  z  X(z) = 𝟐𝝅𝒋 𝒄
𝑿𝟏 (𝒗) 𝑿𝟐 𝒗 𝒗−𝟏 𝒅𝒗
 Where C is a closed contour that encloses the origin and lies within the region of convergence common
to both X1(v) and X1 (1/v).
 EXAMPLE: - Find the z-Transform of this signal, x(n) = n2 u(n)
dx (z)
 SOLUTION: Z{x(n)} = x(z)  Z[nx(n)] = -z  x(z) = Z[n2u(n)] = Z[n{nu(n)}]
dz
d d z −1 1 dz z −1
 x(z) = z −1 dz −1 [Z{nu(n)}]  x(z) = z −1 dz −1 =z
(1−z −1 )2 d (1−z −1 )2
d x d z −1 z −1 (1−z −1 ) d z −1 z −1
 Let z −1 = xx dx = z −1 dz −1 AS: = z −1 dz −1 i.e. nu(n)(1−z −1 )2
(1−x)2 (1−z −1 )2 (1−z −1 )3 (1−z −1 )2
(1−z −1 )2 −z −1 [z(1−z −1 )(−1)] 1−z −1 { 1−z −1 −z −1 (−2)} 1−z −1 +2z −1 1+z −1
= z −1 = z −1 = z −1 = z −1
(1−z −1 )4 (1−z −1 )3 (1−z −1 )3 (1−z −1 )3
𝐳 −𝟏 (𝟏+𝐳 −𝟏 ) 𝐳 (𝐳+𝟏)
= = (ANSWER)
(𝟏−𝐳 −𝟏 )𝟑 (𝐳−𝟏)𝟑

 RATIONAL Z-TRANSFORMS:-
 We have discussed an important family of z-
transforms are those for which X(z) is a
rational function, that is, a ratio of two
polynomials in z-1(or z).
 In this section we discuss some very
important issues regarding the class of
rational z-transforms.
 POLES AND ZEROS:
 The Zeros of a z-transform X(z) are the
values of z for which X(z) = 0.
 The Poles of a z-transform are the values of
z for which X(z) = ∞.

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <Z-TRANSFORM & ITS APPLICATION> [Page 3.9]
𝑴
𝑵(𝒛) 𝒃𝟎 +𝒃𝟏 𝒛−𝟏 +⋯+𝒃𝑴 𝒛−𝑴 𝒌=𝟎 𝒃𝒌 𝒛
−𝒌
 If X( z ) is a rational function, then X(z)= = = 𝑵 𝒂 𝒛−𝒌
𝑫(𝒛) 𝒂𝟎 +𝒂𝟏 𝒛−𝟏 +⋯+𝒂𝑵𝒛−𝑵 𝒌=𝟎 𝒌
 If a0 ≠ 0 and b0 ≠ 0, we can avoid the negative powers of z by factoring out the terms b0z-M and a0z-N as
𝒂 𝒂𝟏 𝑵−𝟏
𝑵(𝒛) 𝒃𝟎 𝒛−𝑴 𝒛𝑵 + 𝟏 𝒛𝑵−𝟏 +⋯+𝒂𝑵/𝒂𝟎 𝒃𝟎 𝒛𝑵 +𝒛 +⋯+𝒂𝑵 /𝒂𝟎
𝒂𝟎 −𝑴+𝑵 𝒂𝟎
follows: X (z) = = 𝒂𝟏 𝑵−𝟏 = 𝒛 𝒂
𝑫(𝒛) 𝒂𝟎 𝒛−𝑵 𝑵
𝒛 + 𝒛 +⋯+𝒂𝑵 /𝒂𝟎 𝒂𝟎 𝒛𝑵 + 𝟏 𝒛𝑵−𝟏 +⋯+𝒂𝑵 /𝒂𝟎
𝒂𝟎 𝒂𝟎

 Since N(z) and D(z) are polynomials in z, they can be expressed in factored form as
𝑴
𝑵(𝒛) 𝒃𝟎 𝒛−𝒛𝟏 𝒛−𝒛𝟐 …(𝒛−𝒛𝑴 ) 𝒌=𝟏 (𝒛−𝒛𝒌)
X (z) = = 𝒛−𝑴+𝑵  X(z) = GzN-M 𝑵 . Where G = b0/a0.
𝑫(𝒛) 𝒂𝟎 𝒛−𝒑𝟏 𝒛−𝒑𝟐 …(𝒛−𝒑𝑴 ) 𝒌=𝟏 (𝒛−𝒑𝒌 )
 Thus X(z) has M finite zeros at z = z1, z2,…, zM (the roots of the numerator polynomial), N finite poles
at z = p1, p2,…, pN (the roots of the denominator polynomial), and |N-M| zeros (if N > M ) or poles (if
N < M ) at the origin z = 0. Poles or Zeros may also occur at z = ∞.
 A Zero exists at z = ∞ if X(∞) = 0 and a Pole exists at z = ∞ if X(∞) = ∞.
 If we count the poles and zeros at zero and infinity, we find that X (z) has
exactly the same number of poles as zeros. We can represent X(z ) graphically
by a pole – zero plot (or pattern) in the complex plane, which shows the
location of Poles by crosses (X) & the location of Zeros by circles (O).
 The multiplicity of multiple order poles or zeros is indicated by a number
close to the corresponding cross or circle. Obviously, by definition, the ROC
of a z-transform should not contain any poles.
 EXAMPLE: - Determine the pole-zero plot for the signal x(n) = an u(n) a > 0
 SOLUTION:- We have X (z) = ∞ 𝑛=−∞ 𝑥(𝑛) 𝑧
−𝑛
= ∞ n
𝑛=−∞ [a u n ] 𝑧
−𝑛
1 𝒛 𝒛
= ∞
𝑛=0 a
n
𝑧 −𝑛 = ∞
𝒏=𝟎 𝐚 𝒛−𝟏 𝐧
=
1− a 𝑧−1
= X(z) = ; ROC: |z| > a
𝒛− 𝐚 𝒛−𝒂
 Thus X (z) has one Zero at z1 = 0 & one Pole at p1 = a. Pole-Zero plot is in Fig.
 Note that the pole p1 = a is not included in the ROC since the z-transform does not converge at a pole.
 EX: - Determine pole-zero plot for the signal x(n) = {an, a ≤ n ≤ M-1 and 0, Otherwise}; Where a>0
(1− a 𝑧−1 )𝑀 𝑍𝑀 − 𝑎𝑀
 SOLUTION: - We have X (z) = ∞
𝑛=−∞ 𝑥(𝑛) 𝑧
−𝑛
X (z) = 𝑀−1 −1 𝑛
𝑛=0 (𝑎𝑧 ) = =
1− a 𝑧−1 𝑍 𝑀 −1 (𝑧−𝑎)
 Since a>0, the equation ZM = aM has M roots at Zk = aej2πk/M, k = 0, 1, 2, ……… M - 1
 The zero z0 = a cancels the pole at z = a.
𝒛−𝒛𝟏 𝒛−𝒛𝟐 …(𝒛−𝒛𝑴−𝟏 )
 Thus X(z) = Which has M-1 zeros and M-1
𝑍 𝑀 −1
poles, located as shown in fig.
 Note that ROC is the entire z-plane except z= 0 because of the M-1
poles located at the origin.
 Pole Location & Time-Domain Behavior of Causal Signals:
 Here we consider the relation between the z-plane location of a pole pair
and the form (shape) of the corresponding signal
in time domain.
 It is based generally on the collection of z-
transform pairs given in Table & the results in the
preceding subsection.
 We deal exclusively with real, causal signals.
 In particular, we see that the characteristic
behavior of causal signals depends on whether the
poles of the transform are contained in the region
|z| < 1, or in the region |z| > 1, or on circle |z| = 1.
 Since the circle |z| = 1 has a radius of 1, it is called
the unit circle.
 If a real signal has a z-transform with one pole,
this pole has to be real.
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <Z-TRANSFORM & ITS APPLICATION> [Page 3.10]
 The only such signal given below is the real exponential having one Zero at z1 = 0 and one Pole at p1 =
𝟏
a on the real axis. x(n) = anu(n)  z  X(z) = ROC: |z| > |a|
𝟏−𝒂𝒛−𝟏
 Fig illustrates the behavior of the signal w.r.t. the location of the pole relative to the unit circle.
 The signal is decaying if the pole is inside the unit circle, fixed if the pole is on the unit circle, and
growing if the pole is outside the unit circle.
 In addition, a negative pole results in a signal that alternates in sign.
 Obviously, causal signals with poles outside the unit circle become unbounded, cause overflow in
digital systems, and in general, should be avoided.
 THE SYSTEM FUNCTION OF A LINEAR TIME-INVARIANT SYSTEM:-
 We demonstrated that the output of a (relaxed) linear time-invariant system to an input sequence x(n)
can be obtained by computing the convolution of x(n) with the unit sample response of the system .
 The convolution property, derived in properties of z- transform section, allows us to express this
relationship in the z-domain as Y(z) = H(z) X(z) ;
 Where Y(z) is the z-transform of the output sequence y(n), X(z) is the z-transform of the input
sequence x(n) and H(z) is the z-transform of the unit sample response h(n).
 If we know h(n) and x(n), we can determine their corresponding z-transforms H(z) and X(z), multiply
them to obtain Y(z), and therefore determine y(n) by evaluating the inverse z-transform of Y(z).
 Alternatively, if we know x(n) and we observe the output y(n) of the system, we can determine the unit
sample response by 1st solving for H(z) from relation & then evaluating inverse z-transform of H(z).
𝒀(𝒛)
 H (z) = 𝑿(𝒛) and then evaluating the inverse z-transform of H(z). As H(z) = ∞ 𝒏=−∞ 𝒉(𝒏)𝒛
−𝒏

 It is clear that H(z) represents the z-domain characterization of a system,

 Whereas h(n) is the corresponding time-domain characterization of the system.
 In other words, H (z) and h(n) are equivalent descriptions of a system in the two domains.
 The transform H (z) is called the System Function.
 EXAMPLE: - Determine the system function and the unit sample response of the system described by
the difference equation y(n) = ½ y(n-1) + 2x(n)
 SOLUTION: - By computing z-transform of difference equation, we obtain Y(z) = ½ z-1Y(z) + 2X(z)
𝒀(𝒛) 𝟐
 Hence the system function is = H(z) = 𝟏 . This system has a pole at z = ½ & a zero at origin.
𝑿(𝒛) 𝟏− 𝒛−𝟏
𝟐
 We obtain the inverse transform h(n) = 2 (½)n u(n). This is the unit sample response of the system.
 We have now demonstrated that rational z-transforms are encountered in commonly used systems and
in the characterization of linear time-invariant systems.
 EXAMPLE: -Find the system function and impulse response of the system described by the difference
𝟏
equation y(n) = 𝟓 y(n-1) + x(n)
 SOLUTION:- Taking z-transform on both sides of the given signal, we get
1 1 1
 Y(z) = z −1 Y(z) + X(z)  Y(z) - z −1 Y(z) = X(z)  Y(z) 1 − z −1 = X(z)
5 5 5
Y(z) 𝟏 -1
 Now the system function is H (z) = = 𝟏 .  h(n) = Z [H(z)]
X(z) 𝟏− 𝐳 −𝟏
𝟓
𝟏 𝐧
 From the Table of page 3.6, we find the inverse z-transform. Thus h(n) = u(n)
𝟓
 EXAMPLE: -Find the system function and impulse response of the system described by the difference
equation y(n) = x(n) + 2x(n-1) - 4x(n-2) + x(n-3).
 SOLUTION:- Given y(n) = x(n) + 2x(n-1) - 4x(n-2) + x(n-3). Taking z-transform on both side, we get
Y(z)
Y(z) = X(z) + 2z −1 X(z) - 4z −2 X(z) + z −3 X(z)  H(z) = X(z) = 1 + 2z −1 - 4z −2 + z −3
 H(z) = h(0) z 0 + h(1) z −1 - h(2) z −2 + h(3) z −3
 From which we can find h(0) = 1; h(1) = 2; h(2) = - 4; h(3) = 1  h(n) = {𝟏, 2, -4, 1}
 PRACTICE PROBLEM: -Find the system function & impulse response of system described by the
1 1
difference equations, (A) y(n) = y(n-1)+2y(n-2)+x(n) & (B) y(n) = 2x(n-1)+ 2 x(n-2)+x(n-3)+ 3 x(n-4)

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <Z-TRANSFORM & ITS APPLICATION> [Page 3.11]
 EXAMPLE:- Determine the pole-zero plot for the system described by difference equation
𝟑 𝟏
y(n) - y(n-1) + y(n-2) = x(n) - x(n-1)
𝟒 𝟖
 SOLUTION:-Taking z-transform on both sides, we get
𝟑 𝟏
 Y(z) - 𝟒 𝐳 −𝟏 Y(z)+ 𝟖 𝐳 −𝟐Y(z) = X(z) - 𝐳 −𝟏X(z)
3 1 −2 Y(z)
 Y(z) 1 − 4 z −1 + z =X(z) 1 − z −1 System function H(z)=
8 X(z)
1 z −1
1−z −1 1− 8z( z−1) 8z( z−1)
 H(z) = 3 1 = 3
z
1 = z
8z 2 −6z +1
= =
1− z −1 + z −2 1− + 2 8z 2 −6z+1 8z 2 −4z−2z+1
4 8 4z 8z 8z 2
8z( z−1) 8z( z−1) 8z( z−1) 𝐳(𝐳−𝟏)
H(z)= 1 1 = 1 = 1 1 = 𝟏 𝟏 ; ROC: z >½
8z z− −2 z− z− (8z−2) z− 8(z− ) 𝐳− 𝐳−
2 2 2 2 4 𝟒 𝟐
1 1
 Zero are at z = 0; Poles are at z = 4, 2. The pole-zero shown in fig. From fig we can observe following:
(i) ROC of the system function includes unit circle. (ii) ROC of system function cannot contain any poles.
 PRACTICE PROBLEM: - Find the pole-zero plot for the system described by difference equation
𝟓 𝟏 1 1
y(n) = 𝟔 y(n-1) - 𝟔 y(n-2) + x(n) - x(n-1) ANSWER: Zeros  0, 1; Poles  2, 3
 THE INVERSE Z - TRANSFORM: -
 Often, we have the z-transform X (z) of a signal and we must determine the signal sequence. The
procedure for transforming from the z-domain to the time domain is called the Inverse z-transform.
 An inversion formula for obtaining x (n) from X (z) can be derived by using the Cauchy integral
theorem, which is an important theorem in the theory of complex variables.
 To begin, we have the z-transform defined as X (z) = ∞ 𝒌=−∞ 𝒙(𝒌)𝒛
−𝒌

 Suppose that we multiply both sides by zn - 1 and integrate both sides over
a closed contour within the ROC of X (z) which encloses the origin.
 Such a contour is illustrated in Fig as above.
 Thus we have 𝒄
𝑿 (z)zn-1dz = 𝒄 ∞ 𝒌=−∞ 𝒙(𝒌)𝒛
𝒏−𝟏−𝒌
𝒅𝒛
Where C denotes the Closed Contour in the ROC of X (z), taken in a
counterclockwise direction.
 Since the series converges on this contour, we can interchange the order
of integration and summation on the right-hand side of above equation.
 Thus it becomes, 𝒄 𝑿 (z)zn-1dz = ∞
𝒌=−∞ 𝒙(𝒌) 𝒄 𝒛
𝒏−𝟏−𝒌
𝒅𝒛
 Now we can invoke the Cauchy integral theorem, which states that,
𝟏
𝒛𝒏−𝟏−𝒌 𝒅𝒛 = {1, k = n and 0, k ≠ n}, Where C is any contour that encloses the origin.
𝟐𝝅𝒋 𝒄
 By applying this, in the right hand side of above equation, that reduces to 2πjx (n) and hence the
𝟏
desired inversion formula x (n) = 𝑿(𝒛)𝒛𝒏−𝟏 𝒅𝒛.
𝟐𝝅𝒋 𝒄
 Although the contour integral provides the desired inversion formula for determining the sequence x(n)
from the z-transform, we shall not use the above directly in our evaluation of inverse z-transforms.
 In our treatment we deal with signals and systems in the z-domain which has rational z-transforms (i.e.,
z-transforms that are a ratio of two polynomials). For such z-transforms we develop a simpler method.
 INVERSION OF THE Z-TRANSFORM
 The inverse z-transform is formally given by x(n) = 𝒄 𝑿 𝒛 𝒛𝒏−𝟏 𝒅𝒛
 Where the integral is a contour integral over a closed path C that encloses the origin and lies within the
region of convergence of X (z). For simplicity, C can be a circle in the ROC of X (z) in the z-plane.
 There are three methods that are often used for the evaluation of the inverse z-transform in practice:-
(A) Residue method
(B) Convolution method
(C) Direct evaluation by contour integration.
(D) Expansion in to a series of terms in the variables z, and z-1. (Long Division method)
(E) Partial-Fraction Expansion Method and table lookup.
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <Z-TRANSFORM & ITS APPLICATION> [Page 3.12]
 The Inverse z-Transform by PARTIAL-FRACTION EXPANSION Method: -
 We attempt to express the function X(z) as linear combination X (z) = α1X1(z) + α2X2(z) +…+ αkXk(z)
 Where X1(z),…, Xk (z) are expressions with inverse transforms x1(n),…, xk(n) available in a table of z-
transform pairs. If such a decomposition is possible, then x(n), the inverse z-transform of X(z ), can
easily be found using the linearity property as x(n) = α1x1(n) + α2x2(n) + …+ αkxk(n)
 This approach is particularly useful if X(z) is a rational function.
 Without loss of generality, we assume that a0 ≠ 1, so that that can be expressed as
𝑵 𝒛 𝒃𝟎 +𝒃𝟏 𝒛−𝟏 +⋯ + 𝒃𝑴 𝒛−𝑴
X(z) = = ------------------- (i)
𝑫 𝒛 𝟏+𝒂𝟏 𝒛−𝟏 +⋯+ 𝒂𝑵 𝒛−𝑵
 If a0 ≠ 1, we can obtain above by dividing both numerator & denominator by a0.
 A rational function of the form equation (i) is called Proper if aN ≠ 0 and M < N.
 An Improper rational function (M ≥ N) can always be written as the sum of a polynomial and a proper
rational function. In general any improper rational function can be expressed as,
N(z) N 1 (z)
X(z) = D(z) = c0 + c1z -1 + …+ cM – N z - (M - N) + D(z) …………… (ii)
 In the above equation (ii) the Inverse z-transform of polynomial can be easily found.
 We focus our attention on the inversion of proper rational functions, since any improper function can
be transformed into a proper function by using eq. (ii).
 Let us consider a rational function of the form of eq. (i).
 We eliminate negative power of z by multiplying both numerator and denominator by zN.
𝐛𝟎 𝐳 𝐍 +𝐛𝟏 𝐳 𝐍−𝟏 +⋯+𝐛𝐌 𝐳 𝐍−𝐌
 This results in X(z) = ……………(iii) . Dividing equation (iii) by z,
𝐳 𝐍 +𝐚𝟏 𝐳 𝐍−𝟏 +⋯+𝐚𝐍
X(z) b 0 z N −1 +b 1 z N −2 +⋯+b M z N −M −1 b 0 z N −1 +b 1 z N −2 +⋯+b M z N −M −1
 We obtain = = ……………(iv)
z z N +a 1 z N −1 +⋯+a N z−p 1 z−p 2 ……(z−p N )
𝐗(𝐳) 𝐂𝟏 𝐂𝟐 𝐂𝐍
 For distinct poles equation (iv) is expanded in the form = + + ⋯+ ……(v)
𝐳 𝐳−𝐩𝟏 𝐳−𝐩𝟐 𝐳−𝐩𝐍
𝐳−𝐩𝐤 𝐗(𝐳)
 We can find the co-efficient, C1, C2, …CN by using formula Ck = k = 1, 2,…, N …(vi)
𝐳 𝒛 = 𝐩𝐤
 If X(z) has a pole of multiplicity l, that is, it contains in denominator the factor (z - pk)l, then expansion
𝐗(𝐳) 𝟏
of the form of equation (v) is no longer valid. Let us assume l = 2 and = ……(vii)
𝐳 𝐳−𝐩𝟐 𝟐 𝐳−𝐩𝟏
𝐗(𝐳) 𝐂𝟏 𝐂𝟐 𝐂𝟑
 The above equation (vii) can be expressed in partial fraction form as = + +
𝐳 𝐳−𝐩𝟏 𝐳−𝐩𝟐 𝐳−𝐩𝟐 𝟐
𝐗(𝐳) 𝐝 𝟐 𝐗(𝐳) 𝟐 𝐗(𝐳)
 Where C1 = 𝐳 − 𝐩𝟏 ; C2 = 𝐳 − 𝐩𝟐 and C3 = 𝐳 − 𝐩𝟐
𝐳 𝒛 = 𝐩𝟏 𝐝𝐳 𝐳 𝒛 = 𝐩𝟐 𝐳 𝒛 = 𝐩𝟐
𝐗(𝐳) 𝐑(𝐳) 𝐂𝐤 𝟏 𝐂𝟐𝐤 𝐂𝐢𝐤 𝐂𝒍𝐤
 In general for a case of l-repeated roots let = = + + ⋯+ + ⋯+
𝐳 𝐳−𝐩𝐤 𝒍 𝐳−𝐩𝐤 𝐳−𝐩𝐤 𝟐 𝐳−𝐩𝐤 𝐢 𝐳−𝐩𝐤 𝒍
X(z) 𝑙
Where i is any term in the partial fraction expansion and R(z) = z − pk
z
𝟏 𝐝𝒍−𝒊 𝒍 𝐗(𝐳)
 Thus to find general term Cik is given as  𝐂 𝒊𝒌 = 𝐳 − 𝐩𝐤
(𝒍−𝒊)! 𝐝𝐳 𝒍−𝒊 𝐳 𝐳 = 𝐩𝐤
𝟏+𝟑𝐳 −𝟏
 EXAMPLE (1) Find the inverse z-transform of X(z) = 𝟏+𝟑𝐳−𝟏 +𝟐𝐳−𝟐 ; ROC : 𝑧 >2
𝐳(𝐳 𝟐 −𝟒𝐳+𝟓)
(2) Find the inverse z-transform of X(z) = for ROC (i) 2 < 𝑧 < 3 (ii) 𝑧 > 3 (iii) 𝑧 < 1
𝐳−𝟑 𝐳−𝟏 (𝐳−𝟐)
1+3z −1
 SOLUTION: - (1) Given X(z) = . First we eliminate the negative power, by multiplying
1+3z −1 +2z −2
2 𝐳 𝟐 +𝟑𝐳 𝐳(𝐳+𝟑) 𝐳(𝐳+𝟑) 𝐳(𝐳+𝟑)
numerator and denominator by z .  X(z) = 𝐳𝟐 +𝟑𝐳+𝟐 = 𝐳𝟐 +𝟐𝐳+𝐳+𝟐 = 𝐳 𝐳+𝟐 +𝟏(𝐳+𝟐) = 𝐳+𝟏 (𝐳+𝟐)
𝐗(𝐳) (𝐳+𝟑)
 Dividing X(z) by z we obtain = …………(1)
𝐳 𝐳+𝟏 (𝐳+𝟐)
𝐗(𝐳) 𝐂𝟏 𝐂𝟐
 The above equation can be written in partial fraction form as = + ……… (2)
𝐳 𝐳+𝟏 𝐳+𝟐
X(z) (z+3) (z+3)
i.e. C1 = z + 1 = z+1 = =2
z 𝑧 = −1 z+1 (z+2) 𝑧 = −1 (z+2) 𝑧 = −1
X(z) (z+3) (z+3)
C2 = z + 2 = z+2 = (z+1) = -1
z 𝑧 = −2 z+1 (z+2) 𝑧 = −2 𝑧 = −2
 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <Z-TRANSFORM & ITS APPLICATION> [Page 3.13]
X(z) 2 1 𝟐𝐳 𝐳 𝑧
 Therefore = -  X(z) = - [As 𝛼 𝑛 u(n)  z  X(z) = ]
z z+1 z+2 𝐳+𝟏 𝐳+𝟐 𝑧−𝛼
2z z z z
 Thus x(n) = Z-1 {X(z)} = Z-1 − = 2 Z-1 - Z-1 = 2(-1)nu(n) - (-2)nu(n)
z+1 z+2 z+1 z+2
 As ROC is 𝑧 >2 the sequence is causal and by inspection we can find x(n) = 2(-1)nu(n) - (-2)nu(n)
z(z 2 −4z+5) 𝐗(𝐳) (𝐳 𝟐 −𝟒𝐳+𝟓)
 SOLUTION: - (2) Given X(z) = . Dividing X(z) by z we obtain =
z−3 z−1 (z−2) 𝐳 𝐳−𝟑 𝐳−𝟏 (𝐳−𝟐)
X(z) C1 C2 C3
 = + + . The value of C1, C2, C3 can be evaluated using equation
z z−1 z−2 z−3
X(z) (z 2 −4z+5) (z 2 −4z+5) 1−4+5 2
 C1 = z − 1 = z−1 = = −2 x −1 = 2 = 1
z z =1 z−3 z−1 (z−2) z = 1 z−3 (z−2) z = 1
X(z) (z 2 −4z+5) (z 2 −4z+5) 4−8+5 1
 C2 = z − 2 = z−2 = = = = -1
z 𝑧 =2 z−3 z−1 (z−2) 𝑧 = 2 z−1 (z−3) 𝑧 = 2 1 x −1 −1
X(z) (z 2 −4z+5) (z 2 −4z+5) 9−12+5 2
 C3 = z − 3 = z−3 = = =2=1
z z =3 z−3 z−1 (z−2) z = 3 z−1 (z−2) z = 3 2.1
𝐗(𝐳) 𝟏 𝟏 𝟏 𝐳 𝐳 𝐳
 Substituting C1, C2, C3, we have = - + . From which X(z) = - +
𝐳 𝐳−𝟏 𝐳−𝟐 𝐳−𝟑 𝐳−𝟏 𝐳−𝟐 𝐳−𝟑
In case (i) when the ROC is 2 < 𝑧 < 3, shown
in figure, the signal x(n) is two sided.
 The poles z = 1 & z = 2 provide the causal
part and the pole z = 3 the anticausal part.
 Thus, x(n) = u(n) - 2nu(n) + 3nu(-n-1)
In case (ii) when ROC is 𝑧 > 3, shown in fig,
the signal x(n) is causal and all the three terms
in eq. are causal terms. Therefore x(n) = u(n) - 2nu(n) + 3nu(n)
In case (iii) when ROC is 𝑧 < 1, shown in fig, the signal x(n) is anti-causal & all the terms in equation
are anti-causal terms. Therefore x(n) = ±u(-n-1) - 2nu(-n-1) + 3nu(-n-1)  x(n) = [-1 + 2n - 3n] u(-n-1)
𝟏
 EXAMPLE-3 Determine the causal signal x(n) having the z-transform X(z) = −𝟏 −𝟏 𝟐 𝟏−𝟐𝐳 𝟏−𝐳
1 z3
 Answer: -Given X(z) = . Multiply numerator & denominator by z  X(z) = 3
1−2z −1 1−z −1 2 z−2 z−1 2
𝐗(𝐳) 𝐳𝟐 𝐂𝟏 𝐂𝟐 𝐂𝟑
 Dividing the above result by z we have = = + +
𝐳 𝐳−𝟐 𝐳−𝟏 𝟐 𝐳−𝟐 𝐳−𝟏 𝐳−𝟏 𝟐
X(z) z2
 C1 = z − 2 = z−2 =4
z z =2 z−2 z−1 2 z =2
1 d 2 X(z) d 2 z2
 C2 = 1! dz z−1 = z − 1
z z =1 dz z−2 z−1 2 z =1
d d
d z2 z−2 z2 − z2 z−2
= = dz dz
[z 2 (1-0) = z 2 ]
dz z−2 z =1 z−2 2
z =1
z−2 2z − z 2 1−2 2− 12 −2− 1
= = = = -3
z−2 2 z =1 1−2 2 1

