Scilab Manual for

Digital Signal Processing

by Mr Vijay P Sompur
Electronics Engineering
Visvesvraya Technological University1

Solutions provided by
Mr. R.Senthilkumar- Assistant Professor
Electronics Engineering
Institute of Road and Transport Technology

June 26, 2020

1 Funded by a grant from the National Mission on Education through ICT,

http://spoken-tutorial.org/NMEICT-Intro. This Scilab Manual and Scilab codes
written in it can be downloaded from the ”Migrated Labs” section at the website
http://scilab.in
1
Contents

List of Scilab Solutions 4

1 Verification of Sampling theorem. 6

2 Impulse response of a given system 11

3 Linear Circular convolution of two given sequences 14

4 Autocorrelation of a given sequence and verification of its

properties. 18

5 Cross correlation of given sequences and verification of its

properties. 21

6 Solving a given difference equation. 24

7 Computation of N point DFT of a given sequence and to

plot magnitude and phase spectrum. 26

8 Linear convolution of two sequences using DFT and IDFT. 30

9 Circular convolution of two given sequences using DFT and

IDFT 33

10 Design and implementation of FIR filter to meet given spec-

ifications. 35

11 Design and implementation of IIR filter to meet given spec-

ifications. 39

2
12 Circular convolution of two given sequences 49

3
List of Experiments

Solution 1.1 Verification of Sampling Theorem . . . . . . . . . 6

Solution 2.2 Program to find impulse response and Frequency
Response of a system . . . . . . . . . . . . . . . . 11
Solution 3.1 Program to Compute the Convolution of Two Se-
quences . . . . . . . . . . . . . . . . . . . . . . . 14
Solution 4.1 Program to Compute the Autocorrelation of a Se-
quence And verfication of Autocorrelation property 18
Solution 5.1 Program to Compute the Crosscorrelation of a Se-
quence And verfication of crosscorrelation property 21
Solution 6.1 Solving Difference Equation Direct Form II Real-
ization . . . . . . . . . . . . . . . . . . . . . . . . 24
Solution 7.1 Program to find the spectral information of discrete
time signal Calculation of DFT and IDFT . . . . 26
Solution 8.1 Linear Convolution using Circular Convolution DFT
IDFT method . . . . . . . . . . . . . . . . . . . . 30
Solution 9.1 Circular Convolution using DFT IDFT method . 33
Solution 10.1 To Design an Low Pass FIR Filter . . . . . . . . . 35
Solution 11.1 To obtain Digital IIR Butterworth low pass filter
Frequency response . . . . . . . . . . . . . . . . . 39
Solution 11.2 To obtain Digital IIR Chebyshev low pass filter Fre-
quency response . . . . . . . . . . . . . . . . . . . 43
Solution 12.1 Program to perform circular convolution of two se-
quences . . . . . . . . . . . . . . . . . . . . . . . 49

4
List of Figures

1.1 Verification of Sampling Theorem . . . . . . . . . . . . . . . 9

1.2 Verification of Sampling Theorem . . . . . . . . . . . . . . . 10

2.1 Program to find impulse response and Frequency Response of

a system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3.1 Program to Compute the Convolution of Two Sequences . . 15

7.1 Program to find the spectral information of discrete time signal

Calculation of DFT and IDFT . . . . . . . . . . . . . . . . . 27

8.1 Linear Convolution using Circular Convolution DFT IDFT

method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

10.1 To Design an Low Pass FIR Filter . . . . . . . . . . . . . . . 36

11.1 To obtain Digital IIR Butterworth low pass filter Frequency

response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
11.2 To obtain Digital IIR Chebyshev low pass filter Frequency
response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

5
 Experiment: 1

Verification of Sampling
theorem.

Scilab code Solution 1.1 Verification of Sampling Theorem

1 // C a p t i o n : V e r i f i c a t i o n o f S a m p l i n g Theorem
2 // [ 1 ] . R i g h t S a m p l i n g [ 2 ] . Under S a m p l i n g [ 3 ] . Over
Sampling
3 clc ;
4 close ;
5 clear ;
6 fm = input ( ’ E n t e r t h e i n p u t s i g n a l f r e q u e n c y : ’ ) ;
7 k = input ( ’ E n t e r t h e number o f C y c l e s o f i n p u t s i g n a l :
’ );
8 A = input ( ’ E n t e r t h e a m p l i t u d e o f i n p u t s i g n a l : ’ ) ;
9 tm =0:1/( fm * fm ) : k / fm ;
10 x = A * cos (2* %pi * fm * tm ) ;
11 figure (1) ;
12 a = gca () ;
13 a . x_location = ” o r i g i n ” ;
14 a . y_location = ” o r i g i n ” ;
15 plot ( tm , x ) ;
16 title ( ’ ORIGINAL SIGNAL ’ ) ;
17 xlabel ( ’ Time ’ ) ;

6
18 ylabel ( ’ A m p l i t u d e ’ ) ;
19 xgrid (1)
20 // S a m p l i n g Rate ( N y q u i s t Rate ) =2∗fm
21 fnyq =2* fm ;
22 // UNDER SAMPLING
23 fs =(3/4) * fnyq ;
24 n =0:1/ fs : k / fm ;
25 xn = A * cos (2* %pi * fm * n ) ;
26 figure (2) ;
27 a = gca () ;
28 a . x_location = ” o r i g i n ” ;
29 a . y_location = ” o r i g i n ” ;
30 plot2d3 ( ’ gnn ’ ,n , xn ) ;
31 plot (n , xn , ’ r ’ ) ;
32 title ( ’ Under S a m p l i n g ’ ) ;
33 xlabel ( ’ Time ’ ) ;
34 ylabel ( ’ A m p l i t u d e ’ ) ;
35 legend ( ’ Sampled S i g n a l ’ , ’ R e c o n s t r u c t e d S i g n a l ’ ) ;
36 xgrid (1)
37 //NYQUIST SAMPLING
38 fs = fnyq ;
39 n =0:1/ fs : k / fm ;
40 xn = A * cos (2* %pi * fm * n ) ;
41 figure (3) ;
42 a = gca () ;
43 a . x_location = ” o r i g i n ” ;
44 a . y_location = ” o r i g i n ” ;
45 plot2d3 ( ’ gnn ’ ,n , xn ) ;
46 plot (n , xn , ’ r ’ ) ;
47 title ( ’ N y q u i s t S a m p l i n g ’ ) ;
48 xlabel ( ’ Time ’ ) ;
49 ylabel ( ’ A m p l i t u d e ’ ) ;
50 legend ( ’ Sampled S i g n a l ’ , ’ R e c o n s t r u c t e d S i g n a l ’ ) ;
51 xgrid (1)
52 //OVER SAMPLING
53 fs = fnyq *10;
54 n =0:1/ fs : k / fm ;
55 xn = A * cos (2* %pi * fm * n ) ;

