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EC : ELECTRONICS AND COMMUNICATIONS

EE : ELECTRICAL ENGINEERING

SIGNALS AND SYSTEMS

INDEX

Pg.
Contents Topics
No.

1. Linear Time Invariant System

Continuous Time Signals and System 1
Classification of signals 1
Useful signal Operations 4
Type of Signals 8
PART(A) :
Classification of Systems 17
TimeDomain Analysis of LTIC Systems 20
System Stability 35
Notes Discrete Time Signals and System 37
Types of Signals 37
Useful Signal Operations 45
Classification of Systems 48
PART(B) :
Time-Domain Analysis of LTID System 49
System Stability 65
List of Formulae 67
LMR (Last Minute Revision) 69
Assignment1 Questions 70
 Pg.
Contents Topics
No.

2. Continuous and Discrete Time Fourier Series and Transform

PART(A) : Continuous Time Fourier Series 75
Orthogonal Signal Space 76
Trigonometric Fourier Series 77
Existence of The Fourier Series: Dirichlet
conditions
80
The Effect of Symmetry 82
Fourier Synthesis of Discontinuous
82
Functions: The Gibbs Phenomenon
Exponential Fourier Series 83
Parseval's Theorem 86
Continuous Time Fourier Transform 87
Notes
Properties of Fourier Transform (FT) 88
Signal Transmission Through LTIC Systems 100
(Only for EE & IN)
Raleigh Energy Theorem 101
Duality 102
Phase and Group Delay 103
PART(B) : Discrete-Time Fourier Series (DTFS) 104
Discrete Time Fourier Transform (DTFT) 106
Nature of Fourier Spectra 106
Properties of DTFT 108
LMR (Last Minute Revision ) 111
Assignment2 Questions 113
3. Definition and Properties of Laplace Transform
Introduction 117
The Unilateral Laplace Transform 120
Properties of Laplace Transform 121
Laplace Transform of Some Elementary Functions 128
Finding Inverse Laplace Transform 129
Notes
Solution of Differential and Intergro-Differential Equation 132
Analysis of LTIC System Using Laplace Transform 133
Bilateral Laplace Transform (BLT) 135
System Realization 140
LMR (Last Minute Revision) 143
Assignment3 Questions 145
 Pg.
Contents Topics
No.

Transform
4. Z
The Z-Transform 149
The Unilateral Z-Transform 151
Properties of Z-Transform 152
Important Theorems 155
Z-Transform of Some Useful Signals 156
Inverse ZTransform 157
Notes
Solving Linear Difference Equation Using Z-Transform 159
Analysis of LTID System 162
Connection Between the Laplace and the Z-Transform 163
The Bilateral Z-Transform 163
System Realization 165
LMR (Last Minute Revision) 167
Assignment4 Questions 169
5. Sampling
Introduction to Sampling 172
Effects of Sampling 174
Sampling Theorem 176
Sampling Theorem for Low Pass Signals 176
Signal Reconstruction : The Interpolation Formula 177
Notes
Sampling Theorem for Band Pass Signals 178
Frequency Relationship 179
The Discrete Fourier Transform (DFT) 180
Fast Fourier Transform (FFT) 187
LMR (Last Minute Revision) 191
Assignment5 Questions 192

1
Test Paper Questions 195

Test Paper2 Questions 201

Practice Problems Questions 207

 Pg.
Contents Topics
No.

SOLUTIONS

Answer Key 220

Assignments
Model Solutions 221

Answer Key 243

Test Papers
Model Solutions 244

Answer Key 254

Practice Problems
Model Solutions 255
 Topic 1 : Linear Time Invariant System

PART(A) : CONTINUOUS TIME SIGNALS AND SYSTEM

Introduction
Signals : A signal is a single valued function of time that represents a physical variable of
interest such as voltage, current, displacement, velocity etc. It can be either real valued
or complex valued.
System : A system is a combination and interconnection of several components to
perform a desired task. Systems may be interconnected with each other to form other
systems. In such cases the component systems are called as subsystems.

CLASSIFICATION OF SIGNALS
There are several classes of signals.
1. Continuous-time and discrete-time signals
2. Analog and digital signals
3. Periodic and aperiodic signals
4. Energy and power signals
5. Deterministic and probabilistic signals
1. Continuous-time and discrete-time signals :

A signal that is specified for every value of time t is a continuous time signal, (fig. (a)
and (b)) and a signal that is specified only at discrete values of time t is a discrete
time signal (fig. (c) and (d)).

f (t) f (t)

t
t
(a) Analog, continuous time (b) Digital, continuous time

f (t) f (t)

t
t

(c) Analog, discrete time (d) Digital, discrete time

Examples of Signals

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Vidyalankar : GATE – EC

2. Analog and digital signals :

Analog signal :
The concept of continuous time is often confused with that of analog. The two are not
the same. The same is true of the concepts of discrete time and digital. A signal
whose amplitude can take on any value in a continuous range is an analog signal
(fig. (a) and (c)). A digital signal, on the other hand, is one whose amplitude can take
on only a finite number of values (fig. (b) and (d)). Signals associated with a digital
computer are digital because they take on only two values (binary signals).

Thus the terms continuous time and discrete time qualify the nature of a signal along
the time (horizontal) axis. The terms analog and digital, on the other hand, qualify the
nature of the signal amplitude (vertical axis). An analog signal can be converted into a
digital signal [analogtodigital (A/D) conversion] through quantization (rounding off).

3. Periodic and aperiodic (non-periodic)signals :

A signal f(t) is said to be periodic if for some positive constant T0

f(t) = f(t + T0) for all t …(1)

The smallest value of T0 that satisfies the periodicity condition (1) is the period of f(t).
A signal for which there is no value of T0 for which equation (1) is satisfied are called
aperiodic (non-periodic) signal.

Note: A periodic signal, by definition, is an everlasting signal.

A signal that does not start before t = 0 is a causal signal.

i.e. f(t) = 0 , t<0

A signal that starts before t = 0 is a noncausel signal. A signal that is zero for all t t 0
is called an anticausal signal.
Note: Observe that an everlasting signal is always noncausal but a noncausal
signal is not necessarily everlasting.

Example
Determine if the signal is periodic, if yes then find its fundamental frequency and
period
1 2 7
1. f1(t) = 2 + 7 cos ⎛⎜ t  T ⎞⎟  3 cos ⎛⎜ t  T2 ⎞⎟  5 cos ⎜⎛ t  T3 ⎞⎟
⎝2 ⎠ ⎝3 ⎠ ⎝6 ⎠

2. f2 (t) 2 cos(2t  T1)  5 sin( πt  T2 )

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.2
 LTI System

Solution :

⎛1 ⎞ ⎛2 ⎞ ⎛7 ⎞
1. f1(t) = 2 + 7 cos ⎜ t  T ⎟  3 cos ⎜ t  T2 ⎟  5 cos ⎜ t  T3 ⎟
⎝2 ⎠ ⎝3 ⎠ ⎝6 ⎠
In order for the signal to be periodic the ratios of their frequencies must be a
rational number here
1 2 7
Z1 , Z2 and Z3
2 3 6
Zfundamental GCD (Z1, Z2 , Z3 ) …(GCD { greatest common divisor)
1 Z 1
(rational) ⇒ ffundamental
6 2π 12 π
1 ⎛ 1 ⎞
Thus, given signal is periodic with fundamental frequency rad/sec ⎜ Hz ⎟
6 ⎝ 12π ⎠
and period (12S seconds)

2. f2 (t) 2 cos(2t  T1)  5 sin( πt  T2 )

The signal is not periodic

ω1 2
∵ ratio of its frequency i.e. is irrational
ω2 π
4. Energy and power signals :
(i) Energy Signals : The signal energy is given by
f
2
E= ∫f (t). dt …(2)
f
in case of complex signal
f

E= ∫ f(t) dt …(3)
f

Note: For signal to be energy signal E must be finite i.e. E < f. Also as t o f, f(t) o 0.

Exemples

1. f(t) = eat u(t)

f f f
⎡ e2at ⎤ 1
E ∫ e 2at .u(t)dt 2at
∫ e dt = ⎢ ⎥ = o Energy signal
f 0 ⎣⎢ 2a ⎦⎥ 0 2a
2. f(t) = G(t)
f
2
E ∫G (t) dt = 1 o Energy signal
f
W W
3. f(t) = A,  dtd
2 2
W /2
W /2
E ∫ A 2 dt = ⎡ A 2 (t)⎤ = A 2 W o Energy signal
⎣ ⎦ W /2
W /2

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Vidyalankar : GATE – EC

(ii) Power Signal :

The signal power is given by

1 T /2 2
T of T ∫ T /2
P lim f (t)dt

We can generalize this definition for a complex signal f(t) as

1 T /2 2
P lim
T of T ∫ T /2
f(t) dt

Observe that the signal power P is the time average (mean) of the signal
amplitude squared, that is, the mean-squared value of f(t). Indeed, the square
root of P is the familiar rms (root mean square) value of f(t).

• Energy signal has zero average power, while power signal has infinite
energy.

• A signal cannot both be energy and a power signal. If it is one, it cannot

be the other. On the other hand, there are signals that are neither energy
nor power signals. The ramp signal is such an example.

• Signals which are either periodic or random are power signals, where as
signals which are both aperiodic and deterministic are energy signals.

5. Deterministic and probabilistic signals :

A signal whose physical description is known completely, either in a mathematical

form or a graphical form, is a deterministic signal. A signal whose values cannot be
predicted precisely but are known only in terms of probabilistic description, such as
mean value, mean squared value, and so on is a random signal.

USEFUL SIGNAL OPERATIONS

(A) Time Shifting

Consider a signal f(t) shown in figure (a) below. To time shift the signal by T seconds,
we replace t with t T which is represented by I(t) = f(t  T). If T is positive, the shift is
to the right (delay) (fig. (b)). If T is negative, the shift is to the left (advance) (fig. (c)).

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.4
 LTI System

(a) Signal f(t) f(t)

0 t

(b) Time delay I(t) = f (t  T)

T 0 t

(c) Time advance f (t + T)

T 0 t

Time shifting a signal

(B) Time Scaling
The compression or expansion of a signal in time is known as time scaling. To time
scale a signal by a factor a, we replace t with at. If a > 1, the scaling results in
compression, and if a < 1, the scaling results in expansion. For example if f(t) were
recorded on a tape and played back at twice the normal recording speed, we would
obtain f (2t) i.e. f(t) compressed by factor of 2.

(a) Signal f(t) f(t)

T1 0 T2 t
I(t) = f(2t)
(b) Time compression

T1 0 T2 t
2 2
⎛t⎞
I(t) = f ⎜ ⎟
(c) Time expansion ⎝ 2⎠

2T1 0 2T2 t
Time Scaling of a Signal
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Vidyalankar : GATE – EC

Note : Observe that in time scaling operation, the origin t = 0 is the anchor point, which
remains unchanged under scaling operation because at t = 0, f(t) = f (at) = f (0).

(C) Time Inversion (Time Reversal)

To time-invert a signal we replace t with t, which can be obtained by taking mirror
image of f(t) about vertical axis.
f(t)
2

2
0 5 t
1

(a) Signal f(t)

I(t) = f(t)
2

2
5 0 t
1

(b) Time inversion

Time inversion (reflection) of a signal

Note : The mirror image of f(t) about the vertical axis is f(t). Recall also that the
mirror image of f(t) about the horizontal axis is f(t).

Combined Operations

Certain complex operations require simultaneous use of more than one of the above
operation. The most general operation involving all the three operations is f (at  b),
which is realized in two possible sequences of operation :

1. Time-shift f(t) by b to obtain f(t  b). Now time-scale the shifted signal f(t  b) by a
(i.e. replace t with at) to obtain f (at  b).
b
2. Time scale f(t) by a to obtain f(at). Now time-shift f(at) by , (i.e. replace t with
a
⎛ b⎞ ⎡ ⎛ b ⎞⎤
⎜ t  ⎟ to obtain f ⎢ a ⎜ t  ⎟ ⎥ f(at  b) ).
⎝ a⎠ ⎣ ⎝ a ⎠⎦

In either case, if a is negative, time scaling involves time inversion.

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.6
 LTI System

Example : f(t)

For given signal f(t) find f(2t  4)

4
2

t
4 0 2
Solution :

Method 1 :
f (t  4)
Step 1 : Time shifting f (t  4) ⇒ f(t)
is delayed by 4 seconds. (Shift left) 4

t
2 4 6
Step 2 : Time scaling f (2t  4) f (2t  4)
⇒ scaling f(t  4) by
factor 2 (compression)
4

t
2 3 4 6
f (2t)
Method 2 :
4
Step 1 : Time scaling f (2t)
⇒ compresession by factor 2 2

t
2 0 1

f (2t  4)
Step 2 : Time shifting f [2 (t2)] = f(2t  4)
⇒ delaying signal f(2t) by 2 seconds 4

0 t
2 3
we get the same result using any of the method

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Vidyalankar : GATE – EC

TYPE OF SIGNALS
(1) Unit Impulse Function :
Mathematically : An impulse is a unit pulse of extremely large magnitude and
infinitesimal duration. As t o 0, magnitude o f, such that area is unity. It is defined
as
G(t) 0 , t z 0
f

∫ G(t) dt = 1
f

Representation :

G(t)

t
0
Important Properties of unit function :
1
• G(at) = G(t)
|a|

• G(t) = G(t)

• f(t) G(t) = x(0) G(t)

f(t) G(t  t0) = x(t0) G(t  t0) …multiplication property

f
• ∫ x(t) G(t) dt = x(0)
f
f

∫ x(t) G(t  a) = x(a) … Sampling property

f

Note: Integral exists, provided f(t) is continuous at the place where impulse occurs.

Example :
Solve
⎡ ⎛ 2 π ⎞⎤
i) ⎢sin ⎜⎝ t  2 ⎟⎠ ⎥ G(t)
⎣ ⎦
f
⎛ πt ⎞
ii) ∫ G(t  2)cos ⎜⎝ 4 ⎟⎠ dt
f

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.8
 LTI System

Solution:

⎛ π⎞ ⎛ π⎞
i) sin ⎜ t 2  ⎟ .G(t) = sin ⎜  ⎟ .G(t) …[multiplication property of G(t)]
⎝ 2⎠ ⎝ 2⎠
=  G(t)
f
⎛ πt ⎞
ii) ∫ G(t  2).cos ⎜⎝ 4 ⎟⎠ dt
f

now G(t  2) is impulse delayed by 2 seconds

? by sampling propertry of G(t)
f
⎛ πt ⎞ ⎛ 2π ⎞
∫ G(t  2).cos ⎜⎝ 4 ⎟⎠ dt = cos ⎜
⎝ 4 ⎠
⎟ …[putting t = 2] = 0
f

(2) Step input :

Mathematically : r = 0, t<0
r = A, tt0
Representation:

0 t

When A = 1, it is called unit step function, denoted by u(t).