2 X(z) 2 z2
 C3 = z−1 = z−1 = -1
z z =1 z−2 z−1 2 z = 1
𝐗(𝐳) 𝟒 𝟑 𝟏
 Substituting C1, C2, C3 values, = - -
𝐳 𝐳−𝟐 𝐳−𝟏 𝐳−𝟏 𝟐
𝟒𝐳 𝟑𝐳 𝐳
 X (z) = – –  x (n) = 4(2)n u(n) - 3u(n) – nu(n)
𝐳−𝟐 𝐳−𝟏 𝐳−𝟏 𝟐
z 3 +z 2
 EXAMPLE - 4 Find the Inverse z-Transform of X(z) = ROC: 𝑧 >3
z−1 z−3
z2 + z
𝐗(𝐳) 𝐳 𝟐 +𝐳 z 2 +z z 2 +z
 ANS - 4:-From the expression = = z 2 −3z−z+3 = z 2 −4z+3 z − 4z + 3 z 2 − 4z + 32
1
𝐳 𝐳−𝟏 𝐳−𝟑
(−) (+) (−)
5z - 3
 Converting the above improper rational function (as M = N) into sum of a constant and a proper
𝐗(𝐳) 𝟓𝐳−𝟑
rational function, we get =1+
𝐳 𝐳−𝟏 𝐳−𝟑

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <Z-TRANSFORM & ITS APPLICATION> [Page 3.14]
𝟓𝐳−𝟑 𝐂𝟏 𝐂𝟐
 The rational expression can be expressed by partial fraction expansion = +
𝐳−𝟏 𝐳−𝟑 𝐳−𝟏 𝐳−𝟑
5z−3 2 5z−3 12
 Where C1 = z − 1 = = -1 and C2 = z − 3 = =6
z−1 z−3 𝑧 = 1 −2 z−1 z−3 𝑧 = 3 2
X(z) 1 6 𝐳 𝟔𝐳
 Therefore =1- +  X(z) = z - +
z z−1 z−3 𝐳−𝟏 𝐳−𝟑
 Taking inverse z-transform on both sides we get x(n) = δ(n+1) – u(n) + 6(3)n u(n)
 EXAMPLE- 5 : Find inverse z-Transform of
𝟏 z −1 1
𝟏+ 𝐳 −𝟏 1+ z 2 +𝑧 z z+ 𝐗(𝐳) 𝐳+𝟏 𝐂𝟏 𝐂𝟐
 SOLUTION:- X(z) = 𝟐
= 2 2
= z 2 +3z+2 = 2
.˙. 𝟐
= 𝐳𝟐 +𝟑𝐳+𝟐 = +
𝟏+𝟑𝐳 −𝟏 +𝟐𝐳 −𝟐 1+3z −1 +2z −2 z+1 (z+2) 𝐳 𝐳+𝟏 𝐳+𝟐
1 1 1 1 1 3
z+ −1+ − 𝟏 z+ −2+ − 𝟑
2 2 2 2
.˙. C1 = (𝑧 + 1) = = = − 𝟐 ; C2 = (𝑧 + 2) = −2+12 = −12 =
z+1 (z+2) 1+2 1 z+1 (z+2) 𝟐
𝑧 = −1 𝑧 = −2
𝐗(𝐳) −𝟏 𝟐𝒛 𝟑 𝒛 𝟏
.˙. = + 𝟐
.˙. x (n) = − 𝟐 −𝟏 𝐧 𝐮(𝐧) + 𝟑 𝟐 −𝟐 𝐧 𝐮(𝐧)
𝐳 𝐳+𝟏 𝐳+𝟐
𝟏 1 n
 PRACTICE PROBLEM:-Find Inverse z-Transform of X(z) = 𝟏 ANS4(n+1) u(n)
𝟏−𝐳 −𝟏 + 𝐳−𝟐 2
𝟒
𝐳𝟑
 EXAMPLE-6Find the causal signal x(n) which is having z-transform as under: X(z) = 𝐳+𝟏 𝐳−𝟏 𝟐
 SOLUTION - 6:- Expanding the given function X(z) in terms of the positive powers of z i.e.
X(z) z2 𝐗(𝐳) 𝐂𝟏 𝐂𝟐 𝐂𝟑
=  = + +
z z+1 z−1 2 𝐳 𝐳+𝟏 𝐳−𝟏 𝐳−𝟏 𝟐
z2 𝟏
 Here, we have C1 = z + 1 F(z) 𝑧 = −1 = =
z+1 𝑧 = −1 𝟒
2 z2 𝟏 d z2 z+1 2z−z 2 𝟑
C3 = z − 1 F(z) 𝑧 =1 = 2
= & C2 = dz = 2
=
z+1 𝑧 =1 𝟐 z+1 𝑧=1 z+1 𝑧 =1 𝟒
1 1 3 1 1 1 𝟏 𝐳 𝟑 𝐳 𝟏 𝐳
 Therefore X(z) = + + ; Hence X(z) = + +
4 z+1 4 z−1 2 z−1 2 𝟒 𝐳+𝟏 𝟒 𝐳−𝟏 𝟐 𝐳−𝟏 𝟐
1 n 3 1 𝟏 𝐧 𝟑 𝟏
 Now inverse z-transform of X(z)  4 (-1) u(n) + u(n) + 2 nu(n)  x(n) = −𝟏 + + 𝐧 u(n)
4 𝟒 𝟒 𝟐
 EXAMPLE-7: - Determine the Inverse z-Transform of the following X(z) by the partial fraction
𝐳+𝟐
expansion method X(z) = 𝟐𝐳𝟐 −𝟕𝐳+𝟑 if the ROCs are (a) 𝑧 > 3, (b) 𝑧 < ½ and (c) ½ < 𝑧 < 3.
𝐗(𝐳) 𝐳+𝟐
 SOLUTION-7:- We desire partial fraction expansion of = 𝐳(𝟐𝐳𝟐 −𝟕𝐳+𝟑)
𝐳
2z 2 − 7z + 3 = 2z 2 − 6z − z + 3 = 2z z − 3 − 1 z − 3 = z − 3 2z − 1 = 2 (z − 1/2)(z − 3)
𝐗(𝐳) 𝐳+𝟐 𝐂𝟏 𝐂𝟐 𝟑 𝐂 X(z) z+2 𝟐
 = 𝟏 = + 𝟏 + 𝐳−𝟑 ; Where C1 = z = 1 =
𝐳 𝟐𝐳 (𝐳− )(𝐳−𝟑) 𝐳 𝐳− z 𝑧 =0 2(z− )(z−3) 𝟑
𝟐 𝟐 2 𝑧 =0
1 X(z) z+2 X(z) z+2 𝟏
 C2 = z − 1 = 2z(z−3) = −1 & C3 = (z − 3) = 1 =
2 z 𝑧 = 𝑧 =
1 z 𝑧 =3 2z(z− ) 𝟑
2 2 2 𝑧 =3
X(z) 𝟐 𝐳 𝐳/𝟑
 Hence, by putting the value of C1, C2¸& C3 and multiply to by z, we obtain X(z) = − 𝟏 + 𝐳−𝟑
z 𝟑 𝐳−
𝟐
1
 Here, the given function X (z) has two poles, p1 = & p2 = 3 and following three inverse transforms.
2

(A) In the region 𝑧 > 3, all poles are Interior, i.e. the signal x(n) is causal, and therefore,
𝟐 𝟏 𝒏 𝟏
x (n) = δ (n) - u(n) + 𝟑 (3)n u(n).
𝟑 𝟐
(B) In the region 𝑧 < ½ , both the poles are Exterior, i.e. x(n) is anti-causal, and hence,
𝟐 𝟏 𝒏 𝟏
x (n) = δ (n) - u(-n-1) - 𝟑 (3)n u(-n-1).
𝟑 𝟐
1 1
(C) In the region 2
𝑧 < l < 3, [ i.e. 𝑧 < 3 anti causal and 𝑧 > ½ causal], the pole p1 = 2
is interior and
𝟐 𝟏 𝒏 𝟏
p2 = 3 is exterior, and hence x(n) = δ(n) - u(n) - (3)n u(-n-1). {For ROC Fig Ref Page- 3.13}
𝟑 𝟐 𝟑

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <Z-TRANSFORM & ITS APPLICATION> [Page 3.15]
𝐳
 EXAMPLE-8: -Determine the Inverse z-Transform of X(z) = 𝟑𝐳𝟐 −𝟒𝐳+𝟏 If the Regions Of Convergence
are (i) 𝑧 > 1, (ii) 𝑧 < 1/3 and (iii) 1/3 < 𝑧 < 1.
𝐗(𝐳) 𝟏
 SOLUTION-8:- Partial fraction expansion of X(z) yields = 𝐳 = 𝟑𝐳𝟐 −𝟒𝐳+𝟏
1
3z 2 − 4z + 1 = 3z 2 − 3z − z + 1 = 3z z − 1 − 1 z − 1 = 3z − 1 (z − 1) = 3 (z − 1) z −
3
𝐗(𝐳) 1 𝟏 𝐂 𝐂𝟐 X(z) 1 𝟏
 = 𝟏 = (𝐳−𝟏) + 𝟏  C1 = (z − 1) = 1 =
𝐳 3 (z−1) 𝐳− 𝐳− z z =1 3 (z− ) 𝟐
𝟑 𝟑 3 z =1
X(z) 1 1 𝟏 𝐗(𝐳) 𝟏/𝟐 −𝟏/𝟐
& C2 = z− 1 = 3(z−1) =-  = (𝐳−𝟏) + 𝟏
z 3 𝑧 = 𝑧 =
1 𝟐 𝐳 𝐳−
3 3 𝟑
𝟏 𝟏
𝐳 𝐳 𝟏 𝐳 𝟏 𝐳
 X (z) = (𝐳−𝟏) - 𝟐 𝟐
𝟏 = - 𝟏 [P1 = 1, P2 = 1/3]
𝐳− 𝟐 (𝐳−𝟏) 𝟐 𝐳−
𝟑 𝟑
(A) When the ROC is 𝑧 > 1, the signal x(n) is causal and both terms are causal.
𝟏 𝟏 𝟏 𝒏 𝟏 𝟏 𝒏
Therefore, x(n) = 𝟐 (1)n u(n) - 𝟐 𝟑 u(n) = 𝟐 𝟏 − 𝟑 u(n)
(B) When the ROC is 𝑧 < 1/3 , the signal x(n) is anti-causal i.e. the inverse gives negative time sequence.
𝟏 𝟏 𝟏 𝒏
Therefore, x(n) = 𝟐 (𝟏)𝐧 − 𝟐 𝟑 u(-n-1)
(C) Here the ROC 1/3 < 𝑧 < 1 is a ring, which implies that the signal x(n) is two-sided. Therefore one of
the terms corresponds to a causal signal and the other to an anti-causal signal, obviously the given ROC
is the overlapping of the region 𝑧 > 1/3 and 𝑧 < 1. The partial fraction expansion remains the same;
however because of ROC, the pole at 1/3 provides the causal part corresponding to positive time and
the pole at 1 is the anticausal part corresponding to negative time. Therefore, x(n) becomes
𝟏 𝟏 𝟏 𝒏
X(n) = - 𝟐 (1)n u(-n-1) - u(n)
𝟐 𝟑
𝐳(𝟏−𝐞−𝐓 )
 EXAMPLE-9 Compute the Inverse z-Transform of x(z) = (𝐳−𝟏)(𝐳−𝐞−𝐓 )
z(1−e −T ) z e −T
 SOLUTION-9: Given that x(z) = (z−1)(z−e −T ) = (z−1)(z−e −T ) − (z−1)(z−e −T )
x(z) 1−e −T A B 1−e −T 1−e −T
 = (z−1)(z−e −T ) = z−1 + z−e −T ; A = (z − 1) x = 1−e −T = 1 and
z (z−1)(z−e −T ) 𝑧=1
1−e −T 1−e −T 1−e −T 1+e −T
B = z − e−T x = = = −(1−e −T ) = -1
(z−1)(z−e −T ) 𝑧=e −T e −T −1 −1+e −T
x(z) A B 1 −1 z −z
Putting values of A & B, = z−1 + z−e −T = z−1 + z−e −T x (z) = z−1 + z−e −T  x(n) = u(n) – e-T u(n)
z
−𝟒 +𝟖𝐳 −𝟏
 EXAMPLE-10: Find Inverse z-Transform of expression by partial fraction method X(z) = 𝟏+𝟔𝐳−𝟏 +𝟖𝐳−𝟐
−𝟒 +𝟖𝐳 −𝟏 A A
 SOLUTION-10: Given that, X(z) = = 1+4z1 −1 + 1+2z2 −1
𝟏+𝟒𝐳 −𝟏 (𝟏+𝟐𝐳 −𝟏 )
−𝟒 +𝟖𝐳 −𝟏 6 −𝟒 +𝟖𝐳 −𝟏 −8
Now, A1 = = − 1/2 = -12 ; And, A2 = = −1 = 8
𝟏+𝟐𝐳 −𝟏 at z −1 =−1/4 𝟏+𝟒𝐳 −𝟏 at z −1 =1/2
−12 8
 H(z) = 1+4z −1 + 1+2z −1  Taking inverse z-transform, we have h(n) = {-12(-4)n u(n) + 8(-2)n u(n)}
𝟐+𝟑 𝐳 −𝟏
 EXAMPLE-11: Find Inverse z-Transform of the expression by X(z) = 𝟏 𝟏
𝟏+𝐳 −𝟏 𝟏+ 𝐳 −𝟏 𝟏− 𝐳 −𝟏
𝟐 𝟒
2+3 z −1 𝐀𝟏 𝐀𝟐 𝐀𝟑
 SOLUTION-11: Given that, X(z) = 1 1 = + 𝟏 + 𝟏
1+z −1 1+ z −1 1− z −1 𝟏+𝐳 −𝟏 𝟏+ 𝐳 −𝟏 𝟏+ 𝐳 −𝟏
2 4 𝟐 𝟒
2+3 z −1 𝟖 2+3 z −1 𝟖 2+3 z −1 𝟑
 A1 = 1 1 = − 𝟓 ; A2 = 1 = ; A3 = 1 =𝟖
1+ z −1 1− z −1 1+z −1 1− z −1 𝟑 1+z −1 1− z −1
2 4 z −1 =−1 4 z −1 =−2 2 z −1 =4
𝟖 𝟖 𝟏𝟒
− 8 8 1 n 14 1 n
 X(z) = 𝟏+𝐳𝟓−𝟏 + 𝟑
𝟏 + 𝟏𝟓
𝟏  Hence x(n) = Z-1[X(z)] = − 5 (−1)n + 3 − 2 + 15 u(n)
𝟏+ 𝐳 −𝟏 𝟏 − 𝐳 −𝟏 4
𝟐 𝟒
𝟏
 EXAMPLE12:-Determine the partial-fraction expansion of the proper function X(z) =
𝟏−𝟏.𝟓𝒛−𝟏 + 𝟎.𝟓𝒛−𝟐

