Chapter2 - Linear Time-Invariant System

241-306 Linear Time-Invariant Systems
1
Chapter 2
Linear Time-Invariant
Systems

2
Outline
1 Discrete-Time LTI Systems : The Convolution
Sum
2 Continuous-Time LTI Systems : The Convolution
Integral
3 Properties of Linear Time-Invariant Systems
4 Causal LTI Systems Described by Differential and
Difference Equation
5 Singularity Functions

3
Representation of discrete-time signals in term of
impulses
Any discrete-time signal can be construct by
discrete-time unit impulse.
We can think as a sequence of shift discrete-time unit
impulse scaling by the value of x[n]
2.1 Discrete-Time LTI Systems : The Convolution
Sum

4
Example
x[−2][n2]=
{x[−2] , n=−2
0, n≠−2
x[2][n−2]=
{x[2], n=2
0, n≠2

5
Therefore, we can write x[n]
x[n]=...x[−3][n3]x[−2][n2]
x[−1][n1]x[0][n]x[1][n−1]
x[2][n−2]x[3][n−3]...
x[n]= ∑
k=−∞
∞
x[k ][n−k ]

6
Example
u[n]= ∑
k=−∞
∞
x[k ][n−k ]=∑
k=0
∞
[n−k ]
u[n]=
{0, n0
1, n≥0

7
Discrete-Time Unit Impulse Response and the
Convolution sum Representation of LTI Systems
The signal x[n] can be represent as a superposition
of scaled versions of a very simple set of elementary
functions namely, shifted unit impulses δ[n-k].
Let hk
[n] denote the response of linear system to the
shifted unit impulse δ[n-k]
[n−k]h[n−k]=hk [n]
x[n] y[n]
δ[n-k] hk
[n]

8
From the superposition of linear system, the response
y[n] of the linear system to the input x[n] is simply the
weighted linear combination of these basic responses.
x[k]δ[n-k] x[k]hk
[n]
δ[n-k] hk
[n]

9
x[-1]δ[n+1] x[-1]h-1
[n]
x[0]δ[n] x[0]h0
[n]
x[1]δ[n-1] x[1]h1
[n]
+
y[n]
y[n]= ∑
k=−∞
∞
x[k ]hk [n]

10
If we know the response of a linear system to the set
of shifted unit impulses, we can construct the
response to an arbitrary input.
Example

11
The response to the signals δ[n+1], δ[n] and δ[n-1] are
h-1
[n] , h0
[n] , h1
[n].

12
The response to the signals x[-1]δ[n+1], x[0]δ[n] and
x[1]δ[n-1] are x[-1]h-1
[n] , x[0]h0
[n] , x[1]h1
[n].

13
The response to the signals x[n] :

14
If the system is time-invariant, then these responses
to the time-shifted unit impulses are all time-shifted
version of each other. Specifically, since δ[n-k] is the
time-shifted version of δ[n], the response hk
[n] is a
time-shifted version of h0
[n]
hk [n]=h0[n−k]
For notational convenience, we will drop the
subscript on h0
[n] and define the unit impulse
response
h[n]=h0[n]

15
The convolution-sum representation for discrete-time
system are linear and time invariant.
The response of LTI system :
y[n]= ∑
k=−∞
∞
x[k ]h[n−k ]
This is referred to as the convolution sum. We will
present the operation of convolution symbolically as:
y[n]=x[n]∗h[n]

16
Example 2.1
Consider an LTI system with impulse response h[n]
and input x[n] below

17
The response of the system is :
y[n]=x[0]h[n−0]x[1]h[n−1]
=0.5h[n]2h[n−1]

18
Example 2.2
Consider again the convolution problem encountered
in example 2.1. The sequence x[k] shown below.

19
The sequence h[n-k], for n fixed and view as a
function of k for several different values of n.

20

21
For any particular value of n, we multiply signal x[k]
and h[n-k] and sum over all values of k
For example, for n<0, we see that x[k]h[n-k] = 0 for n<0
since the nonzero values of x[k] and h[n-k] do not
overlap. Consequently, y[n] = 0 for n<0.