The unit step function, u(t) also proves very useful in specifying a function which has
different mathematical descriptions over different intervals of times, since such a
description often proves clumsy and inconvenient in mathematical treatment. With
the help of unit step function, we can describe such functions by a single expression
that is valid for all t.

Example :
Obtain a single expression for f(t), which is true for all time ‘t’

f(t)

t
0 1 2 4

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Vidyalankar : GATE – EC

Solution :
f(t) can be obtain as a sum of two signals f1(t) and f2(t) as shown below
f1 (t)
(t  1)
1 ⇒ 1

t t
0 1 2 0 1 2

1 f1(t) = (t1) [u(t1)u(t2)]

f2 (t)

t
0 2 4

f2(t) = u(t2)u(t4)
? f(t) = f1(t) + f2(t)
= (t  1) [u (t  1)  u (t  2)] + u (t  2)  u (t  4)
= (t  1) u (t  1)  (t  1) u (t  2) + u (t  2)  u (t  4)
= (t  1) u (t  1)  (t  2 + 1) u (t  2) + u (t  2)  u (t  4)
= (t  1) u (t  1)  (t  2) u (t  2)  u (t  4) …(for all t)

t
• u(t) = ∫ G t) dt
f
d
x Derivative of a unit step is unit impulse u(t) = G(t)
dt
x Derivative of a unit impulse is doublet.
x Derivative of a unit doublet is triplet.

(3) Unit Gate Pulse :

We define a unit gate function rect (t) as a gate pulse of unit height and unit width,
centered at the origin, as illustrated in following figure (a).
Mathematically :
⎧ 1
⎪0 | t | ! 2
⎪
⎪1 1
rect(t) = ⎨ |t|
⎪2 2
⎪ 1
⎪1 | t | 
⎩ 2

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.10
 LTI System

The gate pulse in following figure (b), is the unit gate pulse rect (t) expanded by a
⎛t⎞
factor W and therefore can be expressed as rect ⎜ ⎟ . Observe that W, the
⎝W⎠
⎛t⎞
denominator of the argument of rect ⎜ ⎟ , indicates the width of the pulse.
⎝W⎠
Representation :

⎛t⎞
rect (t) rect ⎜ ⎟
1 1 ⎝ W⎠

1 0 1 t W 0 W t
2 2 2 2
(a) A gate pulse (b)

(4) Ramp Function :

Mathematically : r (t) = 0, t<0
r(t) = Ktu(t), tt0
Representation :

r(t) r(t)

K = slope K = slope

0 t 0 t
t = t1
r(t) = Kt u(t) delayed ramp
r(t) = K(t  t1) u(t  t1)

dr(t)
Derivative of unit ramp( Slope, K = 1) is the unit step signal. i.e. u(t)
dt

(5) Unit Triangle Function

We define a unit triangle function '(t) as a triangular pulse of unit height and unit
width, centered at the origin, as shown in following figure.
Mathematically :

⎧ 1
⎪⎪0 t t
'(t) = 2
⎨
⎪1  2 t 1
t 
⎪⎩ 2

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Vidyalankar : GATE – EC

⎛t⎞
The pulse in following figure (b) is ' ⎜ ⎟ . Here also denominator W of the argument of
⎝W⎠
⎛t⎞
' ⎜ ⎟ indicates the pulse width.
⎝W⎠

Representation :

⎛t⎞
'(t) '⎜ ⎟
⎝ W⎠
1 1

1 0 1 t W 0 W t
2 2 2 2
(a) (b)
A triangle pulse
(6) Sinusoidal input Function :
Mathematically : f(t) = C sin (2SF0t + T)
… where C = amplitude
F0 = Frequency in Hertz
T = Phase in radians
Representation :

f(t)

T0
C
t
0
C

Cleary, this sinusoid repeats every 1/F0 seconds. As a result, there are F0 repetitions
per second. This is the frequency of the sinusoid, and the repetition interval T0 given
by
1 2π
T0 … (Radian frequency, Z0 = 2SF0)
F0 Z

(7) Interpolation Function sinc (t) :

The function sin t/t is the “sine over argument” function denoted by sinc (t). This
function plays an important role in signal processing. It is also known as the filtering
or interpolating function.

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 LTI System

Mathematically :
sin t sin πt
sinc (t) or sinc (t) …(4)
t πt
Inspection of equation (4) shows that
1. sinc (t) is an even function of t.
2. sinc (t) = 0 when sin t = 0 except at t = 0, where it appears indeterminate. This
means that sinc t= 0 for t = rS, r2S, r3S,…
3. Using L’Hopital’s rule, we find sinc (0) = 1.
4. sinc (t) is the product of an oscillating signal sin t (of period 2S) and a
monotonically decreasing function 1/t. Therefore, sinc (t) exhibits sinusoidal
oscillations of period 2S, with amplitude decreasing continuously as 1/t.

Following figure shows sinc(t). Observe that sinc (t) = 0 for values of t that are
positive and negative integral multiples of S.

Representation :

1
1 t
sinc (t)

2S 2S
3S S 0 S 3S t

1
t

A sinc pulse
(8) Random Function :

The amplitude of this function is random and similar to white noise as it randomly
contains all frequencies.
Representation :

f(t)

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Vidyalankar : GATE – EC

(9) The Exponential Function est

One of the most important functions in the area of signals and systems is the
exponential signal est, where s is complex frequency in general, given by
s = V + jZ
Function est encompasses a large class of functions. The following functions are
special cases of est:
1. A constant k = ke0t (s = 0) …(fig. (a))
2. A monotonic exponential eVt (Z = 0, s = V) …(fig. (a))
3. A sinusoid cos Zt (V = 0, s = rjZ) …(fig. (b))

4. An exponentially varying sinusoid eVt cos Zt (s = V r jZ) …(fig. (c and d))

There functions are illustrated in following figures.
Vt
e
V<0 V>0
V=0
V=Z=0
t
t

(a) (b)

V<0 V>0

t t

(c) (d)
Sinusoids of complex frequency V + jZ

The complex frequency s can be conveniently represented on a complex frequency

plane (s plane) as depicted in following figure. The absolute value of the imaginary
part of s is |Z| (the radian frequency), which indicates the frequency of oscillation of
est; the real part V (the neper frequency) gives information about the rate of increase
or decrease of the amplitude of est.

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.14
 LTI System

Left half plane Right half plane

Imaginary axis jZ

Exponentially increasing signals

Exponentially decreasing signals
Real
axis

Complex frequency plane(s-plane)

Note: Thus the s-plane can be differentiated into two parts: the left half-plane (LHP)
corresponding to exponentially decaying signals and the right half-plane
(RHP) corresponding to exponentially growing signals. The imaginary axis
separates the two regions and corresponds to signals of constant amplitude.

Even and Odd Signals

A real valued signal f(t) is called symmetric (even) if
x(t) = f(t)
and is called antisymmetric (odd) if
x(t) = f(t)
For an odd function x(0) = 0

Even and Odd Components of signals

Every signal f(t) can be expressed as a sum of even and odd components
i.e. f(t) = fe(t) + f o(t)

where f e(t) is the even part of f(t)

fo (t) is the odd part of f(t)
The even part xe (t) = ½ [f(t) + f(t)]
The odd part xo (t) = ½ [f(t)  f(t)]

For odd function, f(0) = 0 always

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Vidyalankar : GATE – EC

Example:
1. Find the even/odd part of the signal.

f(t)
1
t
Solution : 0 1

f(t)
1
Step (i) t
1 0
1
1/2 fe (t) [f (t)  f (  t)]
2
Step (ii) t Even component
1 0 1
1
1/2 fo (t) [f (t)  f (  t)]
2
1 Odd component
Step (iii) t
0 1
1/2

Some useful properties of even and odd function

(a) f e(t) and fo (t) are orthogonal functions.
f
i.e. ∫ fe (t)fo (t)dt 0.
f
(b) Energy of signal f(t) is equal to
f f f
E = ∫ | f(t) |2 dt = ∫ fe2 (t)dt  2
∫ fo (t)dt
f f f
Proof :
f f
E = ∫ f 2 (t). dt  ∫ [fe (t)  fo (t)]2 .dt
f f
f
= 2 2
∫ [fe (t) fe (t). fo (t) fo (t)].dt
f
f f f f f
= = 2 2
∫ fe (t) ∫ fo (t). dt
2 2
∫ fe (t)dt   ∫ fe (t). fo (t). dt  ∫ fo (t). dt
f f f f f
f
∵ ∫ fe (t).fo (t).dt 0 as fe(t) and fo(t) are orthogonal Thus
f
f f
E = ∫ fe2 (t)dt  2
∫ fo (t)dt
f f

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 LTI System

(c) Area

Because fe(t) is symmetrical about the vertical axis, it follows that

a a

∫ fe (t)dt 2 ∫ fe (t)dt
a 0
a
It is also clear that ∫ f0 (t)dt 0
a

(d) even function u odd function = odd function

odd function u odd function = even function
even function u even function = even function

CLASSIFICATION OF SYSTEMS
Systems may be classified broadly in the following categories :
1. Linear and Non-linear Systems

A system whose output is proportional to its input is an example of a linear system.

But linearity implies more than this; it also implies additivity property. i.e. if a cause c1
acting alone has an effect e1, and if another cause c2, also acting alone, has an effect
e2, then, with both causes acting on the system, the total effect will be e1 + e2. Thus,
if
c1 o e1 and c2 o e2
then for all c1 and c2
c1 + c2 o e1 + e2
In addition, a linear system must satisfy the homogeneity or scaling property, i.e. if a
cause is increased k-fold, the effect also increases k-fold. Thus, if
coe
then for all real or imaginary k
kc o ke
Both these properties can be combined into one property (superposition), which is
expressed as follows : if
c1 o e1 and c2 o e2

then for all values of constants k1 and k2,

k1c1 + k2c2 o k1e1 + k2e2

This is true for all c1 and c2.

Note : Almost all systems observed in practice becomes non-linear when large
enough signals are applied to them. However, many systems show linear
behavior for small signals or can be linearzed for small signal condition using
Taylor series expansion about the operating point.

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2. Time-Invariant and Time-Varying Parameter Systems :

Systems whose parameters do not change with time are time-invariant (also
constant-parameter) systems. For such a system, if the input is delayed by T
seconds, the output is the same as before but delayed by T.
A system with an input-output relationship described by a linear differential equation
with constant coefficients is a linear time-invariant (LTI) system. If these coefficients
are functions of time, then the system is a linear time-varying system. eg. Rocket.

3. Instantaneous and Dynamic Systems:

If output of the network at some instant t depends only on the input at that instant t,
then such systems are said to be instantaneous or memory less systems. i.e. A
system is said to be instantaneous (or memory less) if its output at any instant t
depends, at most, on the strength of its inputs at the same instant but not on any past
or future values of the inputs. Otherwise, the system is said to be dynamic (or a
system with memory). Networks containing inductive and capacitive elements
generally have infinite memory.
4. Causal and Noncausal Systems :
A causal (also known as a physical or non-anticipative) system is one for which the
output at any instant t0 depends only on the value of the inputs f(t) for t d t0. In order
words, the value of the output at the present instant depends only on the past and
present values of the inputs f(t), not on its future values. If the response starts before
the input, it means that the system knows the input in the future and acts on this
knowledge before the input is applied. Such systems are called non-causal
(or anticipative) system. Any practical system that operates in real time must
necessarily be causal.
Note : It is impossible to implement non-causal system in real time.

5. Lumped-parameter and Distributed-Parameter Systems :

If the physical dimensions of a component are small compared to the wavelength of
the signal propagated, we may assume that the current is constant throughout the
component. This is the assumption made in lumped-parameter systems, where each
component is regarded as being lumped at one point in space. Such an assumption
is justified at lower frequencies (higher wavelength). For such systems, the system
equations require only one independent variable (time) and therefore are described
by ordinary differential equations.
In contrast, for distributed-parameter systems such as transmission lines,
waveguides, antennas, and microwave tubes, the system dimensions cannot be
assumed to be small compared to the wavelengths of the signals; thus the lumped-
parameter assumption breaks down. The signals here are functions of space as well
as of time, leading to mathematical models consisting of partial differential equations.
6. Continuous-Time and Discrete-Time Systems :
Systems whose inputs and outputs are continuous-time signals are continuous-time
systems. On the other hand, systems whose inputs and outputs are discrete-time
signals are discrete-time systems. If a continuous-time signal is sampled, the
resulting signal is a discrete-time signal. We can process a continuous-time signal by
processing its samples with a discrete-time system.

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 LTI System

7. Analog and Digital Systems :

A system whose input and output signals are analog is an analog system; a system
whose input and output signals are digital is a digital system. A digital computer is an
example of a digital (binary) system. Observe that a digital computer is an example of
a system that is digital as well as discrete-time.
Examples:
1. Determine whether following systems are linear or non-linear
dy
(i) 3y(t) + 2 = f(t) (ii)  3ty(t) t 2 f(t)
dt
Solution :
(i) 3y(t) + 2 = f(t)
if inputs f1(t) and f2(t) are applied separately , we have
3y1(t) + 2 = f1(t) …(5)
3y2(t) + 2 = f2(t) …(6)

multiplying (5) by a and (6) by b and adding we get,

3[ay1(t) + by2(t)] + 2(a+b) = [af1(t) + bf2(t)] …(7)

if both input af1(t) and bf2(t) are applied simultaneously, we must get,
3[ay1(t) + by2(t)] + 2 = [af1(t) + bf2(t)] …(8)
we see that equation (7) z equation (8)
? given system is non-linear,
Note: The above system can be linearized if constant ‘2’ is abstent.
dy
(ii)  3ty(t) t 2 f(t)
dt
if inputs f1(t) and f2(t) are applied separately , we have
dy1
 3ty1(t) t 2 f1(t) …(9)
dt
dy 2
 3ty 2 (t) t 2 f2 (t) …(10)
dt
multiplying (9) by a and (10) by b and adding we get,
d
>ay1(t)  by2 (t)@  3t >ay1(t)  by 2 (t)@ t 2 > af1(t)  bf2 (t)@ …(11)
dt
if both input af1(t) and bf2(t) are applied simultaneously, we must get,
d
>ay1(t)  by2 (t)@  3t >ay1(t)  by 2 (t)@ t 2 > af1(t)  bf2 (t)@ …(12)
dt
we see that equation (11) = equation (12) ? given system is linear.

