Scilab Manual for
DIGITAL SIGNAL PROCESSING
by Prof Desai Siddhibahen Deveshbhai
Others
Laxmi Institute Of Technology,sarigam1
Solutions provided by
Prof Desai Siddhibahen Deveshbhai
Others
Laxmi Institute Of Technology,sarigam
April 6, 2023
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 WRITE A SCILAB PROGRAM TO GENERATE COMMON DISCRETE TIME SIGNALS.
7
2 WRITE A SCILAB PROGRAM TO OBSERVE THE EFFECTS OF LOWER SAMPLING RATE AND HIGHER
SAMPLING RATE ON C.T. SIGNAL(SAMPLING THM.)
11
3 WRITE A SCILAB PROGRAM TO COMPUTE LINEARITY PROPERTY OF A GIVEN SIGNAL.
15
4 WRITE A SCILAB PROGRAM TO COMPUTE LINEAR
CONVOLUTION OF TWO SEQUENCES USING BASIC
EQUATION.
18
5 WRITE A SCILAB PROGRAM TO FIND AUTOCORRELATION AND CROSS CORRELATION OF THE GIVEN
SEQUENCES.
22
6 WRITE A SCILAB PROGRAM TO FIND N-POINT DFT
OF THE GIVEN SEQUENCE.
26
7 WRITE A SCILAB PROGRAM TO COMPUTE LINEAR
CONVOLUTION OF TWO SEQUENCES USING DFT BASED
APPROACH.
31
8 WRITE A SCILAB PROGRAM TO COMPUTE CIRCULAR CONVOLUTION OF TWO SEQUECNES USING BA2
SIC EQUATION.
34
9 WRITE A SCILAB PROGRAM TO COMPUTE CIRCULAR CONVOLUTION OF THE TWO SEQUENCES USING DFT BASED APPROACH.
38
10 WRITE A SCILAB PROGRAM TO COMPUTE BLOCK
CONVOLUTION USING OVERLAP ADD METHOD
42
11 WRITE A SCILAB PROGRAM RO COMPUTE BLOCK
CONVOLUTION USING OVERLAP SAVE METHOD.
46
12 WRITE A SCILAB PROGRAM TO FIND FFT USING
DECIMATION IN TIME(DIT) METHOD.
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13 WRITE A SCILAB PROGRAM FOR DESIGNING OF
FIR FILTER FOR LOW PASS, HIGH PASS, BANDPASS
AND BAND REJECT RESPONSES.
54
14 WRITE A SCILAB PROGRAM TO DESIGN DIGITAL
IIR BUTTERWORTH LOW PASS FILTER.
62
15 WRITE A SCILAB PROGRAM TO DESIGN DIGITAL
IIR CHEBYSHEW FILTER.
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3
List of Experiments
Solution 1.01
PROGRAM TO GENERATE COMMON DISCRETE
TIME SIGNALS . . . . . . . . . . . . . . . . . .
7
Solution 2.02 PROGRAM TO OBSERVE THE EFFECTS OF
LOWER SAMPLING RATE AND HIGHER SAMPLING RATE ON CONTINUOUS TIME SIGNAL 11
Solution 3.03 PROGRAM TO COMPUTE LINEARITY PROPERTY OF GIVEN SIGNAL . . . . . . . . . . . .
15
Solution 4.04 PROGRAM TO COMPUTE LINEAR CONVOLUTION OF TWO SEQUENCES USING BASIC
EQUATION . . . . . . . . . . . . . . . . . . . . .
18
Solution 5.05 PROGRAM TO FIND AUTO CORRELATION AND
CROSS CORRELATION OF THE GIVEN SEQUENCES
Solution 6.06 PROGRAM TO FIND N POINT DFT OF THE
GIVEN SEQUENCE . . . . . . . . . . . . . . . .
26
Solution 7.07 PROGRAM TO COMPUTE LINEAR CONVOLUTION OF TWO SEQUENCES USING DFT BASED
APPROACH . . . . . . . . . . . . . . . . . . . .
31
Solution 8.08 PROGRAM TO COMPUTE CIRCULAR CONVOLUTION OF TWO SEQUENCES USING BASIC EQUATION . . . . . . . . . . . . . . . . . .
34
Solution 9.09 PROGRAM TO COMPUTE CIRCULAR CONVOLUTION OF THE TWO SEQUENCES USING
DFT BASED APPROACH . . . . . . . . . . . .
38
Solution 10.10 PROGRAM TO COMPUTE BLOCK CONVOLUTION USING OVERLAP ADD METHOD . . . .
42
Solution 11.11 PROGRAM TO COMPUTE BLOCK CONVOLUTION USING OVERLAP SAVE METHOD . . .
46
4
22
Solution 12.12 PROGRAM TO FIND FFT USING DECIMATION
IN TIME METHOD . . . . . . . . . . . . . . . .
50
Solution 13.13 PROGRAM FOR DESIGNING OF FIR FILTER
FOR LOW PASS HIGH PASS BAND PASS AND
BAND REJECT RESPONSES . . . . . . . . . .
54
Solution 14.14 PROGRAM TO DESIGN DIGITAL IIR BUTTERWORTH LOW PASS FILTER . . . . . . . . . . .
62
Solution 15.15 PROGRAM TO DESIGN DIGITAL IIR CHEBYSHEW
FILTER . . . . . . . . . . . . . . . . . . . . . . .
64
5
List of Figures
6
Experiment: 1
WRITE A SCILAB
PROGRAM TO GENERATE
COMMON DISCRETE TIME
SIGNALS.
Scilab code Solution 1.01 PROGRAM TO GENERATE COMMON DISCRETE TIME SIGNALS
1
2
3
// VERSION : S c i l a b : 5 . 4 . 1
// OS : windows 7
// CAPTION : PROGRAM TO GENERATE COMMON DISCRETE TIME
SIGNALS
4
5 clc ;
6 clear ;
7 close ;
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9 //UNIT IMPULSE SIGNAL
10 L = input ( ’ E n t e r t h e Length= ’ ) ; //SET LIMIT
11 n = - L : L ;
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x =[ zeros (1 , L ) , ones (1 ,1) , zeros (1 , L ) ];
subplot (2 ,4 ,1) ;
a = gca () ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 (n , x )
title ( ’ U n i t I m p u l s e S i g n a l ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
// UNIT STEP SIGNAL
y =[ zeros (1 , L ) , ones (1 , L +1) ];
subplot (3 ,4 ,2) ;
b = gca () ;
b . y_location = ” m i d d l e ” ;
b . foreground = 5;
b . font_color = 5;
b . font_style = 5;
plot2d3 (n , y )
title ( ’ U n i t S t e p S i g n a l ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
//UNIT RAMP SIGNAL
z =[ zeros (1 , L ) ,0: L ] ,
subplot (2 ,4 ,3)
c = gca () ;
c . y_location = ” m i d d l e ” ;
c . foreground = 5;
c . font_color = 5;
c . font_style = 5;
plot2d3 (n , z )
title ( ’ U n i t Ramp S i g n a l ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
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// EXPONENTIALLY INCREASING SEQUENCE
n =0:1:10;
x = exp ( n ) ;
subplot (2 ,4 ,4) ;
d = gca () ;
d . x_location = ” o r i g i n ” ;
d . y_location = ” o r i g i n ” ;
d . foreground = 5;
d . font_color = 5;
d . font_style = 5;
plot2d3 (n , x )
title ( ’ E x p o n e n t i a l l y I n c r e a s i n g S e q u e n c e ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
// EXPONENTIALLY DECREASING SEQUENCE
n =0:1:10;
x = exp ( - n ) ;
subplot (2 ,4 ,5) ;
d = gca () ;
d . x_location = ” o r i g i n ” ;
d . y_location = ” o r i g i n ” ;
d . foreground = 5;
d . font_color = 5;
d . font_style = 5;
plot2d3 (n , x )
title ( ’ E x p o n e n t i a l l y D e c r e a s i n g S e q u e n c e ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
// SINE WAVE
t =0:0.04:1;
x = sin (2* %pi * t ) ;
subplot (2 ,4 ,6) ;
a = gca () ;
a . foreground = 5;
a . font_color = 5;
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a . font_style = 5;
plot2d3 (t , x ) ;
title ( ’ S i n e Wave ’ )
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
// COSINE WAVE
t =0:0.04:1;
x = cos (2* %pi * t ) ;
subplot (2 ,4 ,7) ;
b = gca () ;
b . foreground = 5;
b . font_color = 5;
b . font_style = 5;
plot2d3 (t , x ) ;
title ( ’ C o s i n e Wave ’ )
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
//INPUT :
// E n t e r t h e Length=5
10
Experiment: 2
WRITE A SCILAB
PROGRAM TO OBSERVE
THE EFFECTS OF LOWER
SAMPLING RATE AND
HIGHER SAMPLING RATE
ON C.T. SIGNAL(SAMPLING
THM.)