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 603] <Z-TRANSFORM & ITS APPLICATION> [Page 3.16]
 SOLUTION-12 :-We eliminate negative powers, by multiplying both numerator & denominator by z2.
𝒛𝟐
Thus X(z) = 𝒛𝟐 −𝟏.𝟓𝒛 + 𝟎.𝟓
 The poles of X(z) are p1 = 1 and p2 = 0.5. Consequently, the expansion of the form (distinct poles) is
𝑿 𝒛 𝒛 𝑨𝟏 𝑨𝟐
= = +
𝒛 𝒛−𝟏 (𝒛−𝟎.𝟓) 𝒛−𝟏 𝒛−𝟎.𝟓
 A very simple method to determine A1 & A2 is to multiply the equation by the denominator term (z - l)
(z - 0.5). Thus we obtain z = (z - 0.5) A1 + (z - 1)A2
 Now if we set z = p1 = 1in the above, we eliminate the term involving A2. Hence 1 = (1-0.5)A1
 Thus we obtain the result A1= 2. Next we return to (3.4.18) and set z = p2 = 0.5, thus eliminating the
term involving A1, So we have 0.5 = (0.5-1) A2 And hence A2 = -1.
𝑿 𝒛 𝟐 𝟏
 Therefore, the result of the partial-fraction expansion is = 𝒛−𝟏 - 𝒛−𝟎.𝟓  2u(n) – (½)n u(n)
𝒛
 The example given above suggests that we can determine the coefficients A1, A2,…,AN, by multiplying
both sides of (3.4.15) by each of the terms (z - pk), k = 1, 2,…, N, and evaluating the resulting
expressions at the corresponding pole positions, p1, p2,…,pN.
(𝒛−𝒑𝒌 )𝑿 𝒛 (𝒛−𝒑𝒌 )𝑨𝟏 (𝒛−𝒑𝒌 )𝑨𝑵
 Thus we have, in general, = + …+ Ak + …+
𝒛 𝒛−𝒑𝟏 𝒛−𝒑𝑵
th (𝒛−𝒑𝒌 )𝑿 𝒛
 Consequently, with z = pk, yields the k coefficient as Ak = k = 1, 2,…, N
𝒛 𝒛=𝒑𝒌
𝟏 + 𝒛−𝟏
 EXAMPLE-13: - Determine the partial-fraction expansion of X(z) = 𝟏 − 𝒛−𝟏 + 𝟎.𝟓𝒛−𝟐
 SOLUTION:- To eliminate negative powers of z, we multiply both numerator and denominator by z2.
𝑿 𝒛 𝒛+𝟏 1 1 1 1
 Thus = 𝒛𝟐 −𝒛+𝟎.𝟓 . The poles of X (z) are complex conjugates p1 = 2 + j2 and p2 = 2 - j2
𝒛
𝑿 𝒛 𝒛+𝟏 𝑨 𝑨
 As p1 ≠ p2, we seek an expansion of the form (distinct pole). Thus = = 𝒛 −𝟏𝒑 + 𝒛 −𝟐𝒑
𝒛 𝒛 − 𝒑𝟏 (𝒛 − 𝒑𝟐 ) 𝟏 𝟐
 To obtain A1 and A2, we use the formula, Thus we obtain
𝟏 𝟏 𝟏 𝟏
(𝒛−𝒑𝟏 )𝑿 𝒛 𝒛+𝟏 + 𝐣 +𝟏 𝟏 𝟑 (𝒛−𝒑𝟐 )𝑿 𝒛 𝒛+𝟏 − 𝐣 +𝟏 𝟏 𝟑
𝟐 𝟐
A1 = = 𝒛−𝒑 =𝟏 𝟏 𝟏 𝟏 = 𝟐 + j𝟐 & A2 = = 𝒛−𝒑 =𝟏𝟐 𝟐
𝟏 𝟏 𝟏 = 𝟐 - j𝟐
𝒛 𝒛=𝒑𝟏 𝟐 𝒛=𝒑𝟏 +𝐣 − +𝐣 𝒛 𝒛=𝒑𝟐 𝟏 𝒛=𝒑𝟐 −𝐣 − −𝐣
𝟐 𝟐 𝟐 𝟐 𝟐 𝟐 𝟐 𝟐
 The expansion (distinct pole) and the formula hold for both real and complex poles.
 The only constraint is that all poles be distinct. We also note that A2 = A*1.
 It can be easily seen that this is a consequence of the fact that p2 = p*1.
 Also complex-conjugate poles result in complex-conjugate coefficients in partial-fraction expansion.
𝟎.𝟓(𝟏−𝟎.𝟓𝐳 −𝟏 )
 EXAMPLE-14: - (a) If Y(z) = (𝟏−𝟎.𝟐𝟓𝐳−𝟏 )(𝟏−𝟎.𝟕𝟓𝐳 −𝟏 )(𝟏−𝐳−𝟏 ) find the Steady value of y(n) if it exists.
𝟑𝐳
(b) If X(z) is given by X(z) = 𝐳−𝟏 (𝐳+𝟏)
 SOLUTION-14 (A): - By inspection, we can find that the ROC for Y(z) is z > 1. (1 − z −1 )Y(z) has
no poles on or outside the unit circle, se the condition of the final value theorem satisfied.
 Therefore, the steady state value exists and is given by y(∞) = limz 1 (1 − z −1 )Y(z)
0.5(1−0.5z −1 ) 0.5(1−0.5) 0.5(0.5)
 = limz 1 (1 − z −1 ) (1−0.25z −1 )(1−0.75z −1 )(1−z −1 ) = (1−0.25)(1−0.75) = (0.75)(0.25) = 1.33
3z 3z
 SOLUTION -14 (B): - Given X(z) = ; x(∞) = limz 1 (z − 1)X(z) = limz 1 (z − 1)
z−1 (z+1) z−1 (z+1)
 In this case (z-1) X(z) has a [pole on the unit circle. So, x (∞) is not defined.

-------------- ALL THE BEST -------------------- ALL THE BEST ----------------

 Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School

 A Hand Note on DSP [ETT 603] <FOURIER TRANSFORM: ITS APPLICATIONS PROPERTIES > [Page 4.1]

[CHAPTER-4]
--------- FOURIER TRANSFORM: ITS APPLICATIONS PROPERTIES---------
 INTRODUCTION: -
 Discrete Fourier Transform (DFT)
computes the values of Z- transform
for evenly spaced points around the
unit circle for given sequence.
 If the given sequence to be represented
is of finite duration i.e. it has only a
finite number of non-zero values, then the Transform used is Discrete Fourier Transform (DFT).
 DFT finds its application in DSP including linear filtering, correlation analysis and spectrum analysis.
 To develop the DFT let us first consider the Continuous Time Fourier Transform (CTFT) of a
∞
continuous signal x(t) which is expressed as X(jω) = − ∞ 𝒙(𝒕) 𝒆−𝒋𝝎𝒕 𝒅𝒕 ------------- (1)
 This equation is also called Analysis Equation of CTFT.
𝟏 ∞
 Also Inverse CTFT is given by the following synthesis equations 𝒙 𝒕 = 𝟐𝝅 − ∞ 𝑿(𝒋𝝎) 𝒆 𝒋𝝎𝒕 𝒅𝝎 - (2)
 The Continuous Time Fourier Transform (CTFT) is used for non periodic Continuous Time Signal.
 It produces a continuous spectrum of signal x (t).
∞
 Similarly the DTFT of discrete time signal x(n) is expressed as X(𝒆 𝒋𝝎) = − ∞ 𝒙(𝒏) 𝒆−𝒋𝝎𝒏 ---- (3)
 This is known as the analysis equation of DTFT of a discrete – time signal or sequence.
𝟏 ∞
 Also Inverse DTFT is given by the following synthesis equations 𝒙 𝒏 = 𝟐𝝅 − ∞ 𝑿(𝒆 𝒋𝝎) 𝒆 𝒋𝝎𝒏𝒅𝝎-(4)
 This also produces a continuous spectrum of signal x (n).
 Now DFT of a discrete – time signal x (n) is obtained by performing the sampling operation in both the
time domain and frequency domain. However, the Discrete Time Fourier Transform (DTFT) of a
discrete-time sequence x (n) is obtained by performing the sampling operation in time domain only.
 Now, let x (n) be a finite duration sequence. the N-Point DFT of the sequence x(n) is expressed as
X (k) = 𝐍−𝟏𝒏=𝟎 𝐱 (𝐧) 𝐞
−𝐣𝟐𝛑𝐧𝐤/𝐍
, k = 0, 1, 2, 3, 4 … N-1 ------ (5)
 Also the corresponding Inverse Discrete Fourier Transform (IDFT) is expressed as
𝟏 𝐍−𝟏
x (n) = 𝐗(𝐤)𝐞 𝐣𝟐𝛑𝐧𝐤/𝐍 , n = 0, 1, 2, 3, 4 … N-1 ------ (6)
𝑵 𝒌=𝟎
−𝐣𝟐𝛑/𝐍
 Now let us define WN = 𝐞 , which is called Twiddle Factor.
 So the above equations for DFT and IDFT can be written as
𝐤𝐧 𝟏
X (k) = 𝐍−𝟏
𝒏=𝟎 𝐱(𝐧) 𝐖𝐍 , k = 0, 1,… N-1 – (7) & x (n) = 𝐍−𝟏
𝐗(𝐤)𝐖𝐍 − 𝐤𝐧 , n = 0, 1, … N-1 - (8)
𝑵 𝒌=𝟎
 Here both the indices n and k are ranging from 0 to N-1. The integer n is known as time index since it
denotes the time instant. The integer k denotes discrete frequency and is called frequency Index.
 EXAMPLE - 1: - Find the 4-point DFT of a sequence x(n) = {𝟏, 1, 0, 0}.
 SOLUTION: Let us assume N = L = 4
 We have X(k) = 𝐍−𝟏 𝒏=𝟎 𝐱(𝐧) 𝐞
− 𝐣𝟐𝛑𝐧𝐤/𝐍
, k = 0, 1,… N-1  X(k) = 𝟑𝒏=𝟎 𝐱(𝐧) 𝐞− 𝐣𝟐𝛑𝐧𝐤/𝐍
3 e jθ − e −jθ e jθ + e −jθ
 X(0) = 𝑛=0 x(n) = x(0) + x(1) + x(2) + x(3) = 1+1+0+0 =2 𝐀𝐬 Sin θ = 2j
& Cos θ = 2
3 − jπn/2 𝜋 𝜋
 X(1) = 𝑛=0 x(n) e = x(0) + x(1) e-jπ/2 + x(2) e-jπ + x(3) e-j3π/2 = 1 + cos 2 - 𝑗 sin 2 +0+0= 1 – j
3 − jπn
 X(2) = 𝑛=0 x(n) e = x(0) + x(1) e-jπ + x(2) e-j2π + x(3) e-j3π = 1 + cos 𝜋 - 𝑗 sin 𝜋 = 1 - 1 = 0
3 − j3πn/2 3𝜋 3𝜋
 X(3) = 𝑛=0 x(n) e = x(0) + x(1) e-j3π/2 + x(2) e-j3π + x(3) e-j9π/2 = 1 + cos 2 - 𝑗 sin 2 = 1 + j
 X(k) = {2, 1 - j, 0, 1 + j}
𝟏
 EXAMPLE - 2: - Compute the DFT of the sequence: 𝐱 𝐧 = {𝟒 , 𝐟𝐨𝐫 𝟎 ≤ 𝐧 ≤ 𝟐 ; and 𝟎, 𝐎𝐭𝐡𝐞𝐫𝐰𝐢𝐬𝐞}
 SOLUTION: -From the above expression, the sequence is given by, x(n) = { ¼ , ¼ , ¼ }
 We know that the N-point DFT of the sequence x(n) is expressed as
X(k) = 𝐍−𝟏
𝒏=𝟎 𝐱(𝐧) 𝐞
− 𝐣𝟐𝛑𝐧𝐤/𝐍
, k = 0, 1,… N-1 [AS 𝐞𝐣𝛉 = 𝑪𝒐𝒔𝜽 + 𝒋𝑺𝒊𝒏 𝜽 & 𝐞−𝐣𝛉 = 𝑪𝒐𝒔𝜽 − 𝒋𝑺𝒊𝒏 𝜽]

<DIGITAL SIGNAL PROCESSING>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DSP [ETT 603] <FOURIER TRANSFORM: ITS APPLICATIONS PROPERTIES > [Page 4.2]
1 1 − jω 1
 Therefore, X(k) = 4 [1 + e− jω + e− 2jω ] = e + 1 + e− jω ]
[
𝜔=2πk/N 4 e jω
−
𝜔 =2πk/N
1 − jω jω − jω 1 − jω jω − jω 1
= 4e [e + 1 + e ] = 4e [1 + e + e ] = X(k) = 4 e− jω [1+2cos 𝜔]
𝜔 =2πk/N 𝜔=2πk/N
1 − j2πk/3 2𝜋𝑘
Where 𝝎 = 𝟐𝛑𝐤/𝐍 and N = 3  Therefore, X(k) = 4 e 1 + 2 cos 3
𝟏 − 𝐣𝟐𝛑𝐤/𝟑 𝟐𝝅𝒌
 X(k) = 𝟒 𝐞 𝟏 + 𝟐 𝐜𝐨𝐬 , Where k = 0, 1,… N-1
𝟑
𝟏
 EXAMPLE - 3:- Compute DFT of the sequence: 𝐱 𝐧 = {𝟓 , 𝐟𝐨𝐫 − 𝟏 ≤ 𝐧 ≤ 𝟏 ; and 𝟎, 𝐎𝐭𝐡𝐞𝐫𝐰𝐢𝐬𝐞}
 SOLUTION: - From the above expression
1 𝟏 1
the sequence is given by, x(n) = , ,
5 𝟓 5
 N-point DFT of the Signal x(n) is expressed
as X(k) = ∞ 𝒏=−∞ 𝐱(𝐧) 𝐞
− 𝐣𝛚𝐧
, 𝜔 = 2πk/N
1 jω −jω
 Thus, X(k) = [e + 1 + e ]
5 at ω=2πk/N
1 𝟏 𝟐𝝅𝒌
= 5 [1+2cos ω] = 𝟏 + 𝟐 𝐜𝐨𝐬 ;
𝟓 𝟑
2πk
Where k = 0, 1, … N-1, 𝜔 = and N = 3
N
 EXAMPLE - 4: -
 Derive the DFT of the sample data sequence
x(n) = {1, 1, 2, 2, 3, 3}.
 SOLUTION:
 We know N-point DFT of the Signal x(n) is
X(k) = 𝐍−𝟏𝒏=𝟎 𝐱(𝐧) 𝐞
− 𝐣𝟐𝛑𝐧𝐤/𝐍
, k = 0,1,…N-1
5
 For k = 0  X(0) = 𝑛 =0 x(n)e− j2π(0)n/6
= 5𝑛=0 x(n) = 1 + 1 + 2 + 2 + 3 + 3 = 12
 Similarly, X(1) = 5𝑛=0 x(n)e− j2π(1)n/6
= 5𝑛=0 x(n) e-jπn/3 = 1 + e-jπ/3 + 2e-j2π/3 +2e-jπ
+ 3e-j4π/3 + 3e-j5π/3
= 1 + (0.5 - j0.866) + 2(-0.5- j0.866) + 2(-1) +
3(-0.5- j0.866) + 3(0.5- j0.866) = -1.5 + j2.598
 X(2) = 5𝑛=0 x(n)e− j2π(2)n/6 = 5𝑛=0 x(n) e-2jπn/3 = 1 + e-j2π/3 + 2e-j4π/3 + 2e-j2π + 3e-j8π/3 + 3e-j10π/3
= 1 + (-0.5-j0.866) + 2(-0.5+ j0.866) + 2(1) + 3(-0.5- j0.866) + 3(-0.5+ j0.866) = -1.5 + j0.866
 X(3) = 5𝑛=0 x(n)e− j2π(3)n/6 = 5𝑛=0 x(n) e-jπn = 1 + e-jπ + 2e-j2π + 2e-j3π + 3e-j4π + 3e-j5π
= 1 – 1 + 2(1) + 2(-1) + 3(1) + 3(-1) = 0
 X(4) = 𝑛=0 x(n)e− j2π(4)n/6 = 5𝑛=0 x(n) e-j4πn/3 = 1 + e-j4π/3 + 2e-j8π/3 + 2e-j4π + 3e-j14π/3 + 3e-j20π/3
5

= 1 + (-0.5+j0.866) + 2(-0.5-j0.866) + 2(1) + 3(-0.5+j0.866) + 3(-0.5- j0.866) = - 1.5 - j0.866

 X(5) = 5𝑛=0 x(n)e− j2π(5)n/6 = 5𝑛=0 x(n) e-j5πn/3 = 1 + e-j5π/3 + 2e-j10π/3 + 2e-j5π + 3e-j20π/3 + 3e-j25π/3
= 1 + (-0.5+j0.866) + 2(-0.5+j0.866) + (-1) + 3(-0.5- j0.866) + 3(0.5- j0.866) = - 1.5 - j2.598
 Hence, X(k) = {12, -1.5 + j2.598, -1.5 + j0.866, 0, -1.5 - j0.866, -1.5 - j2.598}

<DIGITAL SIGNAL PROCESSING>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DSP [ETT 603] <FOURIER TRANSFORM: ITS APPLICATIONS PROPERTIES > [Page 4.3]
𝒏𝝅
 EXAMPLE - 5: - Compute the 4 - point DFT of the sequence x(n) = 𝐜𝐨𝐬 𝟒
𝜋 𝜋 3𝜋
 SOLUTION: Given that, N = 4  So x(n) = cos 0 , cos , cos , cos = [1, 0.707, 0, -0.707]
2 4
4
 We know that the N-point DFT of x(n) is expressed as X(k) = 𝐍−𝟏 𝒏=𝟎 𝐱(𝐧) 𝐞 − 𝐣𝟐𝛑𝐧𝐤/𝐍
, k = 0, 1,… N-1
3 − j2πnk /4 3 − jπnk /2
 The DFT is X(k) = 𝑛=0 x(n) e , k = 0, 1, 2, 3  X(k) = 𝑛=0 x(n) e , k = 0, 1, 2, 3
3 3
 For k = 0, we have, X(0) = 𝑛=0 x(n) = 1; For k = 1, we have, X(1) = 𝑛=0 x(n) e− jπ(1)n/2
 X(1) = 1 + 0.707e-π/2 + 0+ (-0.707)e-j3π/2 = 1 + (0.707) (-j) + 0 - (0.707) (j) = 1 - j1.414
 For k = 2, we have, X(2) = ∞𝑛=−∞ x(n) e
− jπ(2)n/2
= ∞𝑛=−∞ x(n) e
− jπn
-jπ -j3π
X(2) = 1 + 0.707e + 0 + (-0.707)e = 1 + 0.707(-1) + 0 + (-0.707)(-1) = 1
 For k = 3, we have, X(3) = 3𝑛=0 x(n) e− jπ(3)n/2 = 1 + 0.707e-3π/2 + 0 + (-0.707)e-j9π/2
Or X(3) = 1 + 0.707(j) + 0 + (-0.707)(-j) = 1 + j1.414  X(k) = [1, 1-j1.414, 1, 1 + j1.414]
 INVERSE DISCRETE FOURIER TRANSFORM (IDFT): -
𝟏
 The IDFT for a sequence is expressed as x(n) = 𝑵 𝐍−𝟏 𝒌=𝟎 𝐗(𝐤) 𝐞
𝐣𝟐𝛑𝐧𝐤/𝐍
, n = 0, 1, 2, 3, 4 … N-1
 EXAMPLE - 6: - Find the IDFT of a sequence Y(k) = {1, 0, 1, 0}.
𝟏
 SOLUTION: We know that the N-point IDFT of x(n) is y(n) = 𝑵 𝐍−𝟏 𝒌=𝟎 𝐘(𝐤) 𝐞
𝐣𝟐𝛑𝐧𝐤/𝐍
, n = 0, 1,… N-1
1 3 1 1
 For n= 0  y(0) = 4 𝑘=0 Y(k)= 4 [y(0) + y(1) + y(2) + y(3)] = 4 [1 + 0 + 1 + 0] = 0.5
1
 Similarly for n = 1, y(1) = 𝑁 3𝑘=0 Y(k) ejπk/2
1 1 1
 y(1) = 4 [Y(0) + Y(1)ejπ/2 + Y(2)ejπ + Y(3)ej3π/2] = 4 [1 + 0 + cos 𝜋 + 𝑗 sin 𝜋 + 0] = 4 [1 + 0 – 1 + 0] = 0
1 3 jπk 1
 For n = 2; y(2) = 4 𝑘=0 Y(k) e = 4 [Y(0) + Y(1) ejπ + Y(2) ej2π + Y(3) ej3π]
1 1
= 4 [1 + 0 + cos 2𝜋 + 𝑗 sin 2𝜋 + 0] = 4 [1 + 0 + 1 + 0] = 0.5
1 3 j6πk/4 1
 For n = 3; y(3) = 4 𝑘=0 Y(k) e = 4 [Y(0) + Y(1)ej3π/2 + Y(2)ej3π + Y(3)ej9π/2]
1 1
= [1 + 0 + cos 3𝜋 + 𝑗 sin 3𝜋 + 0] = [1 + 0 – 1 + 0] = 0  Thus y(n) = {0.5, 0, 0.5, 0}
4 4
 EXAMPLE - 7: - Compute the inverse DFT of a sequence X(k) = {1,2,3,4}.
𝟏
 SOLUTION: We know that the DFT is expressed as, x(n) = 𝑵 𝐍−𝟏
𝒌=𝟎 𝐗(𝐤) 𝐞
𝐣𝟐𝛑𝐧𝐤/𝐍
, n = 0, 1,… N-1
𝟏𝟑 𝐣𝟐𝛑𝐧𝐤/𝐍
 Given that N = 4, Therefore, x(n) = 𝟒 𝒌=𝟎 𝐗(𝐤) 𝐞 , n = 0, 1, 2, 3
1 3 jπ(0)k/2 1
 When n = 0, we have x(0) = 4 𝑘=0 X(k)e = 4 [1 + 2 + 3 + 4] = 5/2
1 3 jπ(1)k/2 1
 When n = 1, we have x(1) = 4 𝑘=0 X(k)e = 4 [1 + 2ejπ/2 + 3ejπ + 4ej3π/2]
1 1 𝟏 𝟏
= 4 [1 + 2(j) + 3(-1) + 4(-j)] = 4(-2 - j2) = - 𝟐 - j𝟐
1 3 jπ(2)k/2 1
 When n = 2, we have x(2) = 4 𝑘=0 X(k)e = 4 [1 + 2ejπ + 3ej2π + 4ej3π]
1 1 𝟏
= 4 [1 + 2(-1) + 3(1) + 4(-1)] = 4 (-2) = - 𝟐
1 3 jπ(3)k/2 1
 When n = 3, we have x(3) = 4 𝑘=0 X(k)e = 4 [1 + 2ej3π/2 + 3ej3π + 4ej9π/2]
1 1 𝟏 𝟏 𝟓 𝟏 𝟏 𝟏 𝟏 𝟏
= 4 [1 + 2(-j) + 3(-1) + 4(j)] = 4(-2 + j2) = - 𝟐 + j𝟐  x(n) = ,− − 𝐣𝟐,− 𝟐,− + 𝐣𝟐
𝟐 𝟐 𝟐
 EXAMPLE - 8:- Find the IDFT of a sequence X(k) = {3, (2+j), 1, (2-j)}.
𝟏
 SOLUTION: The IDFT is expressed as x(n) = 𝑵 𝐍−𝟏
𝒌=𝟎 𝐗(𝐤) 𝐞
𝐣𝟐𝛑𝐧𝐤/𝐍
, 0 ≤ n ≤ N-1
𝟏 𝟑 𝐣𝛑𝐧𝐤/𝟐
 Here, given that N = 4, Therefore, x(n) = 𝟒 𝒌=𝟎 𝐗(𝐤) 𝐞 , 0≤n≤ 3
1 3 0 1
 When n = 0, we have x(0) = 4 𝑘=0 X(k)e = 4[3 + (2+j) + 1 + (2-j)] = 2
1 3 jπk/2 1
 When n = 1, we have x(1) = 4 𝑘=0 X(k)e =4 [3 + (2+j)ejπ/2 + ejπ + (2-j)ej3π/2]
1
= 4 [3 + (2+j)(j) + 1(-1) + (2-j)(-j)] = 0
1 3 jπk 1
 When n = 2, we have x(2) = 𝑘=0 X(k)e = [3 + (2+j)ejπ + ej2π + (2-j)ej3π]
4 4
1
= 4 [3 + (2+j) (-1) + (1) + (2-j) (-1)] = 0