22
For n=0, The product x[k]h[0-k] has only one nonzero
sample with the value 0.5, We conclude that
y[0]= ∑
k=−∞
∞
x[k ]h[0−k ]=0.5

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y[1]= ∑
k=−∞
∞
x[k]h[1−k ]=0.52.0=2.5
The product x[k]h[1-k] has only two nonzero samples
with may be summed to obtain

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y[3]= ∑
k=−∞
∞
x[k ]h[3−k ]=2.0
y[2]= ∑
k=−∞
∞
x[k ]h[2−k ]=0.52.0=2.5
Similarly, for n=2 and n=3, we have
Finally, for n>3, the product x[k]h[n-k] is zero for all k,
from which we conclude that y[n] = 0 for n> 3. The
output values agree with those obtained in example
2.1

25
Output Signal y[n]

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Example 2.3
Consider an input x[n] and a unit impulse response
h[n] given by
x[n]=
n
u[n]
h[n]=u[n]

27
Example of x[k] and h[n-k] as a function of k for
several of n :

28

29

30
For n<0, there is no overlap between the nonzero points
in x[k] and h[n-k]. The product x[k]h[n-k] = 0 for all k.
Consequently, y[n] = 0 for n<0.
Solution

31
y[n]=∑
k=0
n

k
y[n]=∑
k=0
n

k
=
1−
n1
1−
for n≥0
For n≥0,
y[n]=
1−
n1
1−
u[n]
x[k]h[n−k]=
{
k
, 0≤k≤n
0, otherwise

32
1−∑
k=0
N
k
=∑
k=0
N
k
−∑
k=0
N
k1
Prove
1−∑
k=0
N
k
=1−N 1
∑
k=0
N
k
=
1−N 1
1−

33
y[n]=1−
n1
1 u[n]
Output Signal y[n]

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Example 2.4
Consider two sequences x[n] and a unit impulse
response h[n] below. Calculate the convolution of
two signals
x[n]=
{1, 0≤n≤4
0, otherwise
h[n]=
{n
, 0≤n≤6
0, otherwise

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Solution
The convolution is sometimes described in terms of
sliding the sequence h[n-k] past x[k].
In this example, it is convenient to consider five
separate intervals for n.
Interval 1 : For n< 0, there is no overlap between
the nonzero portions of x[k] and h[n-k], and
consequently, y[n] = 0.

36

37
Interval 2 : For 0 ≤ n ≤ 4,
x[k ]h[n−k ]=
{n−k
, 0≤k≤n
0, otherwise
Thus in this interval
y[n]=∑
r=0
n

r
=
1−
n1
1−
y[n]=∑
k=0
n
n−k
Changing variable k to r = n-k

38

39
Interval 3 : For n > 4 but n-6 ≤ 0 (4 < n ≤ 6),
x[k ]h[n−k ]=
{n−k
, 0≤k≤4
0, otherwise
Thus, in this interval
y[n]=
n
∑
k=0
4

−1

k
=
n 1−
−1

5
1−
−1
=

n−4
−
n1
1−
y[n]=∑
k=0
4
n−k

40

41
Interval 4 : For n > 6 but n-6 ≤ 4 (6 < n ≤ 10),
x[k ]h[n−k ]=
{
n−k
, n−6≤k≤4
0, otherwise
Thus, in this interval
y[n]= ∑
r=0
10−n

6−r
=
6 1−
n−11
1−
−1
=

n−4
−
7
1−
y[n]= ∑
k=n−6
4
n−k
Changing variable k to r = k-n+6

42

43
Interval 5 : For n-6 > 4 or n>10, there is no overlap
between the nonzero portions of x[k] and h[n-k],
and consequently, y[n] = 0.