Scilab code Solution 2.02 PROGRAM TO OBSERVE THE EFFECTS
OF LOWER SAMPLING RATE AND HIGHER SAMPLING RATE ON
CONTINUOUS TIME SIGNAL
1
2
3
// VERSION : S c i l a b : 5 . 4 . 1
// OS : windows 7
//CAPTION : PROGRAM TO OBSERVE THE EFFECTS OF LOWER
SAMPLING RATE AND
11
4
// HIGHER SAMPLING RATE ON C . T . SIGNAL (SAMPLING THM
.)
5
6 clc ;
7 clear ;
8 close ;
9 f = input ( ’ E n t e r c o n t i n u o u s t i m e s i g n a l
10
f r e q u e n c y= ’ ) ;
// f =0.1
a = input ( ’ E n t e r c o n t i n u o u s t i m e s i g n a l a m p l i t u d e= ’ ) ;
// a=1
t =0:0.01:100;
x = a * sin (2* %pi * f * t ) ;
subplot (2 ,2 ,1) ;
b = gca () ;
b . x_location = ” o r i g i n ” ;
b . y_location = ” o r i g i n ” ;
b . foreground = 5;
b . font_color = 5;
b . font_style = 5;
plot (t , x ) ;
title ( ’ O r i g i n a l C o n t i n u o u s t i m e S i g n a l ’ ) ;
xlabel ( ’ Time t ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
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25 // S a m p l i n g Rate =2∗max f r e q u e n c y
26 fs1 = input ( ’ E n t e r s a m p l i n g f r e q u e n c y e q u a l t o 2∗ f s = ’ )
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; // f s 1 =0.2
fd = f / fs1 ;
n =0:0.01:100;
x1n = a * sin (2* %pi * f * n / fs1 ) ;
subplot (2 ,2 ,2) ;
b = gca () ;
b . x_location = ” o r i g i n ” ;
b . y_location = ” o r i g i n ” ;
b . foreground = 5;
b . font_color = 5;
b . font_style = 5;
plot2d3 ( ’ gnn ’ ,n , x1n ) ;
12
title ( ’ R e c o n s t r u c t e d s i g n a l w i t h s a m p l i n g f r e q u e n c y
e q u a l 2∗ f s ’ ) ;
39 xlabel ( ’ Time t ’ ) ;
40 ylabel ( ’ A m p l i t u d e ’ ) ;
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41
42
43 // S a m p l i n g Rate <2∗max f r e q u e n c y= A l i a s i n g E f f e c t
44 fs2 = input ( ’ E n t e r s a m p l i n g f r e q u e n c y l e s s t h a n 2∗ f s = ’
) ; // f s 2 =0.1
x2n = a * sin (2* %pi * f * n / fs2 ) ;
subplot (2 ,2 ,3) ;
b = gca () ;
b . x_location = ” o r i g i n ” ;
b . y_location = ” o r i g i n ” ;
b . foreground = 5;
b . font_color = 5;
b . font_style = 5;
plot2d3 ( ’ gnn ’ ,n , x2n ) ;
title ( ’ R e c o n s t r u c t e d s i g n a l w i t h s a m p l i n g f r e q u e n c y
L e s s t h a n 2∗ f s =A l i a s i n g e f f e c t ’ ) ;
55 xlabel ( ’ Time t ’ ) ;
56 ylabel ( ’ A m p l i t u d e ’ ) ;
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// S a m p l i n g Rate >2∗max f r e q u e n c y=P e r f e c t
Reconstuection
fs3 = input ( ’ E n t e r s a m p l i n g f r e q u e n c y G r e a t e r t h a n 2∗
f s = ’ ) ; // f s 3 =1
x3n = a * sin (2* %pi * f * n * fs3 ) ;
subplot (2 ,2 ,4) ;
b = gca () ;
b . x_location = ” o r i g i n ” ;
b . y_location = ” o r i g i n ” ;
b . foreground = 5;
b . font_color = 5;
b . font_style = 5;
plot2d3 ( ’ gnn ’ ,n , x3n ) ;
title ( ’ P e r f e c t R e c o n s t r u c t e d s i g n a l w i t h s a m p l i n g
f r e q u e n c y g r e a t e r t h a n 2∗ f s ’ ) ;
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xlabel ( ’ Time t ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
//INPUT :
// E n t e r c o n t i n u o u s t i m e s i g n a l f r e q u e n c y =0.1
// E n t e r c o n t i n u o u s t i m e s i g n a l a m p l i t u d e =1
// E n t e r s a m p l i n g f r e q u e n c y e q u a l t o 2∗ f s =0.2
// E n t e r s a m p l i n g f r e q u e n c y l e s s t h a n 2∗ f s =0.1
// E n t e r s a m p l i n g f r e q u e n c y G r e a t e r t h a n 2∗ f s =1
14
Experiment: 3
WRITE A SCILAB
PROGRAM TO COMPUTE
LINEARITY PROPERTY OF
A GIVEN SIGNAL.
Scilab code Solution 3.03 PROGRAM TO COMPUTE LINEARITY PROPERTY OF GIVEN SIGNAL
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10
// VERSION : S c i l a b : 5 . 4 . 1
// OS : windows 7
// CAPTION : PROGRAM TO COMPUTE LINEARITY PROPERTY OF
A GIVEN SIGNAL
// I n p u t S e q u e n c e :
x1 ( n )=s i n ( 2 %pi0 . 1 n ) and x2 ( n )=
c o s ( 2 %pi0 . 5 n ) and s y s t e m y ( n ) =0.5 x ( n )
clc ;
clear ;
close ;
n =0:10;
a =2;
b =3;
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x1 = sin (2* %pi *0.1* n ) ; // i n p u t s e q u e n c e 1
x2 = cos (2* %pi *0.5* n ) ; // i n p u t s e q u e n c e 2
x = a .* x1 + b .* x2 ; // h o m o g e n i t y and s u p e r p o s i t i o n −−op1
y =0.5* x ; // s y s t e m y =0.5∗ x
y1 =0.5* x1 ;
y2 =0.5* x2 ;
yt = a .* y1 + b .* y2 ; // h o m o g e n i t y and s u p e r p o s i t i o n −−op2
d =y - yt ; // op2−op1= d i f f e r e n c e
disp ( ’ Output o f System y ( n ) =0.5 x ( n ) i s : ’ ) ;
if ( d ==0)
disp ( ’ System i s l i n e a r ’ ) ;
else
disp ( ’ System i s Non− l i n e a r ’ ) ;
end
// I n p u t S e q u e n c e :
x1 ( n )=s i n ( 2 %pi0 . 1 n ) and x2 ( n )=
c o s ( 2 %pi0 . 5 n ) and s y s t e m y ( n )=s q r t ( x ( n ) )
n =0:10;
a =2;
b =3;
x1 = sin (2* %pi *0.1* n ) ; // i n p u t s e q u e n c e 1
x2 = cos (2* %pi *0.5* n ) ; // i n p u t s e q u e n c e 2
x = a .* x1 + b .* x2 ; // h o m o g e n i t y and s u p e r p o s i t i o n −−op1
y = sqrt ( x ) ; // s y s t e m y=s q r t ( x )
y1 = sqrt ( x1 ) ;
y2 = sqrt ( x2 ) ;
yt = a .* y1 + b .* y2 ; // h o m o g e n i t y and s u p e r p o s i t i o n −−op2
d =y - yt ; // op2−op1= d i f f e r e n c e
disp ( ’ Output o f System y ( n )=s q r t ( x ( n ) ) i s : ’ ) ;
if ( d ==0)
disp ( ’ System i s l i n e a r ’ ) ;
else
disp ( ’ System i s Non− l i n e a r ’ ) ;
end
//OUTPUT:
// Output o f System y ( n ) =0.5 x ( n ) i s :
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// System i s l i n e a r
// Output o f System y ( n )=s q r t ( x ( n ) ) i s :
// System i s Non− l i n e a r
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Experiment: 4
WRITE A SCILAB
PROGRAM TO COMPUTE
LINEAR CONVOLUTION OF
TWO SEQUENCES USING
BASIC EQUATION.
