Lab no.07
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Transcript of Lab no.07
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Designed by : Dawar [email protected]
CECOS College of Engineering and IT March – July 2012
Lab No.07
Analysis of LTI systems (Impulse response, convolution)
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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
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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.
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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
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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]
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CECOS College of Engineering and IT March – July 2012
Impulse response
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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)
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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.
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CECOS College of Engineering and IT March – July 2012
Convolution
Convolve the following two sequences in MATLAB.
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CECOS College of Engineering and IT March – July 2012
Convolution
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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
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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
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CECOS College of Engineering and IT March – July 2012
Convolution (without time adjustment)
*
=
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CECOS College of Engineering and IT March – July 2012
Convolution (after adjusting time)
*
=
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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]
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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]