44
We obtain
y[n]=
{
0, n0
1−n1
1−
, 0≤n≤4

n−4
−
n1
1−
, 4n≤6

n−4
−
7
1−
, 6n≤10
0, n10

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Output Signal y[n]

46
Example 2.5
Consider an input x[n] and a unit impulse response
h[n] given by
x[n]=2
n
u[−n]
h[n]=u[n]
Calculate the convolution of two signals

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Plot of sequence x[k] and h[n-k]

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Note : x[k] is zero for k>0 and h[n-k] is zero for k>n.
The product x[k]h[n-k] always has nonzero samples
along the k-axis.
For n ≥ 0, x[k]h[n-k] has nonzero samples in the
interval k ≤ 0.
y[n]= ∑
k=−∞
0
x[k]h[n−k]= ∑
k=−∞
0
2
k
Changing variable k to r = -k
y[n]= ∑
k=−∞
0
2
k
=∑
r=0
∞
1
2
k
=
1−1/2
∞
1−1/2
=2

49
For n < 0, x[k]h[n-k] has nonzero samples in the
interval k ≤ n.
y[n]= ∑
k=−∞
n
x[k ]h[n−k ]= ∑
k=−∞
n
2
k
Changing variable k to l = -k and m = l+n
y[n]= ∑
l=−n
∞
2
l
=∑
m=0
∞
1
2
m−n
=1
2
−n
∑
m=0
∞
1
2
m
=2
n
⋅2=2
n1

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Output Signal y[n]

51
Problems
1 Let
x[n]=[n]2[n−1]−[n−3]
and
h[n]=2[n1]2[n−1]
Compute and plot each of the following
convolutions :
a) y1
[n] = x[n] *h[n] b) y2
[n] = x[n+2] *h[n]
c) y3
[n] = x[n] *h[n+2]

52
2 Consider an input x[n] and a unit impulse response
h[n] given by
x[n]=1
2
n−2
u[n−2]
h[n]=u[n2]
Determine and plot the output y[n] = x[n] *h[n]

53
3 Let
x[n]=
{1, 0≤n≤9
0, elsewhere
and
h[n]=
{1, 0≤n≤N
0, elsewhere
where N≤9 is an integer. Determine the value
of N, given that y[n] = x[n] *h[n] and y[4] = 5,
y[14] = 0.

54
2.2 Continuous-Time LTI Systems : The Convolution
Integral
The Representation of Continuous-Time Signals
in Terms of impulses
Consider a pulse or “staircase” approximation,
to a continuous time signal x(t).
xt

55
t=
{
1

, 0≤t
0, otherwise
The approximation can be expressed as a linear
combination of delayed pulses. If we define
then, since ΔδΔ
(t) has unit amplitude, we have the
expression
xt= ∑
k=−∞
∞
xk t−k 

56

57

58
xt=lim
0
∑
k=−∞
∞
xk  t−k 
Let Δ approach 0, the approximation becomes
better and better. Therefore,
xt
As Δ→0, the summation approaches an integral.
This can be seen by considering the graphical
interpretation of the equation :

59

60
The product x(τ)δΔ
(t-τ)

61
xt=∫−∞
∞
xt−d 
ut=∫−∞
∞
ut−d =∫
0
∞
t−d 
x(t) equals the limit as Δ→0 of the area under
x(τ)δΔ
(t-τ). Moreover, we know that the limit as Δ→0
of δΔ
(t) is the unit impulse function δ(t).
For example x(t) = u(t)

62
In ideally case, if Δ is very small, we can using
several basic of basic properties of unit impulse. The
integral of this signal from τ = -∞ to τ =+∞ equal :
xt=∫−∞
∞
xt−d 
=∫−∞
∞
xtt−d 
=xt∫−∞
∞
t−d =xt

63

64

65
The Continuous-Time Unit Impulse Response
and the Convolution Integral Representation of
LTI Systems
yt= ∑
k=−∞
∞
xk  hk t
The approximation represent the signal as a
sum of scaled and shifted versions of δΔ
(t). Let
define as a response of an LTI system to
the input δΔ
(t-kΔ). Then
xt
hk  t

66

67

68

69
yt=lim
0
∑
k=−∞
∞
xk  hk t
yt=∫−∞
∞
xhtd 
If Δ is very small :
As Δ→0, the summation approaches an integral.
Therefore,

