Addis Ababa Science and Technology University

College of Electrical & Mechanical Engineering

Electrical & Computer Engineering Department

Digital Signal Processing (EEEg-3151)

Chapter One
Introduction to Digital Signal Processing
Introduction to Digital Signal Processing
Outline
 Introduction
 Limitations of Analog Signal Processing
 Applications of Digital Signal Processing
 Advantages and Disadvantages of DSP
 Sampling of Continuous-time Signals

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Introduction

 Signal processing is a method of modifying one signal into

another signal.
 We encounter many types of signals in various applications
 Electrical signals: voltage, current, magnetic and electric fields,…
 Mechanical signals: velocity, force, displacement,…
 Acoustic signals: sound, vibration,…
 Other signals: pressure, temperature,…

 Most real-world signals are analog

 They are continuous in time and amplitude
 Convert to voltage or currents using sensors and transducers

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Introduction Cont’d……

 Signal processing can be classified into two broad categories:

Analog Signal Processing and Digital Signal Processing

 Analog signal processing can be done using analog circuit

elements such as:
 Resistors, Capacitors, Inductors, Amplifiers,…

 Analog signal processing examples

 Audio processing in FM radios

 Video processing in traditional TV sets

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Limitations of Analog Signal Processing

 Accuracy limitations due to

 Component tolerances

 Undesired nonlinearities

 Limited repeatability due to

 Tolerances

 Changes in environmental conditions

• Temperature

• Vibration

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Limitations of Analog Signal Processing….

 Sensitivity to electrical noise

 Inflexibility to changes

 Limited dynamic range for voltage and currents

 Difficulty of implementing certain operations

 Nonlinear operations

 Time-varying operations

 Difficulty of storing information

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Digital Signal Processing
 Represent signals by a sequence of numbers
 Sampling or analog-to-digital conversions
 Perform processing on these numbers with a digital
processor
 Digital signal processing
 Reconstruct analog signal from processed numbers
 Reconstruction or digital-to-analog conversion

digital digital
signal signal
analog analog
signal A/D D/A signal

7
Digital Signal Processing……..

Forms or Types of Signal Processing:

 Analog input – analog output  Analog Signal Processing

 Digital recording of music

 Analog input – digital output  Mixed Signal Processing

 Touch tone phone dialing

 Digital input – analog output  Mixed Signal Processing

 Text to speech

 Digital input – digital output  Digital Signal Processing

 Compression of a file on computer
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Applications of Digital Signal Processing
 Where is DSP applicable?
 DSP has wide range of applications in every area of
electrical engineering.

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Applications of Digital Signal Processing…..

 Sound applications
 Compression, enhancement, special effects, synthesis, recognition,
echo cancellation,…

 Cell Phones, MP3 Players, Movies, Dictation, Text-to-speech,…

 Communication
 Modulation, coding, detection, equalization, echo cancellation,…

 Cell Phones, dial-up modem, DSL modem, Satellite Receiver,…

 Automotive
 ABS, GPS, Active Noise Cancellation, Cruise Control, Parking,…

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Applications of Digital Signal Processing…..

 Medical
 Magnetic Resonance, Tomography, Electrocardiogram,…

 Military
 Radar, Sonar, Space photographs, remote sensing,…

 Image and Video Applications

 DVD, JPEG, Movie special effects, video conferencing,…

 Mechanical
 Motor control, process control, oil and mineral prospecting,…

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Advantages and Disadvantages of DSP
Advantages:
 Accuracy can be controlled by choosing word length

 Repeatable

 Sensitivity to electrical noise is minimal

 Dynamic range can be controlled using floating point numbers

 Flexibility can be achieved with software implementations

 Non-linear and time-varying operations are easier to implement

 Digital storage is cheap

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Advantages and Disadvantages of DSP…….

Advantages:
 Digital information can be encrypted for security

 Price/performance and reduced time-to-market

Disadvantages:
 Sampling causes loss of information

 A/D and D/A requires mixed-signal hardware

 Limited speed of processors

 Quantization and round-off errors

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Sampling of Continuous-time Signals

 Most discrete-time signals come from sampling a continuous-

time signal, such as speech and audio signals, radar and sonar
data, and seismic and biological signals.

 The process of converting these signals into digital form is

known as analog-to-digital (A/D) conversion.

 The reverse process of reconstructing an analog signal from its

samples is known as digital-to-analog (D/A) conversion.

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Sampling of Continuous-time Signals……..

 A discrete-time signal x(n) is obtained from a continuous-time

signal xa(t) according to the relation:
x(n)  x a (nTs ) ,    n  
where :
Ts : is the sampling period
 The reciprocal of the sampling period is known as the sampling
frequency.
1
fs 
Ts
where :
f s : is the sampling frequency
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Sampling of Continuous-time Signals……..

 A system that implements the above operation is known as an

ideal continuous-time to discrete-time (C/D) converter.

Fig. Ideal continuous-time to discrete-time (C/D) converter

 The sampling operation is generally not invertible.

 This is because many input continuous-time signals can produce
the same output discrete-time signal.

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Sampling of Continuous-time Signals……..

 The overall sampling process can be illustrated by the figure

given below.

Fig. Continuous-to-discrete (C/D) conversion

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Sampling of Continuous-time Signals……..