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2. Determine whether following systems are time invariant or time varying

(i) y(t) = f(t  2) (ii) y(t) = tf(t  2)

Solution:

(i) y(t) = f(t2)

let the input be delayed by T second, then
y1(t) = f(t  T  2) …(13)
now let the output y(t), be delayed by T seconds, then
y2(t) = y(t  T) = f(t  T  2) …(14)
? (13) = (14)
i.e. y1(t) = y2(t)
system is time invariance

(ii) y(t) = tf(t  2)

let the input be delayed by T seconds, then
y1(t) = t f(t  T 2) …(15)
now, let y(t) be delayed by T seconds, then
y2(t) = y(t  T)
= (t  T) f(t  T  2) …(16)
here y1(t) z y2(t)
? system is time varying

TIMEDOMAIN ANALYSIS OF LTIC SYSTEMS

Introduction
Consider linear differential systems with constant co-efficient for which the input f(t) and
the output y(t) are related by linear differential equations of the form

dn y dn1y dy dm f dm1f df
 an1  ...  a1  a 0 y(t) bm  bm1  ...  b1  b0 f(t) …(17)
dtn dtn1 dt dtm dtm1 dt
Using operational notation D to represent d/dt, we can express this equation as
(Dn  an1Dn1  ...  a1D  a 0 )y(t) (bmDm  bm1Dm1  ...  b1D  b0 )f(t)

or Q(D)y(t) P(D)f(t)

where the polynomials Q(D) and P(D) are

Q(D) Dn  an1Dn1  ...  a1D  a 0

P(D) bmDm  bm1Dm1  ...  b1D  b0

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 LTI System

Note : Practical noise considerations, requires m d n.

A system described by Equation (17) is linear and its response can be expressed as the
sum of two components: the zero-input component and the zero-state component.
Therefore,
Total response = zero-input response + zero-state response
The zero-input component is the system response when the input f(t) = 0 so that it is the
result of internal system conditions (such as energy storages, initial conditions) alone. It
is independent of the external input f(t). In contrast, the zero-state component is the
system response to the external input f(t) when the system is in zero state, that is, all
initial conditions are zero.

System Response to Internal Conditions: Zero-Input Response

The zero-input response y0(t) is the solution of equation (17) when the input f(t) = 0 so
that
Q(D)y0(t) = 0

or (Dn + an1 Dn1 +…+ a1D + a0) y0(t) = 0 …(18)

Equation (18) shows that a linear combination of y0(t) and its n successive derivatives is
zero, not at some values of t, but for all t. Such a result is possible if and only if y0(t) and
all its n successive derivatives are of the same form. Only an exponential function e λt
has this property. So let us assume that
y 0 (t) ce λt

Substituting these results in equation (18), we obtain

c(On  an1On1  ...  a1O  a0 )eOt 0

For a nontrivial solution of above equation,

On  an1On1  ...  a1O  a0 0 …(19)

This result means that ceOt is indeed a solution of equation (18), provided that O satisfies
equation (19). Note that the polynomial in equation (19) is identical to the polynomial
Q(D) in equation (18), with O replacing D. Therefore, equation (19) can be expressed as

Q(O) = 0 …(20)

When Q(O) is expressed in factorized form, equation (20) can be represented as

Q(O) (O  O1)(O  O 2 ) ˜ ˜ ˜ (O  On ) 0

We can readily show that a general solution is given by the sum of these n solutions, so
that

y 0 (t) c1eO1t  c 2 eO2 t  ˜ ˜ ˜  cneOnt

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where c1, c2, … , cn are arbitrary constants determined by n constraints (the initial
conditions) on the solution.

The equation
Q(O) = 0
is called the characteristic equation of the system. O1, O2, …, On are the roots of the
characteristic equation; consequently, they are called the characteristic roots,
characteristic values, eigenvalues, and natural frequencies. The exponentials
eOit (i 1,2,....,n) in the zero-input response are the characteristic modes (also known as
modes or natural modes) of the system. The single most important attribute of an LTIC
system is its characteristic modes. Characteristic modes not only determine the zero-
input response but also play an important role in determining the zero-state response.

The zero input response, y0(t) depends upon the nature of the characteristic roots of
equation Q(O) = 0.

Nature of roots of Q(O) = 0 Zero-input response, y0(t)

1. All distinction roots y 0 (t) c1eO1t  c 2 eO2 t  ˜ ˜ ˜cneOnt

2. Some roots repeated ( say r-times ) y 0 (t) (c1  c 2 t  c 3 t 2  ˜ ˜ ˜  cr tr 1)eOt

3. Complex conjugate roots of the form D r jE y 0 (t) ceDt cos(Et  T)

Example

1. Find y0(t), the zero-input component of the response for an LTI system described by
the following differential equation :
(D2  3D  2) y(t) Df(t)
when the initial condition are y 0 (0) 0,y 0 (0) 5.

Solution:
Note that y0(t), being the zero-input component (f(t) = 0), is the solution of (D2 + 3D +2)
y0(t) = 0. The characteristic polynomial of the system is O2 + 3O + 2 = 0. The characteristic
roots of the system are O1 = 1 and O2 = 2. Consequently, the zero-input response is
y 0 (t) c1e t  c 2 e2t …(21)

To determine the arbitrary constants c1 and c2, we differentiate equation (21) to obtain
y 0 (t) c1e t  2c 2 e2t …(22)

Setting t = 0 in equations (21) and (22), and substituting the initial conditions y0(0) = 0
and y 0 (0) 5 we obtain
0 = c1 + c2
5 = c1  2c2

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 LTI System

Solving these two simultaneous equations in two unknowns for c1 and c2 yields
c1 = 5, c2 = 5
Therefore y0(t) = 5et + 5e2t, for t t 0.

2. Similar procedure may be followed for repeated roots. For instance, for a system
specified by
(D2  6D  9)y(t) (3D  5)f(t)

let us determine y0(t), the zero-input component of the response if the initial onditions
are
y0(0) = 3 and y 0 (0) 7

Solution:
The characteristic polynomial is O2 + 6O + 9 = (O + 3)2 = 0, and its characteristic roots are
O1 = 3, O2 = 3 (repeated roots). The zero-input response is given by
y0(t) = (c1 + c2t)e3t
We can find the arbitrary constants c1 and c2 from the initial conditions y0(0) = 3 and
y 0 (0) = 7 following the same procedure in example (1). Which gives c1 = 3 and c2 = 2.

Hence,

y 0 (t) (3  2t)e3t , tt0

3. For the case of complex roots, let us find the zero-input response of an LTI system
described by the equation:
(D2  4D  40) y(t) (D  2)f(t)

with initial conditions y 0 (0) 2 and y 0 (0) 16.78

Solution:
The characteristic polynomial is O2 + 4O + 40 = (O + 2  j6) (O + 2 + j6) = 0. The
characteristic roots are 2 r j6. Since D = 2 and E = 6, the real form solution is [see point
3 in the above table]

y 0 (t) ce 2t cos(6t  T) …(23)

where c and T are arbitrary constants to be determined from the initial conditions y0(0) = 2
and y 0 (0) 16.78. Differentiation of equation (23) yields
y 0 (t) 2ce2t cos(6t  T)  6ce2t sin(6t  T) …(24)

Setting t = 0 in equations (23) and (24), and then substituting initial conditions, we obtain
2 = c cos T
16.78 = 2c cos T  6c sin T

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Solution of these two simultaneous equation, we get

c cos T = 2 …(25)
c sin T = 3.463 …(26)
Squaring and then adding the two sides of the above equations yields
c2 = (2)2 + (3.464)2 = 16 ⇒ c = 4
Next, dividing equation (26) by (25), we get
3.463
and tan T
2
⎛ 3.463 ⎞ π
T tan1 ⎜ ⎟ 
⎝ 2 ⎠ 3
⎛ π ⎞
Therefore y 0 (t) 4e2t cos ⎜ 6t  ⎟
⎝ 3⎠

Practical Initial Conditions and the meaning of 0 and 0+

The input is assumed to start at t = 0, unless otherwise mentioned. Hence, t = 0 is the
reference point. The conditions immediately before t = 0 (just before the input is applied)
are the conditions at t = 0, and those immediately after t = 0 (just after the input is
applied) are the conditions at t = 0+ (compare this with the historical time frame B.C. and
A.D.). In practice, we are likely to know the initial conditions at t = 0 rather than at t = 0+.
The two sets of conditions are generally different, although in some cases they may be
identical.

We are dealing with the total response y(t), which consists of two components; the zero-
input component y0(t) (response due to the initial conditions alone with f(t) = 0) and the
zero-state component resulting from the input alone with all initial conditions zero. At
t = 0, the response y(t) consists solely of the zero-input component y0(t) because the
input has not started yet. Hence the initial conditions on y(t) are identical to those of y0(t).
Thus, y(0) = y0(0), y (0) = y 0 (0 ), and so on. Moreover, y0(t) is the response due to
initial conditions alone and does not depend on the input f(t). Hence, application of the
input at t = 0 does not affect y0(t). This means the initial conditions on y0(t) at t = 0 and 0+
are identical; that is y0(0), y 0 (0 ), … are identical to y 0 (0 ),y 0 (0 ),..., respectively. It is
clear that for y0(t), there is no distinction between the initial conditions at t = 0, 0 and 0+.
They are all the same. But this is not the case with the total response y(t), which consists
of both, the zero-input and the zero-state components. Thus, in general,
y(0) z y(0+), y(0
  ) z y(0
  ), and so on.

The Unit Impulse Response h(t)

h(t) is the system response to an impulse input G(t) applied at t = 0 with all the initial
conditions zero at t = 0. Impulse input G(t) appears momentarily at t = 0, and then it is
gone forever. But in that moment it generates energy storages; that is, it creates nonzero
initial conditions instantaneously within the system at t = 0+. Although the impulse input
G(t) vanishes for t > 0 so that the system has no input after the impulse has been applied,
the system will still have a response generated by these newly created initial conditions.

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 LTI System

Thus impulse response h(t), therefore, must consist of the system’s characteristic modes
for t t 0+.
h(t) = characteristic mode terms t t 0+
This response is valid for t > 0. But what happens at t = 0? At a single moment t = 0,
there can at most be an impulse, so the form of the complete response h(t) is given by
h(t) = A0G(t) + characteristic mode terms tt0
which takes the form
h(t) = bnG(t) + [P(D)yn(t)u(t)]
where bn is the coefficient of the nth-order term in P(D), and yn(t) is a linear combination
of the characteristic modes of the system subject to the following initial conditions:
1)
y(n
n yn (0) ˜ ˜ ˜ yn(n2) (0) 0
(0) 1, and yn (0) y n (0) 

Note : Once we know the system response to an impulse input, we can determine the
system response to an arbitrary input f(t).

Example:
1) Find unit impulse response of LTIC system described by equation
(D + 2) y(t) = (3D + 5) f(t)

Solution:
Here Q(D) = D + 2 and P(D) = 3D + 5 and also bn = b1 = 3 the characteristic equation is
Q (O) = 0 ⇒ O + 2 = 0
⇒ O = 2
? yn(t) = Ce2t subject to yn(0) = 1
⇒ C=1
? yn(t) = e2t
Impulse response

? h(t) = bn G(t) + P(D) [yn(t)] u(t)

= 3 G(t) + [(3D + 5) e2t] u(t)
= 3 G(t) + (5e2t  6e2t) u(t)
= 3G(t)  e2t u(t)
thus, h(t) = 3G(t)  e2t u(t)

Signal Transmission through LTI Systems

In the time domain, a linear system is described in terms of its impulse response, which is
defined as the response of the system (with zero initial conditions) to a unit impulse or
delta function G(t) applied to the input of the system. If the system is time-invariant, then
the shape of the impulse response is the same no matter when the unit impulse is
applied to the system. Let us denote the impulse response of a linear time-invariant
system by h(t). Let this system be subjected to an arbitrary excitation f(t), as in following
figure (a).

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f(t)

Impulse
f(t) response y(t) f(W)
h(t) approximation
(a) Linear System

W t
'W
(b) Approximation of input f(t)

To determine the response y(t) of the system, we begin by first approximating f(t) by a
staircase function composed of narrow rectangular pulses, each of duration 'W, as shown
in above figure (b).

Clearly in the limit this approachest, a delta function weighted by a factor equal to the
height of the pulse times 'W. Consider a typical pulse, shown shaded in above figure (b),
which occurs at t = W. This pulse has an area equal to f(W)'W.

By definition, the response of the system to a unit impulse or delta function G(t), occurring
at t = 0, is h(t). It follows, therefore, that the response of the system to a delta function,
weighted by the factor f(W)'W and occurring at t = W, must be f(W)h(t  W)'W. To find the total
response y(t) at some time t, we apply the principle of superposition. Thus, summing the
various infinitesimal responses due to the various input pulses, we obtain in the limit, as
'W approaches zero.
f
y(t) = ∫f f(W)h(t  W)dW …(27)

This relation is called the convolution integral.

In equation (27), three different time scales are involved: excitation time W, response time
t, and system-memory time t  W. This relation is the basis of time-domain analysis of
linear time-invariant systems. It states that the present value of the response of a linear
time-invariant system is a weighted integral over the past history of the input signal,
weighted according to the impulse response of the system. Thus the impulse response
acts as a memory function for the system.
System Response to External Input : Zero State Response

This is the system response y(t) to an input f(t) when the system is in zero state; that is,
when all initial conditions are zero. Under these conditions, the zero-state response will
be the total response of the system.

The (zero-state) response y(t) to the input f(t) is given by

f
y(t) = ∫f f(W)h(t  W)dW …(28)

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 LTI System

Above equation shows that knowing h(t), we can determine the response y(t) to any
input. Observe the all-pervasive nature of the system’s characteristic modes. The system
response to any input is determined by the impulse response, which in turn is made up of
characteristic modes of the system. The above equation (28) is only valid for LTIC
systems.