7
56 figure (4) ;
57 a = gca () ;
58 a . x_location = ” o r i g i n ” ;
59 a . y_location = ” o r i g i n ” ;
60 plot2d3 ( ’ gnn ’ ,n , xn ) ;
61 plot (n , xn , ’ r ’ ) ;
62 title ( ’ Over S a m p l i n g ’ ) ;
63 xlabel ( ’ Time ’ ) ;
64 ylabel ( ’ A m p l i t u d e ’ ) ;
65 legend ( ’ Sampled S i g n a l ’ , ’ R e c o n s t r u c t e d S i g n a l ’ ) ;
66 xgrid (1)
67 // R e s u l t
68 // E n t e r t h e i n p u t s i g n a l f r e q u e n c y : 1 0 0
69 //
70 // E n t e r t h e number o f C y c l e s o f i n p u t s i g n a l : 2
71 //
72 // E n t e r t h e a m p l i t u d e o f i n p u t s i g n a l : 2

8
Figure 1.1: Verification of Sampling Theorem

9
Figure 1.2: Verification of Sampling Theorem

10
 Experiment: 2

Impulse response of a given

system

Scilab code Solution 2.2 Program to find impulse response and Frequency
Response of a system

1 // C a p t i o n : Program t o f i n d i m p u l s e r e s p o n s e and
2 // F r e q u e n c y R e s p o n s e o f a s y s t e m
3 // y [ n ] = a ∗ y [ n−1]+x [ n ]
4 // Assume y [ n ] = h [ n ] , x [ n ]= d e l t a [ n ]= u n i t i m p u l s e
response
5 // a = 0 . 9
6 // h [ n ] = 0 . 9 ∗ h [ n−1]+ d e l t a [ n ]
7 clc ;
8 clear ;
9 close ;
10 a = 0.9; // c o n s t a n t a = 0 . 9 l e s s t h a n 1
11 h0 = 1;
12 h1 = a ; // f i r s t two v a l u e s o f i m p u l s e r e s p o n s e
13 h = [ h0 , h1 , zeros (1 ,100) ];
14 for i = 1:100
15 h ( i +2) = (( a ) ^( i +1) ) * h ( i +1) ; // i m p u l s e r e s p o n s e

11
Figure 2.1: Program to find impulse response and Frequency Response of a
system

12
16 end
17 [ HW , W ] = frmag (h ,512) ; // f r e q u e n c y r e s p o n s e
18 figure (1)
19 subplot (2 ,1 ,1)
20 a = gca () ;
21 a . x_location = ’ o r i g i n ’ ;
22 a . y_location = ’ o r i g i n ’ ;
23 plot ([1: length ( h ) ] ,h , ’ r ’ ) ;
24 xlabel ( ’ D i s c r e t e Time I n d e x n−−−−> ’ ) ;
25 ylabel ( ’ I m p u l s e R e s p o n s e h [ n]−−−−−> ’ ) ;
26 title ( ’ I m p u l s e R e s p o n s e o f f i r s t o r d e r r e c u r s i v e
system ’ )
27 xgrid (1)
28 subplot (2 ,1 ,2)
29 a = gca () ;
30 a . x_location = ’ o r i g i n ’ ;
31 a . y_location = ’ o r i g i n ’ ;
32 plot ([ mtlb_fliplr ( -2* %pi * W ) ,2* %pi * W (2: $ ) ] ,[
mtlb_fliplr ( abs ( HW ) ) , abs ( HW (2: $ ) ) ])
33 xlabel ( ’ D i s c r e t e F r e q u e n c y i n d e x W−−−−−−> ’ )
34 ylabel ( ’ Magnitude R e s p o n s e | H(W)|−−−−−−> ’ )
35 title ( ’ F r e q u e n c y R e s p o n s e o f a c a u s a l , s t a b l e , LTI
I s t Order R e c u r s i v e System ’ ) ;
36 xgrid (1)

13
 Experiment: 3

Linear Circular convolution of

two given sequences

Scilab code Solution 3.1 Program to Compute the Convolution of Two

Sequences

1 // C a p t i o n : Program t o Compute t h e C o n v o l u t i o n o f Two

Sequences
2 clc ;
3 clear ;
4 close ;
5 x = input ( ’ E n t e r t h e i n p u t S e q u e n c e := ’ ) ;
6 m = length ( x ) ;
7 lx = input ( ’ E n t e r t h e l o w e r i n d e x o f i n p u t s e q u e n c e
:= ’ )
8 hx = lx +m -1;
9 n = lx :1: hx ;
10 h = input ( ’ E n t e r i m p u l s e r e s p o n s e s e q u e n c e := ’ )
11 l = length ( h ) ;
12 lh = input ( ’ E n t e r t h e l o w e r i n d e x o f i m p u l s e
r e s p o n s e := ’ )
13 hh = lh +l -1;

14
Figure 3.1: Program to Compute the Convolution of Two Sequences

15
14 g = lh :1: hh ;
15 nx = lx + lh ;
16 nh = nx + m +l -2;
17 y = convol (x , h )
18 r = nx : nh ;
19 figure (1)
20 subplot (3 ,1 ,1)
21 a = gca () ;
22 a . x_location = ” o r i g i n ” ;
23 a . y_location = ” o r i g i n ” ;
24 plot2d3 ( ’ gnn ’ ,n , x )
25 xlabel ( ’ n===> ’ )
26 ylabel ( ’ Amplitude −−> ’ )
27 title ( ’ I n p u t S e q u e n c e x [ n ] ’ )
28 subplot (3 ,1 ,2)
29 a = gca () ;
30 a . x_location = ” o r i g i n ” ;
31 a . y_location = ” o r i g i n ” ;
32 plot2d3 ( ’ gnn ’ ,g , h )
33 xlabel ( ’ n===> ’ )
34 ylabel ( ’ Amplitude −−−> ’ )
35 title ( ’ I m p u l s e R e s p o n s e S e q u e n c e h [ n ]= ’ )
36 subplot (3 ,1 ,3)
37 a = gca () ;
38 a . x_location = ” o r i g i n ” ;
39 a . y_location = ” o r i g i n ” ;
40 plot2d3 ( ’ gnn ’ ,r , y )
41 xlabel ( ’ n===> ’ )
42 ylabel ( ’ Amplitude −−−> ’ )
43 title ( ’ Output R e s p o n s e S e q u e n c e y [ n ]= ’ )
44 // Example
45 // E n t e r t h e i n p u t S e q u e n c e : = [ 1 , 2 , 3 , 1 ]
46 //
47 // E n t e r t h e l o w e r i n d e x o f i n p u t s e q u e n c e :=0
48 //
49 // E n t e r i m p u l s e r e s p o n s e s e q u e n c e : = [ 1 , 2 , 1 , − 1 ]
50 //
51 // E n t e r t h e l o w e r i n d e x o f i m p u l s e r e s p o n s e :=−1

16
52 //
53 //
54 //−−>y
55 // y =
56 //
57 // 1. 4. 8. 8. 3. − 2. − 1.
58 //

17
 Experiment: 4

Autocorrelation of a given
sequence and verification of its
properties.