Scilab code Solution 4.04 PROGRAM TO COMPUTE LINEAR CONVOLUTION OF TWO SEQUENCES USING BASIC EQUATION
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// VERSION : S c i l a b : 5 . 4 . 1
// OS : windows 7
//CAPTION : PROGRAM TO COMPUTE LINEAR CONVOLUTION OF
TWO SEQUENCES USING BASIC EQUATION
clc ;
clear ;
close ;
// i n p u t s e q u e n c e s
x = input ( ’ E n t e r t h e I n p u t S e q u e n c e x ( n )= ’ ) // x =[1 2 3
1]
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9 m = length ( x ) ;
10 xl = 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= ’ )
// 0
xh = xl +m -1;
n = xl :1: xh ;
subplot (3 ,1 ,1) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,n , x ) ;
title ( ’ I n p u t S e q u e n c e x [ n ] ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
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25 h = input ( ’ E n t e r t h e 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 )= ’ )
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; // h =[1 2 1 −1]
l = length ( h ) ;
hl = 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
S e q u e n c e= ’ ) ; //−1
hh = hl +l -1;
g = hl :1: hh ;
subplot (3 ,1 ,2) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,n , h ) ;
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 ] ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
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42 nx = xl + hl ; // r a n g e o f k
43 nh = xh + hh ; // r a n g e o f n
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44 x =[ x , zeros (1 , l ) ];
45 h =[ h , zeros (1 , m ) ];
46 y = zeros (1 , m +l -1) // n1+n2−1= l e n g t h
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of linear
convolution
for i =1: m +l -1
y ( i ) =0;
for j =1: m +l -1
if (j < i +1)
y ( i ) = y ( i ) + x ( j ) * h (i - j +1) ;
end
end
end
disp ( ’ L i n e a r C o n v o l u t i o n u s i n g E q u a t i o n i s y ( n ) : ’ )
disp ( y ) ;
r = nx : nh ;
subplot (3 ,1 ,3) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,r , y ) ;
title ( ’ Output R e s p o n s e S e q u e n c e o f L i n e a r
Convolution u s i n g Equation y [ n ] ’ );
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
//INPUT :
// E n t e r t h e I n p u t S e q u e n c e x ( n ) =[1 2 3 1 ]
// 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
// E n t e r t h e 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 ) =[1 2 1
−1]
// 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 S e q u e n c e
=−1
//OUTPUT:
// L i n e a r C o n v o l u t i o n u s i n g E q u a t i o n i s y ( n ) :
20
78
// 1 .
4.
8.
8.
21
3.
− 2.
− 1.
Experiment: 5
WRITE A SCILAB
PROGRAM TO FIND
AUTOCORRELATION AND
CROSS CORRELATION OF
THE GIVEN SEQUENCES.
Scilab code Solution 5.05 PROGRAM TO FIND AUTO CORRELATION
AND CROSS CORRELATION OF THE GIVEN SEQUENCES
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9
// VERSION : S c i l a b : 5 . 4 . 1
// OS : windows 7
//CAPTION : PROGRAM TO FIND AUTOCORRELATION AND
CROSSCORRELATION OF THE GIVEN SEQUENCES
clc ;
clear ;
close ;
x = input ( ’ E n t e r t h e I n p u t S e q u e n c e= ’ ) // x =[1 2 3 1 ]
m = length ( x ) ;
xl = 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= ’ )
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// 0
xh = xl +m -1;
n = xl :1: xh ;
subplot (2 ,2 ,1) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,n , x ) ;
title ( ’ I n p u t S e q u e n c e x [ n ] ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
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24 h = input ( ’ E n t e r t h e I m p u l s e r e s p o n s e S e q u e n c e= ’ ) ; // h
=[1 2 1 1 ]
25 l = length ( h ) ;
26 hl = 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
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S e q u e n c e= ’ ) ; // 0
hh = hl +l -1;
g = hl :1: hh ;
subplot (2 ,2 ,2) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,g , h ) ;
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 ] ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
// A u t o c o r r e l a t i o n
y = xcorr (x , x ) ;
disp ( ’ Auto C o r r e l a t i o n Of g i v e n S e q u e n c e y ( n )= ’ ) ;
disp ( y ) ;
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nx = xl + xl ;
nh = xh + xh ;
r = nx : nh ;
subplot (2 ,2 ,3) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,r , y ) ;
title ( ’ Output o f Auto C o r r e l a t i o n o f S e q u e n c e ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
// C r o s s c o r r e l a t i o n
z = xcorr (x , h ) ;
disp ( ’ C r o s s C o r r e l a t i o n o f S e q u e n c e y ( n )= ’ ) ;
disp ( z ) ;
subplot (2 ,2 ,4) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,r , z ) ;
title ( ’ Output o f C r o s s c o r r e l a t i o n o f S e q u e n c e ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
//INPUT :
// E n t e r t h e
// E n t e r t h e
// E n t e r t h e
// E n t e r t h e
=0
I n p u t S e q u e n c e =[1 2 3 1 ]
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
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 ]
lower index of impulse response Sequence
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//OUTPUT:
// Auto C o r r e l a t i o n Of g i v e n S e q u e n c e y ( n )=
// 1 .
5.
11.
15.
11.
5.
1.
// C r o s s C o r r e l a t i o n o f S e q u e n c e y ( n )=
// 1 .
3.
7.
9.
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9.
5.
1.
Experiment: 6
WRITE A SCILAB
PROGRAM TO FIND
N-POINT DFT OF THE
GIVEN SEQUENCE.