Convolution and Its Graphic Interpretation

(i) Convolution Integral
The zero-state response y(t) is given as
f
y(t) = ∫f f(W)h(t  W)dW …(29)

Equation (29) occurs frequency in the physical sciences, engineering, and

mathematics. For this reason this integral is given a special name: the convolution
integral. The convolution integral of two function f1(t) and f2(t) is denoted symbolically
by f1(t) * f2(t) and is defined as
f
f1(t) * f2 (t) { ∫f f1(W)f2 (t  W)dW
Properties :
1) Commutative : x1(t) * x2(t) = x2(t) * x1(t)
2) Associative : [x1(t)* x 2 (t)]* x 3 (t) x1(t) * [x 2 (t) * x 3 (t)]

3) Distributive : x1(t)*[x 2 (t)  x 3 (t)] [x1(t)* x 2 (t)  x1(t)* x 3 (t)]

4) The Shift Property :

If f1(t) * f2(t) = c(t)
then f1(t) * f2(t  T) = c(t  T)
f1 (t  T) * f2(t) = c(t  T)
and
f1(t  T1) * f2(t  T2) = c(t  T1  T2)

5) Convolution with an Impulse:

Convolution of a function f(t) with a unit impulse results in the function f(t) itself.
Therefore
f(t) * G(t) = f(t)

6) The Width Property :

If the durations (widths) of f1(t) and f2(t) are T1 and T2 respectively, then the
duration (width) of f1(t) * f2(t) is (T1 + T2)

Note: In deriving equation (29), we assumed the system to be LTI. There were no
other restrictions either on the system or on the input signal f(t). In practice, most
systems are causal, so that their response cannot begin before the input starts.
Furthermore, most inputs are also causal, which means they start at t = 0.

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Causality restrictions on both signals and systems further simplify the limits of
integration in equation (29). It is important to remember that the integration in
equation (29) is performed with respect to W (not t). If the input f(t) is causal, f(W) = 0
for W < 0. Therefore, f(W) = 0 for W < 0. Similarly, if h(t) is causal, h(t  W) = 0 for
t  W < 0; that is, for W > t. Therefore, the product f(W)h(t W) = 0 everywhere except
over the interval 0 d W d t. Therefore equation (29) reduces to
t
y(t) f(t) * h(t) ∫0 f(W)h(t  W)dW tt0 …(30)
=0 t<0
The lower limit of integration in equation (30) is taken as 0 to avoid the difficulty in
integration that can arise if f(t) contains an impulse at the origin.

Example

For an LTIC system with the unit impulse response h(t) = e2t u(t), determine the
response y(t) for the input
f(t) = etu(t)

f(t) h(t)
1 1
et e2t

0 t 0 t
(a) y(t) (b)
1
et  e2t
et

0 t
e2t

1
(c)

Convolution of f(t) and h(t) in above example

Here both f(t) and h(t) are causal (see above figure). Hence, we need only to perform
the convolution’s integration over the range (0, t) [see equation (30)]. The system
response is therefore given by
t
y(t) = ∫0 f(W)h(t  W)dW tt0
t W 2(t W )
y(t) = ∫0 e e dW tt0
t
y(t) = e2t ∫ eW dW e2t (e t  1) e  t  e 2t tt0 …(31)
0

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 LTI System

Also, y(t) = 0 when t < 0 [see equation (30)]. This result, along with equation (31), yields
y(t) = (e t  e2t )u(t)
The response is depicted in above figure (c).
(ii) Graphical Interpretation of convolution
To have a proper grasp of convolution operation, we should understand the graphical
interpretation of convolution. In addition, graphical convolution allows us to grasp
visually or mentally the convolution integral’s result, which can be of great help in
sampling, filtering, and many other problems.
The procedure for graphical convolution can be summarized as follows:
1. Keep the function f(W) fixed.
2. Visualize the function g(W) as a rigid wire frame, and rotate (or invert) this frame
about the vertical axis (W = 0) to obtain g(W).
3. Shift the inverted frame along the W axis by t0 seconds. The shifted frame now
represents g(t0  W).
4. The area under the product of f(W) and g(t0  W) (the shifted frame) is c(t0), the
value of the convolution at t = t0.
5. Repeat this procedure, shifting the frame by different values (positive and
negative) to obtain c(t) for all values of t.
Example
Find c(t) = f(t) * g(t) for the signals depicted in following figures (a) and (b). Since f(t)
is simpler than g(t), it is easier to evaluate g(t) * f(t) than f(t) * g(t). However, we have
intentionally take the more difficult route and evaluate f(t) * g(t) to clarity some of the
finer points of convolution.
Following figure (a) and (b) shows f(t) and g(t) respectively. Observe that g(t) is
composed of two segments. As a result, it can be described as
t
⎪⎧ 2e segment A, t ! 0
g(t) = ⎨ 2t
⎪⎩2e segment B, t  0
⎧⎪ 2e(t W) segment A, W  W
g(t  W) = ⎨ 2(t W)
⎪⎩2e segment B, W ! t
The segment of f(t) that is used in convolution is f(t) = 1, so that f(W) = 1. Following
figure (c) shows f(W) and g(W).

To compute c(t) for t t 0, we right-shift g(W) to obtain g(t  W), as illustrated in

following figure (d). Clearly, g(t  W) overlaps with f(W) over the shaded interval; that is,
over the range W t 0; Segment A overlaps with f(W) over the interval (0, t), while
Segment B overlaps with f(W) over (t, f). Remembering that f(W) = 1, we have
f
c(t) = ∫0 f(W)g(t  W)dW
t (t W ) f
= ∫0 2e dW  ∫ 2e2(t W)dW = 2(1  e t )  1
t

= 1  2e t tt0

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Vidyalankar : GATE – EC

Following figure (e) shows the situation for t < 0. Here the overlap is over the shaded
interval; that is, over the range W t 0, where only the segment B of g(t) is involved.
Therefore
f f f 2(t W )
c(t) = ∫0 f(W)g(t  W)dW = ∫0 g(t  W)dW = ∫0 2e dW
2t
= e t<0

⎧⎪1  2e2t tt0

c(t) = ⎨ 2t
⎪⎩e t0

Following figure (f) shows a plot of c(t).

g(t)
2
2et
f(t) segment A

1
0 t
0 t segment B
2et 2
(a)
(b)

g(W) f(W) f(W) t t 0

A g(t  W) 1
A

0 W t
0 W
B
B

(c) (d)

g(t  W) t<0
f(W) c(t)
A 1 1

t 0 W t
B 1
(f)
(e)
Convolution of f(t) and g(t)

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.30
 LTI System

Total Response
The total response of a linear system can be expressed as the sum of its zero-input and
zero-state components:
n
O jt
Total Response = ∑ c je  f(t) * h(t)
 
j 1


 zero  state component
zero inputcomponent
Example:

1. Find the total response y(t) for the equation

d2 y dy df
2
3  2y(t)
dt dt dt
if the input f(t) = 10 e3t u(t) and y(0) = 0 and y (0 ) 5

Solution:

[∵ initial condution are given at t = 0, if applies to zero-input response i.e. y0(0) = y(0)

and y (0 )]

i) zero-input response, y0(t)

The given differential equation is
d2 y dy df
2
3  2y(t)
dt dt dt
using ‘D’ operator, we get
D2  3D  2 y(t) Df(t)
to find zero-input response we will consider f(t) = 0
? (D2 + 3D + 2) y0(t) = 0
? characteristic equation is
O2 + 3O + 2 = 0
⇒ (O + 2) (O + 1) = 0
⇒ O1 = 1 and O2 = 2
? y0(t) = C1et + C2e2t
y 0 (t) = C1et  2C2e2t

Putting the values y0(0) = 0 and y 0 (0) 5, we get

C1 + C2 = 0
C1 + 2C2 = 5 …solving we get C1 = 5 and C2 = 5
t 2t
? y0(t) = 5e + 5e o required zero-input response

ii) To find zero-state response, we first have to find impilse response h(t), will be of the
form shown below
? h(t) = bn G(t) + P(D) [yn(t)] u(t)

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Vidyalankar : GATE – EC

where bn = b2 = 0 and yn(t) = k1et + k2e2t subject to y n (0) 1,yn (0) 0

⇒ k1 + k2 = 0
 k1  2k2 = 1 … solving we get k1 = 1, k2 = 1
? yn(t) = et  e2t
? h(t) = P(D) [yn(t)] u(t)
= D[et  e2t]u(t)
h(t) = (et + 2e2t) u(t) impulse response

iii) zero-state response, yzs (t) = h(t) * f(t) = f(t) * h(t)

= f(t) * [et u(t) + 2e2t u(t)]
= f (t) * et u(t) + 2 f(t) * e2t u(t)
t t
=  ∫ 10e 3t .e(t  τ ) dW  2 ∫ 10e 3 W .e2(t W)dW
0 0

…[∵ both f(t) and h(t) are causal, limits are form 0 o t]
t t
= 10e t 2 W 2 W W
∫ e dW  20e ∫ e dW
0 0
t 2t
10e >e2W @0t  20e >eW @0t
=
2 (1)
= 5e [e  1]  20 e2t [et  1]
t 2t

= (5e3t  5et  20 e3t + 20 e2t) u(t)

yzs = (5et + 20e2t  15e3t)u(t) zero-state response
? Total response y(t) = zero-input response + zero-state response
= y0(t) + yzs(t)
(5e t + 5e2t ) u(t)  (5e t  20e2t  15e3t ) u(t) …(32)
   
zero input response zero  stateresponse

Following figure (a) shows the zero-input, the zero-state, and the total response.

y(t) y(t) natural

zero-state

total total

0 0
t t
zero-input forced

(a) (b)

Total response and its components

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.32
 LTI System

Natural and Forced Response

From above example we see that the zero-input response is composed exclusively of
characteristic modes. Also the zero-state response equation (32) contains characteristic
mode terms.
We can now lump together all the characteristic mode terms in the total response, giving
us a component known as the natural response yn(t). The remainder, consisting entirely
of non-characteristic mode terms, is known as the forced response yI(t). The total
response can be expressed in terms of natural and forced components
t
10e
Total response = (  25e 2t
 )  
(15e 3t
) tt0 …(33)
naturalresponse yn (t) forcedresponse y I (t)

Above figure (b) shows the natural, forced, and total response.
Classical Solution of Differential Equations
In the classical method we solve differential equation to find the natural and forced
components rather than the zero-input and zero-state components of the response.
Note : Natural response yn(t) (also known as the homogeneous solution or
complementary solution). The remaining portion of the response consists entirely
of non-characteristic mode terms and is called the system’s forced response yI(t)
(also known as the particular solution.)
The natural response, being a linear combination of the system’s characteristic modes,
has the same form as that of the zero-input response; only its arbitrary constants are
different and forced response yI(t) can be obtained using method of undetermining
coefficients.
Forced Response: The Method of Undetermined Coefficients

Input f(t) Forced Response, yI(t)

1. ]t
e , ] z Oi (i 1,2, ˜ ˜ ˜,n) Ee ]t

2. e ]t , ] O i Ete] t
3. k E
4. cos(Zt + T) E cos(Zt + I)
5. (tr  Dr 1tr 1  ˜ ˜ ˜  D1t  D 0 )e]t (Er tr  Er 1tr 1  ˜ ˜ ˜  E1t  E0 )e]t

Note :
By definition, yI(t) cannot have any characteristic mode terms. If any term appearing in
the right-hand column for the forced response is also a characteristic mode of the
system, the correct form of the forced response must be modified to tiyI(t), where i is the
smallest possible integer that can be used and still can prevent tiyI(t) from having a
characteristic mode term. For example, when the input is e]t, the forced response
(right-hand column) has the form Ee]t. But if e]t happens to be a characteristic mode of
the system, the correct form of the forced response is Ete]t (see pair 2). If te]t also
happens to be a characteristic mode of the system, the correct form of the forced
response is Et2e]t, and so on.

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Vidyalankar : GATE – EC

This method can be used only for inputs having with a finite number of derivatives; this
class of inputs includes a wide variety of the most commonly encountered signals in
practice.

Example:

Find total response (solving same example as above) : natural and forced response
(D2 + 3D + 2) y(t) = Df(t)
f(t) = 10e3t and y(0+) = 0 and y (0 ) = 5

Solution:
Note : here initial condution are given at t = 0+, which applies to the total solution)
i) natural response yn(t) :
The characteristic equation is O2 + 3O + 2 = 0
⇒ (O + 2) (O + 1) = 0
⇒ O1 = 1 and O2 = 2
? yn(t) = k1 et + k2e2t …(34)

ii) Forced response, yI(t) (using method of undetermined co-efficient)

∵ f(t) = 10e
3t
…[is not of the form of characteristic response]
3t
let yI(t) = Ee
Substituting this in given D.E. we get

y I (t)  3y I (t)  2y I (t) f(t)
⇒ 9 Ee3t + 3[3Ee3t] + 2Ee3t = 30e3t
⇒ 2 Ee3t = 30e3t
comparing, we get E = 15
? yI(t) = 15 e3t …(35)
= natural response + forced response
y(t) = yn(t) + yI(t) …[from 34,35]
y(t) = (k1et + k2e2t  15e3t)u(t)
To find k1 and k2, now, we apply initial conditions
y(0+) = 0, y (0  ) = 5
⇒ k1 + k2 = 15
k1  2k2 = 40 …solving we get k1 = 10, k2 = 25
y(t) = (10et + 25e2t  15e3t)u(t) …required total response
which is same as before (refer equation (33))
• advantage of using this method very simple
• disadvantage
 can not be broken into zero-input and zero-state response
 initial condition required are at t = 0+, but generally we know condition at
t = 0, so these conditions must be derived

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.34
 LTI System

Note : In the classical method, the initial conditions are required at t = 0+. The reason is
that because at t = 0, only the zero-input component exists, and the initial conditions at
t = 0 can be applied to the zero-input component only. In the classical method, the zero-
input and zero-state components cannot be separated. Consequently, the initial
conditions must be applied to the total response, which begins at t = 0+.

SYSTEM STABILITY
If, in the absence of an external input, a
Imag
system remains in a particular state (or
condition) indefinitely, then that state is said
to be an equilibrium state of the system Re O < 0 Re O > 0
(zero state). Now suppose an LTI system is
in equilibrium (zero state) and we change Real
this state by creating some nonzero initial stable unstable
conditions. If the system is stable it should
eventually return to zero state. But the
system output generated by initial condtions Marginally
stable
(zero-input response) is made up of its
characteristic modes. For this reason we Characteristic roots location
define stability as follows: a system is and system stability
(asymptotically) stable if, and only if, all its
characteristic modes o 0 as t o f. If any of
the modes grows without bound as t o f,
the system is unstable. If zero-input
response remains bounded (approaches
neither zero nor infinity), approaching a
constant or oscillating with a constant
amplitude as t o f.
For this borderline situation, the system is said to be marginally stable or just stable.
To summarize :
1. An LTIC system is asymptotically stable if, and only if, all the characteristic roots are
in the LHP. The roots may be simple (unrepeated) or repeated.