Scilab code Solution 4.1 Program to Compute the Autocorrelation of a

Sequence And verfication of Autocorrelation property

1 // C a p t i o n : Program t o Compute t h e A u t o c o r r e l a t i o n o f
a Sequence
2 //And v e r f i c a t i o n o f A u t o c o r r e l a t i o n p r o p e r t y
3 clc ;
4 clear ;
5 close ;
6 x = input ( ’ E n t e r t h e i n p u t S e q u e n c e := ’ ) ;
7 m = length ( x ) ;
8 lx = input ( ’ E n t e r t h e l o w e r i n d e x o f i n p u t s e q u e n c e
:= ’ )
9 hx = lx +m -1;
10 n = lx :1: hx ;
11 x_fold = x ( $ : -1:1) ;
12 nx = lx + lx ;
13 nh = nx + m +m -2;
14 r = nx : nh ;

18
15 Rxx = convol (x , x_fold ) ;
16 disp ( Rxx , ’ Auto C o r r e l a t i o n Rxx [ n ] : = ’ )
17 // P r o p e r t y 1 : A u t o c o r r e l a t i o n o f a s e q u e n c e h a s e v e n
symmetry
18 // Rxx [ n ] = Rxx[−n ]
19 Rxx_flip = Rxx ([ $ : -1:1]) ;
20 if Rxx_flip == Rxx then
21 disp ( ’ P r o p e r t y 1 : Auto C o r r e l a t i o n h a s Even
Symmetry ’ ) ;
22 disp ( Rxx_flip , ’ Auto C o r r e l a t i o n t i m e r e v e r s e d
Rxx[−n ] : = ’ ) ;
23 end
24 // P r o p e r t y 2 : C e n t e r v a l u e Rxx [ 0 ] = t o t a l power o f
the sequence
25 Tot_Px = sum ( x .^2) ;
26 Mid = ceil ( length ( Rxx ) /2) ;
27 if Tot_Px == Rxx ( Mid ) then
28 disp ( ’ P r o p e r t y 2 : Rxx [ 0 ] = c e n t e r v a l u e=max . v a l u e=
T o t a l power o f i /p s e q u e n c e ’ ) ;
29 end
30 subplot (2 ,1 ,1)
31 plot2d3 ( ’ gnn ’ ,n , x )
32 xlabel ( ’ n===> ’ )
33 ylabel ( ’ Amplitude −−> ’ )
34 title ( ’ I n p u t S e q u e n c e x [ n ] ’ )
35 subplot (2 ,1 ,2)
36 plot2d3 ( ’ gnn ’ ,r , Rxx )
37 xlabel ( ’ n===> ’ )
38 ylabel ( ’ Amplitude −−> ’ )
39 title ( ’ Auto c o r r e l a t i o n S e q u e n c e Rxx [ n ] ’ )
40 // Example
41 // E n t e r t h e i n p u t S e q u e n c e : = [ 2 , − 1 , 3 , 4 , 1 ]
42 //
43 // E n t e r t h e l o w e r i n d e x o f i n p u t s e q u e n c e :=−2
44 //
45 // Auto C o r r e l a t i o n Rxx [ n ] : =
46 //
47 // 2. 7. 5. 11. 31. 11. 5.

19
 7. 2.
48 //
49 // P r o p e r t y 1 : Auto C o r r e l a t i o n h a s Even Symmetry
50 //
51 // Auto C o r r e l a t i o n t i m e r e v e r s e d Rxx[−n ] : =
52 //
53 // 2. 7. 5. 11. 31. 11. 5.
7. 2.
54 //
55 // P r o p e r t y 2 : Rxx [ 0 ] = c e n t e r v a l u e=max . v a l u e=T o t a l
power o f i / p s e q u e n c e

20
 Experiment: 5

Cross correlation of given

sequences and verification of its
properties.

Scilab code Solution 5.1 Program to Compute the Crosscorrelation of a

Sequence And verfication of crosscorrelation property

1 // C a p t i o n : Program t o Compute t h e C r o s s c o r r e l a t i o n
of a Sequence
2 //And v e r f i c a t i o n o f c r o s s c o r r e l a t i o n p r o p e r t y
3 clc ;
4 clear ;
5 close ;
6 x = input ( ’ E n t e r t h e F i r s t i n p u t S e q u e n c e := ’ ) ;
7 y = input ( ’ E n t e r t h e s e c o n d i n p u t S e q u e n c e := ’ )
8 mx = length ( x ) ;
9 my = length ( y ) ;
10 lx = input ( ’ E n t e r t h e l o w e r i n d e x o f f i r s t i n p u t
s e q u e n c e := ’ )
11 ly = input ( ’ E n t e r t h e l o w e r i n d e x o f s e c o n d i n p u t
s e q u e n c e := ’ )
12 hx = lx + mx -1;
13 n = lx :1: hx ;