Scilab code Solution 6.06 PROGRAM TO FIND N POINT DFT OF
THE GIVEN SEQUENCE
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// VERSION : S c i l a b : 5 . 4 . 1
// OS : windows 7
// CAPTION : PROGRAM TO FIND N−POINT DFT OF THE GIVEN
SEQUENCE
clc ;
clear ;
close ;
N = input ( ’ E n t e r t h e v a l u e o f N= ’ ) ; // n−p o i n t
x = input ( ’ E n t e r t h e i n p u t s e q u e n c e x ( n )= ’ ) ;
subplot (3 ,2 ,1) ;
a = gca () ;
a . foreground = 5;
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12 a . font_color = 5;
13 a . font_style = 5;
14 plot2d3 ( x ) ;
15 title ( ’ I n p u t s e q u e n c e ’ ) ;
16 xlabel ( ’ S a m p l e s n ’ ) ;
17 ylabel ( ’ A m p l i t u d e ’ ) ;
18
19 // c a l c u l a t i o n o f DFT
20 for k =1: N
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y ( k ) =0;
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for n =1: N
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y ( k ) = y ( k ) + x ( n ) .* exp ( - %i *2* %pi *( k -1) *( n -1) / N )
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;
A = real ( y ) ;
B = imag ( y ) ;
end
end
mag = abs ( y ) ;
x1 = atan ( imag ( y ) , real ( y ) ) ;
phase = x1 *(180/ %pi ) ;
disp ( ’ The o u t p u t DFT s e q u e n c e i s : ’ ) ;
disp ( y ) ;
subplot (3 ,2 ,2) ;
a = gca () ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( y ) ;
title ( ’ Output DFT s e q u e n c e ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
//REAL VALUE
disp ( ’ The r e s u l t a n t r e a l v a l u e i s : ’ ) ;
disp ( A ) ;
subplot (3 ,2 ,3) ;
a = gca () ;
a . foreground = 5;
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a . font_color = 5;
a . font_style = 5;
plot2d3 ( A ) ;
title ( ’ R e a l V a l u e ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
//IMAGINARY VALUE
disp ( ’ The r e s u l t a n t i m a g i n a r y v a l u e i s : ’ ) ;
disp ( B ) ;
subplot (3 ,2 ,4) ;
a = gca () ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( B ) ;
title ( ’ I m a g i n a r y V a l u e ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
//MAGNITUDE RESPONSE
disp ( ’ The Magnitude r e s p o n s e i s : ’ ) ;
disp ( mag ) ;
subplot (3 ,2 ,5) ;
a = gca () ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( mag ) ;
title ( ’ Magnitude R e s p o n s e ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
//PHASE RESPONSE
disp ( ’ The Phase r e s p o n s e i s : ’ ) ;
disp ( phase ) ;
subplot (3 ,2 ,6) ;
a = gca () ;
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87 a . foreground = 5;
88 a . font_color = 5;
89 a . font_style = 5;
90 plot2d3 ( phase ) ;
91 title ( ’ Phase R e s p o n s e ’ ) ;
92 xlabel ( ’ S a m p l e s n ’ ) ;
93 ylabel ( ’ Phase ’ ) ;
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95
96
97 //INPUT :
98 // E n t e r t h e v a l u e o f N=4
99 // E n t e r t h e i n p u t s e q u e n c e x ( n ) =[1 2 3 4 ]
100
101 //OUTPUT:
102 // The o u t p u t DFT s e q u e n c e i s :
103
104
// 1 0 .
105
//− 2 . + 2 . i
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//− 2 . − 9 . 7 9 7D−16 i
107
//− 2 . − 2 . i
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109
// The r e s u l t a n t r e a l v a l u e i s :
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111
// 1 0 .
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//− 2 .
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//− 2 .
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//− 2 .
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// The r e s u l t a n t i m a g i n a r y v a l u e i s :
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// 0 .
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// 2 .
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//− 9 . 7 9 7D−16
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//− 2 .
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123
// The Magnitude r e s p o n s e i s :
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// 1 0 .
// 2 . 8 2 8 4 2 7 1
// 2 .
// 2 . 8 2 8 4 2 7 1
// The Phase r e s p o n s e i s :
// 0 .
// 1 3 5 .
//− 1 8 0 .
// − 1 3 5 .
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Experiment: 7
WRITE A SCILAB
PROGRAM TO COMPUTE
LINEAR CONVOLUTION OF
TWO SEQUENCES USING
DFT BASED APPROACH.
Scilab code Solution 7.07 PROGRAM TO COMPUTE LINEAR CONVOLUTION OF TWO SEQUENCES USING DFT BASED APPROACH
1
2
3
4
5
6
7
8
// VERSION : S c i l a b : 5 . 4 . 1
// OS : windows 7
//CAPTION : PROGRAM TO COMPUTE LINEAR CONVOLUTION OF
TWO SEQUENCES USING DFT BASED APPROACH
clc ;
clear ;
close ;
x = input ( ’ E n t e r t h e i n p u t s e q u e n c e x ( n )= ’ ) // x =[1 2 3
1]
m = length ( x ) ;
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9 xl = 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= ’ )
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// 0
xh = xl +m -1;
n = xl :1: xh ;
subplot (3 ,1 ,1) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,n , x ) ;
title ( ’ I n p u t S e q u e n c e x [ n ] ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
h = input ( ’ E n t e r t h e 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 )= ’ )
; // h =[1 2 1 −1]
l = length ( h ) ;
hl = 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
s e q u e n c e= ’ ) ; //−1
hh = hl +l -1;
g = hl :1: hh ;
subplot (3 ,1 ,2) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,n , h ) ;
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 ] ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
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41 nx = xl + hl ; // r a n g e o f k
42 nh = xh + hh ; // r a n g e o f n
43 p = m +l -1;
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x =[ x , zeros (1 ,p - l ) ];
h =[ h , zeros (1 ,p - m ) ];
// d f t − i d f t
XK = fft (x , -1) ; // DFT o f x
HK = fft (h , -1) ; // DFT o f h
YK = XK .* HK ; // M u l t i p l i c a t i o n o f DFT o f x and h
yn = fft ( YK ,1) ; // I n v e r s e DFT f o r o b t a i n i n g yn
disp ( ’ L i n e a r C o n v o l u t i o n by DFT−IDFT Method i s y ( n ) :
’ );
disp ( real ( yn ) ) ;
r = nx : nh ;
subplot (3 ,1 ,3) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,r , yn ) ;
title ( ’ Output R e s p o n s e S e q u e n c e o f L i n e a r
C o n v o l u t i o n by DFT−IDFT Method y [ n ] ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
//INPUT :
// E n t e r t h e
// E n t e r t h e
// E n t e r t h e
−1]
71 // E n t e r t h e
=−1
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i n p u t s e q u e n c e x ( n ) =[1 2 3 1 ]
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
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 ) =[1 2 1
lower index of impulse response sequence
//OUTPUT:
// L i n e a r C o n v o l u t i o n by DFT−IDFT Method i s y ( n ) :
// 1 .
4.
8.
8.
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3.
− 2.
− 1.
Experiment: 8
WRITE A SCILAB
PROGRAM TO COMPUTE
CIRCULAR CONVOLUTION
OF TWO SEQUECNES
USING BASIC EQUATION.
Scilab code Solution 8.08 PROGRAM TO COMPUTE CIRCULAR CONVOLUTION OF TWO SEQUENCES USING BASIC EQUATION
1
2
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5
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7
8
9
// VERSION : S c i l a b : 5 . 4 . 1
// OS : windows 7
//CAPTION : PROGRAM TO COMPUTE CIRCULAR CONVOLUTION
OF TWO SEQUENCES USING BASIC EQUATION
clc ;
clear ;
close ;
x = input ( ’ E n t e r t h e i n p u t s e q u e n c e= ’ ) // x =[1 1 2 2 ]
m = length ( x ) ;
xl = 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= ’ )
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// 0
xh = xl +m -1;
n = xl :1: xh ;
subplot (3 ,1 ,1) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,n , x ) ;
title ( ’ I n p u t S e q u e n c e x [ n ] ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
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21
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23
24 h = input ( ’ E n t e r t h e i m p u l s e r e s p o n s e s e q u e n c e= ’ ) ; // h
=[1 2 3 4 ]
25 l = length ( h ) ;
26 hl = 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
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42
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s e q u e n c e= ’ ) ; // 0
hh = hl +l -1;
g = hl :1: hh ;
subplot (3 ,1 ,2) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,g , h ) ;
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 ] ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
// f o r making l e n g t h o f b o t h s i g n a l s e q u a l
N = max (m , l ) ;
p =m - l ;
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if (p >=0) then
h =[ h , zeros (1 , p ) ];
else
x =[ x , zeros (1 , - p ) ];
end
for i =1: N
y ( i ) =0;
for j =1: N
k =i - j +1;
if (k <=0)
k=N+k;
end
y(i)=y(i)+x(j)*h(k);
end
end
disp ( ’ C i r c u l a r c o n v o l u t i o n by e q u a t i o n i s y [ n ] : ’ ) ;
disp ( y ) ;
nx = xl + hl ;
r = nx : length ( y ) -1;
subplot (3 ,1 ,3) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,r , y ) ;
title ( ’ Output R e s p o n s e S e q u e n c e o f C i r c u l a r
Convolution y [ n ] u s i n g B a s i c Equation ’ );
73 xlabel ( ’ S a m p l e s n ’ ) ;
74 ylabel ( ’ A m p l i t u d e ’ ) ;
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//INPUT :
// E n t e r t h e
// E n t e r t h e
// E n t e r t h e
// E n t e r t h e
=0
i n p u t s e q u e n c e =[1 1 2 2 ]
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
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 3 4 ]
lower index of impulse response sequence
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86
87
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//OUTPUT:
// C i r c u l a r c o n v o l u t i o n by e q u a t i o n i s y [ n ] :
//
15.
