DIGITAL SIGNAL
PROCESSING
Vedat Tavşanoğlu
Response of LTI Systems
The linear operation T[ . ] on a function x(n), i.e.T[x(n)] is
defined as in the following :
(a) T[(n)] = h(n) where (n) is the unit-impulse function;
(b) If T[x1(n)] = y1(n)
and
T[x1(n)] = y1(n)
then
T[a1 x1(n)+a2 x2(n)] = a1 y1(n)+a2 y2(n) (linearity)
Now define the time-invariant (shift-invariant) operation T[.]:
(c) If T[x(n)]=y(n) then T[x(n-k)]=y(n-k) (time-invariance).
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DIGITAL SIGNAL PROCESSING
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Response of LTI Systems
Since any signal x(n) can be expressed in terms of
the unit sample (n) as
x ( n) 

 x(k ) (n  k )
k  
The following are true for an LTI system:
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DIGITAL SIGNAL PROCESSING
3
Response of LTI Systems
 (n)
 (n  k )
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T[.]
T[.]
h(n)  T [ (n)]
h(n  k )  T [ (n  k )]
DIGITAL SIGNAL PROCESSING
Time invariance
4
Response of LTI Systems
T [ x(k ) (n  k )]  x(k )T [ (n  k )]
x(k ) (n  k )
Linearity (scaling)
T[.]
 x ( k ) h( n  k )
Time invariance

T [  x(k ) (n  k )] 

 x(k ) (n  k )
k  
T[.]
k  

 T [ x(k ) (n  k )]
k  
Linearity (additivity)

 x(k )T [ (n  k )]
k  
Linearity (scaling)


 x ( k ) h( n  k )
k  
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DIGITAL SIGNAL PROCESSING
Time invariance
5
Response of LTI Systems
This expression is called the Convolution Sum. With a simple
change of variable we obtain the following:


k  
k  
y( n ) 
 x( k )h( n  k )   x( n  k )h( k )
Given an input x(n) and h(n), the response of the system
to an impulse (n), the above expression tells us how to
evaluate the output y(n). If x(n) and h(n) are of finite
duration then:
length of y(n)= length of x(n)+ length of h(n)-1
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DIGITAL SIGNAL PROCESSING
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Response of LTI Systems
In the case where
x(n)  h(n)  0 for n  0
the lower and upper limits for summation are, respectively, k=0
and n. This can be shown as follows:
Since x(n)=h(n)=0 for n<0 it is evident from the convolution
sum formula that k must satisfy the following:
k 0,nk 0
which yields
k 0,k n
0k n
hence the following:
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DIGITAL SIGNAL PROCESSING
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Response of LTI Systems
n
n
k 0
k 0
y ( n)   x ( k ) h( n  k )   x ( n  k ) h( k )
We can state the following:
The smallest value of the lower limit in the first expression
is zero and the geatest value of the upper limit in the second
expression is n because:
x ( n)  0
for
n0
The geatest value of the upper limit in the first expression is
n and the smallest value of the lower limit in the second
expression is zero because
h( n)  0
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for
DIGITAL SIGNAL PROCESSING
n0
8
Response of LTI Systems

Example:
x(n)
1
Find the output y(n) for the system given below.
1
 
-1
0
1
2
n
h(n)

-2 -1
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
1

1/2
1/4
0
1
2

DIGITAL SIGNAL PROCESSING
0
3
4
y(n)
n
5
9
Response of LTI Systems
Length of x(n)=2, length of h(n)=3 therefore length of y(n)=2+3-1=4
Using the first expression of the convolution sum we should work out the
following four values :
0
y (0)   x(k )h(0  k )  x(0)h(0)  1x1  1
k 0
1
1
1
y (1)   x(k )h(1  k )  x(0)h(1)  x(1)h(0)  1x  1x1  1
2
2
k 0
1
1
y (2)   x(k )h(2  k )  x(0)h(2)  x(1)h(1)  1x1 / 4  1x1  1
4
k 0
1
1 1
y (3)   x(k )h(3  k )  x(1)h(2)  1x 
4 4
k 1
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DIGITAL SIGNAL PROCESSING
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Response of LTI Systems

 

 
x(k)
1
1
k
-1
h(k)
0
1
1
2
1/2
1/4
0
-2 -1
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0
1
2
3
4
k
5
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
Response of LTI Systems
x(k)
1
 


 

h(-k)
0

1
1
2
y(0)=1
k
 

1/2

0
1
2
n
1/4
0
-2 -1
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0
1
2
k
3
4
5
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
Response of LTI Systems
x(k)
1
 


 

0
h(1-k)

1
1
2
y(1)=1.5
k

1/2

 
0
1
2
n
1/4
0
-2 -1
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0

2
k
3
4
5
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
Response of LTI Systems
x(k)
1
1
 


 

0
1
h(2-k)

y(2)=0.75
k
2

1/2

 
0
1
2
3
n
1/4
0
-2
-1
0
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
2
3
k
4
5
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
Response of LTI Systems
x(k)
1
1
 



 

0
1
2
h(3-k)
k

 
y(3)=0.25

1/2

0
1
2
n
3
1/4
-1
0
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
2
3
k
4
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
Response of LTI Systems
x(k)
1
1
 