21
14 x_fold = x ( $ : -1:1) ;
15 y_fold = y ( $ : -1:1) ;
16 nx = lx + ly ;
17 ny = nx + mx + my -2;
18 r = nx : ny ;
19 Rxy = convol (x , y_fold ) ;
20 Ryx = convol ( x_fold , y ) ;
21 disp ( Rxy , ’ C r o s s C o r r e l a t i o n Rxy [ n ] : = ’ )
22 count =1;
23 // P r o p e r t y 1 : c r o s s c o r r e l a t i o n o f a s e q u e n c e h a s
Antisymmetry
24 // Rxy [ n ] = Ryx[−n ]
25 Ryx_flip = Ryx ([ $ : -1:1]) ;
26 for i = 1: length ( Rxy )
27 if ( ceil ( Ryx_flip ( i ) ) == ceil ( Rxy ( i ) ) ) then
28 count = count +1;
29 end
30 end
31 if ( count == length ( Rxy ) ) then
32 disp ( ’ P r o p e r t y 1 : C r o s s C o r r e l a t i o n h a s
AntiSymmetry : Rxy [ n ]=Ryx[−n ] ’ ) ;
33 end
34 // P r o p e r t y 2 :% V e r i f i c a t i o n o f Energy P r o p e r t y o f
Rxy
35 Ex = sum ( x .^2) ;
36 Ey = sum ( y .^2) ;
37 E = sqrt ( Ex * Ey ) ;
38 Mid = ceil ( length ( Rxy ) /2) ;
39 if ( E >= Rxy ( Mid ) ) then
40 disp ( ’ P r o p e r t y 2 : Energy P r o p e r t y o f C r o s s
Correlation verified ’)
41 end
42 subplot (2 ,1 ,1)
43 plot2d3 ( ’ gnn ’ ,n , x )
44 xlabel ( ’ n===> ’ )
45 ylabel ( ’ Amplitude −−> ’ )
46 title ( ’ I n p u t S e q u e n c e x [ n ] ’ )
47 subplot (2 ,1 ,2)

22
48 plot2d3 ( ’ gnn ’ ,r , Rxy )
49 xlabel ( ’ n===> ’ )
50 ylabel ( ’ Amplitude −−> ’ )
51 title ( ’ C r o s s c o r r e l a t i o n S e q u e n c e Rxy [ n ] ’ )
52 // Example
53 // E n t e r t h e F i r s t i n p u t S e q u e n c e : = [ 1 , 2 , 1 , 1 ]
54 // E n t e r t h e s e c o n d i n p u t S e q u e n c e : = [ 1 , 1 , 2 , 1 ]
55 // E n t e r t h e l o w e r i n d e x o f f i r s t i n p u t s e q u e n c e :=0
56 // E n t e r t h e l o w e r i n d e x o f s e c o n d i n p u t s e q u e n c e :=0
57 // C r o s s C o r r e l a t i o n Rxy [ n ] : =
58 // 1. 4. 6. 6. 5. 2. 1.
59 // P r o p e r t y 1 : C r o s s C o r r e l a t i o n h a s AntiSymmetry : Rxy
[ n ]=Ryx[−n ]
60 //
61 // P r o p e r t y 2 : Energy P r o p e r t y o f C r o s s C o r r e l a t i o n
verified

23
 Experiment: 6

Solving a given difference

equation.

Scilab code Solution 6.1 Solving Difference Equation Direct Form II Re-
alization

1 // C a p t i o n : S o l v i n g D i f f e r e n c e E q u a t i o n
2 // D i r e c t Form−I I R e a l i z a t i o n
3 // F i n d i n g o u t t h e Output R e s p o n s e o f t h e f i r s t order
4 // s y s t e m ( F i l t e r )
5 clc ;
6 clear ;
7 close ;
8 x =
[1 ,1/2 ,1/4 ,1/8 ,1/16 ,1/32 ,1/64 ,1/128 ,1/256 ,1/512];
9 b = [3 , -4/3]; // n u m e r a t o r p o l y n o m i a l s
10 a = [1 , -1/3]; // d e n o m i n a t o r p o l y n o m i a l s
11 p = length ( a ) -1;
12 q = length ( b ) -1;
13 pq = max (p , q ) ;
14 a = a (2: p +1) ;
15 w = zeros (1 , pq ) ;
16 for i = 1: length ( x )
17 wnew = x ( i ) - sum ( w (1: p ) .* a ) ;

24
18 w = [ wnew , w ];
19 y ( i ) = sum ( w (1: q +1) .* b ) ;
20 end
21 disp (y , ’ Output R e s p o n s e y [ n ]= ’ ) ;
22 // R e s u l t
23 // Output R e s p o n s e y [ n ]=
24 // 3.
25 // 1.1666667
26 // 0.4722222
27 // 0.1990741
28 // 0.0871914
29 // 0.0394805
30 // 0.0183685
31 // 0.0087270
32 // 0.0042111
33 // 0.0020547

25
 Experiment: 7

Computation of N point DFT

of a given sequence and to plot
magnitude and phase spectrum.

Scilab code Solution 7.1 Program to find the spectral information of dis-
crete time signal Calculation of DFT and IDFT

1 // C a p t i o n : Program t o f i n d t h e s p e c t r a l i n f o r m a t i o n
o f d i s c r e t e time signal
2 // C a l c u l a t i o n o f DFT and IDFT
3 // P l o t t i n g Magnitude and Phase Spectrum
4 clc ;
5 close ;
6 clear ;
7 xn = input ( ’ E n t e r t h e r e a l i n p u t d i s c r e t e s e q u e n c e x
[ n ]= ’ ) ;
8 N = length ( xn ) ;
9 XK = zeros (1 , N ) ;
10 IXK = zeros (1 , N ) ;
11 // Code b l o c k t o f i n d t h e DFT o f t h e S e q u e n c e
12 for K = 0: N -1

26
 Figure 7.1: Program to find the spectral information of discrete time signal
Calculation of DFT and IDFT

13 for n = 0: N -1
14 XK ( K +1) = XK ( K +1) + xn ( n +1) * exp ( - %i *2* %pi * K * n /
N);
15 end
16 end
17 [ phase , db ] = phasemag ( XK )
18 disp ( XK , ’ D i s c r e t e F o u r i e r T r a n s f o r m X( k )= ’ )
19 disp ( abs ( XK ) , ’ Magnitude S p e c t r a l S a m p l e s= ’ )
20 disp ( phase , ’ Phase S p e c t r a l S a m p l e s= ’ )
21 n = 0: N -1;
22 K = 0: N -1;
23 figure (1)
24 subplot (2 ,2 ,1)
25 a = gca () ;
26 a . x_location = ” o r i g i n ” ;
27 a . y_location = ” o r i g i n ” ;
28 plot2d3 ( ’ gnn ’ ,n , xn )
29 xlabel ( ’ Time I n d e x n−−−−> ’ )