// 1 7 .
// 1 5 .
// 1 3 .
37
Experiment: 9
WRITE A SCILAB
PROGRAM TO COMPUTE
CIRCULAR CONVOLUTION
OF THE TWO SEQUENCES
USING DFT BASED
APPROACH.
Scilab code Solution 9.09 PROGRAM TO COMPUTE CIRCULAR CONVOLUTION OF THE TWO SEQUENCES USING DFT BASED APPROACH
1
2
3
4
5
6
7
// VERSION : S c i l a b : 5 . 4 . 1
// OS : windows 7
//CAPTION :PROGRAM TO COMPUTE CIRCULAR CONVOLUTION OF
THE TWO SEQUENCES USING DFT BASED APPROACH
clc ;
clear ;
close ;
x = input ( ’ E n t e r t h e I n p u t s e q u e n c e= ’ ) // x =[1 1 2 2 ]
38
8 m = length ( x ) ;
9 xl = 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= ’ )
// 0
xh = xl +m -1;
n = xl :1: xh ;
subplot (3 ,1 ,1) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,n , x ) ;
title ( ’ I n p u t S e q u e n c e x [ n ] ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
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24 h = input ( ’ E n t e r t h e I m p u l s e r e s p o n s e s e q u e n c e= ’ ) ; // h
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26
=[1 2 3 4 ]
l = length ( h ) ;
hl = 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
s e q u e n c e= ’ ) ; // 0
hh = hl +l -1;
g = hl :1: hh ;
subplot (3 ,1 ,2) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,g , h ) ;
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 ] ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
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40
41 // f o r making l e n g t h
42 N = max (m , l ) ;
o f both s i g n a l s e q u a l
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p =m - l ;
if (p >=0) then
h =[ h , zeros (1 , p ) ];
else
x =[ x , zeros (1 , - p ) ];
end
XK = fft (x , -1) ;
HK = fft (h , -1) ;
YK = XK .* HK ;
y = ifft ( YK ) ;
disp ( ’ C i r c u l a r c o n v o l u t i o n by DFT i s y ( n ) : ’ ) ;
disp ( real ( y ) ) ;
nx = xl + hl ;
r = nx : length ( y ) -1;
subplot (3 ,1 ,3) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,r , y ) ;
title ( ’ Output R e s p o n s e S e q u e n c e o f C i r c u l a r
C o n v o l u t i o n y [ n ] u s i n g DFT ’ ) ;
66 xlabel ( ’ S a m p l e s n ’ ) ;
67 ylabel ( ’ A m p l i t u d e ’ ) ;
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//INPUT :
// E n t e r t h e
// E n t e r t h e
// E n t e r t h e
// E n t e r t h e
=0
I n p u t s e q u e n c e =[1 1 2 2 ]
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
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 3 4 ]
lower index of impulse response sequence
//OUTPUT:
// C i r c u l a r c o n v o l u t i o n by DFT i s y ( n ) :
// 1 5 .
17.
15.
13.
40
41
Experiment: 10
WRITE A SCILAB
PROGRAM TO COMPUTE
BLOCK CONVOLUTION
USING OVERLAP ADD
METHOD
Scilab code Solution 10.10 PROGRAM TO COMPUTE BLOCK CONVOLUTION USING OVERLAP ADD METHOD
1
2
3
4
5
6
7
8
// VERSION : S c i l a b : 5 . 4 . 1
// OS : windows 7
//CAPTION : PROGRAM TO COMPUTE BLOCK CONVOLUTION
USING OVERLAP ADD METHOD
clc ;
clear ;
close ;
x = input ( ’ E n t e r t h e i n p u t s e q u e n c e= ’ ) // x =[1 2 −1 2 3
−2 −3 −1 1 1 2 −1]
m = length ( x ) ;
42
9 xl = 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= ’ )
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// 0
xh = xl +m -1;
n = xl :1: xh ;
subplot (3 ,1 ,1) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,n , x ) ;
title ( ’ I n p u t S e q u e n c e x [ n ] ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
h = input ( ’ E n t e r t h e i m p u l s e r e s p o n s e s e q u e n c e= ’ ) ; // h
=[1 2 3 −1]
l = length ( h ) ;
hl = 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
s e q u e n c e= ’ ) ; // 0
hh = hl +l -1;
g = hl :1: hh ;
subplot (3 ,1 ,2) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,g , h ) ;
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 ] ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
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41 N = m +l -1;
42 h1 =[ h zeros (1 ,l -1) ];
43 n3 = length ( h1 ) ;
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y = zeros (1 , N ) ;
H = fft ( h1 ) ;
for i =1: l : m
if i <=( m +l -1) then
x1 =[ x ( i : i + n3 - l ) zeros (1 , n3 - l ) ];
else
x1 =[ x ( i : m ) zeros (1 , n3 - l ) ];
end
x2 = fft ( x1 ) ;
x3 = x2 .* H ;
x4 = round ( ifft ( x3 ) ) ;
if ( i ==1)
y (1: n3 ) = x4 (1: n3 ) ;
else
y ( i : i + n3 -1) = y ( i : i + n3 -1) + x4 (1: n3 ) ;
end
end
disp ( ’ Output s e q u e n c e u s i n g O v e r l a p add method y ( n ) :
’ );
disp ( y (1: N ) ) ;
nx = xl + hl ;
r = nx : length ( y ) -1;
subplot (3 ,1 ,3) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,r , y ) ;
title ( ’ Output s e q u e n c e u s i n g O v e r l a p add method y [ n ]
’ );
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
//INPUT :
44
// E n t e r
2 −1]
81 // E n t e r
82 // E n t e r
83 // E n t e r
=0
80
84
85
86
87
88
89
90
91
92
93
94
95
t h e i n p u t s e q u e n c e =[1 2 −1 2 3 −2 −3 −1 1 1
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
t h e 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 3 −1]
the lower index of impulse response sequence
//OUTPUT:
// Output s e q u e n c e u s i n g O v e r l a p add method y ( n ) :
//
column
// 1 .
4.
− 8.
//
1 t o 12
6.
3.
5.
8.
column 13 t o 15
// 3 .
− 5.
2.
5.
1.
45
11.
0.
− 16.
Experiment: 11
WRITE A SCILAB
PROGRAM RO COMPUTE
BLOCK CONVOLUTION
USING OVERLAP SAVE
METHOD.
Scilab code Solution 11.11 PROGRAM TO COMPUTE BLOCK CONVOLUTION USING OVERLAP SAVE METHOD
1
2
3
4
5
6
7
8
// VERSION : S c i l a b : 5 . 4 . 1
// OS : windows 7
//CAPTION : PROGRAM TO COMPUTE BLOCK CONVOLUTION
USINF OVERLAP SAVE METHOD
clc ;
clear ;
close ;
x = input ( ’ E n t e r t h e i n p u t s e q u e n c e= ’ ) // x =[1 2 −1 2 3
−2 −3 −1 1 1 2 −1]
m = length ( x ) ;
46
9 xl = 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= ’ )
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// 0
xh = xl +m -1;
n = xl :1: xh ;
subplot (3 ,1 ,1) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,n , x ) ;
title ( ’ I n p u t S e q u e n c e x [ n ] ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
h = input ( ’ E n t e r t h e i m p u l s e r e s p o n s e s e q u e n c e= ’ ) ; // h
=[1 2 3 −1]
l = length ( h ) ;
hl = 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
s e q u e n c e= ’ ) ; // 0
hh = hl +l -1;
g = hl :1: hh ;
subplot (3 ,1 ,2) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,g , h ) ;
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 ] ’ ) ;
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
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41 N = m +l -1;
42 h1 =[ h zeros (1 ,N - m ) ];
43 n3 = length ( h1 ) ;
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y = zeros (1 , N ) ;
x1 =[ zeros (1 , n3 - l ) x zeros (1 , n3 ) ];
H = fft ( h1 ) ;
for i =1: l : N
y1 = x1 ( i : i +(2*( n3 - l ) ) ) ;
y2 = fft ( y1 ) ;
y3 = y2 .* H ;
y4 = round ( ifft ( y3 ) ) ;
y ( i :( i + n3 - l ) ) = y4 ( l : n3 ) ;
end
disp ( ’ Output s e q u e n c e u s i n g o v e r l a p s a v e method Y( n )
: ’ );
disp ( y (1: N ) ) ;
nx = xl + hl ;
r = nx : length ( y ) -1;
subplot (3 ,1 ,3) ;
a = gca () ;
a . x_location = ” o r i g i n ” ;
a . y_location = ” o r i g i n ” ;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot2d3 ( ’ gnn ’ ,r , y ) ;
title ( ’ Output s e q u e n c e u s i n g O v e r l a p s a v e method y [ n
] ’ );
xlabel ( ’ S a m p l e s n ’ ) ;
ylabel ( ’ A m p l i t u d e ’ ) ;
//INPUT :
// E n t e r t h e
2 −1]
72 // E n t e r t h e
73 // E n t e r t h e
74 // E n t e r t h e
=0
75
76
77
i n p u t s e q u e n c e =[1 2 −1 2 3 −2 −3 −1 1 1
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
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 3 −1]
lower index of impulse response sequence
//OUTPUT:
// Output s e q u e n c e u s i n g o v e r l a p s a v e method Y( n ) :
48
78
79
80
81
82
83
84
85
86
//
column
// 1 .