<DIGITAL SIGNAL PROCESSING>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DSP [ETT 603] <FOURIER TRANSFORM: ITS APPLICATIONS PROPERTIES > [Page 4.4]
1 3 jπ3k/2 1
 When n = 3, we have x(3) = 4 𝑘=0 X(k)e = 4 [3 + (2+j)ej3π/2 + ej3π + (2-j)ej9π/2]
1
= 4 [3 + (2+j) (-j) + (-1) + (2-j) (j)] = 1  Hence x(n) = [2, 0, 0, 1]
 EXAMPLE - 9: -
1, for 0 ≤ n ≤ 2
 Find DFT of the sequence x n = for [1] N = 4 & [2] N = 8; Plot 𝑋(𝑘) & ∠X(k).
0, otherwise
 SOLUTION: - From the above expression x(n) = {1, 1, 1, 0}
Given L = 3, so for N = 4, the periodic extension of x(n) can be obtained by adding one zero [Zero Padding]

 Thus x(0) = 1; x(1) = 1; x(2) = 1; x(3) = 0 ; But We have X(k) = 𝐍−𝟏 𝒏=𝟎 𝐱(𝐧) 𝐞
− 𝐣𝟐𝛑𝐧𝐤/𝐍
, k = 0, 1,… N-1
 FOR N = 4 𝟑
X(k) = 𝒏=𝟎 𝐱(𝐧) 𝐞 − 𝐣𝛑𝐧𝐤/𝟐
, k = 0, 1, 2, 3
3
 For k = 0, X(0) = 𝑛=0 x(n) = x(0) + x(1) + x(2) + x(3) = 1+1+1+0 = 3 Thus, 𝑿(𝟎) = 3, ∠X(0) = 0
 For k = 1, X(1) = 3𝑛=0 x(n) e− jπn/2 = x(0) + x(1)e-jπ/2 + x(2) e-jπ + x(3) e-j3π/2
𝜋 𝜋 𝝅
= 1 + cos 2 - 𝑗 sin 2 + cos 𝜋 - 𝑗 sin 𝜋 + 0= 1 – j – 1 = - j ; Therefore, 𝑿(𝟏) = 1, ∠X(1) = − 𝟐
 For k = 2, X(2) = 3𝑛=0 x(n) e− jπn = x(0) + x(1)e-jπ + x(2) e-j2π + x(3) e-j3π
= 1 + cos 𝜋 - 𝑗 sin 𝜋 + cos 2𝜋 - 𝑗 sin 2𝜋 + 0= 1 – 1 +1= 1; Therefore, 𝑿(𝟐) = 1, ∠X(2) = 0
 For k = 3, X(3) = 3𝑛=0 x(n) e− j3πn/2 = x(0) + x(1)e-j3π/2 + x(2) e-j3π + x(3) e-j9π/2
3𝜋 3𝜋 𝝅
= 1 + cos 2 - 𝑗 sin 2 + cos 3𝜋 - 𝑗 sin 3𝜋 + 0= 1 + j-1 = j Therefore, 𝑿(𝟑) = 1, ∠X(3) = 𝟐
𝝅 𝝅
 Thus 𝑿 𝒌 = {3, -j, 1, j}  𝑿(𝒌) = {3, 1, 1, 1} and ∠X(k) = 𝟎, − , 𝟎, ; 𝑿(𝒌) = 𝒂𝟐 + 𝒃𝟐
𝟐 𝟐
 FOR N = 8, x(n)={1,1,1,1,1,1,0,0}
 Given L = 3, thus the periodic
extension of x(n) can be obtained
by adding Five zero.
 Thus, we have x(0) = x(1) = x(2) =
1; x(n) = 0 for 3 ≤ n ≤ 7
 We know that X(k) = N−1 𝑛=0 x(n) e
− j2πnk /N
For N = 8, X(k) = 𝟕𝒏=𝟎 𝐱(𝐧) 𝐞− 𝐣𝛑𝐧𝐤/𝟒 , k = 0, 1, 2,…,7
 For k = 0, X(0) = 7𝑛=0 x(n) = 1 + 1 + 1 + 0 + 0 + 0 + 0 + 0 = 3 ; Therefore, 𝑿(𝟎) = 3, ∠X(0) = 0
 For k = 1, X(1) = 7𝑛=0 x(n) e− jπn/4 = x(0) + x(1)e-jπ/4 + x(2) e-jπ/2
𝝅
= 1 + 0.707 – j0.707 + 0 – j = 1.707 – j1.707 ; Therefore, 𝑿(𝟏) = 2.414, ∠X(1) = − 𝟒
 For k = 2, X(2) = 7𝑛=0 x(n) e− jπn/2 = x(0) + x(1)e-jπ/2 + x(2) e-jπ
𝜋 𝜋 𝝅
= 1 + cos 2 - 𝑗 sin 2 + cos 𝜋 - 𝑗 sin 𝜋 = 1 – j -1= -j ; Therefore, 𝑿(𝟐) = 1, ∠X(2) = − 𝟐
 For k = 3, X(3) = 7𝑛=0 x(n) e− j3πn/4 = x(0) + x(1)e-j3π/4 + x(2) e-j3π/2
3𝜋 3𝜋 3𝜋 3𝜋
= 1 + cos 4 - 𝑗 sin 4 + cos 2 - 𝑗 sin 2
𝝅
= 1 – 0.707 – j0.707 + j = 0.293 + j0.293; Therefore, 𝑿(𝟑) = 0.414, ∠X(3) = 𝟒
 For k = 4, X(4) = 7𝑛=0 x(n) e− jπn = x(0) + x(1)e-jπ + x(2) e-j2π
= 1 + cos 𝜋 - 𝑗 sin 𝜋 + cos 2𝜋 - 𝑗 sin 2𝜋 = 1 – 1 + 1 = 1; Therefore, 𝑿(𝟒) = 1, ∠X(1) = 0
 For k = 5, X(5) = 7𝑛=0 x(n) e− j5πn/4 = x(0) + x(1)e-j5π/4 + x(2) e-j5π/2
5𝜋 5𝜋 5𝜋 5𝜋
= 1 + cos 4 - 𝑗 sin 4 + cos 2 - 𝑗 sin 2 = 1 – 0.707 + j0.707 – j
𝝅
= 0.293 – j0.293 ; Therefore, 𝑿(𝟓) = 0.414, ∠X(5) = − 𝟒
 For k = 6, X(6) = 7𝑛=0 x(n) e− j3πn/2 = x(0) + x(1)e-j3π/2 + x(2) e-j3π/2
3𝜋 3𝜋 𝝅
= 1 + cos 2 - 𝑗 sin 2 + cos 3𝜋 - 𝑗 sin 3𝜋 = 1 + j – 1 = j; Thus, 𝑿(𝟔) = 1, ∠X(6) = − 𝟐

<DIGITAL SIGNAL PROCESSING>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DSP [ETT 603] <FOURIER TRANSFORM: ITS APPLICATIONS PROPERTIES > [Page 4.5]
7 − j7πn/4 -j7π/4 -j7π/2
 For k = 7, X(7) = 𝑛=0 x(n) e = x(0) + x(1)e + x(2) e
7𝜋 7𝜋 7𝜋 7𝜋
= 1 + cos - 𝑗 sin + cos - 𝑗 sin = 1 + 0.707 + j0.707 + j
4 4 2 2
𝝅
= 1.707 + j1.707 ; Therefore, 𝑿(𝟕) = 2.414, ∠X(7) = 𝟒
𝝅 𝝅 𝝅 𝝅 𝝅 𝝅
 Thus 𝑿(𝒌) = {3, 2.414, 1, 0.414, 1, 0.414, 1, 2.414} and ∠X(k) = 𝟎, − , − , , 𝟎 , − , − ,
𝟒 𝟐 𝟒 𝟒 𝟐 𝟒
The plot of 𝑋(𝑘) and ∠X(k) for N = 4 & 8 is shown in figure.

 NOTE: - From figure we can observe that it is difficult to extrapolate the entire frequency spectrum,
with N = 4. That is the resolution of the spectrum is very poor. In order to increase the resolution, we
must increase N. in figure it is possible to extrapolate the frequency spectrum, where the value of N =8.
Thus we concluded that the Zero Padding gives a high density spectrum and provides a better display
version for plotting.
 EXAMPLE - 10: Determine the 8-point DFT of the sequence x(n) = {1, 1, 1, 1, 1, 1, 0, 0}
 SOLUTION: We know X(k) = 𝐍−𝟏 𝒏=𝟎 𝐱(𝐧) 𝐞
− 𝐣𝟐𝛑𝐧𝐤/𝐍
, k = 0, 1,… N-1
𝟕 − 𝐣𝛑𝐧𝐤/𝟒
 For given N = 8; X(k) = 𝒏=𝟎 𝐱(𝐧) 𝐞 , k = 0, 1, 2,…,7
 For k = 0; X(0) = 7𝑛=0 x(n) = x(0) + x(1) + x(2) + x(3) + x(4) + x(5) + x(6) + x(7)
=1+1+1+1+1+1+0+0=6
 For k = 1; X(1) = 7𝑛=0 x(n) e− jπn/4
= x(0) + x(1)e-jπ/4 + x(2) e-jπ/2 + x(3)e-j3π/4 + x(4) e-jπ+ x(5)e-j5π/4 + x(6) e-j3π/2 + x(7)e-j7π/4
= 1 + 0.707 – j0.707 – j - 0.707 – j0.707 – 1 - 0.707 + j0.707 = -0.707 – j1.707
 For k = 2; X(2) = 7𝑛=0 x(n) e− jπn/2
= x(0) + x(1)e-jπ/2 + x(2) e-jπ + x(3)e-j3π/2 + x(4) e-j2π+ x(5)e-j5π/2 + x(6) e-j3π + x(7)e-j7π/2
=1–j–1+j+1–j=1-j
 For k = 3 ; X(3) = 7n=0 x(n) e− j3πn/4
= x(0)+x(1)e-j3π/4+x(2) e-j3π/2+x(3)e-j9π/4+x(4) e-j3π+x(5)e-j15π/4+x(6) e-j9π/2+x(7)e-j21π/4
= 1 - 0.707 – j0.707 + j + 0.707 – j0.707 – 1 + 0.707 + j0.707 = 0.707 + j0.293
 For k = 4 ;X(4) = 7𝑛=0 x(n) e− jπn
= x(0) + x(1)e-jπ + x(2) e-j2π + x(3)e-j3π + x(4) e-j4π+ x(5)e-j5π + x(6) e-j6π + x(7)e-j7π
=1–1+1-1+1–1=0
 For k = 5 ;X(5) = 7n=0 x(n) e− j5πn/4
= x(0)+x(1)e-j5π/4+x(2)e-j5π/2+x(3)e-j15π/4 + x(4) e-j5π+ x(5)e-j25π/4 + x(6) e-j15π/2 + x(7)e-j35π/4
= 1 - 0.707 + j0.707 – j + 0.707 + j0.707 – 1 + 0.707 - j0.707 = 0.707 – j0.293
 For k = 6; X(6) = 7n=0 x(n) e− j3πn/2
= x(0) + x(1)e-j3π/2 + x(2) e-j3π + x(3)e-j9π/2 + x(4) e-j6π+ x(5)e-j15π/2 + x(6) e-j9π + x(7)e-j21π/2
=1+j–1-j+1+j=1+j
 For k = 7; X(7) = 7𝑛=0 x(n) e− j7πn/4
= x(0)+x(1)e-j7π/4+x(2)e-j7π/2+ x(3)e-j21π/4 + x(4) e-j7π+ x(5)e-j35π/4 + x(6) e-j21π/2 + x(7)e-j49π/4
= 1 + 0.707 + j0.707 + j - 0.707 + j0.707 – 1 - 0.707 - j0.707 = -0.707 + j1.707
X(k) = {6, -0.707 - j1.707, 1 - j, 0.707 + j0.293, 0, 0.707 – j0.293, 1 + j, -0.707 + j1.707}
 EXAMPLE - 11: -Find IDFT of the sequence X(k) = {5, 0, 1-j, 0, 1, 0, 1+j, 0}
1
 SOLUTION: We have x(n) = 𝑁 N−1 𝑘=0 X(k) e
j2πnk /N
, n = 0, 1,… N-1
𝟏 𝟕 𝐣𝛑𝐧𝐤/𝟒
 For N = 8, x(n) = 𝟖 𝒌=𝟎 𝐗(𝐤) 𝐞 , n = 0, 1, 2, …,7
1 7 1
 For n = 0; x(0) = 8 𝑘=0 X(k) = 8[5 + 0 + 1-j + 0 + 1 + 0 + 1+j + 0] = 1
1 7 jπk/4 1 1
 For n = 1; x(1) = 8 𝑘=0 X(k) e = 8[5 + 0 + (1-j)(j) + 0 + 1(-1) + 0 + (1+j)(-j) + 0] = 8[6] = 0.75
<DIGITAL SIGNAL PROCESSING>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DSP [ETT 603] <FOURIER TRANSFORM: ITS APPLICATIONS PROPERTIES > [Page 4.6]
1 7 jπk/2 1 1
 For n = 2; x(2) = 8 𝑘=0 X(k) e = 8[5 + 0 + (1-j)(-1) + 0 + 1(1) + 0 + (1+j)(-1) + 0] = 8[4] = 0.5
1 7 j3πk/4 1 1
 For n = 3; x(3) = 8 𝑘=0 X(k) e = 8[5 + 0 + (1-j)(-j) + 0 + 1(-1) + 0 + (1+j)(j) + 0] = 8[2] = 0.25
1 7 jπk 1 1
 For n = 4; x(4) = 8 𝑘=0 X(k) e = 8[5 + 0 + (1-j)(1) + 0 + 1(1) + 0 + (1+j)(1) + 0] = 8[8] = 1
1 7 j5πk/4 1 1
 For n = 5; x(5) = 8 𝑘=0 X(k) e = 8[5 + 0 + (1-j)(j) + 0 + 1(1) + 0 + (1+j)(-j) + 0] = 8[6] = 0.75
1 7 j3πk/2 1 1
 For n = 6; x(6) = 8 𝑘=0 X(k) e = 8[5 + 0 + (1-j)(-1) + 0 + 1(1) + 0 + (1+j)(-1) + 0] = 8[4] = 0.5
1 7 j7πk/4 1 1
 For n = 7; x(7) = 8 𝑘=0 X(k) e = 8[5 + 0 + (1-j)(-j) + 0 + 1(-1) + 0 + (1+j)(j) + 0] = 8[2] = 0.25
x(n) = {1, 0.75, 0.5, 0.25, 1, 0.75, 0.5, 0.25}
 PRACTICE PROBLEMS:
i. Compute 4-point DFT of the sequence x(n) = {0, 2, 4, 6} Ans : 12, -4 + j4, -4, -4 - j4
ii. Compute 8-point DFT of the sequence x(n) = {2, 22, 23, 24} = {2, 4, 8, 16}
Ans: {30, -6.48 - j22, -6 + j12, 10.48 - j6.14, -10, 10.48 + j6.4, -6 - j12, -6.48 + j22}
 RELATIONSHIP OF THE DFT TO OTHER TRANSFORM:-
 RELATIONSHIP TO THE FOURIER TRANSFORM
 The Fourier transform X(ejω) of a finite duration sequence x(n) having length N is given by
X(ejω) = 𝐍−𝟏
𝐧=𝟎 𝐱(𝐧) 𝐞
𝐣𝛚𝐧
…………… (1) Where X(ejω) is a continuous function of ω.
 The discrete Fourier transform of x(n) is given by X(k) = 𝐍−𝟏
𝒏=𝟎 𝐱(𝐧) 𝐞
− 𝐣𝟐𝛑𝐧𝐤/𝐍
, k = 0, 1,… N-1…(2)
 Comparing equation 1 & 2, we find that the DFT of x(n) is a sampled version of the Fourier transform
of the sequence and is given by, X(k) = 𝑿(𝐞 𝐣𝛚 ) 𝝎=𝟐𝛑𝐤/𝐍, k = 0, 1,… N-1 ……… (3)
 RELATIONSHIP TO THE Z- TRANSFORM
𝐍−𝟏 −𝐧
 Let us consider a sequence x(n) of finite duration N with z-transform X(z) = 𝒏=𝟎 𝐱(𝐧) 𝐳 ………(4)
𝟏
 We have x(n) = 𝑵 𝐍−𝟏 𝒌=𝟎 𝐗(𝐤) 𝐞
𝐣𝟐𝛑𝐧𝐤/𝐍
………………………(5)
 Substituting equation 5 in 4,
N−1 𝟏 𝐍−𝟏 j2πnk /N 1 N−1 N−1 n 𝟏−𝐳 −𝐍 𝐍−𝟏 𝑿(𝒌)
X(z) = 𝑛=0 𝑵 𝒌=𝟎 𝐗(𝐤) e z −n = 𝑁 𝑘=0 X(k) 𝑛=0 ej2πk/N z −1 = 𝒌=𝟎 𝟏−𝐞𝐣𝟐𝛑𝐤/𝐍 𝐳 −𝟏
𝑵
 PROPERTIES OF DFT:-
 The properties of DFT can be listed as under:
1) Periodicity 7) Complex conjugate property
2) Linearity 8) Circular convolution
3) Shifting property 9) Circular correlation
4) Time reversal of a sequence 10) Multiplication of two sequences
5) Circular time shift 11) Parseval‫׳‬s Theorem
6) Circular frequency shift
1. PERIODICITY
 This property states that if a discrete-time signal is periodic then its DFT will also be periodic.
 Also, if a signal or sequence repeats its waveform after N number of samples then it is called a periodic
signal or sequence and N is called the period of signal.
 Mathematically, If X(k) is an N-point DFT of x(n), then we have
x(n + N) = x(n) for all values of n & X(k + N) = X(k) for all values of k
2. LINEARITY
 Linearity properties states that if X1(k) and X2(k) are the N-point DFTs of x1(n) and x2(n) respectively,
and a and b are arbitrary constants either real or complex-valued, then we have
𝐷𝐹𝑇
ax1(n) + bx2(n) 𝑁 aX1(k) + bX2(k)
3. SHIFTING PROPERTY
 This property can be explained as follows:
 Let xp(n) is a periodic sequence with period N, which is obtained by extending x(n) periodically i.e.
xp(n) = ∞ 𝑙=−∞ x(n − 𝑙N).
 Now, let us shift the sequence xp(n) by k units to the right.
 Again, let resultant signal be expressed as xp(n) is expressed as xp(n) = xp(n-k) = ∞
𝑙=−∞ x(n − k − 𝑙N)

<DIGITAL SIGNAL PROCESSING>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DSP [ETT 603] <FOURIER TRANSFORM: ITS APPLICATIONS PROPERTIES > [Page 4.7]
 The finite duration sequence x′ n = {xp′ (n), for 0
≤ n ≤ N − 1 ; and 0, Otherwise} can be obtained
from x(n) by a circular shift.
 The circular shift of a sequence can be represented as the index modulo N, x‫(׳‬n) = x[n-k, (mod N)]
4. TIME REVERSAL OF A SEQUENCE
𝐷𝐹𝑇
 This property of DFT states that, If x(n) 𝑁 X(k), then, we have
𝐷𝐹𝑇
x [-n, (mod N)] = x(N - n) 𝑁 X(k, (mod N)) = X(N - k)
 Thus, when the N-point sequence in time is reversed, it is equivalent to reversing the DFT values.
5. CIRCULAR TIME SHIFT
𝐷𝐹𝑇 𝐷𝐹𝑇
−j2πk/N
 This property of DFT states that, If x(n) 𝑁 X(k), then x[n-l, (mod N)] 𝑁 X(k) e
 This means that shifting of the sequence by l units in the time-domain is equivalent to multiplication of
e−j2πk/N in the frequency-domain.
6. CIRCULAR FREQUENCY SHIFT
𝐷𝐹𝑇 𝐷𝐹𝑇
j2π𝑙n/N
 This property of DFT states that, If x(n) 𝑁 X(k), then x(n)e 𝑁 X(k - l, (mod N))
 This means that when the sequence x(n) is multiplied by the complex exponential sequence ej2π𝑙n/N , it
is equivalent to circular shift of the DFT by l units in the frequency domain.
7. COMPLEX CONJUGATE PROPERTY
𝐷𝐹𝑇
 This property of DFT states that, If x(n) 𝑁 X(k), then
𝐷𝐹𝑇 𝐷𝐹𝑇 1 1 ∗
N−1 ∗ j2πkn /N N−1 j2πk(N−n)/N
x*(n) 𝑁 X*(-k,(mod N))=X*(N-k) & X*(k) 𝑁 𝑁 𝑘=0 X (n)e = 𝑁 𝑘=0 X(n)e
𝐷𝐹𝑇
 Hence x*[-n, (mod N)] = x*(N-k) 𝑁 X*(k)
8. CIRCULAR CONVOLUTION: - Circular convolution property states that
𝐷𝐹𝑇 𝐷𝐹𝑇 𝐷𝐹𝑇
If x1(n) 𝑁 X1(k) and x2(n) 𝑁 X2(k) then x1(n) x2(n) 𝑁 X1(k) X2(k)
 Where x1(n) N x2(n) denotes the circular convolution of the sequence x1(n) and x2(n) expressed as
x3(n) = N−1 N−1
𝑚 =0 x1 (m)x2 (n − m, (mod N))= 𝑚 =0 x1 (m)x1 (n − m, (mod N))
9. CIRCULAR CORRELATION
 This property of DFT states that for complex valued sequence x(n) and y(n),
𝐷𝐹𝑇 𝐷𝐹𝑇 𝐷𝐹𝑇
If x(n) 𝑁 X(k) and y(n) 𝑁 Y(k), then rxy(l) 𝑁 Rxy(k) = X(k) Y*(k)
 Where rxy(l) is the (un-normalized) circular cross-correlation sequence, given as
rxy(l) = N−1 ∗
𝑛=0 x(n)y (n − 𝑙, (mod N))
10. MULTIPLICATION OF TWO SEQUENCES: - This property of DFT states that
𝐷𝐹𝑇 𝐷𝐹𝑇 𝐷𝐹𝑇 1
If x1(n) 𝑁 X1(k) and x2(n) 𝑁 X2(k) then x1(n) x2(n) 𝑁 𝑁 X1(k) X2(k)
11. PARSEVAL‫׳‬S THEOREM
 This property of DFT states that for complex valued sequence x(n) and y(n),
𝐷𝐹𝑇 𝐷𝐹𝑇 1
N−1 ∗ N−1 ∗
 If x(n) 𝑁 X(k) and y(n) 𝑁 Y(k), then 𝑛=0 x(n)y (n) =𝑁 𝑘=0 X(k)Y (k)
N−1 1
 If y(n) = x(n), then the above equation reduces to N−1 2
𝑛=0 x(n) = 𝑁 𝑘=0 X(k)
2