27
30 ylabel ( ’ A m p l i t u d e xn−−−−> ’ )
31 title ( ’ D i s c r e t e I n p u t S e q u e n c e ’ )
32 subplot (2 ,2 ,2)
33 a = gca () ;
34 a . x_location = ” o r i g i n ” ;
35 a . y_location = ” o r i g i n ” ;
36 plot2d3 ( ’ gnn ’ ,K , abs ( XK ) )
37 xlabel ( ’ F r e q u e n c y Sample I n d e x K−−−−> ’ )
38 ylabel ( ’ | X(K)|−−−−> ’ )
39 title ( ’ Magnitude Spectrum ’ )
40 subplot (2 ,2 ,3)
41 a = gca () ;
42 a . x_location = ” o r i g i n ” ;
43 a . y_location = ” o r i g i n ” ;
44 plot2d3 ( ’ gnn ’ ,K , phase )
45 xlabel ( ’ F r e q u e n c y Sample I n d e x K−−−−> ’ )
46 ylabel ( ’<X(K) i n r a d i a n s −−−−> ’ )
47 title ( ’ Phase Spectrum ’ )
48 // Code b l o c k t o f i n d t h e IDFT o f t h e s e q u e n c e
49 for n = 0: N -1
50 for K = 0: N -1
51 IXK ( n +1) = IXK ( n +1) + XK ( K +1) * exp ( %i *2* %pi * K * n
/N);
52 end
53 end
54 IXK = IXK / N ;
55 ixn = real ( IXK ) ;
56 subplot (2 ,2 ,4)
57 a = gca () ;
58 a . x_location = ” o r i g i n ” ;
59 a . y_location = ” o r i g i n ” ;
60 plot2d3 ( ’ gnn ’ ,[0: N -1] , ixn )
61 xlabel ( ’ D i s c r e t e Time I n d e x n −−−−> ’ )
62 ylabel ( ’ A m p l i t u d e x [ n]−−−−> ’ )
63 title ( ’ IDFT s e q u e n c e ’ )
64 // Example
65 //
66 // E n t e r t h e r e a l i n p u t d i s c r e t e s e q u e n c e x [ n

28
 ]=[1 ,2 ,3 ,4]
67 //
68 // D i s c r e t e F o u r i e r T r a n s f o r m X( k )=
69 //
70 // 1 0 . − 2 . + 2 . i − 2 . − 9 . 7 9 7D−16 i − 2 . − 2 . i
71 //
72 // Magnitude S p e c t r a l S a m p l e s=
73 //
74 // 10. 2.8284271 2. 2.8284271
75 //
76 // Phase S p e c t r a l S a m p l e s=
77 //
78 // 0. 135. 180. 225.
79 //

29
 Experiment: 8

Linear convolution of two

sequences using DFT and
IDFT.

Scilab code Solution 8.1 Linear Convolution using Circular Convolution

DFT IDFT method

1 // C a p t i o n : L i n e a r C o n v o l u t i o n u s i n g C i r c u l a r
Convolution
2 //DFT−IDFT method
3 clc ;
4 clear ;
5 close ;
6 x = input ( ’ E n t e r t h e i n p u t d i s c r e t e s e q u e n c e := ’ )
7 h = input ( ’ E n t e r t h e i m p u l s e d i s c r e t e s e q u e n c e := ’ )
8 N1 = length ( x ) ;
9 N2 = length ( h ) ;
10 N = N1 + N2 -1; // L i n e a r C o n v o l u t i o n r e s u l t l e n g t h
11 h = [h , zeros (1 ,N - N2 ) ];
12 x = [x , zeros (1 ,N - N1 ) ];
13 // Computing DFT−IDFT

30
Figure 8.1: Linear Convolution using Circular Convolution DFT IDFT
method

31
14 XK = dft (x , -1) ; //N p o i n t DFT o f i / p s e q u e n c e
15 HK = dft (h , -1) ; //N p o i n t DFT o f i m p u l s e s e q u e n c e
16 // M u l t i p l i c a t i o n o f 2 DFT’ s
17 YK = XK .* HK ;
18 // L i n e a r C o n v o l u t i o n r e s u l t
19 yn = dft ( YK ,1) ; //IDFT o f Y(K) ( o / p s e q u e n c e )
20 disp ( real ( yn ) , ’ L i n e a r C o n v o l u t i o n r e s u l t y [ n ] : = ’ )
21 // Example
22 // E n t e r t h e i n p u t d i s c r e t e s e q u e n c e := [ 1 , 2 , 3 ]
23 // E n t e r t h e i m p u l s e d i s c r e t e s e q u e n c e : = [ 1 , 2 , 2 , 1 ]
24 // L i n e a r C o n v o l u t i o n r e s u l t y [ n ] : =
25 //
26 // 1.
27 // 4.
28 // 9.
29 // 11.
30 // 8.
31 // 3.

32
 Experiment: 9

Circular convolution of two

given sequences using DFT and
IDFT

Scilab code Solution 9.1 Circular Convolution using DFT IDFT method

1 // C a p t i o n : C i r c u l a r C o n v o l u t i o n u s i n g DFT−IDFT
method
2 clc ;
3 clear ;
4 close ;
5 L = 4; // Length o f t h e s e q u e n c e
6 N = 4; //N−p o i n t DFT
7 x1 = input ( ’ E n t e r t h e f i r s t d i s c r e t e s e q u e n c e : x1 [ n ]=
’)
8 x2 = input ( ’ E n t e r t h e s e c o n d d i s c r e t e s e q u e n c e : x2 [ n
]= ’ )
9 // Computing DFT
10 X1K = dft ( x1 , -1) ;
11 X2K = dft ( x2 , -1) ;
12 // M u l t i p l i c a t i o n o f 2 DFT’ s
13 X3K = X1K .* X2K ;
14 x3 = dft ( X3K ,1) ; //IDFT o f X3 (K)

33
15 x3 = real ( x3 ) ;
16 disp ( x3 , ’ C i r c u l a r C o n v o l u t i o n r e s u l t : x3 [ n ]= ’ ) ;
17 // Example
18 // E n t e r t h e f i r s t d i s c r e t e s e q u e n c e : x1 [ n ]= [ 2 , 1 , 2 , 1 ]
19 // E n t e r t h e s e c o n d d i s c r e t e s e q u e n c e : x2 [ n ]=
[1 ,2 ,3 ,4]
20 //
21 // C i r c u l a r C o n v o l u t i o n r e s u l t : x3 [ n ]=
22 //
23 // 14.
24 // 16.
25 // 14.
26 // 16.

34
 Experiment: 10

Design and implementation of

FIR filter to meet given
specifications.