4.
− 8.
//
// 3 .
1 t o 12
6.
3.
5.
8.
5.
column 13 t o 15
− 5.
2.
1.
49
11.
0.
− 16.
Experiment: 12
WRITE A SCILAB
PROGRAM TO FIND FFT
USING DECIMATION IN
TIME(DIT) METHOD.
Scilab code Solution 12.12 PROGRAM TO FIND FFT USING DECIMATION IN TIME METHOD
1
2
3
4
5
6
7
// VERSION : S c i l a b : 5 . 4 . 1
// OS : windows 7
// CAPTION : PROGRAM TO FIND FFT USING DECIMATION IN
TIME( DIT ) METHOD
clc ;
clear ;
close ;
x = input ( ’ E n t e r I n p u t s e q u e n c e= ’ ) ; // x =[0 1 2 3 4 5
6 7]
o r x= [ 0 1 2 3 ]
N = length ( x ) ;
8
9
10 s = log2 ( N ) ; // c o m p u t i n g a r r a y
11
12 // f o r s e q u e n c e s i z e 8
50
size
13 if ( N ==8) then
14 stage1 =1;
15
16 x =[ x (1) x (5) x (3) x (7) x (2) x (6) x (4) x (8) ]; //
17
18
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41
stage1
for stage =1: s
for index =0:(2^ stage ) :( N -1) ; // s e r i e s o f
b u t t e r f l y f o r each s t a g e
for n =0:( stage1 -1) ;
// c r e a t e b u t t e r l f y
and s a v e r e s u l t
pos = n + index +1; // i n d e x o f d a t a s a m p l e
pow =(2^( s - stage ) ) * n ; // p a r t o f power o f
complex m u l t i p l i e r
w = exp (( -1* %i ) *(2* %pi ) * pow / N ) ; // c o m p l e x
multiplier
a = x ( pos ) + x ( pos + stage1 ) .* w ; // 1 s t p a r t
of butterfly creating operations
b = x ( pos ) -x ( pos + stage1 ) .* w ; // 2 nd p a r t o f
butterfly creating operation
x ( pos ) = a ; // s a v i n g c o m p u t a t i o n o f 1 s t
half
x ( pos + stage1 ) = b ; // s a v i n g c o m p u t a t i o n
of second part
end
end
stage1 =2* stage1 ; // c o m p u t i n g n e x t s t a g e
end
y=x;
disp ( ’FFT o f t h e g i v e n i n p u t s e q u e n c e i s y ( n )= ’ ) ;
disp ( y ) ;
// f o r s e q u e n c e s i z e −4
else
stage1 =1;
x =[ x (1) x (3) x (2) x (4) ]; //
for stage =1: s
51
stage1
for index =0:(2^ stage ) :( N -1) ; // s e r i e s o f
b u t t e r f l y f o r each s t a g e
for n =0:( stage1 -1) ;
// c r e a t e b u t t e r l f y
and s a v e r e s u l t
pos = n + index +1; // i n d e x o f d a t a s a m p l e
pow =(2^( s - stage ) ) * n ; // p a r t o f power o f
complex m u l t i p l i e r
w = exp (( -1* %i ) *(2* %pi ) * pow / N ) ; // c o m p l e x
multiplier
a = x ( pos ) + x ( pos + stage1 ) .* w ; // 1 s t p a r t
of butterfly creating operations
b = x ( pos ) -x ( pos + stage1 ) .* w ; // 2 nd p a r t o f
butterfly creating operation
x ( pos ) = a ; // s a v i n g c o m p u t a t i o n o f 1 s t
half
x ( pos + stage1 ) = b ; // s a v i n g c o m p u t a t i o n
of second part
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end
end
stage1 =2* stage1 ; // c o m p u t i n g n e x t s t a g e
end
y=x;
disp ( ’FFT o f t h e g i v e n i n p u t s e q u e n c e y ( n )= ’ ) ;
disp ( y ) ;
end
//INPUT :
// E n t e r I n p u t s e q u e n c e =[0 1 2 3 4 5 6 7 ]
//OUTPUT:
// FFT o f t h e g i v e n i n p u t s e q u e n c e i s y ( n )=
//
//
column 1 t o 3
28.
− 4. + 9.6568542 i
52
− 4. + 4. i
72
73
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81
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83
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86
87
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91
//
//
column 4 t o 5
− 4. + 1.6568542 i
//
//
− 4.
column 6 t o 7
− 4. − 1.6568542 i
//
− 4. − 4. i
column 8
// − 4 . − 9 . 6 5 6 8 5 4 2 i
// INPUT :
// E n t e r I n p u t s e q u e n c e =[0 1 2 3 ]
//OUTPUT:
//FFT o f t h e g i v e n i n p u t s e q u e n c e y ( n )=
//
6.
− 2. + 2. i
− 2.
53
− 2. − 2. i
Experiment: 13
WRITE A SCILAB
PROGRAM FOR
DESIGNING OF FIR FILTER
FOR LOW PASS, HIGH PASS,
BANDPASS AND BAND
REJECT RESPONSES.