 

0
1
k
2

h(4-k)
  
y(4)=0
1/2

0
1
2
3
n
4
1/4
-1
0
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
2
3
k
4
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16
Response of LTI Systems
y (n)   (n)  1.5 (n  1)  0.75 (n  2)  0.25 (n  3)

 
1.5
y(n)
1

-2 -1
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3/4

1/4
0
1
2
3
4
n
5
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Response of LTI Systems
MATLAB program and plots of x(n),h(n) and y(n)
close all
x=[1 1 0 0 0];
h=[1 1/2 1/4 0 0];
y=conv(x,h);
stem(x);
title('x(n)');
figure,stem(h);
title('h(n)');
figure,stem(y(1:5));
title('y(n)');
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DIGITAL SIGNAL PROCESSING
18
Frequency Response of LTI
Systems
In the following we will find the response of an LTI
system to the sinusoidal input signal
x(n)  A cos( 0 n   )
x(n)  A cos( 0 n   )
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h(n)
DIGITAL SIGNAL PROCESSING
y(n)
19
Frequency Response of LTI
Systems
x(n) can be written as:
A j ( 0 n  ) A  j ( 0 n  )
x ( n)  e
 e
2
2
A j j0n
A  j  j0n
 ( e )e
 ( e )e
2
2
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DIGITAL SIGNAL PROCESSING
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Frequency Response of LTI
Systems
x(n)  e
j0n
*  j0n
 e
where * is complex-congugate of . We can further write
x(n)  x1 (n)   x2 (n)
*
Now let us first compute the response to x1(n).
Using the convolution sum,we can write:
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DIGITAL SIGNAL PROCESSING
21
Frequency Response of LTI
Systems
y1 (n) 


 x ( n  k ) h( k )   h( k ) e
k  

1
 [  h( k ) e
j ( n  k )
k  
 j k
]e
j n
k  
Now defining
j
H (e ) 

 h( k ) e
 j k
k  
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DIGITAL SIGNAL PROCESSING
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Frequency Response of LTI
Systems
which is the Fourier transform of the system impulse
response h(n), yields
j
y1 (n)  H (e ) e
jn
It is evident from above that the response y2(n) to the
input x2(n) can be obtained by replacing  by -:
y 2 ( n )  H (e
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 j
)e
DIGITAL SIGNAL PROCESSING
 jn
23
Frequency Response of LTI
Systems
On the other hand since h(k) are real-valued we can write:
H (e
 j

 h( k ) e
)
jk
j
 H (e )
*
k  
where ( )* denotes complex-congugate.
Now express H(ej) in the polar form:
j
j
H (e ) | H (e ) | e
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DIGITAL SIGNAL PROCESSING
j  ( )
24
Frequency Response of LTI
Systems
Also considering that:
| H (e
 j
j
) || H (e ) |
 ( )   ( )
we obtain
j
j (n  ( ))
j
 j (n  ( ))
y1 (n) | H (e ) | e
y2 (n) | H (e ) | e
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DIGITAL SIGNAL PROCESSING
25
Frequency Response of LTI
Systems
Since the system is linear the response to
x(n)  x1 (n)   x2 (n)
*
is given as:
y (n)  y1 (n)   y2 (n)
*
A
j
j j (n  ( ))
 j  j (n  ( ))
y ( n )  | H (e ) | [e e
e e
]
2
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DIGITAL SIGNAL PROCESSING
26
Frequency Response of LTI
Systems
j
y (n)  A | H (e ) |[cos(n     ( ))]
Example: An LTI system is given by the impulse response
h(n)  {0.5, 1, 0.5}
The input to this system is given by:

2
x(n)  cos( n)  cos( n  )  cos( n  )  cos( n)
6
3
6
2
3
3
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