Scilab code Solution 10.1 To Design an Low Pass FIR Filter

1 // C a p t i o n : To D e s i g n an Low P a s s FIR F i l t e r
2 clc ;
3 clear ;
4 close ;
5 wp = input ( ’ E n t e r t h e p a s s band e d g e ( r a d )= ’ ) ;
6 ws = input ( ’ E n t e r t h e s t o p band e d g e ( r a d )= ’ ) ;
7 ks = input ( ’ E n t e r t h e s t o p band a t t e n u a t i o n ( dB )= ’ ) ;
8 // I f 43<Ks<54 c h o o s e hamming window .
9 //To s e l e c t N, o r d e r o f f i l t e r .
10 N = (2* %pi *4) ./( ws - wp ) ; // k=4 f o r Hamming window .
11 N = ceil ( N ) ; //To round− o f f N t o t h e n e x t i n t e g e r .
12 wc =( wp +( ws - wp ) /2) ./ %pi
13 // To o b t a i n FIR f i l t e r I m p u l s e R e s p o n s e ’ wft ’
14 //And FIR F i l t e r F r e q u e n c y r e s p o n s e ’ wfm ’
15 [ wft , wfm , fr ]= wfir ( ’ l p ’ ,N +1 ,[ wc /2 ,0] , ’hm ’ ,[0 ,0]) ;

35
Figure 10.1: To Design an Low Pass FIR Filter

36
16 figure (1)
17 a = gca () ;
18 a . x_location = ” o r i g i n ” ;
19 a . y_location = ” o r i g i n ” ;
20 a . data_bounds = [0 , -150;1 ,50];
21 plot (2* fr ,20* log10 ( wfm ) , ’ r ’ )
22 xlabel ( ’ N o r m a l i z e d D i g i t a l F r e q u e n c y w−−−> ’ )
23 ylabel ( ’ F r e q u e n c y R e s p o n s e i n dB H( jw )= ’ )
24 title ( ’ F r e q u e n c y R e s p o n s e o f FIR LPF ’ )
25 xgrid (1)
26 // R e s u l t
27 // E n t e r t h e p a s s band e d g e ( r a d )= 0 . 3 ∗ %pi
28 // E n t e r t h e s t o p band e d g e ( r a d )= 0 . 4 5 ∗ %pi
29 // E n t e r t h e s t o p band a t t e n u a t i o n ( dB )= 50
30 //N = 54.
31 //−−>wc
32 // wc = 0.375
33 //−−>d i s p ( wft , ’ I m p u l s e R e s p o n s e o f FIR LPF= ’)
34 // I m p u l s e R e s p o n s e o f FIR LPF=
35 // column 1 t o 7
36 // 0.0003609 − 0.0007195 − 0.0010869 1 . 5 7 5D
−18 0.0016485 0.0015927 − 0.0010883
37 // column 8 t o 14
38 // − 0 . 0 0 3 5 7 0 3 − 0 . 0 0 1 7 0 0 9 0.0038764
0 . 0 0 6 1 8 9 6 − 5 . 9 6 5D−18 − 0 . 0 0 9 0 2 0 8 − 0 . 0 0 8 2 5 1 6
39 // column 15 t o 21
40 // 0.0053105 0.0164428 0.0074408 −
0.0162551 − 0.0251602 1 . 1 9 1D−17 0.0359480
41 // column 22 t o 28
42 // 0.0334760 − 0.0225187 − 0.0756838 −
0.0394776 0.1111441 0.2931653 0.375
43 // column 29 t o 35
44 // 0.2931653 0.1111441 − 0.0394776 −
0.0756838 − 0.0225187 0.0334760 0.0359480
45 // column 36 t o 42
46 // 1 . 1 9 1D−17 − 0 . 0 2 5 1 6 0 2 − 0 . 0 1 6 2 5 5 1
0.0074408 0.0164428 0.0053105 − 0.0082516
47 // column 43 t o 49

37
48 // − 0 . 0 0 9 0 2 0 8 − 5 . 9 6 5D−18 0.0061896
0.0038764 − 0.0017009 − 0.0035703 − 0.0010883
column 50 t o 55
49 // 0.0015927 0.0016485 1 . 5 7 5D−18 −
0.0010869 − 0.0007195 0.0003609

38
 Experiment: 11

Design and implementation of

IIR filter to meet given
specifications.

Scilab code Solution 11.1 To obtain Digital IIR Butterworth low pass
filter Frequency response

1 // C a p t i o n : To o b t a i n D i g i t a l I I R B u t t e r w o r t h low
pass f i l t e r
2 // F r e q u e n c y r e s p o n s e
3 clc ;
4 clear ;
5 close ;
6 fp = input ( ’ E n t e r t h e p a s s band e d g e ( Hz ) = ’ ) ;
7 fs = input ( ’ E n t e r t h e s t o p band e d g e ( Hz ) = ’ ) ;
8 kp = input ( ’ E n t e r t h e p a s s band a t t e n u a t i o n ( dB ) = ’ )
;
9 ks = input ( ’ E n t e r t h e s t o p band a t t e n u a t i o n ( dB ) = ’ )
;
10 Fs = input ( ’ E n t e r t h e sampling rate samples / sec = ’ )
;

39
Figure 11.1: To obtain Digital IIR Butterworth low pass filter Frequency
response

40
11 d1 = 10^( kp /20) ;
12 d2 = 10^( ks /20) ;
13 d = sqrt ((1/( d2 ^2) ) -1) ;
14 E = sqrt ((1/( d1 ^2) ) -1) ;
15 // D i g i t a l f i l t e r s p e c i f i c a t i o n s ( r a d / s a m p l e s )
16 wp =2* %pi * fp *1/ Fs ;
17 ws =2* %pi * fs *1/ Fs ;
18 disp ( wp , ’ D i g i t a l P a s s band e d g e f r e q i n r a d / s a m p l e s
wp= ’ )
19 disp ( ws , ’ D i g i t a l S t o p band e d g e f r e q i n r a d / s a m p l e s
ws= ’ )
20 // Pre w a r p i n g
21 op =2* Fs * tan ( wp /2) ;
22 os =2* Fs * tan ( ws /2) ;
23 disp ( op , ’ Analog P a s s Band Edge Freq . i n r a d / s e c op= ’
)
24 disp ( os , ’ Analog S t o p band Edge Freq . i n r a d / s e c o s= ’
)
25 N = log10 ( d / E ) / log10 ( os / op ) ;
26 oc = op /(( E ^2) ^(1/(2* N ) ) ) ;
27 N = ceil ( N ) ; // r o u n d e d t o n e a r e s t i n t e g e r
28 disp (N , ’ I I R F i l t e r o r d e r N = ’ ) ;
29 disp ( oc , ’ C u t o f f F r e q u e n c y i n r a d / s e c o n d s OC = ’ )
30 [ pols , gn ] = zpbutt (N , oc ) ;
31 disp ( gn , ’ Gain o f Analog I I R B u t t e r w o r t h LPF Gain = ’ )
32 disp ( pols , ’ P o l e s o f Analog I I R B u t t e r w o r t h LPF P o l e s
=’)
33 HS = poly ( gn , ’ s ’ , ’ c o e f f ’ ) / real ( poly ( pols , ’ s ’ ) ) ;
34 disp ( HS , ’ T r a n s f e r f u n c t i o n o f Ananlog I I R
B u t t e r w o r t h LPF H( S )= ’ )
35 z = poly (0 , ’ z ’ )
36 Hz = horner ( HS ,(2* Fs *( z -1) /( z +1) ) )
37 num = coeff ( Hz (2) )
38 den = coeff ( Hz (3) )
39 Hz (2) = Hz (2) ./ den (3) ;
40 Hz (3) = Hz (3) ./ den (3) ;
41 disp ( Hz , ’ T r a n s f e r f u n c t i o n o f D i g i t l a I I R
B u t t e r w o r t h LPF H( Z )= ’ )