Scilab code Solution 13.13 PROGRAM FOR DESIGNING OF FIR FILTER FOR LOW PASS HIGH PASS BAND PASS AND BAND REJECT
RESPONSES
1
2
3
4
5
// VERSION : S c i l a b : 5 . 4 . 1
// OS : windows 7
//CAPTION : PROGRAM FOR DESIGNING OF FIR LOW PASS ,
HIGH PASS , BAND PASS AND BAND REJECT RESPONSES
// F i l t e r l e n g t h =5 , Order o f
Rectangular
54
f i l t e r =4 , Window=
6 clc ;
7 clear ;
8 close ;
9 xdel ( winsid () ) ;
10 fc1 = input ( ’ E n t e r t h e a n a l o g
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c u t o f f F r e q u e n c y i n Hz= ’
) ; // 250
fc2 = input ( ’ E n t e r a n a l o g h i g h e r c u t o f f f r e q u e n c y i n
Hz= ’ ) ; // 600
fs = input ( ’ E n t e r t h e a n a l o g s a m p l i n g F r e q u e n c y i n Hz=
’ ) ; // 2 0 0 0
M = input ( ’ E n t e r o r d e r o f f i l t e r = ’ ) ; // 4
w1 =(2* %pi ) *( fc1 / fs ) ;
w2 =(2* %pi ) *( fc2 / fs ) ;
// FIR LOW PASS FILTER
disp ( ’ DESIGNING OF FIR LOW PASS FILTER : ’ ) ;
disp ( w1 , ’ D i g i t a l C u t o f f F r e q u e n c y i n r a d i a n s . c y c l e s
/ samples ’ );
wc1 = w1 / %pi ;
disp ( wc1 , ’ N o r m a l i z e d d i g i t a l c u t o f f f r e q u e n c y i n
c y c l e s / samples ’ );
[ wft , wfm , fr ]= wfir ( ’ l p ’ ,M +1 ,[ wc1 /2 ,0] , ’ r e ’ ,[0 ,0]) ;
disp ( wft , ’ I m p u l s e R e s p o n s e o f LPF FIR f i l t e r : h ( n )= ’ )
;
// p l o t t i n g m a g n i t u d e r e s p o n s e o f LPF f i l t e r
subplot (2 ,4 ,1) ;
a = gca () ;
a . thickness =2;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot (2* fr , wfm ) ;
title ( ’ Magnitude R e s p o n s e o f FIR LPF ’ ) ;
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 ’ ) ;
ylabel ( ’ Magnitude | H(w) | ’ ) ;
xgrid (1)
subplot (2 ,4 ,2) ;
a = gca () ;
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a . thickness =2;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot ( fr * fs , wfm ) ;
title ( ’ Magnitude R e s p o n s e o f FIR LPF ’ ) ;
xlabel ( ’ Analog f r e q u e n c y i n Hz f ’ ) ;
ylabel ( ’ Magnitude | H(w) | ’ ) ;
xgrid (1)
// FIR HIGH PASS FILTER
disp ( ’ DESIGNING OF FIR HIGH PASS FILTER : ’ )
disp ( w1 , ’ D i g i t a l C u t o f f F r e q u e n c y i n r a d i a n s . c y c l e s
/ samples ’ );
wc1 = w1 / %pi ;
disp ( wc1 , ’ N o r m a l i z e d d i g i t a l c u t o f f f r e q u e n c y i n
c y c l e s / samples ’ );
[ wft , wfm , fr ]= wfir ( ’ hp ’ ,M +1 ,[ wc1 /2 ,0] , ’ r e ’ ,[0 ,0]) ;
disp ( wft , ’ I m p u l s e R e s p o n s e o f HPF FIR f i l t e r : h ( n )= ’ )
;
// p l o t t i n g m a g n i t u d e r e s p o n s e o f HPF f i l t e r
subplot (2 ,4 ,3) ;
a = gca () ;
a . thickness =2;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot (2* fr , wfm ) ;
title ( ’ Magnitude R e s p o n s e o f FIR HPF ’ ) ;
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 ’ ) ;
ylabel ( ’ Magnitude | H(w) | ’ ) ;
xgrid (1)
subplot (2 ,4 ,4) ;
a = gca () ;
a . thickness =2;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
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104
plot ( fr * fs , wfm ) ;
title ( ’ Magnitude R e s p o n s e o f FIR HPF ’ ) ;
xlabel ( ’ Analog f r e q u e n c y i n Hz f ’ ) ;
ylabel ( ’ Magnitude | H(w) | ’ ) ;
xgrid (1)
// FIR BAND PASS FILTER
disp ( ’ DESIGNING OF FIR BAND PASS FILTER : ’ )
disp ( w1 , ’ D i g i t a l l o w e r c u t o f f f r e q u e n c y i n r a d i a n s
c y c l e s / samples ’ );
disp ( w2 , ’ D i g i t a l h i g h e r c u t o f f f r e q u e n c y i n r a d i a n s
c y c l e s / samples ’ );
wc1 = w1 / %pi ;
wc2 = w2 / %pi ;
disp ( wc1 , ’ N o r m a l i z e d d i g i t a l l o w e r c u t o f f f r e q u e n c y
i n c y c l e s / samples ’ );
disp ( wc2 , ’ N o r m a l i z e d d i g i t a l h i g h e r c u t o f f f r e q u e n c y
i n c y c l e s / samples ’ );
[ wft , wfm , fr ]= wfir ( ’ bp ’ , M +1 ,[ wc1 /2 , wc2 /2] , ’ r e ’
,[0 ,0]) ;
disp ( wft , ’ I m p u l s e r e s p o n s e o f Bandpass F i l t e r FIR
f i l t e r : h ( n )= ’ ) ;
// p l o t t i n g t h e m a g n i t u d e R e s p o n s e o f HPF FIR f i l t e r
subplot (2 ,4 ,5) ;
a = gca () ;
a . thickness =2;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot (2* fr , wfm ) ;
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 ’ ) ;
ylabel ( ’ Magnitude | H(W) | ’ ) ;
title ( ’ Magnitude R e s p o n s e o f FIR BPF ’ ) ;
xgrid (1) ;
subplot (2 ,4 ,6) ;
a = gca () ;
a . thickness =2;
a . foreground = 5;
57
105 a . font_color = 5;
106 a . font_style = 5;
107 plot ( fr * fs , wfm ) ;
108 xlabel ( ’ Analog F r e q u e n c y i n Hz f ’ ) ;
109 ylabel ( ’ Magnitude | H(w) | ’ ) ;
110 title ( ’ Magnitude r e s p o n s e o f FIR BPF ’ ) ;
111 xgrid (1) ;
112
113 // FIR BAND REJECT FILTER
114 disp ( ’ DESIGNING OF FIR BAND REJECT FILTER : ’ )
115 disp ( w1 , ’ D i g i t a l l o w e r c u t o f f f r e q u e n c y i n r a d i a n s
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c y c l e s / samples ’ );
disp ( w2 , ’ D i g i t a l h i g h e r c u t o f f f r e q u e n c y i n r a d i a n s
c y c l e s / samples ’ );
wc1 = w1 / %pi ;
wc2 = w2 / %pi ;
disp ( wc1 , ’ N o r m a l i z e d d i g i t a l l o w e r c u t o f f f r e q u e n c y
i n c y c l e s / samples ’ );
disp ( wc2 , ’ N o r m a l i z e d d i g i t a l h i g h e r c u t o f f f r e q u e n c y
i n c y c l e s / samples ’ );
[ wft , wfm , fr ]= wfir ( ’ s b ’ , M +1 ,[ wc1 /2 , wc2 /2] , ’ r e ’
,[0 ,0]) ;
disp ( wft , ’ I m p u l s e r e s p o n s e o f B a n d r e j e c t F i l t e r FIR
f i l t e r : h ( n )= ’ ) ;
// p l o t t i n g t h e m a g n i t u d e R e s p o n s e o f HPF FIR f i l t e r
subplot (2 ,4 ,7) ;
a = gca () ;
a . thickness =2;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot (2* fr , wfm ) ;
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 ’ ) ;
ylabel ( ’ Magnitude | H(W) | ’ ) ;
title ( ’ Magnitude R e s p o n s e o f FIR BRF ’ ) ;
xgrid (1) ;
subplot (2 ,4 ,8) ;
a = gca () ;
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a . thickness =2;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot ( fr * fs , wfm ) ;
xlabel ( ’ Analog F r e q u e n c y i n Hz f ’ ) ;
ylabel ( ’ Magnitude | H(w) | ’ ) ;
title ( ’ Magnitude r e s p o n s e o f FIR BRF ’ ) ;
xgrid (1) ;
//INPUT :
// E n t e r t h e a n a l o g c u t o f f F r e q u e n c y i n Hz=250
