In the name of Allah
Chapter 2
First, we will see the syntax of three main functions that will be used in this
exercise called:
conv
filter
lsim
conv:
The MATLAB function
y[n] =
conv computes the sum of:
+∞
∑ h[k ] x[n − k ]
k = −∞
Let x[ n ] and h[ n ] be discrete finite-length sequences. If x[ n ] is
nonzero only on the interval of [ sx, fx ] and h[ n ] is nonzero only on
the interval of[ sh, fh ], then LX = length( x ) = ( fx – sx + 1) and LH =
length( h ) = ( fh – sh + 1).You can easily verify that y[ n ] can be
nonzero only on the interval of [ sx + sh, fx + fh ] and LY = length( y )
= LX + LH – 1. The syntax of the command is y = conv( x, h ) where x
and y are the vectors containing the values of the two signals x
and h. conv does not return the indices of the y[ n ]; which makes
sense because the interval of x and h are not inputted to conv.
Example:
Suppose that x[ n ] and y[ n ] are as shown in the following figures :
We know that the x[ n ] * y[ n ] must be:
In MATLAB this is implemented as follow:
x = [ 1, 2, 3 ]
y = [ 1, 2 ]
conv( x, y ) = [ 1, 4, 7, 6 ]
Note that the indices are not inputted and they could be
implemented by another vector which holds the indices.
filter:
The filter command computes the output of a causal, LTI system for a given
Input when the system is specified by a linear constant-coefficient difference
Equation as y = filter (b, a, x). The vectors a and b contain the coefficients of
y and x in the equation. The vector x contains the value of the input signal.
As like the conv function, the indices of x is not specified (note that the
system is causal and time invariant ). The output y, has the same size as x[n].
This is clear, because for determining y in n = n0, usually, x[ n = n0] should
be specified.
Example:
Y[ n ] = x[ n ] + 2x[ n –1 ] then a = [ 1 ], b = [ 1, 2 ], now if x = [ 1 ],
then y = [ 1 ] but if x = [ 1, 0, 0 ] then y = [1, 2, 0 ].
lsim:
The function lsim can be used to simulate the output of continuous-time,
causal LTI systems described by linear constant-coefficient differential
equations as y = lsim( b, a, x, t ). The vectors a and b contain the coefficients
of y and x in the equation. The t and x have the same size, the vector t contains
the time samples for the input and output, x contains the values of the
input x( t ) at each time in t, and y contains the simulated values of the
output y( t ) at each time in t. The accuracy of the simulated values
depends upon how well x and t represent the true function x(t).
Hint:
Basically, lsim interpolates the pair t, x in much the same way as does plot.
For instance, consider the plot produced by the following code:
>> t = [0, 1, 2, 5, 8, 9, 10];
>> x = [0, 0, 0, 3, 0, 0, 0];
>> plot(t, x)
The function lsim( b, a, x, t ) will consider x( t ) to be on the interval of 0 ≤ t ≤ 10
equal to
Example:
y′( t ) + ½ y( t ) = x( t ) then a = [ 1 0.5 ], b = [ 1 ]. Now we want to
simulate the step response of this system. t = [ 0:10 ], x = ones( 1, 11 ).
y = lsim( b, a, x, t ). Note that in this special case, at each value of t,
the step response computed by lsim is essentially identical to the true
step response because the input signal is precise.
☺see also:
impulse(b, a) and
step(b, a).
Computer-based exercises:
1- Commutative property: Interchange the INPUT and INPULSE RESPONSE
you know that x[n]*h[n] = h[n]*x[n] by commutative property. That
means the input and impulse response signals can be interchanged, so
h can be supposed to be input and x be impulse response. First argue for
yourself that x[ n ] * h[ n – n0 ] = x[ n – n0 ] * h[ n ]. After that show this
equality for these signals, n0 = 2 :
1 n = 0
−1 n = 2

h[ n] = 
n=4
3
0
otherwise
3

x[ n] = 1
0

n =1
n=2
otherwise
Use subplot and stem to plot both convolutions in a single figure.
Name your mfile: p2_1.m
2- Consider an input x[ n ] and impulse response h[ n ] given by:
x[ n ] = ((½) ^ ( n – 2 ) ) u[ n – 2 ]
h[ n ] = u [ n + 2 ]
If you wish to compute the x[n]*h[n] using conv, you must deal
appropriately with the infinite length of both x[ n ] and h[ n ].
Assume x[ n ] is zero for n ≥ 25 and h[ n ] is zero for n ≥ 14 and
now compute x[n]*h[n]. Argue that only a portion of the output of
conv will be valid. Specify which values in the output are valid and
which are not.
Use stem to plot the convolution and your program should give the
interval in which values are valid as output. Name your mfile: p2_2.m
3- Use filter function to solve the exercise No. 2-32 in the textbook.
Name your mfile: p2_3.m
4- The pulse response of the continuous-time LTI systems
You know that the impulse function can be approximated by a pulse
with width ∆ and height 1 / ∆. Think about an RC circuit which R = 1K,
and C = 1mf ( i.e. τ = 1 ) connected in series with a voltage source.
Output is the voltage of Capacitor. Use impulse function to get impulse
response, and for ∆ = 0.1 and 0.2 Get the pulse response with lsim function
and plot all three response on one figure using subplot. Perhaps functions
zeros and ones can be useful. Name your mfile: p2_4.m
Deadline: 82/8/11
No paper is required for computer exercises. Just Zip them and
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