This document describes analysis of linear time-invariant (LTI) systems through impulse response and convolution. It defines LTI systems and explains that their behavior is characterized by impulse response. The input and output of an LTI system are related by convolution. Impulse response is the output of an LTI system when the input is a unit impulse. Given impulse response, the output can be found for any input using convolution. Difference equations are also used to represent LTI systems. The document provides examples of finding impulse responses and performing convolution in MATLAB. It discusses properties of convolution like associativity and commutativity. Tasks are given on finding impulse responses and convolving various signals.

1. Lab No.07
Analysis of LTI systems (Impulse
response, convolution)
Designed by : Dawar Awan
dawar@cecos.edu.pk
CECOS College of Engineering and IT
March – July 2012

2. LTI systems
 Systems that are both linear and time invariant are
called LTI systems.
 The behavior of LTI systems is completely characterized
by their impulse response.
 The input and output of an LTI system is related by
convolution sum/integral.
CECOS College of Engineering and IT
March – July 2012

3. Impulse response
 Impulse response ‘ h[n] ’, is the output of an LTI system,
when the input is a unit impulse.
 Given the impulse response, we can find the output for
any input using convolution.
CECOS College of Engineering and IT
March – July 2012

4. Difference equation
 A very common representation of LTI systems is in the
form of difference equation
 The general difference equation is
K=M
K=N
K=0
K=0
∑ ak y[n-k] = ∑ bk x[n-k]
 example : for , y[n] – 5/6y[n-1] + 1/6y[n-2] = 1/3x[n-1]
a0=1, a1=-5/6, a2=1/6,
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and bo=0, b1=1/3
March – July 2012

5. Impulse response
 Plot the impulse response of the following difference
equation .
y[n] – 5/6y[n-1] + 1/6y[n-2] = 1/3x[n-1]
CECOS College of Engineering and IT
March – July 2012

6. Impulse response
CECOS College of Engineering and IT
March – July 2012

7. Practice
1- Plot the impulse response of the following systems
I. y[n] = x[n] + y[n-1]
(Accumulator)
II. y[n] – y[n-1] = 1/7 (x[n] + x[n-1] + x[n-2] + x[n-3]
+ x[n-4] + x[n-5] + x[n-6])
(Moving average system)

8. Convolution
 y[n] = x[n] * h[n] =
 here y[n] is the output signal, x[n] is the input signal,
and h[n] is the impulse response of the LTI system.
 in MATLAB use the instruction ‘ y=conv(x,h) ‘ to perform
convolution.
CECOS College of Engineering and IT
March – July 2012

9. Convolution
 Convolve the following two sequences in MATLAB.
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March – July 2012

10. Convolution
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March – July 2012

11. Convolution
NOTE
 MATLAB assumes that both the convolving signals are
starting from zero index, hence the time/sample no. of
the output signal is not correct always.
 You have to adjust the time axis of the output signal
keeping in view some rules related to convolution.
CECOS College of Engineering and IT
March – July 2012

12. Convolution
Hint :
length of y = length of (x) + length of (h) - 1 and the starting
index for y will be the sum of starting indices of x and h
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March – July 2012

13. Convolution (without time adjustment)
*
=
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March – July 2012

14. Convolution (after adjusting time)
*
=
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March – July 2012

15. Task
1- Convolution is associative. Given the three signals x1[n], x2[n], and
x3[n] as:
x1= [ 3,1,1]
x2= [ 4,2,1]
x3= [ 3,2,1,2,3]
Show that (x1*x2)*x3 = x1*(x2*x3)
2- Convolution is commutative. Given x and as:
x=[1,3,2,1]
h=[1,1,2]
Show that x* h = h*x
3- Determine the output of the LTI system with:
h[n] = 2δ[n] + δ[n‐1] + 2δ[n‐2] + 4δ[n‐3] + 3δ[n‐4]
x[n] = δ[n] + 4δ[n‐1] +3δ[n‐2] + 2δ[n‐3]
CECOS College of Engineering and IT
March – July 2012

16. Task
4- Convolve the signal x=[1,2,3,4,5,6] with an impulse delayed by two
samples. Plot the original signal and the result of convolution.
5- Find the output signal y[n] for any range of ‘n’ using convolution,
for the following cases.
a) h[n] = 5(-1/2)n u[n]
b) h[n] = (0.5)2n u[-n]
c) h[n] = 2n u[-n-1]
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,
,
,
x[n] = (1/3)n u[n]
x[n] = u[n]
x[n] = u[n] – u[n - 10]
March – July 2012