 It relates the energy in finite duration sequence x(n) to the power in the frequency components X(k).
 PROPERTIES OF THE DFT: -
SN PROPERTIES TIME DOMAIN FREQUENCY DOMAIN
1. PERIODICITY x(n) = X(n + N) X(k) = X(k + N)
2. LINEARITY a1x1(n) + a2x2(n) a1X1(k) + a2X2(k)
3. TIME REVERSAL OF A SEQUENCE x(N - n) X(N - k)
4. CIRCULAR TIME SHIFT x((n - l))N X(k) e-j2πkl/N
5. CIRCULAR FREQUENCY SHIFT x(n) ej2πln/N X((k - l))N
6. CIRCULAR CONVOLUTION x1(n) x2(n) X1(k) X2(k)
7. CIRCULAR CORRELATION x1(n) y* (-n) X(k) Y* (k)
1
8. MULTIPLICATION OF TWO SEQUENCES x1(n) x2(n) [X1(k) X2(k)]
𝑁
<DIGITAL SIGNAL PROCESSING>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DSP [ETT 603] <FOURIER TRANSFORM: ITS APPLICATIONS PROPERTIES > [Page 4.8]
9. COMPLEX CONJUGATE PROPERTY x* (n) X* (N - k)
N−1 ∗ 1 N−1 ∗
10. PARSEVAL‫׳‬S THEOREM 𝑛=0 x(n)y (n) 𝑁 𝑘=0 X(k)Y (k)

 Comparison Between CIRCULAR CONVOLUTION & LINEAR CONVOLUTION:-

 The linear convolution of two sequences x(n) of L number of samples and h(n) of M number of
samples produce a result y(n) which contains N = L + M – 1 samples.
 In the same case of circular convolution the situation is entirely different.
 If x(n) contains L number of samples and h(n) has M number of samples and that L > M, then we
perform circular convolution between the two using N = Max (L, M), by adding L – M number of zero
samples to the sequence h(n), so that both sequence are periodic with N.
 Linear convolution can be used to find the response of a filter.
 Circular convolution cannot be used to find the response of a linear filter without zero padding.
 The methods used to find the circular convolution of two sequences are
(1) Concentric Circle method (2) Matrix multiplication method
 Methods To Evaluate CIRCULAR CONVOLUTION of Two Sequences:-
 CONCENTRIC CIRCLE METHOD
 Given two sequences x1(n) & x2(n), the circular convolution of two sequences x3(n) = x1(n) x2(n)
can be found by using the following steps.
1. Graph N samples of x1(n) as equally spaced points around an outer circle in counterclockwise way.
2. Start as the same point as x1(n) graph N samples of x2(n) as equally spaced points around an inner
circle in clockwise direction.
3. Multiply corresponding samples on the two circles and sum the products to produce output.
4. Rotate the inner circle one sample at a time in counterclockwise direction and go to step 3 to obtain
the next value of output.
5. Repeat step No. 4 until the inner circle first sample lines up with the first sample of the exterior
circle once again.
 MATRIX MULTIPLICATION METHOD
 In this method, the circular convolution of two sequences x1(n) & x2(n) can be obtained by representing
the sequences in matrix form as shown below: -

 The sequence x2(n) is repeated via circular shift of samples and represented in N x N matrix form.
 The sequence x1(n) is represented as column matrix.
 The multiplication of these two matrices gives the sequences x3(n).
 EXAMPLE – 1: -
 Find the circular convolution of two finite duration sequences x1(n) = {1, -1, -2, 3, 1}; x2(n) = {1, 2, 3}
 SOLUTION (By Circular Convolution Method)
 To find circular convolution, both sequences must be of same length.
 Therefore we append two zeros to the sequence x2(n) and use concentric
circle method to find circular convolution.
 We have x1(n) = {1, -1, -2, 3, 1} and x2(n) = {1, 2, 3, 0, 0}
 Graph all points of x1(n) on the outer circle in counterclockwise direction.
 Starting at same point as x1(n) graph all points of x2(n) on the inner circle
in clockwise direction. Multiply corresponding samples on the circle and
add to obtain, y(0) = 1(1) + 0(-1) + 0(-2) + 3(3) + 2(-1) = 8
 Rotate the inner circle in counterclockwise direction by one sample, multiply corresponding samples.

<DIGITAL SIGNAL PROCESSING>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DSP [ETT 603] <FOURIER TRANSFORM: ITS APPLICATIONS PROPERTIES > [Page 4.9]

 y(1) = 1(2) + 1(-1) + 0(-2) + 3(0) + 3(-1) = - 2 ;  y(2) = 3(1) + 2(-1) + 1(-2) + 0(3) + 0(-1) = - 1
 y(3) = 0(1) + 3(-1) + 2(-2) + 1(3) + 0(-1) = - 4 ;  y(4) = 0(1) + 0(-1) + 3(-2) + 2(3) + 1(-1) = - 1
 y(n) = {8, -2, -1, -4, -1}
 SOLUTION (By Matrix Method)
 Given that x1(n) = {1, -1, -2, 3, 1} & x2(n) = {1, 2, 3} {We Can Inter change Two Signals}
 By adding two zeros to the
sequences x2(n), we bring the
length of the sequence x2(n) to 5.
 Now x2(n) = {1, 2, 3, 0, 0}
 The matrix form can be written
by substituting N = 5 in above
matrix method equation, we get
 By representing the sequence x2(n) in N x N matrix form and x1(n) in column matrix form and multiply
to get y(n).  y(n) = {8, -2, -1, -4, -1}
 EXAMPLE - 2
 Find the circular convolution of two finite duration
signal x1(n) = {1, 2, 2, 1} & x2(n) = {1, 2, 3, 1} using
(1) Concentric circle method (2) Matrix
multiplication method.
 SOLUTION: - (A)By Concentric Circle Method

 y(0) = 1(1) + 2(1) + 2(3) + 1(2) = 11 ;  y(1) = 1(2) + 2(1) + 2(1) + 1(3) = 9
 y(2) = 1(3) + 2(2) + 2(1) + 1(1) = 10 ;  y(3) = 1(1) + 2(3) + 2(2) + 1(1) = 12
 y(n) = {11, 9, 10, 12}
a) By Matrix Multiplication Method
 Represent the sequence x2(n) in N x N matrix form & x1(n) in column matrix form i.e. substitute N = 4.

 Substitute the sequence values and multiply as shown fig above.  y(n) = {11, 9, 10, 12}
 EXAMPLE - 3
 Given the sequences x1(n) = {1, 2, 3, 4}; x2(n) = {1, 1, 2, 2}. Find x3(n) such that X3(k) = X1(k) X2(k).
 SOLUTION
 As x3(n) = IDFT [X3(k)] = IDFT [X1(k) X2(k)] = x1(n) x2(n)
<DIGITAL SIGNAL PROCESSING>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DSP [ETT 603] <FOURIER TRANSFORM: ITS APPLICATIONS PROPERTIES > [Page 4.10]
 So we find x3(n) by circular convolving x1(n) and x2(n).
 Given that x1(n) = {1, 2, 3, 4} and x2(n) = {1, 1, 2, 2}
 Representing x2(n) as N x N matrix form and x1(n) in column
matrix and multiplying we have,  y(n) = {15, 17, 15, 13}
 EXAMPLE – 4: - Perform the circular convolution of the
following sequences x(n) = {1, 1, 2, 1} and h(n) = {1, 2, 3, 4}. Using DFT and IDFT method
 SOLUTION : - We know X3(k) = X1(k) X2(k)
X1(k) = 𝐍−𝟏
𝒏=𝟎 𝐱 𝟏 (𝐧) 𝐞
− 𝐣𝟐𝛑𝐧𝐤/𝐍
, k = 0, 1,… N-1 ; But given x1(n) = {1, 1, 2, 1} and N = 4
3
 X1(0) = 𝑛 =0 x1 (n) = 1+1+2+1 = 5 ; X1(1) = 3𝑛 =0 x1 (n) e− jπn/2 = 1 – j – 2 + j = -1
3
 X1(2) = 𝑛 =0 x1 (n) e − jπn
= 1 – 1 + 2 - j = 1; X1(3) = 3𝑛 =0 x1 (n) e− j3πn/2 = 1 + j – 2 - j = -1
DFT {x1 (n)}  X1(k) = (5, -1, 1, -1)
𝐍−𝟏 − 𝐣𝟐𝛑𝐧𝐤/𝐍
X2(k) = 𝐱 𝟐 (𝐧) 𝐞
𝒏=𝟎 , k = 0, 1,… N-1 ; But given x2(n) = {1, 2, 3, 4} and N = 4
3
 X2(0) = 𝑛 =0 x2 (n) = 1+2+3+4=10
 X2(1) = 3𝑛 =0 x2 (n) e− jπn/2 = 1 + 2(–j) + 3(-1) + 4(j) = -2 + j2
 X2(2) = 3𝑛 =0 x2 (n) e− jπn = 1 + 2(-1) + 3(1) + 4(-1) = -2
 X2(3) = 3𝑛 =0 x2 (n) e− j3πn/2 = 1 + 2(j) + 3(-1) + 4(-j) = -2 - j2
DFT {x2 (n)}  X2(k) = (10, -2 + j2, -2, -2 - j2)
𝟏
X3(k) = X1(k) X2(k) = {50,2–j2, -2,2+j2}; But IDFT{x3(k)}=𝑵 𝐍−𝟏
𝒌=𝟎 𝐗 𝟑 (𝐤) 𝐞
𝐣𝟐𝛑𝐧𝐤/𝐍
, n = 0, 1,… N-1
1 3 1
x3(0) = 4 𝑛=0 X 3 (k) = 4 (50 + 2 – j2 – 2 + 2 + j2) = 13
1 4 jπk/2 1
x3(1) = 4 𝑘=0 X 3 (k) e = 4 [50 + (2 – j2)j + (– 2) (-1) + (2 + j2) (-j)] = 14
1 4 jπk 1
x3(2) = 4 𝑘=0 X 3 (k) e = 4 [50 + (2 – j2) (-j) + (– 2) (-1) + (2 + j2) (j)] = 12
 x3(n) = {13, 14, 11, 12}
 EXAMPLE - 5
 Consider the sequence x1(n) = {0, 1, 2, 3, 4} and x2(n)
= {0, 1, 0, 0, 0}. Find y(n) so that Y(k) = X1(k) X2(k).
 SOLUTION The sequence y(n) can be obtained by
circular convolution of x1(n) and x2(n). Using Matrix
approach we have  y(n) = {4, 0, 1, 2, 3}
 TASK -1: - Perform the circular convolution of the following sequences,
i. x1(n) = {1, 1, 2, 1}; x2(n) = {1, 2, 3, 4}. Ans: - {13, 14, 11, 12}
ii. x1(n) = {1, 2, 3, 1}; x2(n) = {4, 3, 2, 2}. Ans: - {17, 19, 22, 19}
 TASK 2: - Find circular convolution of the following sequences, x(n) = {1, 1, 1, 2}; h(n) = {1, 2, 3, 2}
using DFT and IDFT method. Ans: - {10, 11, 10, 09}
 TASK 3: - Find circular convolution of the following sequences, x1(n) = {2,1,0,0,5}; x2(n) = {2,2,1,1}
 TASK 4: - Find circular convolution of the following sequences, x1(n) = {1, 2, 2, 1}; x2(n) = {1,2,3,1}
 TASK 5: - Find circular convolution of the following sequences, x1(n) = {1, 1, -1, 2}; x2(n) = {2,0,1,1}
 TASK 6: - Find circular convolution of the following sequences, x1(n) = {1, 2, 3, 4}; x2(n)= {4, 3,2,1}
 TASK 7: - Find circular convolution of the following sequences, x1(n) = {1,-1, 2, 3}; x2(n)= {0, 1,2,3}
∞ −𝒋𝝎𝒕
𝟏 ∞
CTFT X(jω) = − ∞ 𝒙(𝒕) 𝒆 𝒅𝒕 ICTFT 𝒙 𝒕 = 𝑿(𝒋𝝎) 𝒆 𝒋𝝎𝒕 𝒅𝝎
𝟐𝝅 − ∞
𝒋𝝎 ∞ −𝒋𝝎𝒏
𝟏 ∞
DTFT X(𝒆 ) = −∞
𝒙(𝒏) 𝒆 IDTFT 𝒙 𝒏 = 𝑿(𝒆 𝒋𝝎) 𝒆 𝒋𝝎𝒏 𝒅𝝎
𝟐𝝅 − ∞
N-Point 𝐍−𝟏 −𝐣𝟐𝛑𝐧𝐤/𝐍 𝐍−𝟏 𝐤𝐧
X(k) = 𝒏=𝟎 𝐱(𝐧)𝐞  X(k) = 𝒏=𝟎 𝐱(𝐧) 𝐖𝐍
DFT
N-Point 𝟏 𝐍−𝟏 𝐣𝟐𝛑𝐧𝐤/𝐍 𝟏 𝐍−𝟏 − 𝐤𝐧
x(n) = 𝒌=𝟎 𝐗(𝐤)𝐞 = 𝒌=𝟎 𝐗(𝐤)𝐖𝐍
IDFT 𝑵 𝑵

-------------- ALL THE BEST -------------------- ALL THE BEST ----------------

<DIGITAL SIGNAL PROCESSING>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 604] [Page 5.1]
[CHAPTER-5]
------ FAST FOURIER TRANSFORM ALGORITHM & DIGITAL FILTERS------
 INTRODUCTION: -
 We had discussed the Discrete
Fourier Transform (DFT) and
the way it can be used to
perform linear filtering.
 In this chapter we study about a set of algorithms known as Fast Fourier Transform (FFT).
 The FFT is a highly efficient procedure for computing the DFT of a finite series and requires less
number of computations than that of direct evaluation of DFT.
 It reduces the computations by taking advantage of the fact that the calculation of the coefficients of the
DFT can be carried out iteratively. Due to this, FFT computation technique is used in digital spectral
analysis, filter simulation, autocorrelation and pattern recognition.
 The FFT is based on decomposition and breaking the transform into smaller transforms and combining
them to get the total transform. FFT reduces the computation time required to compute a DFT and
improves the performance by a factor 100 or more over direct evaluation of the DFT.
 DIRECT EVALUATION OF THE DFT: -
 The DFT of a sequence can be evaluated using the formula
X(k) = 𝐍−𝟏
𝐧=𝟎 𝐱(𝐧)𝐞
−𝐣𝟐𝛑𝐧𝐤/𝐍
0 ≤ k ≤ N-1 …….(01)
Substituting WN = 𝐞−𝐣𝟐𝛑/𝐍, We have X(k) = 𝐧=𝟎 𝐱(𝐧)𝐖𝐍 𝐧𝐤 0 ≤ k ≤ N-1
𝐍−𝟏
…….(02)
= N−1
n=0 Re x(n) + j Im x(n) Re WN nk + j Im WN nk …….(03)

= N−1
n=0 Re x(n) Re WN nk − N−1
n=0 Im x(n) Im WN nk
nk nk
+ j N−1
n=0 Im x(n) Re WN + N−1
n=0 Re x(n) Im WN …….(04)
 From Eq.(3) we see that to evaluate one value of X (k), the number of complex multiplications required
is N. Therefore to evaluate all N value of X(k), the number of complex multiplications required is N2.
 In the same way, to evaluate all N value of X(k), the total number of additions required is N(N-1).
 From Eq.(4), we can observe that the computation of X(k) for each k requires 4N real multiplications.
 Therefore, to evaluate X(k) for all k from 0 to N - 1 requires 4N2 real multiplications.
 Each of the four sums of N terms requires N - 1 real two-input additions, and to combine the sum to get
the real part and imaginary part requires two more.
 Therefore to evaluate X(k) for each k requires 4 (N - 1) + 2 real additions.
 For all values of k a total number of real additions N (4N - 2) required for direct evaluation of the DFT.
 The above results are obtained by assuming the value of WN nk as always complex, even though for
some values of kn, it equals to 1, -1, j or -j. the direct evaluation of the DFT is basically inefficient
because it does not use the symmetry and periodicity properties of the twiddle factor WN·
 These two are Symmetry Property: WN k+N/2 = −WN k & Periodicity Property: WN k+N = WN k ...(05)
 THE FAST FOURIER TRANSFORM : -
 The fast Fourier transform algorithms exploit the two basic properties of the twiddle factor shown in
𝑵
Eq. (5) & reduces the number of complex multiplications required to perform DFT from N2 to log2N.
𝟐
6
 For ex if N = 1024, this implies about 5000 instead of 10 multiplications - a reduction factor of 200.
 FFT algorithm is based on the fundamental principle of decomposing the computation of Discrete
Fourier Transform of a sequence of length N into successively smaller Discrete Fourier Transforms.
<FAST FOURIER TRANSFORM (FFT)>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [Page 5.2]
[ETT 604]
 There are two classes of FFT algorithms. That is Decimation-In-Time & Decimation-In-Frequency.
 In Decimation-In-Time, sequence for which we need DFT is successively divided into sequences &
DFTs of these subsequences are combined in a certain pattern to obtain required DFT of entire signal.
 In the Decimation-In-Frequency approach, the frequency samples of the DFT are decomposed into
smaller and smaller subsequences in a similar manner.
 DECIIMATION.IN-TIME ALGORITHM : -
 This algorithm is also known as Radix-2 DIT FFT algorithm which means the number of output points
N can be expressed as a power of 2, that is, N = 2M, where M is an integer.
 Let x(n) is an N-point sequence, where N is assumed to be a power of 2.
 Decimate or break this sequence into two sequences of length N /2, where one sequence consisting of
the Even-indexed values of x( n) and the other of Odd-indexed values of x(n).
𝑁 𝑁
 That is xe(n) = x(2n), n = 0, 1, …, − 1 AND Xo(n) = x(2n+1), n = 0, 1, …, −1 …….(06)
2 2
𝐍−𝟏 𝐧𝐤
 The N-point DFT of x(n) can be written as X(k) = 𝐧=𝟎 𝐱(𝐧)𝐖𝐍 , k = 0, 1, …, N-1 …….(07)
 Separating x (n) into Even and Odd indexed values of x(n),we obtain
nk nk
X(k) = N−1 n=0 x(n)WN + N−1
n=0 x(n)WN
(Even ) (Odd )
𝑁 𝑁
−1 2nk −1
= 2
n=0 x(2n)WN + 2
n=0 x(2n + 1)WN (2n+1)k
𝑁 𝑁
−1 2nk k −1
= 2
n=0 x(2n)WN + WN 2
n=0 x(2n + 1)WN 2nk …….(08)
𝑁 𝑁
−1 2nk −1
 Substituting Eq.(6) in Eq.(8) we have, X(k) = 2
n=0 xe (n) WN + WN k 2
n=0 xo (n)WN 2nk …….(09)
2 j2π
 But we can write WN 2 = e−j2π/N = e− N /2 = WN/2  WN 2 = WN/2 …….(10)
 Substituting Eq.(10) in Eq.(9) we get
𝑁 𝑁
−1 nk k −1
X(k) = 2
n=0 xe (n) WN/2 + WN 2
n=0 xo (n)WN/2 nk …….(11)  X(k) = 𝐗 𝐞 (𝐤) + 𝐖𝐍 𝐤 𝐗 𝐨 (𝐤)…(12)
𝑁 𝑁
−𝑝𝑜𝑖𝑛𝑡 𝐷𝐹𝑇 𝑜𝑓 𝑒𝑣𝑒𝑛 −𝑝𝑜𝑖𝑛𝑡 𝐷𝐹𝑇 𝑜𝑓 𝑜𝑑𝑑
2 2
𝑖𝑛𝑑𝑒𝑥𝑒𝑑 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 𝑖𝑛𝑑𝑒𝑥𝑒𝑑 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒
𝑁 𝑁
 Each of the sums in Eq. (1l) is an 2
-point DFT, the First sum being the 2
-point DFT of the Even-
𝑁
indexed sequence and the Second being the -point DFT of Odd-indexed sequence.
2
𝑁
 Although index k ranges from k = 0, 1..., N - 1, each of the sums is computed only for k = 0, 1..., -1,
2
𝑁
since Xe(k) and Xo(k) are periodic in k with period 2 .
 After two DFTs are computed, they are combined according to Eq. (12) to get N-point DFT of X (k).
𝑁
 So Eq. (12) holds for the values of k = 0, 1, ... , - l. For k ≥ N/2, 𝐖𝐍 𝐤+𝐍/𝟐 = −𝐖𝐍 𝐤 …….(13)
2
𝐍
𝑵 𝑵 𝑁 𝑁
 Now X(k) for k ≥ N/2 is given by X(k) = 𝐗 𝐞 𝐤 − 𝟐 - 𝐖𝐍 𝐤−𝟐 𝐗 𝐨 𝐤 − 𝟐 for k = 2 , 2 +1, …, N-1...(14)
 Let us find the number of complex multiplications and complex additions required to compute Eq.(12).
 For direct evaluation of DFT we know that number of complex multiplications required is equal to N2.
𝑵 𝑁 2
 In same way to calculate - Point DFT of Xe(k) we need complex multiplications & to compute
𝟐 2
𝑁 2 𝑁 2
Xo(k) we need another complex multiplications. That is we need 2 complex multiplications.
2 2
𝑁
 Then the two - point DFTs are combined to get X(k). For this we need N complex multiplications.
2
𝑵 𝟐 𝑵 𝟐 𝐍𝟐
 Thus total number of complex multiplications required for computing X(k) is + +N=N+
𝟐 𝟐 𝟐
𝑵 𝑵 𝑵 𝑵 𝐍𝟐
 Similarly the total number of complex Additions required is −𝟏 + +𝟏 +N=
𝟐 𝟐 𝟐 𝟐 𝟐