41
42 [ Hw , w ] = frmag ( Hz ,256) ;
43 figure (1)
44 plot (2* w * %pi ,20* log10 ( abs ( Hw ) ) ) ;
45 xlabel ( ’ D i g i t a l F r e q u e n c y w−−−> ’ )
46 ylabel ( ’ Magnitude i n dB 20 l o g | H(w) |= ’ )
47 title ( ’ Magnitude R e s p o n s e o f I I R LPF ’ )
48 xgrid (1)
49 // R e s u l t
50 // E n t e r t h e p a s s band e d g e ( Hz ) = 1 5 0 0
51 //
52 // E n t e r t h e s t o p band e d g e ( Hz ) = 2 0 0 0
53 //
54 // E n t e r t h e p a s s band a t t e n u a t i o n ( dB ) = −1
55 //
56 // E n t e r t h e s t o p band a t t e n u a t i o n ( dB ) = −3
57 //
58 // E n t e r t h e s a m p l i n g r a t e samples / s e c = 8000
59 //
60 // D i g i t a l P a s s band e d g e f r e q i n r a d / s a m p l e s wp=
61 //
62 // 1.1780972
63 //
64 // D i g i t a l S t o p band e d g e f r e q i n r a d / s a m p l e s ws=
65 //
66 // 1.5707963
67 //
68 // Analog P a s s Band Edge Freq . i n r a d / s e c op=
69 //
70 // 10690.858
71 //
72 // Analog S t o p band Edge Freq . i n r a d / s e c o s=
73 //
74 // 16000.
75 //
76 // I I R F i l t e r o r d e r N =
77 //
78 // 2.
79 //

42
 80 // C u t o f f F r e q u e n c y i n r a d / s e c o n d s OC =
81 //
82 // 16022.769
83 //
84 // Gain o f Analog I I R B u t t e r w o r t h LPF Gain =
85 //
86 // 2 . 5 6 7D+08
87 //
88 // P o l e s o f Analog I I R B u t t e r w o r t h LPF P o l e s =
89 //
90 // − 1 1 3 2 9 . 8 0 9 + 1 1 3 2 9 . 8 0 9 i − 1 1 3 2 9 . 8 0 9 −
11329.809 i
91 //
92 // T r a n s f e r f u n c t i o n o f Ananlog I I R B u t t e r w o r t h LPF
H( S )=
93 //
94 // 2 . 5 6 7D+08
95 // −−−−−−−−−−−−−−−−−−−−−−−−−
96 // 2
97 // 2 . 5 6 7D+08 + 2 2 6 5 9 . 6 1 8 s + s
98 //
99 // T r a n s f e r f u n c t i o n o f D i g i t l a I I R B u t t e r w o r t h LPF
H( Z )=
100 //
101 // 2
102 // 0.2933099 + 0.5866197 z + 0.2933099 z
103 // −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
104 // 2
105 // 0.1715734 + 0.0016661 z + z
106 //

Scilab code Solution 11.2 To obtain Digital IIR Chebyshev low pass filter
Frequency response

43
Figure 11.2: To obtain Digital IIR Chebyshev low pass filter Frequency re-
sponse

44
1 // C a p t i o n : To o b t a i n D i g i t a l I I R Chebyshev low p a s s
filter
2 // F r e q u e n c y r e s p o n s e
3 clc ;
4 clear ;
5 close ;
6 fp = input ( ’ E n t e r t h e p a s s band e d g e ( Hz ) = ’ ) ;
7 fs = input ( ’ E n t e r t h e s t o p band e d g e ( Hz ) = ’ ) ;
8 kp = input ( ’ E n t e r t h e p a s s band a t t e n u a t i o n ( dB ) = ’ )
;
9 ks = input ( ’ E n t e r t h e s t o p band a t t e n u a t i o n ( dB ) = ’ )
;
10 Fs = input ( ’ E n t e r t h e s a m p l i n g r a t e samples / sec = ’ )
;
11 d1 = 10^( kp /20) ;
12 d2 = 10^( ks /20) ;
13 d = sqrt ((1/( d2 ^2) ) -1) ;
14 E = sqrt ((1/( d1 ^2) ) -1) ;
15 // D i g i t a l f i l t e r s p e c i f i c a t i o n s ( r a d / s a m p l e s )
16 wp =2* %pi * fp *1/ Fs ;
17 ws =2* %pi * fs *1/ Fs ;
18 disp ( wp , ’ D i g i t a l P a s s band e d g e f r e q i n r a d / s a m p l e s
wp= ’ )
19 disp ( ws , ’ D i g i t a l S t o p band e d g e f r e q i n r a d / s a m p l e s
ws= ’ )
20 // Pre w a r p i n g
21 op =2* Fs * tan ( wp /2) ;
22 os =2* Fs * tan ( ws /2) ;
23 disp ( op , ’ Analog P a s s Band Edge Freq . i n r a d / s e c op= ’
)
24 disp ( os , ’ Analog S t o p band Edge Freq . i n r a d / s e c o s= ’
)
25 N = acosh ( d / E ) / acosh ( os / op ) ;
26 oc = op /(( E ^2) ^(1/(2* N ) ) ) ;
27 N = ceil ( N ) ; // r o u n d e d t o n e a r e s t i n t e g e r
28 disp (N , ’ I I R F i l t e r o r d e r N = ’ ) ;
29 disp ( oc , ’ C u t o f f F r e q u e n c y i n r a d / s e c o n d s OC = ’ )
30 [ pols , gn ] = zpch1 (N ,E , op ) ;