DIGITAL SIGNAL PROCESSING


27
Frequency Response of LTI
Systems
close all
close all
n=0:1:24
x1=cos((pi/6)*n);
x2=cos((pi/3)*n-(pi/6));
x3=cos((pi/2)*n);
x4=cos((2*pi/3)*n+(pi/6));
x=x1+x2+x3+x4;
stem(x(1:25));
h=[1/2 1 1/2];
y1=conv(x1,h);
figure,stem(y1(1:25));
y2=conv(x2,h);
figure,stem(y2(1:25));
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y3=conv(x3,h);
figure,stem(y3(1:25));
y4=conv(x4,h);
figure,stem(y4(1:25));
y=conv(x,h);
figure,stem(y(1:25));
fx=fft(x,101);
fy=fft(y,101);
fx=abs(fx);
fy=abs(fy);
f=0:.02*pi:2*pi;
figure,plot(f,fx);
figure,plot(f,fy);
DIGITAL SIGNAL PROCESSING
28
Difference Equations
Consider the following digital filter which is a realization
of the difference equation:
y (n)  ay (n  1)  bx(n)
x(n)
The hardware implementation
of this equation will contain
two registers,namely Ry and Ra
to store the past output y(n-1)
and the multiplier coefficient a,
respectively.
b
T
+
y(n)
a
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DIGITAL SIGNAL PROCESSING
29
Difference Equations
MATLAB program implementing the above difference
equation and the plot of output y(n) for x(n)=0.
close all
y=1
y1=0
a=.5
x=0
for k = 1:10
y=a*y+x
x1(k)=x;
y1(k)=y;
end
figure,stem(y1),
title('y(n)')
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DIGITAL SIGNAL PROCESSING
30
Difference Equations
MATLAB program and the plot of output y(n) for x(n)=u(n).
close all
y=0
y1=0
a=.5
x=1
for k = 1:10
y=a*y+x
x1(k)=x;
y1(k)=y;
end
figure,stem(x1),title('x(n)')
figure,stem(y1),title('y(n)')
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DIGITAL SIGNAL PROCESSING
31
Difference Equations
MATLAB program and the plot of output y(n)
for x(n)= cos((pi/6)*n)
y=0
a=.5
for k = 1:24
x=cos((pi/6)*k);
y=a*y+x;
y1(k)=y;
end
figure,stem(y1),title('y(n)')
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DIGITAL SIGNAL PROCESSING
32
Difference Equations
Example: Suppose a person opens a morgage account
with an annual interest rate of 0.04% and borrows
$100,000 to be paid back in monthly instalments in a
period of 25 years.
(a)Find the necessary amount for monthly morgage
payments.
(a)Write down the difference equation for the remaining
morgage amount.
(a) For the monthly morgage payments x we can write:
(1  x)12  1  0.04
x  (1  .04)
1 / 12
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 1  0.00327374
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33
Difference Equations
MATLAB program for the remaining morgage
clear all
in(1)=100000;
i=1;
x=523.896;
y(1)=100000;
while i<320
in(i+1)=x;
y(i+1)=1.00327374*y(i)-in(i+1);
i=i+1;
%pause
end
stem(y)
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DIGITAL SIGNAL PROCESSING
34
Difference Equations
x(n)
T
a2
b0
b1
b2
T
+
y(n)
a1
y (n)  b0 x(n)  b1 x(n  1)  b2 x(n  2)  a1 y (n  1)  a2 y (n  2)
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DIGITAL SIGNAL PROCESSING
35
Difference Equations
The general form for an Mth-order linear constant-coefficient
difference equation can be given as follows:
N
M
i 0
i 0
y (n)   bi x(n  1)  ai y (n  1)
Case 1: ai=0, the system is called NONRECURSIVE or FINITE IMPULSE
RESPONSE (FIR) or FEEDFORWARD SYSTEM.
Case 2: ai=0 for at least one value of i, the system is called RECURSIVE or
INFINITE IMPULSE RESPONSE (IIR) or FEEDBACK SYSTEM.
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DIGITAL SIGNAL PROCESSING
36
Digital Filters
close all
%Plotting of continuous-time (CT) signals
t=0:1/50:1;
xa=(exp).^t+cos(2*pi*6*t);
plot(t(1:51),xa(1:51));
title(‘xa(t)=(exp).^t+cos(2*pi*6*t)');
xa1=(exp).^t;
figure,plot(t(1:51),xa1(1:51));
title(‘xa1(t)=(exp).^t)');
xa2=cos(2*pi*6*t);
figure,plot(t(1:51),xa2(1:51));
title(‘xa2(t)=cos(2*pi*6*t)');
%Sampling of CT and plotting of discrete-time signals
t=0:1/50:1;
figure,stem(x);
title(‘x(n)=(1.02)^n+cos(2pi/8)n)');
figure,stem(x1);
title(‘x1(n)=(1.02)^n)');
figure,stem(x2);
title(‘x2(n)=cos(2pi/8)n)');
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DIGITAL SIGNAL PROCESSING
37
Digital Filters
%Running Average Filter Impulse Response h(n)
h=(1/7)*[1 1 1 1 1 1 1];
%Plotting of h(n)
figure,stem(h)
title('h(n)=(u(n)-u(n-8))');
%Application of h(n) to the input x(n),namely conv of h(n) with x(n)
y=conv(h,x);
%Plotting of the output y(n)
figure,stem(y);
title('y(n)');
%Fourier Transforms of x(n),h(n) and y(n)
fx=fft(x,201);
fx=abs(fx);
fh=fft(h,201);
fh=abs(fh);
fy=fft(y,201);
fy=abs(fy);
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DIGITAL SIGNAL PROCESSING
38
Digital Filters
%Plotting of the Magnitudes of Fourier Transforms
f=0:pi/100:2*pi;
figure,plot(f,fx);
title('Fourier Transform of the Input x(n), the Frequency Spectrum of the
Input Signal');
figure,plot(f,fh);
title('Fourier Transform of h(n), the Frequency response of Filter');
figure,plot(f,fy);
title('Fourier Transform of the Output y(n),the Frequency Spectrum of the
Output Signal');
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DIGITAL SIGNAL PROCESSING
39
Digital Filters
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DIGITAL SIGNAL PROCESSING
40
Digital Filters
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DIGITAL SIGNAL PROCESSING
41
Digital Filters
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Digital Filters
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DIGITAL SIGNAL PROCESSING
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