<FAST FOURIER TRANSFORM (FFT)>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 604] [Page 5.3]
2
 The direct evaluation of X(k) requires N Complex multiplications and N(N - 1) complex additions.
𝑁
 When we decompose x(n) into two subsequences of length and compute X(k) using Eq.(12) the
2
number of computations are reduced by a factor of 2. If we again decompose the sequence Xe(k) and
Xo(k) into two subsequences, then the amount of computation again be cut in half.
 Now let us take N = 8. Then Xe(k) and Xo(k) are 4-point (N/2) DFTs of Even-indexed sequence xe(n)
and Odd-indexed sequence xo(n) respectively, where
xe(0) = x(0); xe(1) = x(2); xe(2) = x(4); xe(3) = x(6);
xo(0) = x(1); xo (1) = x(3); xo (2) = x(5); xo (3) = x(7)
k
 From Eq.(12) and Eq.(14) we have X(k) = Xe (k) + W8 Xo (k) for 0 ≤ k ≤ 3
𝐤−𝟒
= 𝐗 𝐞 (𝐤 − 𝟒) + 𝐖𝟖 𝐗 𝐨 (𝐤 − 𝟒) for 4 ≤ k ≤ 7 …….(15)
 By substituting different values of k we get
X(0) = Xe (0) + W8 0 Xo (0); X(4) = X e (0) - W8 0 Xo (0)
X(1) = Xe (1) + W8 1 Xo (1); X(5) = X e (1) - W8 1 Xo (1)
X(2) = Xe (2) + W8 2 Xo (2); X(6) = X e (2) - W8 2 Xo (2)
X(3) = Xe (3) + W8 3 Xo (3); X(7) = X e (3) - W8 3 Xo (3) …….(16)
 From above equation, we find X(0) & X(4), X(1) & X(5), X(2) & X(6), X(3) & X(7) have same inputs.
 X(0) is obtained by multiplying Xo(0) with W8 0 and adding the
product to Xe(0). Similarly X(4) is obtained by multiplying Xo(0)
with W8 0 and subtracting the product from Xe(0).
 This operation can be represented by a butterfly diagram as in Fig.
 Now the values X(k) for k = 1,2,3,4,5,6,7 can be obtained
 An 8-point DFT flow graph can be constructed from two 4-point DFTs as shown in Fig. 5.2.
 From Fig. 5.2 we can find that initially the sequence xo(n) is
shuffled into even indexed sequenced xe(n) & odd-indexed
sequence xo(n) & then transformed to give Xe(k) and Xo(k).
 For k = 0, 1, 2, 3 the values Xe(k) and Xo(k) are combined
according to Eq. (16) and using butterfly structure shown in
Fig. 5.1 the 8-point DFT is obtained.
N
 The inputs to the butterfly are separated by 2 samples i.e. 4
samples and the powers of the twiddle factors associated in
this set of butterflies are in natural order.
N
 Now we apply the same approach to decompose each of 2
samples DFT. This can be done by dividing the sequence xe(n) and xo(n) into two sequences consisting
of even and odd members of the sequences.
N N
 The - point DFTs can be expressed as a combination of – point DFTs.
2 4
N N
 That is, Xe(k) for 0 ≤ k ≤ − 1 can be written as 𝐗 𝐞 (𝐤) = 𝐗 𝐞𝐞 (𝐤) + 𝐖𝐍 𝟐𝐤 𝐗 𝐞𝐨 (𝐤) for 0 ≤ k ≤ −1
2 2
𝐍 𝐍 N N
= 𝐗 𝐞𝐞 𝐤 − 𝟒 - 𝐖𝐍 𝟐(𝐤−𝐍/𝟒) 𝐗 𝐞𝐨 𝐤 − 𝟒 for ≤k≤ −1 ……. (17)
4 2
N N
 Where Xee (k) is the - point DFT of the even member of xe(n) and Xeo (k) is the 4 - point DFT of the
4
N
odd members of xe(n). In the same way 𝐗 𝐨 (𝐤) = 𝐗 𝐨𝐞 (𝐤) + 𝐖𝐍 𝟐𝐤 𝐗 𝐨𝐨 (𝐤) for 0 ≤ k ≤ −1
2
𝐍 𝐍 N N
= 𝐗 𝐨𝐞 𝐤 − 𝟒 - 𝐖𝐍 𝟐(𝐤−𝐍/𝟒) 𝐗 𝐨𝐨 𝐤 − 𝟒 for ≤k≤ −1 …….(18)
4 2
N N
 Where Xoe (k) is the 4 - point DFT of the even member of xo(n) and Xoo (k) is the - point DFT of the
4
odd members of xo(n).
<FAST FOURIER TRANSFORM (FFT)>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 604] [Page 5.4]
 For N = 8 the sequence xe(n) can be divided into even and odd indexed sequences as
xee(0) = xe(0); xee(1) = xe(2) & xeo(0) = xe(1); xeo(1) = xe(3)
 Now from Eq.(17) we have Xe (0) = Xee (0) + W8 Xeo (0) ; Xe (1) = Xee (1) + W8 2 Xeo (1)
0

Xe (2) = Xee (0) - W8 0 Xeo (0) ; Xe (3) = Xee (1) - W8 2 Xeo (1) …(19)
 Where, Xee (k) is 2 point DFT of even member of xe(n) and Xeo(k) is of odd members of xe(n).
 Similarly the sequence xo(n) can be divided into even and odd member sequences as
xoe(0) = xo(0); xoe(1) = xo(2) & xoo(0) = xo(1); xoo(1) = xo(3)
 Now from Eq.(17) we have Xo (0) = Xoe (0) + W8 Xoo (0) ; Xo (1) = Xoe (1) + W8 2 Xoo (1)
0

Xo (2) = Xoe (0) - W8 0 Xoo (0) ; Xo (3) = Xoe (1) - W8 2 Xoo (1) … (19)
 Where Xoe (k) is the 2 point DFT of even
member of xo(n) and Xoo(k) is the 2-point
DFT of odd members of xo(n).
 Fig.5.3 shows the resulting flow graph the
four-point DFTs of Fig.5.2 are evaluated
as in Eq.(19) and Eq.(20).
 From Fig. 4.3 we find that the input
sequence is again reordered, the input
samples to each butterfly are separated by
N
samples i.e., 2 samples and there are two
4
sets of butterflies. In each set of butterflies the twiddle factor exponents are same and separated by two.
N
 For the more general case, We could proceed by decomposing the 4 - point transforms in Eq. (17) and
N
Eq.(18) into 8 - point transforms and continue until we are left with only 2-point transforms.
 Each decomposition is called a stage, and the total number of stages is given by M = log2(N).
 In each stage we compute N values. Each value computed from a butterfly equation requires one
complex multiplication. Therefore each stage requires a total of N complex multiplications.
 Since the number of stages is equal to log2N, the entire FFT requires a total of Nlog2N multiplications.
 But we know W0 = 1; WN/2 = -1 and W±N/4 = -i about half of the multiplications are not really needed.
N
 Therefore the entire FFT requires log2N complex multiplications. Each butterfly equation requires
2
one complex addition. Thus the entire number complex additions are given by N log2N.
 For an 8-point DFT the number of stages required is three.
 So far we have seen the decomposition for stage 3 and
stage 2. For stage 1 the two point DFT can be easily found
by adding and subtracting the input sequences as the
twiddle factor associated with first stage is W80 = 1.
 That is the first stage involves no multiplication but
addition and subtraction. Now we have
Xee(0) = xee(0) + xee(1) = xe(0) + xe(2) = x(0) + x(4)
Xee(1) = xee(0) - xee(1) = xe(0) - xe(2) = x(0) + x(4) ….(21)
 The algorithm has been called Decimation In Time since at
each stage; input sequence is divided into smaller
sequences, i.e., input sequences are decimated at each stage.
 BIT-REVERSAL:
 In DIT algorithm we can find that for the output sequence to
be in a natural order (i.e. X(k), k = 0, 1, …, N-1) the input
sequence has to be stored in a shuffled order.
<FAST FOURIER TRANSFORM (FFT)>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 604] [Page 5.5]
 For an 8 - point DIT algorithm the Input sequence is x(0), x(4), x(2), x(6), x(5), x(3), and x(7).
 We can see that when N is a Power of 2, the input sequence must be stored in blt-reversal order for the
output to be computed in a natural order. For N = 8 the bit-reversal process is shown in above table 5.1.
 BASIC OPERATION:
 The basic computation block in the diagram is the "butterfly" in
which two inputs are combined to give two outputs.
 If we use m to represent the stage and p and q to represent nodes in
the stage, each butterfly can be represented as shown in Fig. where
power k of WN is variable & depends on the position of butterfly.
 The nodes p and q represent memory locations. At the input nodes p and q the input values Xm(p) and
Xm(q) are stored. After the outputs Xm+1(p) and Xm+1(q) are calculated, the same memory location is
used to store the new values in place of the input values.
 An algorithm that uses the same locations to store both the input and output sequences is called an in-
place algorithm. The FFT algorithm reduces the number of computations.
 The total number of complex
multiplications required for
N
calculating DIT-FFT = log2N and
2
the total number of complex
additions for evaluating a DFT
using DIT-FFT is N log2N.
 A comparison of number of complex
multiplications required for direct
evaluation of the DFT ^ the number
needed for FFT is given in table 5.2.
 SUMMARY OF STEPS OF RADIX - 2 DIT-FFT ALGORITHM : -
1. The number of input samples N = 2M, where, M is an integer.
2. The input sequence is shuffled through bit-reversal.
3. The number of stages in the flow graph is given by M = log2N.
𝑁
4. Each stage consists of butterflies.
2
5. Inputs/outputs for each butterfly are separated by 2m-1 samples, where m represents the stage index, i.e,
for first stage m = 1 and for second stage m = 2 so on.
𝑁
6. The number of complex multiplications is given by log2N.
2
7. The number of complex additions is given by N log2N.
𝐍
8. The twiddle factors are a function of the stage index m & is given by K = 𝟐𝐦𝐭 ; t = 0,1,2,.., 2m-1 – 1...(22)
The number of sets or sections of butterflies in each stage is given by the formula 2M-m.
9. The Exponent Repeat Factor (ERF), which is the number of times the exponent sequence associated
with m is repeated is given by 2M-m.
 EXAMPLE: Find the DFT of a sequence x(n) = {1, 2, 3, 4, 4, 3, 2, 1} using DIT algorithm.
 Solution: The twiddle factors associated with the flow graph are
1
𝐖𝟖 𝟎 = 1; 𝐖𝟖 𝟏 = e−j2π/8 = e−jπ/4 = 0.707 - j0.707
2
𝐖𝟖 𝟐 = e−j2π/8 = e−jπ/2 = -j ;
3
𝐖𝟖 𝟑 = e−j2π/8 = e−j3π/4 = -0.707 - j0.707
 The basic operation is,
<FAST FOURIER TRANSFORM (FFT)>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 604] [Page 5.6]

PRACTICE PROBLEM: Find the 8-point DFT of sequence x(n) = {1, 1, 1, 1, 1, 0, 0, 0} using DIT-FFT
algorithm. [ANS: {5, -j2.414, 1, -0.414j, 0.414j, 1, j2.414}]
 DECIMATION-IN-FREQUENCY ALGORITHM : -
 DIT algorithm is based on the decomposition of the DFT computation by forming smaller and smaller
subsequences of the sequence x(n).
 In DIF algorithm the output sequence X(k) is divided into smaller and smaller subsequences.
 In this algorithm the input sequence x(n) is partitioned into two sequences each of length N/2 samples.
N
 The first sequence x1(n) consists of first 2 samples of x(n) and the second sequence x2(n) consists of
N
the last 2 samples of x(n) i.e., x1(n) = x(n), n = 0, 1, 2,.. N/2-1; x2(n) = x(n+ N/2), n = 0, 1, 2,..N/2-1
 If N = 8 the first sequence x1(n) has values for 0 ≤ n ≤ 3 and x2(n) has values for 4 ≤ n ≤ 7.
𝐍
−𝟏
 The N-point DFT of x(n) can be written as, X(k) = 𝟐
𝐧=𝟎
𝐱(𝐧)𝐖𝐍 𝐧𝐤 + 𝐍
𝐍 𝐱(𝐧)𝐖𝐍
𝐧𝐤
𝐧=
𝟐
N N
−1 nk (n+N/2)k −1 nk
= 2
n=0 x1 (n) WN + N
n=0 x2 (n)WN = 2
n=0 x1 (n) WN + WN Nk /2 N
n=0 x2 (n)WN
nk

𝐍
−𝟏
= 𝟐
𝐧=𝟎
𝐱 𝟏 (𝐧) 𝐖𝐍 𝐧𝐤 + 𝐞−𝐣𝛑𝐤 𝐍
𝐧=𝟎 𝐱 𝟐 (𝐧)𝐖𝐍
𝐧𝐤

N
−1
 When k is Even, e −jπk
= 1  X(2k) = 2
n=0 x1 n + x2 (n) WN 2nk
N 𝐍
−1 −𝟏
= 2
n=0 x1 n + x2 (n) WN/2 nk = 𝟐
𝐧=𝟎
𝐟(𝐧)𝐖𝐍/𝟐 𝐧𝐤 --- (25a); Where f(n) = 𝐱 𝟏 𝐧 + 𝐱 𝟐 (𝐧) --- [25b]
N N
 Eq. (25a) is the 2 -point DFT of the 2 -point sequence f(n) obtained by adding the First-half and the last-
half of the input sequence.
N
−1
 When k is Odd, e −jπk
= -1  X(2k+1) = 2
n=0 x1 n − x2 (n) WN (2k+1)n
N 𝐍
−1 −𝟏
= 2
n=0 x1 n − x2 (n) WN n WN/2 nk = 𝟐
𝐧=𝟎
𝐠(𝐧)𝐖𝐍/𝟐 𝐧𝐤 ---(26a); Where g(n)=[x1 n − x2 (n)]WN n [26b]
N
 Eq. (26) is the 2 -point DFT of the sequence g(n) obtained by subtracting the Second-half of the input
sequence from the first half and then multiplying the resulting sequence with WN n .
 From Eq.(25a) and Eq.(26a) we find that the Even and Odd
N
samples of the DFT can be obtained from the -point
2
DFTs of f(n) and g(n) respectively.
 The Eq.(25b) and Eq.(26b) can be represented by a
butterfly as shown in Fig. 5.10.
 This is the basic operation of DIF algorithm.
<FAST FOURIER TRANSFORM (FFT)>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 604] [Page 5.7]
 From Eq.(4.25), for n = 8, we have
X(0) = 3n=0 x1 n + x2 (n) = 3n=0 f(n) = f(0) + f(1) + f(2) + f(3)
X(2) = 3n=0 x1 n + x2 (n) W8 2n = 3n=0 f(n)W8 2n = f(0) + f(1) 𝐖𝟖 𝟐 - f(2) - f(3) 𝐖𝟖 𝟐
4 8
W8 4 = ej2π/8 = ejπ = −1; W8 8 = ej2π/8 = ej2π = 1
X(4) = 3n=0 x1 n + x2 (n) W8 4n = 3n=0 f(n)W8 4n = 3n=0 f(n)(−1)n = f(0) - f(1) + f(2) - f(3)
X(6) = 3n=0 x1 n + x2 (n) W8 6n = 3n=0 f(n)(−W8 2 )n = f(0) - f(1) 𝐖𝟖 𝟐 + f(2) + f(3) 𝐖𝟖 𝟐
 From Eq. (4.26) we have
X(1) = 3n=0 x1 n − x2 (n) W8 n = 3n=0 g(n) = g(0) + g(1) + g(2) + g(3)
X(3) = 3n=0 x1 n − x2 (n) W8 3n = 3n=0 g(n)W8 2n = g(0) + g(1) 𝐖𝟖 𝟐 - g(2) - g(3) 𝐖𝟖 𝟐
X(5) = 3n=0 x1 n − x2 (n) W8 5n = 3n=0 g(n)W8 4n = 3n=0 g(n)(−1)n = g(0) - g(1) + g(2) - g(3)
X(3) = 3n=0 x1 n − x2 (n) W8 7n = 3n=0 g(n)(−W8 2 )n = g(0) - g(1) 𝐖𝟖 𝟐 + g(2) + g(3) 𝐖𝟖 𝟐
 We have seen that the Even-indexed samples of X(k) can be obtained from the 4-point DFT of the
sequence f(n) where f(n) = x1 n + x2 (n) n = 0, 1, 2, …, N/2-1.
i.e. f(0) = x1 0 + x2 (0) ; f(1) = x1 1 + x2 (1)
f(2) = x1 2 + x2 (2) ; f(3) = x1 3 + x2 (3)
 The Odd-indexed samples of X(k) can be
obtained from the 4-point DFT of the sequence
g(n) where g(n)= [x1 n − x2 (n)]W8 n
i.e. g(0) = x1 0 − x2 (0) W8 0
g(1) = x1 1 − x2 (1) W8 1
g(2) = x1 2 − x2 (2) W8 2
g(3) = x1 3 − x2 (3) W8 3
 Using the above information and the butterfly structure shown in fig.5.10 we can draw the flow graph
of 8-point DFT shown in fig 5.11.
N
 Now each 2 -point DFT can be computed by combining the first half and the last half of the input points
N N
for each of the 2 -point DFTs and then computing 4 -point DFTs.
 For the 8-point DFT example the resultant flow graph is shown in Fig. 5.12. The 2-point DFT can be
found by adding and subtracting the input points. The Fig. 5.12 can be further reduced as in. Fig 5.13.

 The complete flow graph of 8-point DFT using DIF

algorithm is shown in Fig. 5.14.
 From Fig. 5.14 we observe that for DIF algorithm the
input sequence is in natural order, while the output
sequence is in bit reversal order, whereas the reverse is
true for the DIT algorithm.
 From Fig. 5.13 we see no of complex multiplications
N
required to find X(k) is 2 1og2N & additions is N1og2N.

<FAST FOURIER TRANSFORM (FFT)>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 604] [Page 5.8]
 Thus the total number of computations is same for DIT and
DIF algorithms. The basic computational block in the
diagram is the "butterfly" shown in Fig. 5.15.
 Like DIT algorithm, DIF algorithm also is an in-place
algorithm where the same locations are used to store both the
input and output sequences.
 Summary of Steps for Radix - 2 DIF-FFT Algorithm : -
1. The number of input samples N = 2M, where, M is number of stages.
2. The input sequence is in natural order.
3. The number of stages in the flow graph is given by M = log2N.
N
4. Each stage consists of butterflies.
2
5. Inputs/outputs for each butterfly are separated by 2M-m samples, where m represents the stage index i.e.,
for first stage m = 1 and for second stage m = 2 so on.
N
6. The number of complex multiplications is given by log2N.
2
7. The number of complex additions is given by N log2N.
𝐍𝐭
8. Twiddle factor exponents are a function of stage index m & is given by k = 𝟐𝐌−𝐦+𝟏 , t = 0,1,2,..2M-m – 1
9. The number of sets or sections of butterflies in each stage is given by the formula 2m-1.
10. The Exponent Repeat Factor (ERF), which is the number of times the exponent sequence associated
with m repeated is given by 2m-1.
 Differences and Similarities between DIT and DIF Algorithms : -
 DIFFERENCE
1. For Decimation-In-Time (DIT), the input is bit-reversed while the output is in natural order. Whereas,
for Decimation-In-Frequency (DIF) the input is in natural order while the output is bit reversed order.
2. The DIF butterfly is slightly different from the DIT wherein DIF the complex multiplication takes
place after the add-subtract operation.
 SIMILATITIES: Both algorithms require N log2 N operations to compute the DIF. Both algorithms
can be done in-place and both need to perform bit reversal at some place during the computation.
EXAMPLE: Find the DFT of a sequence x(n) = {1, 2, 3, 4, 4, 3, 2, 1}
using DIF algorithm.
 Solution: The twiddle factors associated with the flow graph are: -
W8 0 = 1, W8 1 = 0.707 – j0.707; W8 2 = - j, W8 3 = - 0.707 - j0.707
 The basic operation is,

<FAST FOURIER TRANSFORM (FFT)>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 604] [Page 5.9]
1; 0 ≤ n ≤7
EXAMPLE: Find 8-point DFT of sequence x(n) = by using DIT & DIF algorithms.
0 ; Otherwise
 Solution: The twiddle factors associated with butterflies can be found as,
W8 0 = 1; W8 1 = 0.707 - j0.707; W8 2 = -j; W8 3 = -0.707 - j0.707

DIT Algorithm X(k) = {8, 0, 0, 0, 0, 0, 0, 0}   DIF Algorithm: X(k) = {8, 0, 0, 0, 0, 0, 0, 0}

EXAMPLE: Compute 4-point DFT of a sequence x(n) = {0, 1, 2, 3} using DIT, DFT algorithm.
 SOLUTION: The twiddle factors associated with butterflies are W4 0 = 1; W4 1 = e−j2π/4 = -j
 DIT ALGORITHM:
Input S1 Output
0 0+2=2 2+4=6
1 0 – 2 = -2 -2 + (-j) (-2) = -2 + 2j
2 1+3=4 2 – 4 = -2
3 1 – 3 = -2 -2 - (-j) (-2) = -2 - 2j
X(k) = {6, -2 + 2j , -2, -2 - 2j}
 DIF ALGORITHM:
Input S1 Output
0 0+2=2 2+4=6
1 1+3=4 2 – 4 = -2
2 0 – 2 = -2 -2 + 2j
3 (1 – 3) (-j) = 2j -2 - 2j
X(k) = {6, -2 + 2j , -2, -2 - 2j}
EXAMPLE: Compute the Eight-point DFT of the sequence x(n) = {0.5, 0.5, 0.5, 0.5, 0, 0, 0, 0} by
using the radix-2 DIT & DIF algorithm.
 Solution: Twiddle factors areW8 0 = 1; W8 1 =0.707- j0.707; W8 2 = -j; W8 3 = -0.707-j0.707

X(k) = {2, 0.5 - j1.207, 0, 0.5 – j0.207, 0, 0.5 + j0.207, 0, 0.5 + j1.207} [Same ans in both Method]
<FAST FOURIER TRANSFORM (FFT)>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 604] [Page 5.10]
EXAMPLE: Find DFT of a sequence EXAMPLE: Find DFT of a sequence
x(n) = {1,-1, 1, -1} by DIT algorithm. x(n) = {1, 0, 0, 1} by DIF algorithm.
 SOLUTION: Twiddle factors are  SOLUTION: Twiddle factors are 
0
W4 = 1; W4 = -j1
W4 0 = 1; W4 1 = -j