// E n t e r a n a l o g h i g h e r c u t o f f f r e q u e n c y i n Hz=600
// E n t e r t h e a n a l o g s a m p l i n g F r e q u e n c y i n Hz=2000
// E n t e r o r d e r o f f i l t e r =4
//OUTPUT:
// DESIGNING OF FIR LOW PASS FILTER :
// D i g i t a l C u t o f f F r e q u e n c y i n r a d i a n s . c y c l e s /
samples
//
0.7853982
// N o r m a l i z e d d i g i t a l c u t o f f f r e q u e n c y i n c y c l e s /
samples
//
0.25
// I m p u l s e R e s p o n s e o f LPF FIR f i l t e r : h ( n )=
// 0 . 1 5 9 1 5 4 9
0.1591549
0.2250791
0.25
0.2250791
//DESIGNING OF FIR HIGH PASS FILTER :
// D i g i t a l C u t o f f F r e q u e n c y i n r a d i a n s . c y c l e s /
samples
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//
0.7853982
// N o r m a l i z e d d i g i t a l c u t o f f f r e q u e n c y i n c y c l e s /
samples
// 0 . 2 5
// I m p u l s e R e s p o n s e o f HPF FIR f i l t e r : h ( n )=
// − 0 . 1 5 9 1 5 4 9
0.1591549
− 0.2250791
0.75
− 0.2250791
−
//DESIGNING OF FIR BAND PASS FILTER :
// D i g i t a l l o w e r c u t o f f f r e q u e n c y i n r a d i a n s c y c l e s /
samples
// 0 . 7 8 5 3 9 8 2
// D i g i t a l h i g h e r c u t o f f f r e q u e n c y i n r a d i a n s c y c l e s
/ samples
// 1 . 8 8 4 9 5 5 6
// N o r m a l i z e d d i g i t a l l o w e r c u t o f f f r e q u e n c y i n
c y c l e s / samples
// 0 . 2 5
// N o r m a l i z e d d i g i t a l h i g h e r c u t o f f f r e q u e n c y i n
c y c l e s / samples
// 0 . 6
// I m p u l s e r e s p o n s e o f Bandpass F i l t e r FIR f i l t e r : h (
n )=
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//− 0 . 2 5 2 7 0 3 9
0.2527039
0.0776516
0.35
0.0776516
−
//DESIGNING OF FIR BAND REJECT FILTER :
// D i g i t a l l o w e r c u t o f f f r e q u e n c y i n r a d i a n s c y c l e s /
samples
// 0 . 7 8 5 3 9 8 2
// D i g i t a l h i g h e r c u t o f f f r e q u e n c y i n r a d i a n s c y c l e s
/ samples
// 1 . 8 8 4 9 5 5 6
// N o r m a l i z e d d i g i t a l l o w e r c u t o f f f r e q u e n c y i n
c y c l e s / samples
// 0 . 2 5
// N o r m a l i z e d d i g i t a l h i g h e r c u t o f f f r e q u e n c y i n
c y c l e s / samples
// 0 . 6
// I m p u l s e r e s p o n s e o f B a n d r e j e c t F i l t e r FIR f i l t e r :
h ( n )=
// 0 . 2 5 2 7 0 3 9 − 0 . 0 7 7 6 5 1 6
0.2527039
61
0.65
− 0.0776516
Experiment: 14
WRITE A SCILAB
PROGRAM TO DESIGN
DIGITAL IIR
BUTTERWORTH LOW PASS
FILTER.
Scilab code Solution 14.14 PROGRAM TO DESIGN DIGITAL IIR BUTTERWORTH LOW PASS FILTER
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// VERSION : S c i l a b : 5 . 4 . 1
// OS : windows 7
//CAPTION : PROGRAM TO DESIGN BUTTERWORTH LOW PASS
FILTER
clc ;
clear ;
close ;
xdel ( winsid () ) ;
fc = input ( ’ E n t e r c u t o f f f r e q i n Hz f c= ’ ) ; // 1 0 0 0
fs = input ( ’ E n t e r s a m p l i n g f r e q i n Hz f s = ’ ) ; // 1 0 0 0 0
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N = input ( ’ E n t e r o r d e r o f B u t t e r w o r t h f i l t e r N= ’ ) ; // 4
Fp =2* fc / fs ;
[ Hz ]= iir (N , ’ l p ’ , ’ b u t t ’ ,[ Fp /2 ,0] ,[0 ,0]) ;
[ Hw , w ]= frmag ( Hz ,256) ;
subplot (2 ,1 ,1) ;
a = gca () ;
a . thickness =2;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot (2* w , abs ( Hw ) ) ;
title ( ’ Magnitude R e s p o n s e o f I I R LPF ’ ) ;
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 ’ ) ;
ylabel ( ’ Magnitude | H(w) | ’ ) ;
xgrid (1) ;
subplot (2 ,1 ,2) ;
a = gca () ;
a . thickness =2;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot (2* w * fs , abs ( Hw ) ) ;
title ( ’ Magnitude R e s p o n s e o f I I R LPF ’ ) ;
xlabel ( ’ Analog F r e q u e n c y i n Hz f ’ ) ;
ylabel ( ’ Magnitude | H(w) | ’ ) ;
xgrid (1) ;
//INPUT :
// E n t e r c u t o f f f r e q i n Hz f c =1000
// E n t e r s a m p l i n g f r e q i n Hz f s =10000
// E n t e r o r d e r o f B u t t e r w o r t h f i l t e r N=4
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Experiment: 15
WRITE A SCILAB
PROGRAM TO DESIGN
DIGITAL IIR CHEBYSHEW
FILTER.
Scilab code Solution 15.15 PROGRAM TO DESIGN DIGITAL IIR CHEBYSHEW
FILTER
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// VERSION : S c i l a b : 5 . 4 . 1
// OS : windows 7
//CAPTION : PROGRAM TO DESIGN DIGITAL CHEBYSHEV I I R
FILTER
clc ;
clear ;
close ;
wp = input ( ’ E n t e r t h e D i g i t a l P a s s Band e d g e F r e q u e n c y
= ’ ) ; // 0 . 2 ∗ %pi
ws = input ( ’ E n t e r t h e D i g i t a l S t o p Band e d g e F r e q u e n c y
= ’ ) ; // 0 . 6 ∗ %pi
t = input ( ’ S a m p l i n g I n t e r v a l = ’ ) ; // 1
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del1 = input ( ’ E n t e r t h e P a s s Band R i p p l e= ’ ) ; // 0 . 8
del2 = input ( ’ E n t e r t h e S t o p Band R i p p l e= ’ ) ; // 0 . 2
disp ( wp , ’Wp= ’ ) ;
disp ( ws , ’Ws= ’ ) ;
del = sqrt (((1/ del2 ) ^2) -1) ;
disp ( del , ’ D e l t a= ’ ) ;
epsilon = sqrt (((1/ del1 ) ^2) -1) ;
disp ( epsilon , ’ E p s i l o n= ’ ) ;
N =( acosh ( del / epsilon ) ) /( acosh ( ws / wp ) ) ;
N = ceil ( N ) ;
disp (N , ’N= ’ ) ;
wc = wp /((((1/ del1 ) ^2) -1) ^(1/(2* N ) ) ) ;
[ pols , gn ]= zpch1 (N , epsilon , wp ) ;
hs = poly ( gn , ’ s ’ , ’ c o e f f ’ ) / real ( poly ( pols , ’ s ’ ) ) ;
z = poly (0 , ’ z ’ ) ;
hz = horner ( hs ,((2/ t ) *(( z -1) /( z +1) ) ) ) ;
hw = frmag ( hz (2) , hz (3) ,512) ; // f r e q . r e s p o n s e f o r 512
points
w =0: %pi /511: %pi ;
a = gca () ;
a . thickness =2;
a . foreground = 5;
a . font_color = 5;
a . font_style = 5;
plot ( w / %pi , abs ( hw ) ) ;
xgrid (1) ;
title ( ’ Magnitude R e s p o n s e o f D i g i t a l Chebyshew LPF
IIR F i l t e r ’ );
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 ’ ) ;
ylabel ( ’ Magnitude i n db ’ ) ;
//INPUT :
// E n t e r t h e D i g i t a l P a s s Band e d g e F r e q u e n c y =0.2∗ %pi
// E n t e r t h e D i g i t a l S t o p Band e d g e F r e q u e n c y =0.6∗ %pi
// S a m p l i n g I n t e r v a l =1
// E n t e r t h e P a s s Band R i p p l e =0.8
// E n t e r t h e S t o p Band R i p p l e =0.2
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//OUTPUT:
//Wp=
// 0 . 6 2 8 3 1 8 5
//Ws=
// 1 . 8 8 4 9 5 5 6
// D e l t a=
// 4 . 8 9 8 9 7 9 5
// E p s i l o n=
//
0.75
//N=
2.
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