 First, the continuous-time signal is multiplied by a periodic

sequence of impulses to form a sampled signal.
 The periodic sequence of impulses is given by:

s a (t )    t  nT 
n  
s

 And the sampled signal is given by:

x s (t )  x a (t ) * s a (t )

 x s (t )   x nT  t  nT 
n  
a s s

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Sampling of Continuous-time Signals……..
 Then, the sampled signal is converted into a discrete-time signal
by mapping the impulses that are spaced in time by Ts into a
sequence x(n) where the sample values are indexed by the
integer variable n:

x(n)  xa nTs 

Fig. An example that illustrates the conversion process

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Sampling of Continuous-time Signals……..

 Consider a continuous-time sinusoidal signal given by:

xa (t )  cos m t   cos2f m t 
 The sampled discrete-time sinusoidal signal is given by:

x(n)  xa (nTs )  cos m nTs   cosm n

 Thus, the continuous-time angular frequency  m and discrete-
time angular frequency  m can be related as:

 m   mTs ,  m : continuous - time angular frequency

 m : discrete - time angular frequency

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Sampling of Continuous-time Signals……..
Nyquist Sampling Theorem:
 Let x a (t ) be a bandlimited signal with:
Xa ( f )  0 , for f  f m
 Then x a (t ) can be uniquely recovered from its
samples x(n)  x a (nTs ) only if:

fs  2 fm  f N
 The maximum frequency content of the continuous-time signal
fm is called the Nyquist frequency and the minimum sampling
frequency fN is known as the Nyquist rate.

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Sampling of Continuous-time Signals……..

Example-1:
Consider the continuous-time signal given by:

xa (t )  3 cos50t   5 cos100t   10 cos300t 

a. Find the Nyquist rate for xa(t).

b. What will be the resulting discrete-time signal, if xa(t) is

sampled at:

i. the Nyquist rate.

ii. a rate of twice the Nyquist rate.

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Sampling of Continuous-time Signals……..

Solution:
a. The maximum frequency of the continuous-time signal is
obtained as:
m 300
fm   fm   150 Hz
2 2
The Nyquist rate is then:

f N  2 fm  f N  2 *150  300 Hz

b. The discrete-time signal obtained by sampling the above

continuous-time signal can be calculated as follows:

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Sampling of Continuous-time Signals……..

Solution:

i. When the sampling frequency is equal to the Nyquist

rate:
f s  f N  f s  300 Hz
 1   2   3 
x(n)  3 cos n   5 cos n   10 cos n 
 fs   fs   fs 
 50   100   300 
 x(n)  3 cos n   5 cos n   10 cos n
 300   300   300 
 n   n 
 x(n)  3 cos   5 cos   10 cos n 
 6   3
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Sampling of Continuous-time Signals……..

Solution:

ii. When the sampling frequency is equal to twice the

Nyquist rate:
f s  2 f N  f s  2 * 300 Hz  600 Hz
 1   2   3 
x(n)  3 cos n   5 cos n   10 cos n 
 fs   fs   fs 
 50   100   300 
 x(n)  3 cos n   5 cos n   10 cos n
 600   600   600 
 n   n   n 
 x(n)  3 cos   5 cos   10 cos 
 12   6   2
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Sampling of Continuous-time Signals……..

Example-2:
Consider the discrete-time signal given by:

 n 
x(n)  cos 
 8 
Find two different continuous-time signals that would produce
the above discrete-time signal when sampled at a frequency of
fs=10 kHz.
Solution:
• Consider a continuous-time sinusoidal signal given by:
xa (t )  cos m t   cos2f m t 
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Sampling of Continuous-time Signals……..

Solution:
• Sampling the above continuous-time sinusoidal signal with a
sampling frequency of fs results in the discrete-time signal
given by:
 fm 
x(n)  xa (nTs )  cos 2 n 
 fs 
• However, for any integer k, we have:

 fm   f m  kf s 
cos 2 n   cos 2 n 
 fs   fs 

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Sampling of Continuous-time Signals……..

Solution:
• Therefore, any sinusoid with a frequency f  f mkf s will
produce the same discrete-time signal when sampled with a
sampling frequency fs.
• For the given discrete-time signal, we have:
fm  1
2   fm  f s  625 Hz
fs 8 16
• Therefore, two continuous-time signals that produce the given
discrete-time signal are:
x1 (t )  cos1250t  and x2 (t )  cos21250t 
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Exercise

1. Describe some application areas of digital signal processing in

the field of:
a. Communication Engineering
b. Military
2. Consider the continuous-time signal given by:
xa (t )  cos650t   2 sin720t 
a. What is the Nyquist rate for xa(t) ?

b. If xa(t) is sampled at a rate of twice the Nyquist rate, what

will be the resulting discrete-time signal?

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Exercise……..

3. Consider the continuous-time signal given by:

xa (t )  sin200t 
What will be the resulting discrete-time signal x(n), if the
above continuous-time signal is sampled with a sampling
1
period of Ts  seconds.
400

4. Find two different continuous-time signals that will produce

the discrete-time signal x(n)  cos(0.15n) when sampled
with a sampling frequency of 8 kHz.

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Exercise……..

5. Consider the discrete-time signal given by:

 n 
x(n)  cos  ,   n  
 4
The above discrete-time signal was obtained by sampling the
following continuous-time signal at a sampling rate of 1000
samples/sec.

xa (t )  cos m t  ,   t  

What are the two possible positive values of  m that would

have resulted in the discrete-time signal x(n)?

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Exercise……..

6. Consider a continuous-time signal given by:

xa (t )  sin(20t )  cos(40t )

The above discrete-time signal is sampled with a sampling

period Ts to obtain the discrete-time signal:

 n   2n 
x(n)  sin   cos 
 5  5 
Determine two possible values of Ts that would have resulted in
the discrete-time signal x(n)?

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