2. An LTIC system is unstable if, and only if, either one or both of the following
conditions exist:
(i) at least one root is in the RHP, (ii) there are repeated roots on the imaginary axis.

3. An LTIC system is marginally stable if, and only if, there are no roots in the RHP, and
there are some unrepeated roots on the imaginary axis.

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Vidyalankar : GATE – EC

Characteristic Root Characteristic Root

Location Zero-Input Response Zero-Input Response
Location

0 0 0 0 t
t
(a) (b)

0 0 t 0 0 t
(d)
(c)

0 0 0 0 t
t

(f)
(e)

0 0 t 0 0 t

(g) (h)
Location of characteristic roots and the corresponding characteristic modes

Alternative form for System Stability

Consequently, for an asymptotically stable system

f
∫f | h(W) | dW  K 2  f
These result lead to the formulation of an alternative definition of stability known as
bounded-input, bounded-output (BIBO) stability: a system is BIBO stable if, and only if, a
bounded input produces a bounded output. Observe that an asymptotically stable system
is always BIBO stable. However, a marginally stable system is BIBO unstable.

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.36
 LTI System

PART(B) : DISCRETE TIME SIGNALS AND SYSTEM

Introduction

A typical discrete-time signal is a sequence of numbers specified at uniform interval of

time. This signal may be viewed as a function of time t where signal values are specified
at t = kT or may be viewed as a function of k (k, integer) as shown in the following figure.
f[k] or f(kT)

2 1 5 10 k

2T T 5T 10T t

A discrete-time signal
The representation f[k] is more convenient and will be followed throughout this book.

Advantages of Digital Signal Processing

1. Digital filters have a greater degree of precision and stability.
2. Digital filters are more flexible.
3. A greater variety of filters can be realized by digital systems.
4. Very low frequency filters, if realized by continuous-time systems, require
prohibitively bulky components. Such is not the case with digital filters.
5. Digital signals can be stored easily on magnetic tapes or disks without deterioration
of signal quality.
6. More sophisticated signal processing algorithms can be used to process digital
signals.
7. Digital filters can be time shared, and therefore can serve a number of inputs
simultaneously.
8. Using integrated circuit technology, they can be fabricated in small packages
requiring low power consumption.

TYPES OF SIGNALS

1. Discrete-Time Impulse Function, G[k]

The discrete-time counterpart of the continuous-time impulse function G(t) is G[k],
defined by
⎧1 k 0
G[k] = ⎨
⎩0 k z 0

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Vidyalankar : GATE – EC

This function, also called the unit impulse sequence, is shown in following figure.
Unlike its continuous-time counterpart G(t), this is a very simple function without any
mystery.

G [k]
1

0 k

Discrete-time impulse function

2. Discrete-Time Unit Step Function, u[k]

The discrete-time counterpart of the unit step function u(t) is u[k] defined by
⎧ 1 for k t 0
u[k] = ⎨
⎩0 for k  0
If we want a signal to start at k = 0 (so that it has a zero value for all k < 0), we need
only multiply the signal with u[k]. The following figure shows u[k]

u [k]
1

2 0 1 2 3 4 5 6 k
A discrete-time unit step function u[k]

3. Discrete-Time Exponential, Jk

The discrete-time exponential can be expressed in two forms as

eOk = Jk (J = eO or O = ln J)
In the study of discrete-time signals and systems, unlike the continuous-time case,
the form Jk proves more convenient than the form eOk.
Nature of Jk : The signal eOk grows exponentially with k if Re O > 0 (O in RHP), and
decays exponentially if Re O < 0 (O in LHP). It is constant or oscillates with constant
amplitude if Re O = 0 (O on the imaginary axis). Clearly, the location of O in the
complex plane indicates whether the signal eOk grows exponentially, decays
exponentially, or oscillates with constant frequency [refer following figure (a)]. A
constant signal (O = 0) is also an oscillation with zero frequency. A similar criterion for
determining the nature of Jk from the location of J in the complex plane [refer
following figure (b)].

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.38
 LTI System

LHP RHP Exponentially

Im Im
increasing

Exponentially

Exponentially
decreasing

increasing
Exponentially
1 decreasing 1
Re Re

O-plane J-plane

(a) (b)
The O-plane, the J-plane and their mapping
• The imaginary axis in the O-plane maps into the unit circle in the J-plane.
• The left-half plane in the O-plane maps into the inside of the unit circle.
• The right-half of the O-plane maps into the outside of the unit circle in the J-plane,
as depicted in above figure (b).
This fact means that the signal Jk grows exponentially with k if J is outside the unit
circle (|J| > 1), and decays exponentially if J is inside the unit circle (|J| < 1).The signal
is constant or oscillates with constant amplitude if J is on the unit circle (|J| = 1).
1 (0.8)k
1 (0.8)k

0 1 2 3 4 5 6 7 8
0 1 2 3 4 5 6 k
k

1

1
(0.5)k (1.1)k

0 1 2 3 4 5 6 k 0 1 2 3 4 5 6 k

Discrete-time exponentials Jk
GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.39
Vidyalankar : GATE – EC

4. Discrete-Time Exponential, ej:k

A general discrete-time exponential ej:k (also called phasor) is a complex valued

function of k, we shall plot the values of ej:k in the complex plane for various values
of k, as illustrated in following figure. The function f[k] = ej:k takes on values ej0, ej:,
ej2:, ej3:, … at k = 0, 1, 2, 3, … , respectively.

Locus of ej:k k=3 Locus of ej:k

k=2
:
: k=1 1
:
k=0 k=0
:
1 : k=1
:
k=2
k=3

(a) (b)
Note that
e jΩk re jT , r = 1, and T = k:

The locus of ej:k may be viewed as a phasor rotating counterclockwise at a uniform

speed of : radians per unit sample interval. The exponential ej:k, on the other hand,
takes on values ejT = 1, ej:, ej2:k, ej3:, … at k = 0, 1, 2, 3, …, as depicted in above
figure (b). Therefore, ej:k may be viewed as a phasor rotating clockwise at a uniform
speed of : radians per unit sample interval.

Using Euler’s formula, we can express an exponential ej:k in terms of sinusoids of the
form cos (:k + T), and vice versa.

ej:k = (cos :k + j sin :k)

ej:k = (cos :k  j sin :k)

5. Discrete-Time sinusoid, cos(:k + T)

A general discrete-time sinusoid can be expressed as C cos (:k + T), where C is the
amplitude, : is the frequency (in radians per sample), and T is the phase (in radians).
⎛ S S⎞
Following figure shows a discrete-time sinusoid cos ⎜ k  ⎟ .
⎝ 12 4⎠

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.40
 LTI System

1 ⎛S S⎞
cos ⎜ k ⎟
⎝ 12 4⎠

33 21 9 0 3 15 27
k

⎛S S⎞
A discrete-time sinusoid cos ⎜ k ⎟
⎝ 12 4⎠

• Some Peculiarities of Discrete-Time Sinusoids

There are three unexpected properties of discrete-time sinusoids which
distinguish them from their continuous-time relatives.
a. In the continuous-time case, the period of a sinusoid can take on any value;
integral, fractional, or even irrational. The discrete-time signal, in contrast, is
specified only at integral values of k. Therefore, the period must be an
integer (in terms of k) or an integral multiple of T (in terms of variable t).
b. A continuous-time sinusoid is always periodic regardless of the value of its
frequency Z. But a discrete-time sinusoid cos :k is periodic only if : is 2S
⎛Ω ⎞
times some rational number ⎜ is a rational number ⎟ .
⎝ 2S ⎠
c. A continuous-time sinusoid cos Zt has a unique waveform for each value of
Z. In contrast, a sinusoid cos :k does not have a unique waveform for each
value of :. In fact, discrete-time sinusoids with frequencies separated by
multiples of 2S are identical. Thus, a sinusoid cos :k = cos (: + 2S) k = cos
(: + 4S) k = …
• Not All Discrete-Time Sinusoids Are Periodic
A discrete-time signal f[k] is said to be N0-periodic if
f[k] = f [k + N0] …(36)
for some positive integer N0. The smallest value of N0 that satisfies equation (36)
is the period of f[k].
If a signal cos :k is to be N0-periodic, then
Cos :k = cos : (k + N0) = cos (:k + :N0)
This result is possible only if :N0 is an integral multiple of 2S; that is,
:N0 = 2Sm m integer
Ω m
…(37)
S N0
Because both m and N0 are integers, equation (37) implies that the sinusoid cos
:k is periodic only if Ω is a rational number.
2S

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Vidyalankar : GATE – EC

Example :
State with reasons if the following sinusoids are periodic. If periodic, find the period.
⎛ 3S ⎞
(i) cos ⎜ k ⎟ (ii) cos( Sk)
⎝ 7 ⎠
Solution:

⎛ 3S ⎞ 3S
(i) cos ⎜ k ⎟ comparing with standard form cos :k, we get : =
⎝ 7 ⎠ 7
Ω 3S 1 3 ⎛ m⎞
= ˜ = ⎜ which is of the form N ⎟
S 7 2S 14 ⎝ 0 ⎠
Thus, this sinusoid is periodic with N0 (period) equal to 14.

(ii) cos Sk
Similarly, here : = S
Ω S 1
? (irrationational)
S 2( S )2 2 S
? The given sinusoid is aperiodic

• Non-uniqueness of Discrete-Time Sinusoid Waveforms

A continuous-time sinusoid cos Zt has a unique waveform for every value of Z in

the range 0 to f. Increasing Z results in a sinusoid of ever increasing frequency.
Such is not the case for the discrete-time sinusoid cos :k because
cos (: r 2Sm) k = cos (:k r 2Smk)
Now, if m is an integer, mk is also an integer, and the above equation reduces to
cos (: r 2Sm) k = cos :k , m integer
This result shows that a discrete-time sinusoid of frequency : is indistinguishable
from a sinusoid of frequency : plus or minus an integral multiple of 2S. This
statement certainly does not apply to continuous-time sinusoids.
This result means that discrete-time sinusoids of frequencies separated by
integral multiples of 2S are identical. Thus the fundamental range of frequencies
is from S to S.
Thus cos (8.7Sk + T) = cos (0.7 Sk + T)
cos (9.6Sk + T) = cos (0.4Sk + T)

This result shows that a sinusoid cos (:k + T) can always be expressed as cos
(:fk + T), where S d :f < S (the fundamental frequency range). The graphical
explanation is shown in the figure below, which shows that the given discrete
time sinusoid may be sample of two continuous time sinusoids.

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.42
 LTI System

6 7 8 9 10
0 1 2 3 4 5 k

Physical explanation of non-uniqueness of Discrete-time sinusoid waveforms

Further Reduction in the Frequency Range of Distinguishable Discrete-Time

Sinusoids

The frequencies in the range (0 to S) can be expressed as frequencies in the range
(0 to S) with opposite phase.
cos (9.6Sk + T) = cos (0.4Sk + T) = cos (0.4Sk  T)
This result shows that a sinusoid of any frequency : can always be expressed as a
sinusoid of a frequency |:f|, where |:f| lies in the range 0 to S.

Example :

Consider sinusoids of frequencies : equal to (a) 1.6 S (b) 2.5 S. Each of these sinusoids
is equivalent to a sinusoid of some frequency |:f| in the range 0 to S. Determine |:f|.

Solution :

(a) The frequency 1.6S = 2S  0.4S, and :f = 0.4S. Therefore, a sinusoid of

frequency 1.6S can be expressed as a sinusoid of frequency |:f| = 0.4S.

(b) 2.5S = 2S + 0.5S, and :f = 0.5S. Therefore, a sinusoid of frequency 2.5S can be
expressed as a sinusoid of frequency |:f| = 0.5S.

6. Exponentially Varying Discrete-Time Sinusoid, Jk cos (:k + T)

This is a sinusoid cos (:k + T) with an exponentially varying amplitude Jk. It is

obtained by multiplying the sinusoid cos (:k + T) by an exponential Jk. Following
⎛S S⎞ ⎛S S⎞
figure shows (0.9)k cos ⎜ k  ⎟ , and (1.1)k cos ⎜ k  ⎟ . Observe that if |J| < 1, the
⎝6 3⎠ ⎝6 3⎠
amplitude decays, and if |J| > 1, the amplitude grows exponentially.

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.43
Vidyalankar : GATE – EC

5 ⎛S S⎞
k (0.9) k cos ⎜ k ⎟
(0.9) ⎝6 3⎠

(a)
13 7 0 5 11
k

5
⎛S S⎞
(1.1) k cos ⎜ k ⎟
5 ⎝6 3⎠
k
(1.1)

13 7 0 5 11 k (b)

5

Examples of Exponentially Varying Discrete-Time Sinusoids

Energy And Power For Discrete Time Signals

• Energy of a signal
The energy of the discrete time signals is given by
f
E = ∑ | f[k] |2
k f
For the signal to be energy signal, E < f (i.e. it must be finite). Thus for energy signal
f[k]o 0 as ko f

• Power of a signal
The power of discrete time signal is given by
N
1
Pavg = lim
k of 2N  1
∑ | f[k] |2
k N

For the signal to be power signal, Pavg < f. (i.e. it must be finite)

Example :

1. f[k] = ak u[k]
f f f
2
E = ∑ f[k] ∑ a2k ∑ (a2 )k
k f k 0 k 0
1
= o Energy signal
1  a2

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.44
 LTI System

2. f[k] = u[k]
f f
E = ∑ | f[k] |2 = ∑ | u[k] |2 = ¥ o Power signal
k f k 0

N N
or we define EN = ∑ | f[k] |2 = ∑ | f[k] |2
k 0 k 0

= N+1
then E = Lt EN = ¥
Nof

N
1
To find power = Lt
Nof 2N  1
∑ | f[k] |2
k 0

N N
1 1
P = Lt
Nof 2N  1
∑ u2 = Lt
Nof 2N  1
∑ 12
k 0 k 0

N1 N1 1
= Lt Lt =
Nof 2N  1 Nof 2N  1 2

USEFUL SIGNAL OPERATIONS

Signal operations discussed for continuous-time systems also apply to discrete-time
systems with some modification in time scaling.
(A) Time Shifting
A signal f[k] may be shifted in time by replacing the independent variable k by km,
where m is an integer. If m is a positive integer, the time shift results in a delay of the
signal by m units of time. If m is a negative integer, the time shift results in an
advance of the signal by m units of time, as shown in following figure.

f[k] 1
k
(0.9)
(a)

0 3 6 8 10 15
k

1 fd[k] = f [k  5]
k5
(0.9)
(b)

0 8 10 12 15
k
Time-Shifting Property
GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.45
Vidyalankar : GATE – EC

(B) Time Inversion (or Reversal)

We can show that to time invert a signal f[k], we replace k with k. This operation
rotates the signal about the vertical axis, as shown in following figure.
f[k] 1
k
(0.9)
(a)

0 3 6 8 10 15

1 fr [k] = f [k]
k
(0.9)
(b)
k

10 6 3 0 6 15 k
Time inversion of a signal
(C) Time Scaling
Time Compression: Decimation or Downsampling
Consider a signal
fc[k] = f[2k]

The signal fc[k] is the signal f[k] compressed by a factor 2. Observe that fc[0] = f[0],
fc[1] = f[2], fc[2] = f[4], and so on. This fact shows that fc[k] is made up of even
numbered samples of f[k]. The odd numbered samples of f[k] are missing as shown
in following figure (b). This operation loses part of the data, and that is why such time
compression is called decimation or downsampling. In the continuous-time case, time
compression merely speeds up the signal without loss of data. In general, f[mk]
(m integer) consists of only every m th sample of f[k].