70

71
ht=h0t
yt=∫−∞
∞
xht−d 
yt=xt∗ht
If the system is also time-invariant. The response of
the LTI system to unit impulse δ(t-τ) is hτ
(t) = h0
(t-τ)
Let h0
(t) is the response of the unit impulse δ(t)
The response of LTI system :
This is referred to as the convolution integral. We will
present the operation of convolution symbolically as:

72
Example 2.6
xt=e
−at
ut, a0
ht=ut
Let x(t) be the input to an LTI system with impulse
response h(t), where
and
Find the response of LTI system to x(t)

73
xht−=
{e−a 
, 0≤t
0, otherwise
Solution
For t<0, the product of x(τ) and h(t-τ) is zero,
and consequently, y(t) is zero. For t>0,
=∫
0
t
e−a 
d =−
1
a
1−e−at

yt=∫−∞
∞
xht−d 

74

75

76
The response to input x(t)

77
Example 2.7
Find the convolution of the following two signals
xt=
{1, 0tT
0, otherwise
ht=
{t , 0t2T
0, otherwise

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xht−=
{t− , 0t
0, otherwise
Solution
For t<0 and for t>3T, x(τ)δΔ
(t-τ)=0 for all values of
τ and consequently y(t) = 0.
For interval 0 < t < T
yt=∫−∞
∞
xht−d 
=∫
0
t
t−d =
1
2
t2

79

80
xht−=
{t− , 0T
0, otherwise
For interval T < t < 2T
yt=∫−∞
∞
xht−d 
=∫
0
T
t−d =Tt−
1
2
T2

81

82
xht−=
{t− , t−2TT
0, otherwise
For interval 2T < t < 3T
yt=∫−∞
∞
xht−d 
= ∫
t−2T
T
t−d =−
1
2
t2
Tt−
3
2
T2

83

84
For interval 3T < t

85
yt=
{
0, t0
1
2
t2,
0tT
Tt−
1
2
T
2
, Tt2T
−
1
2
t2
Tt
3
2
T2
, 2Tt3T
0, 3Tt

86
The response to input x(t)

87
Example 2.8
xt=e
2t
u−t
ht=ut−3
Find the response of the system to input signal x(t).

88
yt=∫−∞
t−3
e
2
d =
1
2
e
2t−3
yt=∫−∞
0
e
2
d =
1
2
Solution
The signal x(τ) and h(t-τ) have regions of
nonzero overlap, regardless of the value t. When
t-3 ≤ 0, the product of x(τ) and h(t-τ) is nonzero for
-∞ < τ < t-3, and the convolution integral becomes
For t-3 ≥ 0, the product x(τ)h(t-τ) is nonzero for
-∞ < τ < 0, so that the convolution integral is

89

90
The response to input x(t) Ex. 2.8

91
Problems
1 Determine and sketch the convolution of the
following two signals :
xt=
{
t1, 0≤t≤1
2−t , 1t≤2
0, elsewhere
ht=t22t1

92
2 Suppose that
xt=
{1, 0≤t≤1
0, elsewhere
and h(t) = x(t/α) where 0 < α ≤ 1.
a) Determine and sketch y(t) = x(t)*h(t).
b) If dy(t)/dt contains only three
discontinuities, what is the value of α?

93
3 Let
xt=ut−3−ut−5
and
ht=e
−3t
ut
a) Compute y(t) = x(t)*h(t).
b) Compute g(t) = (dx(t)/dt)*h(t).
c) How is g(t) related to y(t)?

