Designed by : Dawar Awandawar@cecos.edu.pk

CECOS College of Engineering and IT March – July 2012

Lab No.07

Analysis of LTI systems (Impulse response, convolution)

CECOS College of Engineering and IT March – July 2012

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.

LTI systems

CECOS College of Engineering and IT March – July 2012

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

Difference equation

A very common representation of LTI systems is in the

form of difference equation

The general difference equation is

∑ 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, and bo=0, b1=1/3

K=0

K=NK=M

K=0

CECOS College of Engineering and IT March – July 2012

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

Impulse response

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)

CECOS College of Engineering and IT March – July 2012

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

Convolution

Convolve the following two sequences in MATLAB.

CECOS College of Engineering and IT March – July 2012

Convolution

CECOS College of Engineering and IT March – July 2012

Convolution

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.

NOTE

CECOS College of Engineering and IT March – July 2012

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

CECOS College of Engineering and IT March – July 2012

Convolution (without time adjustment)

*

=

CECOS College of Engineering and IT March – July 2012

Convolution (after adjusting time)

*

=

CECOS College of Engineering and IT March – July 2012

Task1- 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

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] , x[n] = (1/3)n u[n] b) h[n] = (0.5)2n u[-n] , x[n] = u[n] c) h[n] = 2n u[-n-1] , x[n] = u[n] – u[n - 10]