Time Expansion

Consider a signal
⎡k ⎤
fe [k] = f ⎢ ⎥ …(38)
⎣2⎦
The signal fe[k] is the signal f[k] expanded by a factor 2. According to equation (38),
fe[0] = f[0], fe[1] = f[1/2], fe[2] = f[1], fe[3] = f[3/2], fe[4] = f[2], fe[5] = f [5/2], fe[6] = f[3],
and so on. Now, f[k] is defined only for integral values of k, and is zero (or undefined)
for all fractional values of k. Therefore, for fe[k], its odd numbered samples fe[1], fe[3],
fe[5], …are all zero (or undefined), as depicted in following figure (c).

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.46
 LTI System

Interpolation

In the time-expanded signal in following figure (c), the missing odd numbered

samples can be reconstructed from the non-zero valued samples using some

suitable interpolation formula. We may use a realizable interpolation, such as a linear

interpolation, where fi[1] is taken as the mean of fi[0] and fi[2]. Similarly, fi[3] is taken

as the mean of fi[2] and fi[4], and so on. This process of time expansion and inserting

the missing samples using an interpolation is called interpolation or upsampling.

f[k]

2 4 6 8 10 12 14 16 18 20 k
(a) Original Signal

fc[k] = f[2k]

2 4 6 8 10 k

(b) Decimation (Downsampling)

⎡k⎤
f e [k ] f
⎢⎣ 2 ⎥⎦

2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 k
(c) Expansion

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.47
Vidyalankar : GATE – EC

fi[k]

2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 k
(d) Interpolation (Upsampling)

CLASSIFICATION OF SYSTEMS
(A) Static Vs Dynamic Systems
A discrete time system is called static or memory less if its output at any instant
depends on the present input and not on past or future values of input. In any other
case, the system is said to be dynamic or to have memory.
e.g. y[k] = a f[k] static
k
y[k] = ∑ f[k  m] dynamic
m 0

(B) Time Invariant Vs Time Variant Systems

A system is said to be time invariant if the input  output characteristics do not
change with time. If they change with time, then the system is time variant.

Example:
i) y[k] = f[k]  f[k  1]
? y[k  m] = f[k  m]  f[k  m  1] o Time invariant

ii) y[k] = k f[k]

? y[k  m] = (k  m) f[k  m] = k f[k  m]  m f[k  m]o Time variant

(C) Linear Vs Non Linear System

A linear system is one that satisfies superposition principle. A system H is linear if
and only if
H [a1 f1[k] + a2 f2[k]] = a1 H[f1[k]] + a2 H[f2[k]]
for any arbitrary sequence f1[k] and f2[k] and any arbitrary constants a1 and a2. A
relaxed system that does not satisfy the principle of superposition is called nonlinear.
If y[k] z 0 for the input f[k] = 0, then the system is nonlinear [A linear system with 0
input produces 0 output].

y[k] = f[k2] o linear

y[k] = f2[k] o non linear

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.48
 LTI System

(D) Causal Vs Non Causal System

A system is said to be causal if the output of the system at any time k depends only
on present and past inputs but does not depend on future inputs. The output of a
causal system satisfies the equation y[k] = ƒ(f[k], f[k  1], f[k  2] . . . . . .). If the
system does not satisfy this condition, it is called non causal

Example :

y[k] = f[k]  f[k  1] causal

y[k] = f[k2] non causal

Following are the conditions for Causal System :

x Signal is right handed
x Output does not depends on further input.

(E) Stable Vs Unstable System

An arbitrary relaxed system is said to be bounded inputbounded output (stable) if

and only if every bounded input produces a bounded output. In other words,

Impulse response should be absolutely summable.

i.e. 6 | h[k]| < f
i.e. 6 | h[k] | should be finite

(F) Invertible Vs Non Invertible System

A system is said to be invertible if there is a one correspondence between its input
and output signals. i.e. if we know the output sequence y[k] ¥ d k d ¥ of an
invertible system, we can find its input f[k] ¥ d k d ¥.
Example:
y[k] = a f[k] o invertible
y[k] = f2[k] o non invertible

TIME-DOMAIN ANALYSIS OF LTID SYSTEM

Introduction

The procedure for analysis is parallel to that for continuous-time systems, with minor
differences. A general nth-order difference equation, using advance operator form:

y[k  n]  an1y [k  n  1]  ˜ ˜ ˜  a1y[k  1]  a 0 y[k]

bm f[k  m]  bm1f[k  m  1]  ˜ ˜ ˜  b1f[k  1]  b0 f[k] …(39)

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.49
Vidyalankar : GATE – EC

Causality Condition

The left-hand side of equation (39) consists of the output at instants k + n, k + n  1, k + n  2,

and so on. The right-hand side equation (39) consists of the input at instants k + m,
k + m  1, k + m  2, and so on. For a causal system the output cannot depend on future
input values. The causality requires m d n. Therefore, we replace k by k  n throughout the
equation yields the alternative form (the delay operator form) of equation (39) for m = n
y[k]  an1y [k  1]  ˜ ˜ ˜  a1y[k  n  1]  a0 y[k  n]
bn f[k]  bn1f[k  1]  ˜ ˜ ˜  b1f[k  n  1]  b0 f[k  n] …(40)

We shall designate form (39) the advance operator form, and form (40) the delay
operator form.

Initial Conditions and Iterative Solution of Difference Equations

Equation (40) can be expressed as

y[k] an1y [k  1]  an2y [k  2]  ˜ ˜ ˜  a0y [k  n]  bn f[k]  bn1f[k  1]  ˜ ˜ ˜  b0 f[k  n]

This equation shows that y[k], the output at the kth instant, is computed from 2n + 1
pieces of information. These are the past n values of the output: y[k  1], y[k  2] ,…,
y[k  n], the past n values of the input: f [k  1], f [k  2],…, f [k  n], and the present value
of the input f[k]. If the input is causal, then f[1] = f[2] = … = f[n] = 0, and we need only
n initial conditions y[1], y[2],…, y[n]. This result allows us to compute iteratively or
recursively the output y[0], y[1], y[2], y[3], … , and so on.

Example
Solve iteratively
y[k]  0.5y[k  1] f[k]

with initial condition y[1] = 16 and causal input f[k] = k2 (starting at k = 0).

Solution :
This equation can be expressed as
y[k] 0.5y[k  1]  f[k] …(41)
If we set k = 0 in this equation, we obtain
y[0] 0.5y[ 1]  f[0]
0.5(16)  0 8
Now, setting k = 1 in equation (41) and using the value y[0] = 8 (computed in the first step)
and
f[1] = (1)2 = 1, we obtain
y[1] = 0.5(8) + (1)2 = 5

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.50
 LTI System

Next, setting k = 2 in equation (41) and using the value y[1] = 5

(computed in the previous step)
and f[2] = (2)2, we obtain
y[2] = 0.5(5) + (2)2 = 6.5
Continuing in this way iteratively, we obtain
y[3] = 0.5(6.5) + (3)2 = 12.25
y[4] = 0.5(12.25) + (4)2 = 22.125
The output y[k] is depicted in figure below.

y[k]
12.25
8
6.5
5

0 1 2 3 4 5 k

Iterative solution of a difference equation

Iterative method can be applied to a difference equation in delay or advance operator

form. The iterative procedure cannot be applied to continuous-time systems. It is
interesting to note that the hardware realization generates the solution precisely in this
(iterative) fashion.

Operational Notation

In difference equations it is convenient to use operational notation similar to that used in

differential equations for the sake of compactness. For discrete-time systems we shall
use the advance operator E to denote advancing of the sequence by one time unit. Thus

Ef [k] { f [k +1]
E2f [k] { f [k + 2]
………………..
En f [k] { f [k + n]
A general nthorder difference equation (39), for m = n can be expressed as

(En  an1En1    a1E  a 0 )y[k] (bnEn  bn1En1    b1E  b0 )f[k] …(42)

or Q[E] y[k] = P[E] f[k]

where Q[E]and P[E] are nth-order polynomial operators

Q[E] = En  an1En1    a1E  a 0
P[E] = bnEn  bn1En1    b1E  b0

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Vidyalankar : GATE – EC

System response to Internal Conditions: The zero-Input response

The zero-input response y0[k] is the solution of equation (42) with f[k] = 0; that is,
Q[E]y0[k] = 0 …(43)
This equation states that a linear combination of y0[k] and advanced y0[k] is zero not for
some values of k, but for all k. Such situation is possible if and only if y0[k] and advanced
y0[k] have the same form. Only an exponential function Jk has this property.
Therefore, the solution of equation (43) must be of the form
y0[k] = cJk
Substitution of these yields

c( J n  an1J n1    a1J  a 0 )Jk = 0

For a nontrivial solution of this equation

( Jn  an1J n1    a1J  a 0 ) = 0
Q[J] = 0 …(44)
k
Thus cJ is indeed the solution provided J satisfies equation (44)
The polynomial Q[J] is called the characteristic polynomial of the system, and Q[J] = 0 is
the characteristic equation of the system. Moreover, J1, J2, …Jn, the roots of the
characteristic equation, are called characteristic roots or characteristic values (also
eigenvalues) of the system. The exponentials Jki (i 1,2,…,n) are the characteristic modes
or natural modes of the system. Characteristic modes not only determine the zero-input
response but also plays an important role in determining the zero-state response.

The zero input response, y0[k] depends upon the nature of the characteristic roots of equation
Q(J) = 0.
Nature of roots of Q(J) = 0 Zero-input response, y0[k]
1. All distinction roots y 0 [k] c1e J1k  c 2 e J 2k  ˜ ˜ ˜  cne Jnk
2. Some roots repeated ( say r-times ) y 0 [k] (c1  c 2k  c 3k 2  ˜ ˜ ˜  cr kr 1)e Jk
3. Complex conjugate roots of the form y 0 [k] ce Jk cos(Ek  T)
D r jE = |J|erjE

Example

1. For an LTID system described by the difference equation

y[k  2]  0.6y[k  1]  0.16y[k] 5f[k  2]
25
find zero-input response if the initial condition are y[1] = 0 and y[2] = ,
4

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.52
 LTI System

Solution:
The system equation in operational notation is
(E2  0.6E  0.16)y[k] 5E2 f[k]
The characteristic polynomial is
J 2  0.6J  0.16 ( J  0.2)( J  0.8)
The characteristic equation is
(J + 0.2) (J  0.8) = 0
The characteristic roots are J1 = 0.2 and J2 = 0.8. The zero-input response is
y 0 [k] = c1(0.2)k  c 2 (0.8)k …(45)

To determine arbitrary constants c1 and c2, we set k = 1 and 2 in equation (45) then
substitute
25
y0[1] = 0 and y0[2] = to obtain
4
5 ⎫ 1
0 5c1  c 2 ⎪ c1
4 ⎪ 5
⎬ ⇒
25 25 ⎪ 4
25c1  c2 c2
4 16 ⎪⎭ 5
Therefore
1 4
y 0 [k] (0.2)k  (0.8)k
5 5

2. A similar procedure may be followed for repeated roots. For instance, for a system
specified by the equation
(E2  6E  9) y[k] (2E2  6E) f[k]

determine y 0 [k], the zero-input component of the response if the initial condition are
1 2
y 0 [ 1]  and y 0 [ 2]  .
3 9
Solution :

The characteristic polynomial is J2 + 6J + 9 = (J + 3)2, and we have a repeated

characteristic root at J = 3. The characteristic modes are (3)k and k(3)k.

Hence, the zero-input response is

y0[k] = (c1 + c2k) (3)k
We can determine the arbitrary constants c1 and c2 from the initial conditions following
the procedure in example (1).

We find that c1 = 4 and c2 = 3 so that

y0[k] = (4 + 3k) (3)k

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.53
Vidyalankar : GATE – EC

3. For the case of complex roots, let us find the zero-input response of an LTID system
described by the equation
(E2  1.56E  0.81) y[k] (E  3)f[k]
When the initial conditions are y0[1] and y0[2] = 1.
Solution :
The characteristic polynomial is (J2  1.56J + 0.81) = (J  0.78  j0.45) (J  0.78 + j0.45).
S
rj
The characteristic roots are 0.78 r j0.45; that is, 0.9e Thus, | J | = 0.9 and E = S / 6.
6.

and the zero-input response, refer above table point (3).