94
2.3 Properties of Linear Time-Invariant Systems
The representation of continuous-time and discrete-
time LTI systems in the terms of their impulse
responses.
yt=∫−∞
∞
xht−d =xt∗ht
y[n]= ∑
k=−∞
∞
x[k ]h[n−k ]=x[n]∗h[n]
Convolution sum
Convolution integral

95
The Commutative Property
x[n]∗h[n]=h[n]∗x[n]= ∑
k=−∞
∞
h[k ]x[n−k ]
xt∗ht=ht∗xt=∫−∞
∞
hxt−d 
Continuous-time
Discrete-time

96
The Distributive Property
x[n]∗h1[n]h2[n]=x[n]∗h1[n]x[n]∗h2[n]
Discrete-time
Continuous-time
xt∗h1 th2t=xt∗h1txt∗h2t

97
y1t=xt∗h1t
y2t=xt∗h2 t
yt=xt∗h1t
xt∗h2t
yt=xt∗h1t
xt∗h2t

98
Example 2.10
Find the convolution of the following two sequences
x[n]=1
2
n
u[n]2
n
u[−n]
h[n]=u[n]
Note : The sequence x[n] is nonzero along the
entire time axis. Direction evaluation of such a
convolution is very difficult. Instead, we may use the
distributive property to express as the sum of two
simpler convolution problem.

99
Let
x1[n]=1
2
n
u[n] x2[n]=2
n
u[−n]and
Using the distributive property
y[n]= y1[n] y2[n]
Where
y1[n]=x1[n]∗h[n] y1[n]=x2 [n]∗h[n]and

100
From example 2.3 with α=1/2
Output Signal y1
[n]
y[n]=1−n1
1 u[n]

101
From example 2.5
Output Signal y2
[n]

102
Output Signal y[n]

103
The Associative Property
x[n]∗h1[n]∗h2[n]=x[n]∗h1[n]∗h2[n]
Discrete-time
Continuous-time
xt∗h1 t∗h2t=xt∗h1t∗h2t

104
Commutative

105
LTI Systems with and without Memory
h[n]=K [n]
y[n]=Kx[n]
A system is memoryless if its output at any time
depends on only the value of input at that same
time. The only way that this can be true for a
discrete-time LTI system is if h[n] = 0 for n≠0.
The impulse response has the form
Where K=h[0] is constant, and the convolution sum
reduces to :

106
ht=K t
yt=Kxt
For continuous-time LTI systems, the system is
memoryless if h(t) = 0 for t≠0. The memoryless
system has the form :
For some constant K and has the impulse response

107
If K = 1, then these system become identity
systems, with output equal to input and with unit
impulse response equal to the unit impulse :
x[n]=x[n]∗[n] xt=xt∗tand
with reduce to the sifting properties of the
discrete-time and continuous-time unit impulse :
x[n]= ∑
k=−∞
∞
x[k ][n−k]
xt=∫−∞
∞
xt−d 

108
Invertibility of LTI Systems
ht∗h1t=t
h[n]∗h1[n]=[n]
The system is invertible, then an inverse system
exists that, when connected in series with the
original system, produces an output equal to the
input to the first system.
Continuous-time
Discrete-time

109

110
Example 2.11
Find the inverse system of
yt=xt−t0
ht=t−t0
xt−t0=xt∗t−t0
Solution
The impulse response of the system is
Therefore,

111
h1 t=tt0
ht∗h1t=t−t0∗tt0=t
To invert the system, all output is required to shift
back. We take
then

112
Example 2.12
Find the inverse of the LTI system with impulse
response
h[n]=u[n]
y[n]= ∑
k=−∞
∞
x[k]u[n−k ]
solution
The response of this system

113
y[n]= ∑
k=−∞
n
x[k ]
since u[n-k] is 0 for n-k < 0 and 1 for n-k ≥0
This system is a summer or accumulator that
computes the running sum of all the values of the
input up to the present time. Its inverse is
y[n]=x[n]−x[n−1]
which is a first difference operation

114
h1 [n]=[n]−[n−1]
h[n]∗h1[n]=u[n]∗{[n]−[n−1]}
=u[n]∗[n]−u[n]∗[n−1]
=u[n]−u[n−1]
=[n]
The impulse response of the inverse system is
Check the inverse system :

115
Causality for LTI Systems
The property of causality : The output of the causal
system depend only on the present and the past
values of the input to the system.
For a discrete-time LTI system to be causal, y[n]
must not depend on x[k] for k > n.
y[n]= ∑
k=−∞
∞
x[k ]h[n−k ]
This is to be true if all of the coefficients h[n-k] must
be zero for k>n.