X(k) = {0, 0, 4, 0} X(k) = {2, 1 + j, 0, 1 - j}

EXAMPLE: Evaluate & compare 8-point for following sequences using DIT-FFT algorithm.
1, for − 3 ≤ n ≤ 3 1for 0 ≤ n ≤ 6
(A) x1(n) = & (B) x2(n) =
0 Otherwise 0 Otherwise
1 for − 3 ≤ n ≤ 3
 SOLUTION (A): Given that x1(n) =
0 otherwise
 The sequence x1(n) for N = 8 is x1(n) = {1, 1, 1, 1, 0, 1, 1, 1}
 The twiddle factors are  W8 0 = 1; W8 1 = 0.707 - j0.707; W8 2 = -j; W8 3 = -0.707 - j0.707

X1(k) = {7, 1, -1, 1, -1, 1, -1, 1}

 SOLUTION (B): Given that x2(n) = {1, 1, 1, 1, 1, 1, 1, 0}. The sequence x2(n) can be obtained by
shifting the sequence x1(n) right 3 times. Thus by Time Shifting property of DFT we have the relation,
X2(k) = 𝐞−𝐣𝟔𝛑𝐤/𝟖 X1(k) = e−j3πk/4 X1(k)
X2(0) = X1(0) e0 = 7 X2(5) = X1(5) e−j15π/4 = 0.707 + j0.707
X2(1) = X1(1) e−j3π/4 = -0.707 - j0.707 X2(6) = X1(6) e−j9π/2 = j
X2(2) = X1(2) e−j3π/2 = -j X2(7) = X1(7) e−j21π/4 = -0.707 + j0.707
−j9π/2
X2(3) = X1(3) e = 0.707 - j0.707
−j3π
X2(4) = X1(4) e =1
X2(k) = {7, -0.707 - j0.707, -j, 0.707 - j0.707, 1,
0.707 + j0.707, j, -0.707 + j0.707}
EXAMPLE: Find FFT of x(n) = {1, 0, 0, 0, 0, 0, 0, 0}
 SOLUTION: Twiddle factors are 
W8 0 = 1; W8 1 = 0.707 - j0.707;
W8 2 = -j; W8 3 = -0.707- j0.707
X(k) = {1, 1, 1, 1, 1, 1, 1, 1}

<FAST FOURIER TRANSFORM (FFT)>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 604] [Page 5.11]
EXAMPLE: An 8-point sequence is given by EXAMPLE: Find the 8-point DFT of the given
x(n) = {2, 2, 2, 2, 1, 1, 1, 1}. Compute 8-point sequence x(n) = {0, 1, 2, 3, 4, 5, 6, 7} using DIF,
DFT of x(n) by radix-2 DIT FFT. radix-2, FFT algorithm.
 Solution:  Solution:

X(k) = {12, 1 – j2.414, 0, 1 – j0.4142, 0, 1+ X(k) = {28, - 4 + j9.565, - 4 + j4, - 4 + j1.656,

j0.4142, 0, 1 + j2.4142} -4, - 4 - j1.656, - 4 - j4, - 4 - j9.565}
EXAMPLE: Find DFT of Signal x(n) by DIT- EXAMPLE: Find DFT of the signal
𝐧𝛑
FFT algorithm x(n) = {1, -1, -1, -1, 1, 1, 1, -1} x (n) = 𝐜𝐨𝐬 , N = 4 by DIF FFT algorithm.
𝟐
 SOLUTION: 𝐧𝛑
 Solution: Given that x(n) = 𝐜𝐨𝐬 ;
𝟐
 For 4-Point DFT, x(n) = {1, 0, -1, 0}

X (k) = {0, -1.414 + j3.414, 2 – j2, 1.414 – j0.586,  Thus the from the butterfly diagram the 4-Pont
4, 1.414 + j0.586, 2 + j2, -1.414 - j3.414} DFT of the given sequence is X(k) = {0, 2, 0, 2}
EXAMPLE: Given x(n) = 2n & N = 8, fine PRACTICE PROBLEM: Find the 8-point
X(k) by DIF-FFT algorithm. DFT of sequence x(n) = {1, 2, 2, 1, 1, 2, 2, 1}
 Solution: Given x(n) = 2n using DIF-FFT algorithm.
 x(n) = {1, 2, 4, 8, 16, 32, 64, 128} Answer: - {12, 0,.2 - j2, 0, 0, 0, -2 + j2, 0}

A (i) {0, 1+0.414j, 0, 1+2.414j, 4, 1-2.414j, 0, 1-0.414j}

(ii) {12, 0, 0, 0, 4, 0, 0, 0}
(iii) {2, 0, 0, 0, 2, 0, 0, 0}
(iv) {16, 0, -4, 0, 0, 0, -4, 0}}
B (i) {10, -2+j2, -2, -2-j2}
X(k) = {255, 48.63 + j166, -51 + j102, (ii) {0, 2-j2, 0, 2+j2}
-78.63 + j46.05, -85 – 78.63 – j46.05, (iii) {3, 2-j, -3, 2+j}
-51 – j102, 48.63 –j166 } (iv) {6, -2+j2, -2, -2-j2}

<FAST FOURIER TRANSFORM (FFT)>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 604] [Page 5.12]
 INTRODUCTION TO DIGITAL FILTERS
 Basically a digital filter is a linear time invariant (LTI) system. The term Finite Impulse Response
(FIR) and Infinite Impulse Response (IIR) are used to distinguish filter types.
 The FIR filters are of non-recursive type, whereby the present output sample depends on the present
input sample and previous input samples; whereas the IIR filters are recursive type, whereby the
present output sample depends on present input, past input samples and output samples.
 The impulse response h(n) for a realizable filter is h(n) = 0 for n ≤ 0 and for stability it must satisfy
the condition ∞ 𝐧=𝟎 𝐡(𝐧) < ∞.
𝐌
∞ −n 𝐛𝐤 𝐳 −𝐤
 IIR digital filters have the transfer function of the form H(z) = n=0 h(n)z = 𝟏+ 𝐤=𝟎
𝐍 −𝐤
𝐤=𝟏 𝐚𝐤 𝐳
 The design of an IIR filter for given specifications is to find filter coefficient ak s and bk s of above eqn.
 FREQUENCY SELECTIVE FILTERS
 A filter, which rejects unwanted frequencies from the input signal and allows the desired frequencies.
 The range of frequencies of signal that are passed through the filter is called pass band.
 The range of frequencies that are blocked is called stop band.
 The filters are of different types,
1) Low Pass Filter 2) High Pass Filter 3) Band Pass Filter 4) Band Reject Filter
1) LOW PASS FILTER: The magnitude response of
an ideal low pass filter allows low frequencies in the
pass band 0 < Ω < ΩC to pass, whereas the higher
frequencies in the stop band Ω > ΩC are blocked.
 The frequency ΩC between the two bands is the
cutoff frequency, where magnitude H(jΩ) = 1 2.
In practice it is impossible to obtain the ideal
response. The practical response of a low pass filter is shown in solid line in fig.
2) HIGH PASS FILTER: The high pass
filter allows high frequencies above Ω > ΩC
and rejects the frequencies between Ω = 0
and Ω = ΩC. The magnitude response of an
ideal & practical HPF is shown.
 BAND PASS FILTER: It allows only a
band of frequencies Ω1 to Ω2 to pass and
stops all other frequencies. The ideal and
practical responses of BPF are shown in fig.
3) BAND REJECT FILTER: It rejects all the
frequencies between Ω1 and Ω2 and allows
remaining frequencies. The magnitude response of
an ideal and practical filter in shown in fig.
 Design Of Digital Filters From Analog Filters
 The common technique used for designing IIR
digital filters known as indirect method, involves
first designing an analog prototype filter and then transforming the prototype to a digital filter.
 For the given specification, the derivation of the digital filter transfer function requires the three steps.
Step-1 Map the desired digital filter specifications into those for an equivalent analog filter.
Step-2 Derive the analog transfer function for the analog prototype.
Step-3 Transform the transfer function of analog prototype to equivalent digital filter transfer function.
<FAST FOURIER TRANSFORM (FFT)>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 604] [Page 5.13]
 DIGITAL VERSUS ANALOG FILTERS
SN ANALOG FILTER DIGITAL FILTER
Analog filter process analog inputs and A digital filter processes and generates
1)
generates analog output digital data.
Analog filters are constructed from active or A digital filter consists of elements like
2)
passive electronic components adder, multiplies and delay unit.
Analog filter is described by a differential Digital filter is described by a difference
3)
equation. equation.
The frequency response of an analog filter can The frequency response can be changed by
4)
be modified by changing the components. changing the filter co-efficients.

 ADVANTAGES OF DIGITAL FILTERS:

1) Unlike analog filter, the digital filter performance is not influenced by component ageing, temperature
and power supply variations.
2) A digital filter is highly immune to noise and possesses considerable parameter stability.
3) Digital filters afford a wide variety of shapes for the amplitude and phase responses.
4) There are no problems of input or output impedance matching with digital filters.
5) Digital filters can be operated over a wide range of frequencies.
6) The coefficients of digital filter can be programmed and altered any time to obtain the desirec
characteristics. Multiple filtering is possible only in digital filter.
 DISADVANTAGES OF DIGITAL FILTERS:
1) The quantization error arises due to finite word length in the representation of signals and parameters.
 ANALOG LOWPASS FILTER DESIGN
M
N(s) ai si
 The most general form of analog filter transfer function is, H(s) = D(s) = 1+ i=0N i
i=0 b i s
∞
 Where H(s) is the laplace transform of the impulse response h(t), that is H(s) = −∞
h(t)e−st dt
 And N ≥ M must be satisfied. For a stable analog filter, the poles of H(s) lie in left half of the s-plane.
 INTRODUCTION TO FINITE IMPULSE RESPONSE <FIR> FILTERS
 In many digital processing applications FIR filters are preferred over their IIR counterparts.
 The following are the main ADVANTAGES OF FIR Filter over IIR filter.
1) FIR filters are always stable.
2) FIR filters with exactly linear phase can be easily be designed.
3) FIR filters can be realized in both recursive and non-recursive structure.
4) FIR filters are free of limit cycle oscillation, when implemented on a finite word length digital system.
5) Excellent design methods are available for various kinds of FIR filters.
 The DISADVANTAGES OF FIR Filters are
1) The implementation of narrow transition band FIR filters are very costly, as it requires considerably
more arithmetic operations and hardware components such as multipliers, adders and delay elements.
2) Memory requirement and execution time are very high.
 LINEAR PHASE FIR FILTERS
 The transfer function of a FIR causal filter is given by H(z) = 𝐍−𝟏
𝐧=𝟎 𝐡(𝐧)𝐳
−𝐧
𝐍−𝟏
 Where h(n) is impulse response of the filter. Fourier transform of h(n) is 𝑯 𝒆𝒋𝝎 = 𝐧=𝟎 𝐡 𝐧 𝐞−𝒋𝝎𝐧
 This is periodic in frequency with period 2π. 𝑯(𝒆𝒋𝝎) = ± 𝑯(𝒆𝒋𝝎) 𝐞𝒋𝜽(𝝎)
 Where 𝐻(𝑒 𝑗𝜔 ) is magnitude response and θ(ω) is phase response.

<FAST FOURIER TRANSFORM (FFT)>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 A Hand Note on DIGITAL SIGNAL PROCESSING [ETT 604] [Page 5.14]
−𝜽(𝝎) −𝒅𝜽(𝝎)
 We define the Phase delay and Group delay of a filter as, τp = and τg =
𝝎 𝒅𝝎
 For FIR filters with linear phase we can define θ(ω) = -αω; -π ≤ ω ≤ π
 Where α is a constant phase delay in samples. Substituting θ(ω) in above equation, we have τp = τg = α,
 Which means that α is independent of frequency. We can write, 𝐍−𝟏 𝐧=𝟎 𝐡(𝐧)𝐞
−𝐣𝛚𝐧
= ± 𝐇(𝐞𝐣𝛚 ) 𝐞𝐣𝛉(𝛚)
 Which gives us, N−1 jω
n=0 h(n) cos ωn = ± H(e ) cos θ(ω) & −
N−1
n=0 h(n) sin ωn =± H(ejω ) sin θ(ω)
 By taking ratio of above two equations, we obtain
𝐍−𝟏
𝐧=𝟎 𝐡(𝐧) 𝐬𝐢𝐧 𝛚𝐧 𝐬𝐢𝐧 𝛂𝛚
𝐍−𝟏 𝐡(𝐧) 𝐜𝐨𝐬 𝛚𝐧 = 𝐜𝐨𝐬 𝛂𝛚 θ(ω) = -αω]
𝐧=𝟎
𝐍−𝟏
 After simplifying this equation, we have 𝐧=𝟎 𝐡(𝐧) 𝐬𝐢𝐧(𝛂 − 𝐧)𝛚 = 0.
𝑵−𝟏
 The above equation will be zero when h(n) = h(N – 1 - n) and α= 𝟐
𝑵−𝟏
 FIR filters have constant phase & group delays when impulse response is symmetrical about α = .
𝟐
 QUESTION AND ANSWERS:
1. What are the difference types of filters based on impulse response?
 Based on impulse response, filters are of two types. 1. IIR Filter 2. FIR Filter
 The IIR filters are recursive type, whereby the present output sample depends on the present input, past
input samples and output samples. The FIR filters are of non-recursive type whereby the present output
sample depends on the present input sample and previous input samples.
2. What are the difference types of filters based on frequency response?
 The filters can be classified based on frequency response.
 They are 1. Low Pass Filter 2. High Pass Filter 3. Band Pass Filter 4. Band Reject Filter
3. What is the most general form of IIR filter?
𝐌
𝐛𝐤 𝐳 −𝐤
 The most general form of IIR filter can be written as, H(z) = 𝟏+ 𝐤=𝟎
𝐍 −𝐤
. Where at least one of the
𝐤=𝟏 𝐚𝐤 𝐳
ak is nonzero and all the roots of the denominator are not cancelled by the roots of the numerator.
4. Distinguish between FIR and IIR filters.
SN FIR FILTER IIR FILTER
It can be easily designed to have perfectly linear
1) These filters do not have linear phase
phase.
FIR filters can be realized recursively & non-
2) IIR filters are easily realized recursively
recursively.
Greater flexibility to control shape of their Less flexibility, usually limited to
3)
magnitude response. specific kind of filters.
Errors due to round off noise are less severe in The round off noises in IIR filters is
4)
FIR filters, mainly because feedback is not used. more
5. What are the techniques of designing FIR filters?
 There are three well-known methods for designing FIR filters with linear phase.
 These are 1. Windows method 2. Frequency sampling method 3. Optimal or minimal design.
6. What is the reason that FIR filter is always stable?
 FIR filter is always stable because all its poles are at the origin.
7. What are the properties of FIR filter?
1) FIR filter is always stable
2) A realizable filter can always be obtained
3) FIR filter has a linear phase response.

-------------- ALL THE BEST -------------------- ALL THE BEST ----------------

<FAST FOURIER TRANSFORM (FFT)>  Prepared by Er. PARAMANANDA GOUDA, Dept of ETC, UCP Engg School
 Previous Year Semester Question of DIGITAL SIGNAL PROCESSING [6TH ETC - ETT 603] P a g e | 1 |

VI / SEM / E & TC / 2019 (W) [03-12-19, EX-REG]

DIGITAL SIGNAL PROCESSING

Full Marks: 80 Sub Code – ETT 603 Time: 3 Hours
Answer any FIVE Questions including Q. No. 1 & 2
The figures in the right-hand margin indicate marks
[GROUP-A]
1. Answer the following: [10 X 2]
a) Define sampling Theorem.
b) Differentiate between discrete signal and digital signal.
c) What is an LTI system?
d) Define ROC.
e) Write down any two properties of Z- transform.
f) What is the condition for system stability?
g) What is Zero padding?
h) Distinguish between DFT& DTFT.
i) Define Twiddle factor.
j) Define causal & non causal system
[GROUP-B]
2. Answer any FIVE questions: [5 X 5]
a) Determine the system described by y (n) = [x (n) + 1/x(n) ] is linear or non linear, where x (n)
& y (n) are input and output respectively.
b) Write down the properties of Z-transform.
c) Find the 4-point DFT of the sequence x(n)= {𝟏, 2, 1, 1}
d) Sketch the block diagram representation of discrete time system described by the input, output
output relation y(n) = 2y(n-1) + 3y(n+1) + x(n) + [1/2x(n+1)] + [1 /4x(n-2)], where x(n )is
the input sequence & y(n) is the output sequence.
e) Determine the Power and Energy of Unit step signal.
f) Compare the advantage of digital signal processing over analog signal processing.
g) Explain the properties of recursive and non recursive discrete time system
[GROUP-C]
3. Define the term signal and signal processing, Explain the digital signal processing system with
neat block diagram.
4. Determine the Z-transform and ROC of the signal: x(n)= a n u(n) + b n u( - n - 1)
5. Determine the circular convolution of the sequences x1 (n) = {𝟐, 1, 2, 1} & x2 (n) = {𝟏, 2, 3, 4}
6. Find 8-point DFT of the sequence is given by x(n)={ 𝟐, 1, 2, 1, 1, 2, 1, 2} by radix-2 DIT-FFT.
7. Write down the properties of DFT.

-------------- ALL THE BEST -------------------- ALL THE BEST ----------------

Collected By:-
Er. Paramananda Gouda
(Dept. of ETC, UCP Engg School)
 Previous Year Semester Question of DIGITAL SIGNAL PROCESSING [6TH ETC - ETT 603] P a g e | 2 |

[DIGITAL SIGNAL PROCESSING] VI / SEM / ETC / 2019 (S) [28-06-19, REG]

 Previous Year Semester Question of DIGITAL SIGNAL PROCESSING [6TH ETC - ETT 603] P a g e | 3 |

VI / SEM / E & TC / 2018 (W) [11-12-18, EX-REG]

DIGITAL SIGNAL PROCESSING

Sub Code – ETT-603
Full Marks: 70
Time: 3 Hours
Answer any FIVE Questions
The figures in the right-hand margin indicate marks
1) [2 +5 + 7]
a) What is multichannel signal?
b) What are advantages of DSP over ASP?
c) Classify different signal and explain briefly.
2) [2 +5 + 7]
a) What is shifting operation on discrete time signal?
b) What are the elementary of discrete time signal and explain?
c) Classify discrete time system and explain them briefly.
3) [2 +5 + 7]
a) What is quantization?
b) Determine the system function & unit sample response of the system? y(n) = ½ y(n-1)+2x(n)
c) Show the graphical representation of the signal, If x(n) = {2, 1, 0, 2, 1, 3}, find x( - n - 2)
4) [2 +5 + 7]
a) What is ROC?
b) Discuss various properties of Z-transform.
c) Compute the Z-transform and ROC of x(n) = 2n u(n) + 3n u( - n - 1).
5) [2 +5 + 7]
a) What is signal processing? Give any two application of it?
b) Determine the inverse z-transform of the sequence using long division method
𝟏 + 𝟐𝐳 −𝟏
X(z) = ; if x(n) is causal.
𝟏 − 𝟐𝐳 −𝟏 + 𝐳 −𝟐
c) Compute 4-point DFT of the following sequence, x(n) = {0, 1, 2, 3}
6) [2 +5 + 7]
a) Find the Z-transform and ROC of x(n) = {2, 3, 0, 1, 3}
b) Verify whether the following systems are time variant or time invariant
(i) Y(n) = X (− n) (ii) Y(n) = X(n) + X(n −1)
c) Determine the circular convolution of the given sequence using concentric circle method:
X(n) = {1, 2, 3, 0} and Y(n) = {1, 2, 1, 1} [2 +5 + 7]
7)
a) Draw the basic butterfly diagram for DIF-FFT.
b) Compute poles, zeros and system response of the following: Y(n) = 2 y(n - 1) + 3x(n)
c) Determine the DFT of the sequence using DIT-FFT algorithm: x(n) = {1, 2, 3, 4}

-------------- ALL THE BEST ------------------------- ALL THE BEST ----------------

 Previous Year Semester Question of DIGITAL SIGNAL PROCESSING [6TH ETC - ETT 603] P a g e | 4 |

VI / SEM / E & TC / 2018 (S) [10-05-18, REG]

DIGITAL SIGNAL PROCESSING

Sub Code – ETT-603
Full Marks: 70
Time: 3 Hours
Answer any FIVE Questions
The figures in the right-hand margin indicate marks
1) [2 +5 + 7]
a) Verify whether y(n)  x(n / 2) is a time invariant system.
b) Determine the value of power and energy of x(n) = Sin (nπ/4).
n / 2,0  n  5
c) Find the convolution of x(n)    & h(n)   (n)   (n  1)   (n  2)   (n  3) .
0, elsewhere 
2) [2 +5 + 7]
a) Define correlation.
b) Determine the z-transform and ROC of the signal x (n) = {2, 4, 5, 7, 0, 1} with the starting
index of the sequence is equal to -2.
c) Find the z-transform and ROC of x(n)  (n  0.5)(1 / 3) n u(n)
3) [2 +5 + 7]
a) State sampling theorem.
b) If the signal x (n) = {1, 2, 6, 4, 3 ,7, 5} then find x(2n), x(n/2), x(n+2), x(3-n), 3x(n-1)
c) Find the IDFT of the sequence x(n) = {1, 1-2j, -1 , 1+2j}
4) [2 +5 + 7]
a) What are the properties of frequency response of an LTI system?
b) What is twiddle factor and define zero padding with example.
c) Plot the pole-zero pattern and determine the stability of the system : -
y(n)  0.7 y(n  1)  0.1y(n  2)  2 x(n)  x(n  2)
5) [2 +5 + 7]
a) Define the pole and zero of a system function.
b) Define Linear Convolution. State its properties.
c) Find the DFT of a sequence x (n) = {1 for 0 ≤ n ≤2; 0 otherwise} for N = 4.
6) [2 +5 + 7]
a) What do you mean by time domain aliasing?
b) Differentiate between analog and digital filter.
1  3z 1
c) Find the Inverse z transform of X ( Z )  .
1  3z 1  2 z  2
7) [2 +5 + 7]
a) How many complex additions and multiplications are required for a 16 bit sample in DIT-FFT
algorithm?
b) Compute 4 point DFT of a sequence x (n) = {0, 1, 2, 3} using DIF-FFT algorithm.
c) Define circular convolution. Find the circular convolution of {1, 2, 2, 1} and {1, 2, 3, 1} using
circular convolution.
-------------- ALL THE BEST ------------------------- ALL THE BEST ----------------
 Previous Year Semester Question of DIGITAL SIGNAL PROCESSING [6TH ETC - ETT 603] P a g e | 5 |

VI / SEM / E & TC / 2017(W), [11-12-17, BACK]

DIGITAL SIGNAL PROCESSING

Sub Code – ETT-603
Full Marks: 70
Time: 3 Hours
Answer any FIVE Questions
The figures in the right-hand margin indicate marks
1. (a) What is Sampling Theorem? [2]
(b) Classify Different Signals. [5]
(c) Explain basic elements of Digital Signal Processing. [7]

2. (a) What is Quantization? [2]

𝟏
(b) y(n) = 2x(n) + 𝒙 (𝒏−𝟏) is Linear or Nonlinear System verify. [5]
(c) Draw block diagram of Discrete Time System. [7]

3. (a) What is ROC? [2]

(b) Determine Z-Transform and ROC of Signal x(n) = an u(n). [5]
𝟏+𝟑𝐳 −𝟏
(c) Find Inverse Z-Transform of x(z) = 𝟏+𝟑𝐳−𝟏 +𝟐𝐳−𝟐 [7]

4. (a) Write down properties of Convolution. [2]

(b) Explain recursive and non-recursive discrete time system. [5]
n
(c) Determine solution of difference equation y(n) = 5/6 y(n-1) -1/6 y(n-2) + x(n) for x(n) = 2 u(n).