⎛S ⎞
y 0 [k] c(0.9)k cos ⎜ k  T ⎟
⎝6 ⎠
To determine the arbitrary constants c and T, we set k = 1 and 2 in this equation and
substituting the initial conditions y 0 [ 1] 2,y 0 [ 2] 1, we obtain
c ⎛ S ⎞ c ⎡ 3 1 ⎤
2 cos ⎜   T ⎟ ⎢ cos T  sin T ⎥
0.9 ⎝ 6 ⎠ 0.9 ⎣ 2 2 ⎦
c ⎛ S ⎞ c ⎡1 3 ⎤
1 cos ⎜   T ⎟
2 ⎢ cos T  sin T ⎥
(0.9) ⎝ 3 ⎠ 0.81 ⎣ 2 2 ⎦
3 1
or c cos T  c sin T 2
1.8 1.8
1 3
c cos T  c sin T 1
1.62 1.62
These are two simultaneous equation in two unknowns c cos T and c sin T. Solution of
these equation yields
c cos T = 2.308
c sin T =  0.397
Dividing c sin T by c cos T yields
0.397 0.172
tan T
2.308 1
T tan1(0.172) 0.17rad
substituting T = 0.17 radian in c cos T = 2.308 yields c = 2.34, therefore
⎛S ⎞
y 0 [k] 2.34(0.9)k cos ⎜ k  0.17 ⎟ kt0
⎝6 ⎠
Observe that here we have used radian unit for both E and T. We also could have used
the degree unit, although it is not recommended. The important consideration is to be
consistent and to use the same units for both E and T.
The Unit Impulse Response h[k]
h[k] is the system response to input G[k], which is zero for k > 0. We know that when the
input is zero, only the characteristic modes can be sustained by the system. Therefore,
h[k] must be made up of characteristic modes for k > 0. At k = 0, it may have some
nonzero value A0, so that h[k] can be expressed as
h[k] = A0G[k] + yn[k]

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.54
 LTI System

b0
where, yn [k] is a linear combination of the characteristic modes and A0 =
a0
Moreover, because h[k] is causal, we must multiply yn [k] by u[k]. Therefore,
b0
h[k] = G[k]  yn [k]u[k] …(46)
a0

The n unknown coefficients in yn[k] (on the right-hand side) can be determined from the
knowledge of n values of h[k].

Example:
Determine the unit impulse response h[k] for a system specified by the equation
y[k]  0.6y[k  1]  0.16y[k  2] 5f[k]

Solution:

This equation can be expressed in the advance operator form as

y[k  2]  0.6y[k  1]  0.16y[k] 5f[k  2]

or (E2  0.6E  0.16) y[k] 5E2 f[k] …(47)

The characteristic polynomial is

J 2  0.6 J  0.16 ( J  0.2)( J  0.8)
The characteristic modes are (0.2)k and (0.8)k. Therefore

yn [k] c1(0.2)k  c 2 (0.8)k

Also, from equation (47), also we have a0 = 0.16 and b0 = 0. Therefore, according to
equation (46)

h[k] [c1(0.2)k  c 2 (0.8)k ]u[k] …(48)

To determine c1 and c2, we need to find two values of h[k] which can be found out
iteratively, we get h[0] = 5 and h[1] = 3. Now setting k = 0 and 1 in equation (48) and
using the fact that h[0] = 5 and h[1] = 3, we obtain

5 c1  c 2 ⎫ c1 1
⎬ ⇒
3 0.2c1  0.8c 2 ⎭ c2 4
Therefore
h[k] [(0.2)k  4(0.8)k ] u[k]

System Response to External Input : The zero-State Response

The zero-state response y[k] is the system response to an input f[k] when the system is in
zero state. Here we follow the procedure parallel to that used in the continuous-time case

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Vidyalankar : GATE – EC

Therefore
f
y [k] = ∑ f[m]h[k  m] …(49)
m f

The summation on the righthand side is known as the convolution sum of f[k] and h[k],
and is represented symbolically by f[k] * h[k]
f
f[k] h[k] ∑ f[m]h[k  m]
m f
Properties of the Convolution Sum
The structure of the convolution sum is similar to that of the convolution integral.
Moreover, the properties of the convolution sum are similar to those of the convolution
integral.
1. The Commutative Property
f1[k] f2 [k] f2 [k] f1[k]

2. The Distributive Property

f1[k] (f2 [k]f3 [k]) f1[k] f2 [k]f1[k] f3 [k]

3. The Associative Property

f1[k] (f2 [k] f3 [k]) (f1[k] f2 [k]) f3 [k]

4. The Shifting Property

If
f1[k] f2 [k] c[k]
then+
f1[k  m] f2 [kn] c[kmn]

5. The Convolution with an Impulse

f[k] G[k] f[k]

6. The Width Property

If f1[k] and f2[k] have lengths of m and n elements respectively, then the length of c[k]
is m + n  1 elements.

Causality and ZeroState Response

In deriving equation (49), we assumed the system to be LTI. There were no other
restrictions on either the input signal or the system. In practice, almost all of the input
signals are causal, and a majority of the systems are also causal. These restrictions
further simplify the limits of the sum in equation (49).
Therefore, equation (49) in this case reduces to
k
y[k]  ∑ f[m]h[k  m] …(50)
m 0

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.56
 LTI System

Example

Find the (zerostate) response y[k] of an LTID system described by the equaiton
y[k + 2]  0.6y[k + 1]  0.16y[k] = 5f [k +2] …(51)
if the input f[k] = 4ku[k].
Solution:
In order to find zero state response, first we have to find impulse response h[k].
Step 1 : To find h[k]
The given DE for zero-input, becomes
y[k + 2]  0.6y [k + 1]  0.16 y[k] = 0
? the characteristic polynomial is
J2  0.6J  0.16 = 0
Solving, we get
J = 0.8, 0.2
b
? h[k] = 0 f[k]  [c1(0.8)k  c 2 (0.2)k ]u[k]
a0
as b0 = 0
? h[k] = [c1(0.8)k  c 2 (0.2)k ]u[k] …(52)

c1 and c2 can be found out by knowing two values of h[k] say h[0] and h[1]
now from equation (51), we get
h[k + 2]  0.6 h [k + 1]  0.16 h[k] = 5G[k + 2]
putting k = 2, we get
h[0] = 5G[0]
= 5
and putting k = 1, we get
h[1]  0.6h [0] = 0
6
? h[1] = .5 3
10
using values of h[0] and h[1] and from (52), we get
c1 + c2 = 5
0.8c1  0.2c2 = 3
? c1 = 4, c2 = 1
? h[k] = [4(0.8)k + (0.2)k] u[k] …impulse response
?zero-state response for input (4)ku[k] is given by
y[k] = h[k] * f[k] = f[k] * h[k]
= (4)k u[k] * {4(0.8)k u[k] + (0.2)k u[k]}
= (0.25)k u[k] * 4(0.8)k u[k] + (0.25)k u[k] * (0.2)k u[k]
k k
= 4 ∑ (0.25)m (0.8)(k m)  ∑ (0.25)m (0.2)(k m)
m 0 m 0
k m k m
⎛ 0.25 ⎞ ⎛ 0.25 ⎞
= 4(0.8)k ∑ ⎝ 0.8 ⎠
⎜ ⎟  ( 0.2)k
∑ ⎜⎝ 0.2 ⎟⎠
m 0 m 0

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⎡ (0.25)k 1  (0.8)k 1 ⎤ k ⎡ (0.25)

k 1
 (0.2)k 1 ⎤
= 4(0.8)k ⎢ k ⎥  (0.2) ⎢ k ⎥
⎣ (0.8) (0.25  0.8) ⎦ ⎣ (0.2) (0.25  0.2) ⎦
= 7.27(0.25)k 1  7.27(0.8)k 1  2.22(0.25)k 1  2.22(0.2)k 1
= 1.8175(4)k  5.81(0.8)k  0.55(4)k  0.444(0.2)k
y[k] = [1.26(4)k  5.81(0.8)k  0.444(0.2)k ]u[k]

Graphical Procedure for the Convolution Sum :

The steps in evaluating the convolution sum are parallel to those followed in evaluating
the convolution integral. The convoluton sum of causal signals f[k] and g[k] is given by
k
c[k]  ∑ f[m]g[km]
m 0

The convolution operation can be performed as follows:

1. Invert g[m] about the vertical axis (m = 0) to obtain g[m] following figure (d).
Following figure (e) shows both f[m] and g[m].
2. Time shift g[m] by k untis to obtain g[k  m]. For k > 0, the shift is to the right (delay);
for k < 0, the shift is to the left (advance). Following figures ((f) and (g)) show g[k m]
for k > 0 and for k < 0, respectively.

3. Next we multiply f[m] and g[k  m] and add all the products to obtain c[k]. The
procedure is repeated for each value of k over the range  f to f.

Example
Find c[k] = f[k] * g[k]
where f[k] and g[k] are depicted in following Figures (a) and (b), respectively.

Solution:
We are given
f[k] = (0.8)k and g[k] = (0.3)k
Therefore
f[m] = (0.8)m and g[k  m] = (0.3)k  m
Following figure (f) shows the general situation for k t 0. The two functions f[m] and
g[k  m] overlap over the interval 0 d m d k. Therefore
k
c[k] ∑ f[m]g[k m]
m 0
k
= ∑ (0.8)m (0.3)k m
m 0
k m
⎛ 0.8 ⎞
= (0.3)k ∑ ⎜⎝ 0.3 ⎟⎠ = 2 ⎡⎣(0.8)k 1  (0.3)k 1 ⎤⎦ k t 0
m 0

GATE/EC/SSA/SLP/Module_5/Ch.1_Notes /Pg.58
 LTI System

For k < 0, there is no overlap between f[m] and g[k  m], as shown in following figure (g)
so that
c[k] = 0 k<0
and
? c[k] = 2[(0.8)k + 1 (0.3)k + 1] u[k]
f [k] g [k]
1 1
(0.8)k
(a) (0.3)k (b)

0 1 2 3 4 ko 0 1 2 3 4 ko

f [m] g [m]
1 m
(0.8)
(c) (d)
(0.3)m

0 1 2 3 4 mo 4  3  2 1 0 mo

g [m] 11 f [m]

(e)

4 3 2 1 mo

g [km] (f)
1
f [m]
k>0

0 k mo

c [k]
g [km] 1
f [m] k<0 1
(g) (h)

k 0 mo 0 1 2 3 4 5 ko

Graphical understanding of convolution of f[k] and g[k]

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An Alternative Form of Graphical Procedure : The Sliding Tape Method

This algorithm is convenient when the sequences f[k] and g[k]are short or when they are
available only in graphical form. This method is explained with the help of the following
example
Example
Using the sliding tape method, convolve the two sequences f[k] and g[k] depicted in
following
Figures (a) and (b), respectively.
f[k]

c[k]

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 LTI System

Solution:
• write the sequences f[k] and g[k] in the slots of two tapes: f tape and g tape above figure (c).
• Now leave the f tape stationary (to correspond to f[m]) and invert g tape about origin,
corresponding to the g[m], above figure (d).
• Shift the inverted g tape by k slots, multiply values on two tapes in adjacent slots, and
add all the products to find c[k]. Above figures (d, e, f, g, h, i and j) show the cases for
k = 0, 1, 2, 3, 4, 5, and 6, respectively.
For the case of k = 0, for example, above figure (d)
c[0] = 0 u 1 = 0
For k = 1, above figure (e)
c[1] = (0 u 1) + (1 u 1) = 1
Similarly,
c[2] = (0 u 1) + (1 u 1) + (2 u 1) = 3
c[3] = (0 u 1) + (1 u 1) + (2 u 1) + (3 u 1) = 6
c[4] = (0 u 1) + (1 u 1) + (2 u 1) + (3 u 1) + (4 u 1) = 10
c[5] = (0 u 1) + (1 u 1) + (2 u 1) + (3 u 1) + (4 u 1) + (5 u 1) = 15
c[6] = (0 u 1) + (1 u 1) + (2 u 1) + (3 u 1) + (4 u 1) + (5 u 1) = 15
Above figure (j) shows that c[k] = 15 for k t 5. Moreover, the two tapes are
nonoverlaping for k < 0, so that c[k] = 0 for k < 0. Above figure (k) shows the plot of c[k].
An Array Form of Graphical Procedure
The convolution sum can also be obtained from the array formed by sequences f[k] and
g[k]. In this numerical convolution can also be perfomed from the arrays using the sets
f[0], f[2],…, and g[0], g[1], g[2],…, as depicted in following figure below. The ijth element
(element in the ith row and jth column) is given by g[i] f[j].

f[k] c [0] c [1] c[2] ••

f[0] f[1] f[2] f[3] •••

g[k]
g [0] f [0] g [0] f [1] g [0] f [2] g [0] f [3] g [0] •••

g [1] f [0] g [1] f [1] g [1] f [2] g [1] f [3] g [1] •••

g [2] f [0] g [2] f [1] g [2] f [2] g [2] f [3] g [2] •••

g [3] f [0] g [3] f [1] g [3] f [2] g [3] f [3] g [3] •••

• • • • •
• • • • •
• • • • •

Array method for computing convolution sum

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Vidyalankar : GATE – EC

We add the elements of the array along its diagonals to produce c[k] = f[k] * g[k]. For
example, if we sum the elements corresponding to the first diagonal of the array, we
obtain c[0]. Similarly, if we sum along the second diagonal, we obtain c[1], and so on.
The process in shown above.

Example :
Convolution array method find convolution of given signal
x1[k] [1, 1, 1, 1, 1, 1, 1,]
n
x 2 [k] [1, 1, 1, 1, 1, 1, 1,]
n
Solution :
[Since signal are of finite duration, we can use sliding tape method or array method]
here we will demonstrate array method

? y[k] = x1[k] * x2[k]

= [1, 2, 3, 4, 5, 6, 7, 6, 5, 4, 3, 2, 1]
n
Total Response
The total response of an LTID system can be expressed as a sum of the zero input and
zerostate components:
n
Total response ∑ c j Jkj  f[k] h[k]
 
j 1


 Zero  state component
Zero input component
Refering examples solved in section zero-input response and zero-state response, we get
Total response 0.2(0.2)k  0.8(0.8)k 1.26(4)k  0.444(0.2)k  5.81(0.8)k …(53)
   
Zero input component Zero  state component

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 LTI System

Natural and Forced Response

The zeroinput component is made up of characteristic modes exclusively, but the

characteristic modes also appear in the zerostate response. When all the characteristic
mode terms in the total response are lumped together, the resulting component is the
natural response. The remaining part of the total response that is made up of
noncharactericstic modes is the forced response.

? For above example

Total response 0.644(0.2)k  6.61(0.8)k 1.26(4)k kt0 …(54)
   
Natural response Forced response

Classical solution of Linear Difference Equations

If yn[k] and y I [k] denote the natural and the forced response respectively, then the total
response is given by

Total response yn [k]  y I [k]

 
mod es nonmodes

Because the total response yn[k] + yI[k] is a solution of the system equaiton (42), we have
Q[E] (yn[k] + yI[k]) = P[E]f[k]
But since yn[k] is made up of characteristic modes,
Q[E]yn [k] = 0
?we get Q[E]yI [k] P[E]f[k] …(55)

The natural response is a linear combination of characteristic modes. The arbitraty

constants (multipliers) are determined from suitable auxiliary conditions usually given as
y[0], y[1], ...., y[n  1].