116
y[n]= ∑
k=−∞
n
x[k ]h[n−k ]
y[n]=∑
k=0
∞
h[k ]x[n−k ]
h[n]=0 for n0
This then requires that the impulse response of a
causal discrete-time LTI system satisfy the condition
The convolution sum become
or

117
ht=0 for t0
yt=∫−∞
t
xht−d 
=∫
0
∞
h xt−d 
For a continuous-time LTI system
The convolution integral become

118
Stability for LTI Systems
A system is stable if every bounded input
produces a bounded output.
To determine conditions under which LTI systems
are stable, consider an input x[n] that is bounded in
magnitude :
∣x[n]∣B for all n
The output magnitude is :
∣y[n]∣=
∣∑
k=−∞
∞
h[k ]x[n−k ]
∣

119
Since the magnitude of the sum of a set of
numbers is no large than the sum of the
magnitudes of numbers.
∣y[n]∣≤ ∑
k=−∞
∞
∣h[k ]∣∣x[n−k]∣
|x[n-k]| < B for all values of k and n.
∣y[n]∣≤B ∑
k=−∞
∞
∣h[k ]∣

120
We can conclude that if
∑
k=−∞
∞
∣h[k ]∣≤∞
then y[n] is bounded in magnitude, and hence,
the system is stable.

121
In continuous-time
∣yt∣=∣∫−∞
∞
h xt−d ∣
≤∫−∞
∞
∣h∣∣xt−∣d 
≤B ∫−∞
∞
∣h∣d 
∫−∞
∞
∣h∣d ∞

122
Example 2.13
Find the stability of the system that is a pure time
shift in either continuous-time or discrete-time.
discrete-time
∑
n=−∞
∞
∣h[n]∣= ∑
n=−∞
∞
∣[n−n0]∣=1
The system is stable.

123
continuous-time
The system is stable.
∫−∞
∞
∣h∣=∫−∞
∞
∣−t0d ∣=1
For accumulator system
∑
n=−∞
∞
∣u[n]∣=∑
n=0
∞
∣u[n]∣=∞
The system is unstable.
discrete-time

124
continuous-time
The system is unstable.
ht=∫−∞
t
∣d ∣=ut
yt=∫−∞
t
xd 
Continuous-time accumulator
The impulse response
and
∫−∞
t
∣u∣d =∫
0
∞
d =∞

125
The Unit Step Response of LTI System
The unit step signal is another signal that is also
used quite often in describing the behavior of the LTI
system . Let s[n] or s(t) is the response (output) of
the unit step input x[n] = u[n] or x(t) = u(t)
Discrete-time
s[n]=u[n]∗h[n]
The unit step response of LTI system is

126
s[n]= ∑
k=−∞
n
h[k ]
h[n]=s[n]−s[n−1]
From Example 2.12, the response is
and h[n] can be recovered from s[n]
The step response of a discrete-time LTI system is
the running sum of its impulse response.
Conversely, the impulse response of a discrete-time
LTI system is the first difference of its step
response.

127
Continuous-time
st=∫−∞
t
hd 
ht=
d st
dt
=x' t
The unit step response of LTI system is
The unit impulse response is the first derivative of
the unit step response.

128
2.4 Causal LTI Systems Described by Differential and
Difference Equation
In continuous-time system, the input and output are
related through a linear constant-coefficient
differential equation.
In discrete-time system, the input and output are
related through a linear constant-coefficient
difference equation.

129
Linear Constant-Coefficient Differential Equation
∑
k=0
N
ak
d
k
yt
dtk
=∑
k=0
M
bk
d
k
xt
dtk
yt=
1
a0
∑
k=0
M
bk
d
k
xt
dt
k
A general Nth-order linear constant-coefficient
differential equation.
When N= 0 :
(*)

130
∑
k=0
N
ak
d
k
yt
dt
k
=0
The solution y(t) consists of two parts – a particular
solution to (*) slide 123 plus a solution to
homogeneous differential equation.