5. (a) Define a stable system. [2]

(b) Describe parallel connection of systems. [5]
(c) Find circular convolution of two finite duration sequence
x1(n) = {1, -1, -2, 3, -1} and x2(n) = {1, 2, 3}. [7]

6. (a) Relate DFT to Z-Transform. [2]

(b) Describe different properties of DFT. [5]
(c) Find DFT of a sequence x (n) = {1, 0, 0}. [7]

7. (a) What are the advantages of FIR Filters? [2]

(b) Compute DFT of a sequence x (n) = {1, -1, 1, -1} using DIT Algorithm. [5]
(c) What are different steps required for radix-2 DIF-FFT algorithm? [7]

-------------- ALL THE BEST -------------------- ALL THE BEST ----------------

Collected By:-
Er. Paramananda Gouda
(Dept. of ETC, UCP Engg School)
 Previous Year Semester Question of DIGITAL SIGNAL PROCESSING [6TH ETC - ETT 603] P a g e | 6 |

VI / SEM / E & TC / 2017(S) <APR, REG>

DIGITAL SIGNAL PROCESSING
Sub Code – ETT-603
Full Marks: 70
Time: 3 Hours
Answer any FIVE Questions
The figures in the right-hand margin indicate marks
1. (a)What is the difference between Domestic and Random signal? [2]
(b) What is signal processing? Draw block diagram & explain digital signal processing system [5]
(c) x (n) = e2nu(n). Determine the signal is Energy signal or Power signal. [7]
2. (a) Determine x(n) = u(n+1) is a casual signal or non-casual signal. [2]
1, for n = -1, 0, 2, 3
(b) If x(n) = -1, for n = -2, 1
0, otherwise; Then find out x (n+2), x (-n), x (-n+3), x (-n-1). [5]
(c) Define linear & non-linear system & prove y(n)=2x(n)+1/x(n-1) is linear or non-linear system.
3. (a) What are the necessary conditions for stable system? [2]
(b) Draw and explain principle of Analog to Digital Converter. [5]
(c) Find out the Convolution between two signals [7]
1, for n = -2, 0, 1
x(n) = 2, for n = -1
0, elsewhere and h(n) = δ(n) - δ(n-1) - δ(n-2) - δ(n+1).
4. (a) Write down the properties of convolution. [2]
(b) Find the z-transform and ROC of a system x(n) = an u(n) - bn u(n). [5]
(c) Write down all the properties of z-transform with proof. [7]
5. (a) Write down all methods are used for find out inverse z-transform. [2]
(b) Determine the Inverse z-transform of X(z) = 1/1 - 1.5z -1 + 0.5z -2
Where (i) ROC : |z| > 1 (ii) ROC : |z| < 0.5 (iii) ROC : 0.5 < |z| < 1 [5]
(c) Find out the forced response of the system described by the equation:
y(n) = 0.6y(n - 1) – 0.08y(n - 2) + x(n) [7]
6. (a) Define DFT. [2]
(b) Explain the relation of DFT to the other transform. [5]
(c) Find out the 4-point DFT of x(n) = (-1)n. [7]
7. (a) Write down the periodicity and time reversal properties of DFT. [2]
(b) Derive an expression for DFT in Radin-2 DFT-FFT and justify in case of DIT-FFT the total
sequence is contain equal no. of even part and odd part. [5]
(c) Find the 8-point DFT of the sequence given by x (n) = {2, 2, 2, 2, 1, 1, 1, 1} by using Radix -2
DIT-FFT algorithm. [7]
-------------- ALL THE BEST -------------------- ALL THE BEST ----------------

Collected By:-
Er. Paramananda Gouda
(Dept. of ETC, UCP Engg School)
 Previous Year Semester Question of DIGITAL SIGNAL PROCESSING [6TH ETC - ETT 603] P a g e | 7 |

VI / SEM / E & TC / 2016(W), <DEC, BACK>

DIGITAL SIGNAL PROCESSING
Sub Code – ETT-603
Full Marks: 70
Time: 3 Hours
Answer any FIVE Questions
The figures in the right-hand margin indicate marks
1. (a) Define Zero Padding. [2]
(b) Discuss the properties of the DFT. [5]
(c) Find the 4 point DFT of the sequence x (n) = {1, 0, 1, 0} [7]

2. (a) Define ROC. [2]

(b) Write down the advantages of Digital Signal Processing over analog signal processing. [5]
(c) By using partial fraction expansion method, find the inverse z transform of,
𝑧(𝑧 2 − 4𝑧 +5)
X(z) = [7]
𝑧−1 𝑧−2 (𝑧−3)

3. (a)What is z transform? [2]

(b) Discuss various properties of z-transform. [5]
(c) Compute 4 point DFT of the following sequences using DIT algorithm: x(n) = {4, 3, 2, 1} [7]

4. (a) Define DFT. [2]

(b) Determine z-transform & ROC of the finite duration signals: x(n) = {1, 0, 3, -1, 2} [5]
(c) Obtain the circular convolution of the following sequencers. [7]
x1(n) = {1, 2, 2, 1} & x2(n) = {1, 2, 3, 1}

5. (a) What is circular convolution? [2]

(b) Discuss the algorithm of DIT - FFT. [5]
(c) Compute a 4 point DFT of the sequences x (n) = {1, 0, 0, 1} using DIF algorithm. [7]

6. (a) Draw the basic butterfly diagram for DIF – FFT. [2]
(b) Differentiate between linear and circular convolution. [5]
(c) Compute the z-transform and ROC of x (n) = 2n u (n) [7]

7. (a) State the applications of FFT algorithm. [2]

(b) Draw the reduced flow graph for 4 point DIF – FFT. [5]
(c) Determine the Convolution of the two sequences. [7]
x1 (n) = {2, 1, 0, 0, 5} & x2 (n) = {2, 2, 1, 1}

-------------- ALL THE BEST -------------------- ALL THE BEST ----------------

Collected By:-
Er. Paramananda Gouda
(Dept. of ETC, UCP Engg School)
 Previous Year Semester Question of DIGITAL SIGNAL PROCESSING [6TH ETC - ETT 603] P a g e | 8 |

VI / SEM / E & TC / 2016(S) <APR, REG>

DIGITAL SIGNAL PROCESSING
Sub Code – ETT-603
Full Marks: 70
Time: 3 Hours
Answer any FIVE Questions
The figures in the right-hand margin indicate marks
1. (a) Define Time variant and Time invariant system [2]
(b) Explain the different properties of z transform. [5]
𝟏
(c) Determine the causal signal x(n) having the z transform x(z) = using partial
(𝟏− 𝟐𝒛−𝟏 )(𝟏− 𝒛−𝟏 )𝟐
fraction expansion method. [7]

2. (a) What is Twiddle Factor? [2]

(b) Find H(z), the system function for the following:
i. y(n) - 3y(n-1) + 2y(n-2) = x(n) – x(n-1)
ii. y(n) = x(n) + x(n-1) - 2x(n-2) + 3x(n-3) [5]
𝒏𝝅
(c) Find the z transform and ROC of the sequence x (n) = 2n.𝐬𝐢𝐧 . u (n) [7]
𝟒

3. (a) Define Zero Padding. [2]

(b) Compute the Convolution y (n) correlation r (n) for the given signals:
x1(n) = {1, 2, 3, 4} & x2(n) = {1, 2, 3, 4} [5]
(c) Find the Circular Convolution of two finite duration sequences
x1(n) = {1, 1, -1, 2} & x2(n) = {2, 0, 1, 1} [7]

4. (a) What is the need of signal processing and give any two applications. [2]
(b) Find the step response of the following differential equation:
y (n) - 5y (n-1) + 6y (n-2) = x(n) [5]
𝒛 (𝟏− 𝒆−𝑻 )
(c) Compute the inverse z transform of x (z) = [7]
(𝒛 − 𝟏)(𝒛 − 𝒆−𝑻 )

5. (a) Draw the basic butterfly diagram for DIT – FFT and DIF-FFT. [2]
(b) Find the z transform and ROC of x (n) = (0.4)n u(n) + (0.3)n u(n-4) [5]
(c) Compute a 4 point DFT of the sequence x (n) = {0, 2, 4, 6} [7]

6. (a) Define Periodic and Aperiodic signals. [2]

(b) Verify whether the following systems are linear or non-linear:
i. y(nT) = x(nT+T) + x(nT - T) ii. y(n) = x(n+7) [5]
(c) Determine the DFT of the sequence x(n) = {1, 2, -1, 1} using DIT-FFT algorithm. [7]

7. (a) What is Discrete Fourier transform? [2]

(b) Determine the IDFT of X (K) = {1, 0, 1, 0} [5]
(c) Using property find the z-transform and ROC of x(n) = n. u(n-1) where, x(n) is causal. [7]

-------------- ALL THE BEST -------------------- ALL THE BEST ----------------

 Previous Year Semester Question of DIGITAL SIGNAL PROCESSING [6TH ETC - ETT 603] P a g e | 9 |

VI / SEM / E & TC / 2014 (S)

DIGITAL SIGNAL PROCESSING
Sub Code – THEORY-4
Full Marks: 80
Time: 3 Hours
Answer any FIVE Questions
The figures in the right-hand margin indicate marks
1. (a) Define signal and system. [2]
(b) What are the different types of signal representation? Explain with examples. [6]
(c) Compute the convolution of the discrete time signals,
x(n) = {1, 0, 2, 5, 4}, h(n) = {1, -1, 1, -1} [8]
2. (a) Define periodic and aperiodic signals. [2]
(b) Verify whether the following systems are causal or non-causal.
𝟏
y1(n) = x(2n), y2(n) = x(n+1) + [6]
𝐱(𝐧−𝟏)
(c) State the advantages of digital signal processing over analog signal processing. [8]
3. (a) State the difference (any four) between FIR and IIR systems. [2]
(b) Show the graphical representation of the signals,
x(n-2), x(n+3), x(-n-2), and x(-n+3) where x(n) = {1, 2, 1, 2, 1} [6]
(c) Determine the impulse response of the causal system,
y(n) + y(n-1) + 2y(n-2) = x(n-1) + 2x(n-2) [8]
4. (a) Define z-transform and where it is used? [2]
(b) Find the system function of the system described by the difference equation
y(n) = x(n) + 2x(n-1) – 4x(n-2) + x(n-3) [6]
(c) Find the z-transform and ROC of the sequence x(n) = (4)n u(-n-1) [8]
5. (a) Define poles and zeros of a system. [2]
(b) State the time shifting and time reversal property of z-transform. [6]
𝟏
𝟏+ 𝐳−𝟏
𝟐
(c) Find the inverse z-transform of X(z) = [8]
𝟏+𝟑𝐳 −𝟏 +𝟐𝐳 −𝟐

6. (a) Name number of complex multiplications & additions required to compute N point DFT. [2]
(b) Find the circular convolution of two finite duration sequence
x1(n) = {1, 2, 3, 4} & x1(n) = {4, 3, 2, 1} [6]
(c) Compute the four point DFT of the sequence x(n) = {1, 1, 1, 1} [8]
7. (a) Define twiddle factor. [2]
(b) State the difference between analog filter and digital filter. [6]
(c) Determine the DFT of the sequence x(n) = {1, 2, 1, 1, 0, 1, 1, 1} using DIT-FFT algorithm. [8]

-------------- ALL THE BEST -------------------- ALL THE BEST ----------------

Collected By:-
Er. Paramananda Gouda
(Dept. of ETC, UCP Engg School)
 Previous Year Semester Question of DIGITAL SIGNAL PROCESSING [6TH ETC - ETT 603] P a g e | 10 |

VI / SEM / E & TC / 2013 (S)

DIGITAL SIGNAL PROCESSING
Sub Code – THEORY-4
Full Marks: 80
Time: 3 Hours
Answer any FIVE Questions including Q. No. 1 & 2
The figures in the right-hand margin indicate marks
[GROUP-A]
8. Answer the following: [10 X 2]
k) What is signal processing?
l) What are the major classifications of digital signals?
m) Distinguish between energy and power signal.
n) What is LTI system? Give one example.
o) Why is folding of a signal required in convolution of two signals?
𝐝𝐲(𝐭)
p) Is the system described by the differential equation + 5y(t) + 2 = x(t) is linear?
𝐝𝐭
q) Define Fourier transform pair.
r) What are twiddle factors of the DFT?
s) State direct and inverse z-transform.
t) What is circular convolution?
u) Write one advantages of FIR filter over IIR filter.
[GROUP-B]
9. Answer any FIVE questions: [5 X 5]
a) What are the various types of realization structure for FIR and IIR filter?
b) Give the frequency response characteristics of Butterworth filter.
c) State and explain symmetry property of DFT.
d) Find the z-transform of u(n) = cosω0n for n ≥ 0.
e) State and explain time reversal and time shifting properties of Z-transform.
f) Give the frequency response characteristics of Butterworth filters.
g) Discuss the radix-4 FFT algorithm.
[GROUP-C]
n
10. Given x(n) = 2 and N = 8. Find x (k) using DIF-FFT algorithm. [10]
𝟏 −𝐧
11. Find the discrete-time Fourier transform of x (n) = . u(-n-1) [10]
𝟐
12. Determine the z-transform of the sequence given by,
2n , n < 0
1 n
x(n) = , n = 0,2,4 [10]
2
1 n
, n = 1,3,5
3
13. Write short notes on any TWO: [5 X 2]
a) What is signum function? Give its plot.
b) Discuss causal and non-causal LTI systems with example.
c) State and explain sampling theorem.
-------------- ALL THE BEST -------------------- ALL THE BEST ----------------
 Previous Year Semester Question of DIGITAL SIGNAL PROCESSING [6TH ETC - ETT 603] P a g e | 11 |

VI / SEM / E & TC / 2012 (S)

DIGITAL SIGNAL PROCESSING
Sub Code – THEORY-4
Full Marks: 80
Time: 3 Hours
Answer any FIVE Questions including Q. No. 1 & 2
The figures in the right-hand margin indicate marks
[GROUP-A]
1. Answer the following: [10 X 2]
a) What is discrete-time signal?
b) Define multichannel and multidimensional signal.
c) State sampling theorem and why it is essential?
d) Distinguish between causal and non-causal system with examples.
e) Define recursive discrete time system.
f) State convolution theorem.
g) Find the z-transform of the sequence x(n) = {-1, 3, 3, 1, 0, 2}
h) What do you mean by N-point DFT?
i) What are digital filters?
j) Define poles and zeroes of a discrete time system.
[GROUP-B]
2. Answer any FIVE questions: [6 X 5]
a) What are the advantages of digital signal processing over analog signal processing?
𝐧
b) Determine whether the system is time-invariant or time variant of the system. y(n) = x
𝟐
c) Write the expression to determine energy and power of discrete time signals.
𝟏, 𝐧 = −𝟐, 𝟎, 𝟏
d) Find the convolution sum of the signals x(n) = 𝟐, 𝐧 = −𝟏
𝟎 𝐞𝐥𝐬𝐞𝐰𝐡𝐞𝐫𝐞
e) Determine the z-transform of the signal x(n) = anu(n)
f) Explain the linear property of Discrete Time Fourier Transform.
𝟏
𝐟𝐨𝐫 𝟎 ≤ 𝐧 ≤ 𝟐
g) Compute DFT of the given signal sequence x(n) = 𝟑
𝟎 𝐞𝐥𝐬𝐞𝐰𝐡𝐞𝐫𝐞
[GROUP-C]
−𝟒+𝟖𝒛
3. Find the Inverse z-transformation X(z) = [10]
𝟏+𝟔𝒛−𝟏 +𝟖𝒛−𝟐

4. Express the signal x(n) = {1, -2, 3, 0, 1, -5, 2, 1} in even and odd signal. [10]
𝟓
5. Find out the impulse response of the system y(n) - y(n-1) + y(n-2) = x(n)- x(n-1) [10]
𝟐
2
6. (a) Determine the linearity of the system described by the input output equation y(n) = x(n ) [5]
(b) Differentiate between continuous valued and discrete valued signals. [5]
7. Describe DFT and FFT algorithm and write the computational formula. [10]

-------------- ALL THE BEST -------------------- ALL THE BEST ----------------

 Previous Year Semester Question of DIGITAL SIGNAL PROCESSING [6TH ETC - ETT 603] P a g e | 12 |

VI / SEM / E & TC / 2011 (S)

DIGITAL SIGNAL PROCESSING
Sub Code – THEORY-4
Full Marks: 80
Time: 3 Hours
Answer any FIVE Questions including Q. No. 1 & 2
The figures in the right-hand margin indicate marks
[GROUP-A[]
1. Answer the following: [10 X 2]
a) Name the basic elements of a digital signal processing system.
b) What is a static system?
c) What is discrete time signal?
d) What do you mean by symmetric signal?
e) Define unit ramp signal.
f) Define continuous time and discrete time.
g) What do you mean by DFT?
h) What do you mean by quantization of a sinusoidal signal?
i) What are digital filter?
j) What is the advantage of FFT algorithm over DFT algorithm?
[GROUP-B]
2. Answer any FIVE questions: [6 X 5]
a) Write the advantage of digital signal processing.
b) Discuss the properties of a discrete time sinusoidal signals.
c) Compute DFT of the four point sequence x(n) = {0, 1, 2, 3}
d) Write the properties of z-transform.
e) Find the z-transform of x(n) where x(n) = 3 cos 2000 μt + 5 sin 6000 μt + 10 cos 12000 μt.
What is the Nyquist rate for this signal?
[GROUP-C]
3. Explain poles and zeros. Determine the pole zero plot for the signal X(n) = anU(n). [10]
4. With a neat block diagram explain different parts of digital to analog converter. [10]
5. State and explain inverse z-transform. Determine the inverse z-transform of
𝟏
X(z) = [10]
𝟏−𝟏.𝟓𝒛−𝟏 +𝟎.𝟓𝒛−𝟐
6. Design a linear phase FIR filter using windows. [10]
7. Write the algorithm for divide and conquer approach for computation of discrete Fourier
transform. [10]
-------------- ALL THE BEST -------------------- ALL THE BEST ----------------

Collected By:-
Er. Paramananda Gouda
(Dept. of ETC, UCP Engg School)
 Previous Year Semester Question of DIGITAL SIGNAL PROCESSING [6TH ETC - ETT 603] P a g e | 13 |

VI / SEM / E & TC / 2010 (S)

DIGITAL SIGNAL PROCESSING
Sub Code – ETT-603
Full Marks: 70
Time: 3 Hours
Answer any FIVE Questions including Q. No. 1 & 2
The figures in the right-hand margin indicate marks
1. Answer the following: [10 X 2]
a) Define discrete time signal and system.
b) Show the graphical representation of 4(n+2).
c) Write four operations that are performed on discrete time signals.
d) Compute the convolution if x(n) = h(n) = 1, 0, 1}
e) What is the z transform of unit impulse?
f) Define energy and power signal.
g) What is the advantage of digital signal processing?
h) What is the basic element of DSP?
i) Define FIR and IIR system.
j) What is meant by Radix – 2 FFT?
2. Answer any Six questions: [6 X 5]
a) Compare three properties of continuous & discrete time sinusoidal signal.
b) Discuss the basic parts of analog to digital (AD) converter.
c) Define convolution. Compute the convolution of following signals
x(n) = {1, 0, 2, 5, 4}, h(n) = {1, -1, 1, -1}
d) Determine the z transform and ROC of signal x(n) = {cos ω0 n}u(n).
e) Describe the properties of z transform.
f) Determine the pole and zero for the signal x(n) = n an u(n)
g) Determine if the following systems are causal or non causal.
1
i. y(n) = x(n) + (ii) y(n) = x(n2)
x(n)
h) Find the circular convolution of the two sequence. x1(n) = {1, -1, 2, 3}; x2(n) = {0, 1, 2, 3}
3. Find the inverse z transform
𝑧(𝑧 2 −4𝑧+5) 1+3𝑧 −1
i. X(z) = for ROC 𝑧 > 3 (ii) X(z) = 𝑍 >2 [5]
𝑧−3 𝑧−1 (𝑧−2) 1+3𝑧 −1 +2𝑧 −2

4. (a) Determine time variant and time invariant system. [4]

(b) Determine whether the following system is time invariant: [6]
i. y(n) = n x(n) (ii) y(n) = cos x(n) (iii) y(n) = ex(n)
5. (a) Write the properties of DFT. [5]
(b) Find the DFT of the sequence x(n) = {1, 1, 0, 0} [5]
6. (a) State and explain sampling theorem. [5]
(b) An analog signal is represented by xa (t) = sin (480πt) + 3sin(720πt). What is the Nyquist rate.

7. (a) What is the relationship of DFT with z-transform? [5]

(b) Compute 4-point DFI of a sequence x(n) = {0, 1, 2, 3} using Radix-2 algorithm. [5]

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