Forced Response
We have shown that the forced response yI[k] satisfies the system Equation (55)
Q[E]yI [k] P[E]f[k]
To determine the forced response, we shall use a method of undetermined coefficients,
the same method used for the continuoustime system. These coefficients can be
determned by substituting yI[k] in Eq. (55) and equating the coefficients of similar terms.

Input f[k] Forced Response yI[k]

k k
1. r , r z Ji (i = 1, 2, ..., n) cr
2. rk ,r = Ji ckrk
3. cos (Ek + T ) c cos (Ek + I)
4. ⎛ m i⎞ k ⎛ m i⎞ k
⎜ ∑ D ik ⎟ r ⎜ ∑ c ik ⎟ r
⎝ i 0 ⎠ ⎝ i 0 ⎠

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Note: By definition, yI[k] cannot have any characteristic mode terms. If any terms shown
in the righthand column for the foced response should also be a characteristic mode of
the system, the correct form of the forced response must be modified to ki y I [k] , where i
is the smallest integer that will prevent ki y I [k] from having a characteristic mode term.
For example, when the input is rk, the forced response in the righthand column is of the
form crk. But rk happens to be a natural mode of the system, the correct form of the
forced response is ckrk (see Pair 2).

Example :

Find total response of

y[k + 2]  0.6y[k + 1]  0.16y [k] = 5f [k +2]
if input f[k] = 4k u[k] and initial conditions are y[0] = 5.994, y[1] = 4.8442

Solution:

[∵ initial conditions are given at y[0] and y[1], they are valid for total response]
The given equation expressed in advance operator ‘E’ form is
(E2  0.6 E  0.16) y[k] = 5E2 f[k]
?characteristic equation is
J2  0.6J  0.16 = 0
J = 0.8, 0.2
?natural response is given by
yn[k] = c1(0.8)k + c2(0.2)k …(56)
k
Now let forced response yI[k] be of the form E(0.25) substituting in given DE, we get
? [E(0.25)(k  2)  0.6E(0.25)k 1  0.16E(0.25)k ] 5(0.25)k  2

? E(0.25)k (0.25)2  0.6E(0.25)k (0.25)  0.16E(0.25)k 5(0.25)k (0.25)2

? 0.0625 E(0.25)k  0.15E(0.25)k  0.16E(0.25)k 0.3125(0.25)k

0.2475E 0.3125
E = 1.26
? yI[k] = 1.26 (0.25)k …(57)
? Total response
y[k] = yn[k] + yI[k]
y[k] = c1(0.8)k + c2(0.2)k  1.26 (0.25)k …(58)
now substituting initial conditions we get
c1 + c2 = 7.254
and 0.8c1  0.2c2 = 5.1592
solving we get,
c1 = 6.61, c2 = 0.644

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 LTI System

Substituting this in equation (58), we get

⎡ ⎛ 1⎞
k ⎤
? y[k] = 6.61(0.8)k  0.644(0.2)k 1.26(4)k … ⎢(0.25)k ⎜ ⎟ 4 k ⎥
    ⎢⎣ ⎝4⎠ ⎥⎦
natural response forcedresponse

Same as before the response we got in equation (54)

SYSTEM STABILITY
Just as in a continuous time system, we define a discretetime system to be
asymptotically stable if, and only if, the zeroinput response approaches zero as k o f.
If the zeroinput response grows without bound as k o f, the system in unstable.
To be more general, let J be complex so that
k
J = J e jE and Jk J e jEk

Since the magnitude of e jEk is always unity regardless of the value of k, the magnitude of
k
Jk is J .
Therefore
if J 1, Jk o 0 as k o f
k
if |J| > 1, J of as k o f
and if |J| = 1, |J|k = 1 for all k

1m Unstable
Marginally

|J| J
E
1 1 Re o
Stable

Characteristic roots location and system stability.

It is clear that a system is asymptotically stable if and only if

J i 1 i = 1, 2, ...., n.

To summarize:
1. An LTID system is asymptotically stable if and only if all the characteristic roots are
inside the unit circle. The roots may be simple or repeated.

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2. An LTID system is unstable if and only if either one or both of the following conditions
exist:
(i) at least one root is outside the unit circle; (ii) there are repeated roots on the unit
circle.
3. An LTID system is marginally stable if and only if there are no roots outside the unit
circle and there are some unrepeated roots on the unit circle.

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 LTI System

System Response to Bounded Inputs

As in the case of continuoustime systems, asymptotically stable discretetime systems

have the property that every bounded input produces a bounded output if.
f
∑ h[k] K 2 f …(59)
k f

A system is said to be stable in the sense of having bounded output for every bounded
input
(BIBO stability) if and only if its impulse response h[k] satisfied equation. (59).

LIST OF FORMULAE

• Signal energy
For Continuous signal :
f
2
E= ∫ | f(t) | dt
f

For Discrete signal :

f
E= ∑ | f[k] |2
k f

• Signal Power
For Continuous signal :
7
1
| f(t) |2 dt
T of T ∫
Pavg = lim
 T /2

For Discrete signal :

1
1
Pavg = lim
Nof 2N  1
∑ | f[k] |2
k N

x Convolution :
For Continuous signal :
f f
y(t) = f1 t f2 t = ∫ f1 O f2 t  O dO = ∫ f1 t  O f2 O dO
f f
i) f (t) G (t) = f (t)
ii) f (t) G (t  t0) = f (t  t0)
iii) f (t  t1) G (t  t2) = f (t  t1  t2)
iv) G (t  t1) G (t  t2) = G (t  t1  t2)

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For Discrete signal :

f
y[k] f[k] h[k] ∑ f[m]h[k  m]
m f

(i) f1[k] f2 [k] f2 [k] f1[k]

(ii) f1[k] (f2 [k]f3 [k]) f1[k] f2 [k]f1[k] f3 [k]

(iii) f1[k] (f2 [k] f3 [k]) (f1[k] f2 [k]) f3 [k]

(iv) f1[k  m] f2 [kn] c[kmn]

(v) f[k] G[k] f[k]

x Miscellaneous :
k 1
(i) Sk ∑ am 1  a  a 2  ...  ak 1
m 0

1  ak
(ii) Sk [ For G.P. with common ratio as ‘a’ such that | a | < 1 ]
1 a
k 1
(iii) ∑ am k, a 1
m 0

ak  1
, az1
a 1
1  ak
(iv) if |a| < 1, Sk
1 a
• Shifting Property of impulse function
f
i) ∫f t G t f 0
f
f
ii) ∫f t G ta f a
f

iii) f(t) G(t) = f (0) G(t)

iv) f (t) G(t  t0) = f (t0) G(t  t0)

• Any signal can be expressed in its even and odd components

1
Even component: fe(t) = f(t)  f( t)
2
1
Odd component: fo(t) = f(t)  f( t)
2

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 LTI System

LMR (LAST MINUTE REVISION)

• Signal cannot be energy signal and power signal simultaneously.
• All periodic signals are power signal, but converse is not true.
x Derivative of unit ramp o unit step
Derivative of unit step o unit impulse
Derivative of unit impulse o unit doublet
t
x unit step u(t) = ∫G t dt
f

x A causal signal are one which are zero for time less than zero
i.e. for continuous f(t) = 0 , for t < 0
for discrete f[k] = 0 , for k < 0
x All memoryless systems are causal.
x System with memory tend to be unstable.
x An initially relaxed system is a causal system.
x For linear systems, zero input yields zero output.
x A memoryless system depends for its output only at that time when the input is
present. If the output is the same as the input it is called identity system.
x The discrete time unit step is the running sum of the unit sample or impulse
k
u[k] = ∑ G>k]
m f
The running sum is zero for k < 0 and 1 for k t 0 from the definition of unit impulse.
The discrete time unit step can be written in terms of unit sample as
f
u[k] = ∑ G>k  m]
m 0

• Some useful parties of convolution integral are :

(i) Commutative (ii) Distributive (iii) associative (iv) width property
x Nonperiodic functions such as unit ramp and unit step are called singularity or
generalized functions.
x If impulse response is not absolutely integrable, the system is unstable.
• For system to be absolutely stable all its characteristic modes must o 0 as t o f
i.e. (i) for LTIC system, all characteristic roots must lie in the LH-side of S-plane
(ii) for LTID system, all characteristic roots must lie in side unit circle of Z-plane
• Classical solution of differential/difference equation given total response in terms of
natural and forced response.
x Homogeneous solution is often referred to as natural response and forced response
is also called particular integral.
x A system is said to be relaxed if the energy storing elements alias memory elements
are set to zero before excitation.
x Linearity indicates additivity, scaling homogeneity (i.e. satisfies superposition theorem).



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Vidyalankar : GATE – EC

ASSIGNMENT  1

Duration : 45 mins Marks : 30

Q1 to Q6 carry one mark each

1. The following circuit is :

R

V C V0

(A) Memory less system (B) Causal system

(C) Static system (D) Dynamic system

2. For a continuous time system, to be BIBO stable ______.

(A) the impulse response of the system should be absolutely integrable
(B) the impulse response of the system should be absolutely summable
(C) the o/p of the system depends upon present and past inputs
(D) the o/p of the system depends upon present, past and future inputs

⎛ 3 ⎞
3. Find the signal x ⎜ 1  t⎟
⎝ 2 ⎠
x(t)
1

t
0 1 2
(A) (B)
1 1

t t
2/3 2/3 2/3 4/3

(C) (D)
1 1

t t
3/2 3/2 2/3 2/3

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 LTI System

4. A periodic function of half wave symmetry is necessarily __________.

(A) an even function (B) an odd function
(C) neither odd nor even (D) both odd and even

5. About the fourier series expansion of a periodic function it can be said that
(A) Even functions have only a constant and cosine terms in their FS
expansion.
(B) Odd functions have only sine terms in their FS expansion.
(C) Functions with half wave symmetry contain only odd harmonics.
(D) All of above

6. A periodic signal has power P/4 equal to average energy per period then RMS
value of signal is given by __________

P
(A) (B) P/4
2
P
(C) P/2 (D)
4

Q7 to Q18 carry two marks each

7. x(s) y1(s)
G1(s) G2(s) y2(s)

For the above system

s 1
G1(s) =
1  2s 1  s 2
Find G 2 (s) for the given system to be invertible.
s 1
(A) 2
(B)
1  2s  s 1  2s  s2

1  2s  s2
(C) (D) 1  2s  s2
s

⎡ 4Sk ⎤
8. If x[k] = cos ⎢ ⎥ , this sequence will be periodic after N0 samples. The value of
⎣ 21 ⎦
N0 is
(A) 12 (B) 24
(C) 21 (D) 2

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9. Consider a discrete time signal x[k] defined by

⎧1 2 d k d 2
x[k] = ⎨ then y[k] = x[3k  2] is :
⎩0 | k |! 2

⎧1 k 0,1 ⎧1 k 1
(A) y[k] = ⎨ (B) y[k] = ⎨
⎩1 otherwise ⎩1 k 1
⎧1 k 0,1
(C) y[k] = ⎨ (D) none of these
⎩0 otherwise
10. Match the following :
1
i) Unit step function (a) SG( Z) 
jZ
2a sin Za
ii) Exponential signal (b)
Za
1
iii) Signum function (c)
a  jZ
2
iv) Gate function (d)
jZ
(A) (i)  (a), (ii)  (d), (iii)  (c), (iv)  (b)
(B) (i)  (d), (ii)  (c), (iii)  (a), (iv)  (b)
(C) (i)  (b), (ii)  (c), (iii)  (a), (iv)  (d)
(D) (i)  (a), (ii)  (c), (iii)  (d), (iv)  (b)

11. Gate function is represented as _______ for t1  t 2 .

(A) (B)

u(t  t1) u(t  t1)

u(t  t2) u(t  t2)

(C) (D) None of these

u(t  t2)

u(t  t1)

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 LTI System

12. Consider the signals as shown below

x1(t) x2(t)
1 1

t t
0 1 0 1
The energy in combined signal (i.e. of addition of above two signals) is given by _____
(A) 2 (B) 3.5
(C) 2.75 (D) 1

⎛ Ec ⎞
13. Consider a signal x(t) with energy ⎜ ⎟ then time scaling by a factor 2 (i.e.
⎝ 2⎠
doubling) will change energy to ___________.

(A) Ec/2 (B) Ec

(C) Ec/4 (D) 2Ec

14. For the signal given below :

⎛2 ⎞ ⎛1 S⎞ ⎛1 S⎞ ⎛1 S⎞
y t 2sin ⎜ t ⎟  4sin ⎜ t  ⎟  6sin ⎜ t  ⎟  8sin ⎜ t  ⎟
⎝3 ⎠ ⎝4 4⎠ ⎝3 5⎠ ⎝2 7⎠
The common period of y(t) is given by __________
(A) 12S (B) 8S
(C) 24S (D) 16S
f
2
15. The value of integral ∫ 4t G t  1 dt is
0
(A) 4 (B) 4
(C) 4/3 (D) None of these

16. Match the following :

List I List II
(1) ycc t  2tyc t x t (i) causal and instantaneous but not
linear
(2) y t x t 3 (ii) causal and dynamic
(3) y t 2 t 1 x t (iii) causal and instantaneous but time
varying
(iv) causal and instantaneous but time
invariant
(A) 1  (i), 2  (ii), 3  (iii) (B) 1  (ii), 2  (i), 3  (iii)
(C) 1  (ii), 2  (iii), 3  (iv) (D) 1  (i), 2  (ii), 3  (iv)

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17. The signal energy in range 1 d Z d 1 rad/s for x(t) = et u(t) is __________.

(A) 1/2 (B) 1/6

(C) 1 (D) ¼

18. Two statements are made regarding Response of LTI system described by
difference equation

(1) The input terms of difference equation completely determine the forced
response

(2) Initial conditions satisfy the total response to yield the constants in the
natural response.

(3) The input terms of difference equation completely determines the natural
response
Choose the correct options :
(A) 1  True, 2  True, 3  False
(B) 1  False, 2  True, 3  True
(C) 1  False, 2  True, 3  False
(D) 1  True, 2  False, 3  False



GATE/EC/SSA/SLP/Module_5/Ch.1_Assign /Pg.74