131
Fine the relationship of input voltage vs
and output voltage
vc
in RC circuit
Example 1.8

132
By KVL, we have
vs t=vRtvct=Ritvct
it=C
dvct
dt
The current in this circuit is
Replace i(t)
vs t=RC
dvct
dt
vct
or
dvc t
dt

1
RC
vct=
1
RC
vs t

133
Example 2.14
Consider the system :
dyt
dt
2yt=xt
Find the solution of the system when the input
signal is
xt=Ke
3t
ut
where K is a real number :

134
The complete solution of the differential equation
consists of the sum of a particular solution yp
(t),
and a homogeneous solution, yh
(t).
yt=y pt yh t
yh
(t) is a solution of the homogeneous differential
equation
dyt
dt
2yt=0 (*)

135
y pt=Ye
3t
3Ye3t
2Ye3t
=Ke3t
The particular solution for an exponential input
signal is a signal of the same form as the input.
We hypothesize a solution for t > 0 of the form
Y is a number that we must determine.
Substituting yp
(t) and x(t) into the differential
equation yields.
3Y 2Y =K
Y =
K
5

136
y pt=
K
5
e
3t
, t0
yh t=Ae
st
so that
To determine yh
(t) ,We hypothesize a solution of
the form :
Substituting yh
(t) and x(t) into the differential
equation (*).
Ase
st
2 Ae
3t
=Ae
st
s2=0

137
yt=Ae
−2t

K
5
e
3t
, t0.
from this equation, we see that we must take
s =-2 and that Ae-2t
is a solution to eq. (*) for
any A. The solution of the differential
equation for t > 0 is :
from the condition of initial rest, at t = 0, y(t) = 0
0=A
K
5
A=−
K
5

138
yt=
K
5
[e
3t
−e
−2t
]
yt=
K
5
[e
3t
−e
−2t
]ut
Thus, for t > 0,
while for t<0, y(t) = 0, because of the condition
of initial rest.

139
Linear Constant-Coefficient Difference Equation
∑
k=0
N
ak y[n−k ]=∑
k=0
M
bk x[n−k ]
∑
k=0
N
ak y[n−k ]=0
A general Nth-order linear constant-coefficient
The solution y[n] consists of two parts – a particular
solution to (**) plus a solution to homogeneous
(**)

140
y[n]=
1
a0
{∑
k=0
M
bk x[n−k]−∑
k=1
N
ak y[n−k ]
}
y[n]=∑
k=0
M
bk
a0
x[n−k ]
The output at time n expresses in the term of previous
values of the input and output. (recursive equation)
When N = 0 (nonrecursive equation)

141
h[n]=
{
bn
a0,
0≤n≤M
0, otherwise
The impulse response of nonrecursive equation

142
Example 2.15
y[n]−
1
2
y[n−1]=x[n]
y[n]=x[n]
1
2
y[n−1]
Consider the difference equation
rewrite in the form
We need the previous value of the output, y[n-1], to
calculate the current value.

143
x[n]=K [n]
y[0]=x[0]
1
2
y[−1]=K
y[1]=x[1]
1
2
y[0]=
1
2
K
Consider the input
In this case, since x[n] = 0 for n ≤ -1, the condition
of initial rest implies that y[n] = 0 for n ≤ -1.

144
y[n]=x[n]
1
2
y[n−1]=1
2
n
K
h[n]=1
2
n
u[n]
y[2]=x[2]
1
2
y[1]=1
2
2
K
⋮ ⋮
The impulse response is

145
Problems
1 Consider a causal LTI system whose input
x(t) and output y(t) are related by the
differential equation
d yt
dt
4 yt=xt
The system also satisfies the condition of
initial rest.
a) If x(t) = e(-1+3j)t
u(t), what is y(t) ?

146
2 Consider a causal LTI system whose input x[n]
and output y[n] are related by the difference
equation
y[n]=x[n]
1
4
y[n−1]
Determine y[n] if x[n] = δ[n-1].