45
31 disp ( gn , ’ Gain o f Analog I I R Chebyshev Type−I LPF
Gain = ’ )
32 disp ( pols , ’ P o l e s o f Analog I I R Chebyshev Type−I LPF
Poles =’)
33 HS = poly ( gn , ’ s ’ , ’ c o e f f ’ ) / real ( poly ( pols , ’ s ’ ) ) ;
34 disp ( HS , ’ T r a n s f e r f u n c t i o n o f Ananlog I I R Chebyshev
Type−I LPF H( S )= ’ )
35 z = poly (0 , ’ z ’ )
36 Hz = horner ( HS ,(2* Fs *( z -1) /( z +1) ) )
37 num = coeff ( Hz (2) )
38 den = coeff ( Hz (3) )
39 Hz (2) = Hz (2) ./ den (3) ;
40 Hz (3) = Hz (3) ./ den (3) ;
41 disp ( Hz , ’ T r a n s f e r f u n c t i o n o f D i g i t l a I I R Chebyshev
LPF H( Z )= ’ )
42 [ Hw , w ] = frmag ( Hz ,256) ;
43 figure (1)
44 plot (2* w * %pi ,20* log10 ( abs ( Hw ) ) ) ;
45 xlabel ( ’ D i g i t a l F r e q u e n c y w−−−> ’ )
46 ylabel ( ’ Magnitude i n dB 20 l o g | H(w) |= ’ )
47 title ( ’ Magnitude R e s p o n s e o f I I R LPF ’ )
48 xgrid (1)
49 // R e s u l t
50 // E n t e r t h e p a s s band e d g e ( Hz ) = 1 5 0 0
51 //
52 // E n t e r t h e s t o p band e d g e ( Hz ) = 2 0 0 0
53 //
54 // E n t e r t h e p a s s band a t t e n u a t i o n ( dB ) = −1
55 //
56 // E n t e r t h e s t o p band a t t e n u a t i o n ( dB ) = −3
57 //
58 // E n t e r t h e s a m p l i n g r a t e samples / s e c = 8000
59 //
60 // D i g i t a l P a s s band e d g e f r e q i n r a d / s a m p l e s wp=
61 //
62 // 1.1780972
63 //
64 // D i g i t a l S t o p band e d g e f r e q i n r a d / s a m p l e s ws=

46
 65 //
66 // 1.5707963
67 //
68 // Analog P a s s Band Edge Freq . i n r a d / s e c op=
69 //
70 // 10690.858
71 //
72 // Analog S t o p band Edge Freq . i n r a d / s e c o s=
73 //
74 // 16000.
75 //
76 // I I R F i l t e r o r d e r N =
77 //
78 // 2.
79 //
80 // C u t o f f F r e q u e n c y i n r a d / s e c o n d s OC =
81 //
82 // 17642.912
83 //
84 // Gain o f Analog I I R Chebyshev Type−I LPF Gain =
85 //
86 // 1 . 1 2 3D+08
87 //
88 // P o l e s o f Analog I I R Chebyshev Type−I LPF P o l e s =
89 //
90 // − 5 8 6 7 . 8 6 1 + 9 5 6 9 . 6 9 2 7 i − 5 8 6 7 . 8 6 1 − 9 5 6 9 . 6 9 2 7 i
91 //
92 // T r a n s f e r f u n c t i o n o f Ananlog I I R Chebyshev Type−I
LPF H( S )=
93 //
94 // 1 . 1 2 3D+08
95 // −−−−−−−−−−−−−−−−−−−−−−−−−
96 // 2
97 // 1 . 2 6 0D+08 + 1 1 7 3 5 . 7 2 2 s + s
98 //
99 // T r a n s f e r f u n c t i o n o f D i g i t l a I I R Chebyshev LPF H(
Z )=
100 //

47
101 // 2
102 // 0.1971055 + 0.3942111 z + 0.1971055 z
103 // −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
104 // 2
105 // 0.3409008 − 0.4562766 z + z

48
 Experiment: 12

Circular convolution of two

given sequences

Scilab code Solution 12.1 Program to perform circular convolution of two

sequences

1 // C a p t i o n : Program t o p e r f o r m c i r c u l a r c o n v o l u t i o n
o f two s e q u e n c e s
2 clc ;
3 clear ;
4 close ;
5 x1 = input ( ’ E n t e r t h e f i r s t d i s c r e t e s e q u e n c e := ’ )
6 x2 = input ( ’ E n t e r t h e s e c o n d d i s c r e t e s e q u e n c e := ’ )
7 m = length ( x1 ) ; // l e n g t h o f f i r s t s e q u e n c e
8 n = length ( x2 ) ; // l e n g t h o f s e c o n d s e q u e n c e
9 //To make l e n g t h o f x1 and x2 a r e e q u a l
10 if (m > n )
11 for i = n +1: m
12 x2 ( i ) =0;
13 end
14 elseif (n > m )
15 for i = m +1: n
16 x1 ( i ) =0;
17 end

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18 end
19 N = length ( x1 ) ;
20 x3 = zeros (1 , N ) ; // c i r c u l a r convolution result
i n i t i a l i z e d to zero
21 a (1) = x2 (1) ;
22 for j = 2: N
23 a ( j ) = x2 (N - j +2) ;
24 end
25 for i = 1: N
26 x3 (1) = x3 (1) + x1 ( i ) * a ( i ) ;
27 end
28 X (1 ,:) = a ;
29 // C a l c u l a t i o n o f c i r c u l a r c o n v o l u t i o n
30 for k =2: N
31 for j = 2: N
32 x2 ( j ) = a (j -1) ;
33 end
34 x2 (1) = a ( N ) ;
35 X (k ,:) = x2 ;
36 for i = 1: N
37 a ( i ) = x2 ( i ) ;
38 x3 ( k ) = x3 ( k ) + x1 ( i ) * a ( i ) ;
39 end
40 end
41 disp ( x3 , ’ C i r c u l a r C o n v o l u t i o n R e s u l t x3 [ n ]= ’ )
42 // Example
43 // E n t e r t h e f i r s t d i s c r e t e s e q u e n c e := [ 2 , 1 , 2 , 1 ]
44 // E n t e r t h e s e c o n d d i s c r e t e s e q u e n c e := [ 1 , 2 , 3 , 4 ]
45 // C i r c u l a r C o n v o l u t i o n R e s u l t x3 [ n ]=
46 // 14. 16. 14. 16.
47 //

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