147
Block Diagram Representation of First-Order
Systems Described by Differential and Difference
Equations
An important property of systems described by linear
constant-coefficient difference and differential
equation is that they can be represented in term of
block diagram interconnections of elementary
operations.

148
2 The representations can be of considerable
value for simulation or implementation of the
system.
The advantage of the block diagram :
1 Provide a pictorial representation which can
add to out understanding of the behavior and
properties of these systems.

149
Basic Block Diagram
for discrete-time
systems

150
y[n]=−ay[n−1]bx[n]

151
Basic Block Diagram
Continuous-Time system

152
yt=−
1
a
dyt
dt

b
a
xt

153
Pictorial representation of an integrator
yt=∫−∞
t
[b x−ay]d 

154
Problems
1 Consider the cascade of the two systems
S1
and S2
, as depicted below :
S1
S2
y[n]x[n] w[n]
S1
: causal LTI
w[n]=
1
2
w[n−1]x[n]
S2
: causal LTI
y[n]= y[n−1] w[n]

155
The difference equation relating x[n] and y[n] is :
y[n]=−
1
2
y[n−2]
3
4
y[n−1]x[n]
a) Determine α and β.
b) Show the impulse response of the cascade
connection of S1
and S2

156
2.5 Singularity Functions
The Unit Impulse as an Idealized Short Pulse
xt=xt∗t
t=t∗t
The unit impulse δ(t) is the impulse of the identity
system
If we take x(t) = δ(t)

157
r t=t∗ t
Let δΔ
(t) is the rectangular pulse and let

158
Example 2.16
Consider the LTI system :
d yt
dt
2yt=xt
The response of this system to δΔ
(t), rΔ
(t), rΔ
(t)*δΔ
(t)
and rΔ
(t)*rΔ
(t) for several values of Δ.

159

160

161
From the figure, we
need a smaller
value of Δ in order
for the responses to
be the same as
impulse response.

162
Defining the Unit Impulse through Convolution
xt=xt∗t
From the point of view of linear systems analysis,
we may define the unit impulse as that signal
which, when applied to an LTI system, yields the
impulse response. We define δ(t) as the signal
for any x(t). In this sense, signals, such as, δΔ
(t),
rΔ
(t), etc, which correspond to short pulse with Δ→0,
all behave like a unit impulse.

163
All the properties of the unit impulse that we need
can be obtained from the operational definition.
Example : Let x(t) = 1 for all t, then
1=xt=xt∗t=t∗xt
=∫−∞
∞
 xt−d =∫−∞
∞
d 

164
Unit Doublets and Other Singularity Function
yt=
dxt
dt
dxt
dt
=xt∗u1t
Consider the LTI system :
The unit impulse response of this system is the
derivative of the unit impulse, which is called the unit
doublet u1
(t).
(*)

165
We take (*) in slide 147 as the operational
definition of u1
(t). We can define u2
(t), the
second derivative of δ(t), as the impulse response of
an LTI system :
d
2
xt
dt
2
=xt∗u2t
We see that
d
2
xt
dt
2
=
d
dt d xt
dt =xt∗u1t∗u1t

166
u2t=u1t∗u1t
uk t=u1 t∗⋯∗u1 t
k times
therefore
For kth derivative of δ(t), the impulse response is
the kth derivative of the input and the system can
be obtain as:

167
For example, Consider the short pulse δΔ
(t).
This pulse behaves like an impulse as Δ→0.
Consequently, we would expect its derivative to
behave like a doublet as Δ→0.

168
yt=∫−∞
t
xd 
ut=∫−∞
t
d 
xt∗ut=∫−∞
t
xd 
Consider the system
Let δ(t) is the input to the system, the unit step is the
impulse response of this system
We have the operational definition of u(t) :

169
u−2t=ut∗ut=∫−∞
t
ud 
u−2t=tut
We define the system that consist of a cascade
of two integrators

170
u−k t=ut∗⋯∗ut
k times
=∫−∞
t
u−k−1d 
We can define higher order integrals of as the
impulse response of cascades of integrators :

171

Chapter2 - Linear Time-Invariant System

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