Meril Final List to Be Given

8/6/2019 Meril Final List to Be Given

1/64

Expt 1:Basic Matlab Programs

Program 1

Generate unit impulse and unit step signal

clc;

clearall;close all;

n=-10:20;

i=(n==0);subplot(2,2,1);stem(n,i);

title('impulse response');

xlabel('range');

ylabel('Amplitude');s=(n>=0);

subplot(2,2,2);

stem(n,s);

title('step response');xlabel('range');

ylabel('amplitude');


2/64

Program 2

Plot the magnitude and phase response of fourier transform of sequence

x(n)= 0.5n u(n)

clc;clearall;

close all;

w=-pi:0.1:pi;b=[1];

a=[1 0.5];

h=freqz(b,a,w);

x=real(h);y=imag(h);

subplot(2,2,3);

stem(w/pi,x);

title('real part');subplot(2,2,4);

stem(w/pi,y);title('imaginary part');

mag=abs(h);

ph=angle(h);

subplot(2,2,1);plot(w/pi,mag);

title('mag response');

subplot(2,2,2);plot(w/pi,ph);

title('phase response');


3/64


4/64

Program 3

Find the Convolution of two sequences

x(n)={1,0,1,2,-1,3,2}

h(n)={1,1,2,2,1,1}

clc;

clearall;

close all;x=input('enter the value x(n)');

h=input('enter the value h(n)');

y=conv(x,h);

stem(y);xlabel('n');


title('convolution');

enter the value x(n)[1 0 1 2 -1 3 2]

enter the value h(n)[1 1 2 2 1 1]


5/64

Program 4

Generate a complex exponential sequence of the form exp(-1/12)+j*(pi/6)n

clc;clearall;

close all;

m1=input('enter the value for m1');m2=input('enter the value for m2');

n=-m1:m2;

y=((-1/12)+j*(pi/6));p=y*n;

x=exp(p);

stem(n,x);

xlabel('n');ylabel('Amplitude');

enter the value for m110

enter the value for m223


6/64

Program 5

Consider 50 points of a 10 Hz sinusoid sampled at fs=50 Hz and scaled by T.

Compute the 50 point FFT of the sequence and plot the spectrum.

clc;clearal;

close all;

t=input('enter the value of t');n=0:49;

x=t*sin(((2*pi*10)/50)*n);

figure(1);

stem(n,x);figure(2);

y=fft(x,50);

stem(n,abs(y));

figure(3);stem(n,angle(y));

enter the value of t.02


7/64


8/64

Program 6

Compute 32 point DFT X(K) of length 32 sinusoidal sequence x(n) of frequency 10

Hz with a sampling rate of 64 Hz.

clc;

clearal;

close all;n=0:31;

x=sin(((2*pi*10)/64)*n);

figure(1);

stem(n,x);title('input sigmal');

figure(2);

y=fft(x,32);

stem(n,abs(y));title('32-point dft');


9/64


10/64

Program 7

Verify that convolution in time domain is equal to multiplication in frequency

domain.

x1=input ('enter the seq1');

x2=input('enter the seq2');l1=length(x1)

l2=length(x2)

y=conv(x1,x2);figure(1);

stem(y);

xlabel('n');

ylabel('amplitude');title('covoluted sequence');

y1=fft(x1,(l1+l2-1));

y2=fft(x2,(l1+l2-1));

x=y1.*y2;x1=ifft(x,(l1+l2-1));

figure(2);stem(x1);xlabel('n');


title('recovered sequence');

enter the seq1[1 2 3 4]

enter the seq2[1 1 1]


11/64


12/64

Program 8

Find the DFT and IDFT of a sequence without using the formula

clc;

clearall;

close all;xn= input('enter the values of x');

N=input ('enter the value of N');

a=exp(-j*2*pi/N);fork=1:N

X(k)=0;

forn=1:N

X(k)=X(k)+ xn(n)*a^((n-1)*(k-1));end

end

display(X)

forn=1:Nx(n)=0;

fork=1:Nx(n)=x(n)+ X(k)*a^(-(n-1)*(k-1));

end

end

display(x/N)

enter the values of x [1 2 3 4]

enter the value of N 4

X =

10.0000 -2.0000 + 2.0000i -2.0000 - 0.0000i -2.0000 - 2.0000i

ans =

1.0000 - 0.0000i 2.0000 - 0.0000i 3.0000 - 0.0000i 4.0000 + 0.0000i


13/64

Program 9

Determine the first 41 samples of the impulse response and step response of the

causal LTI system described by

y(n)+.7y(n-1)-.45y(n-2)-.6y(n-3) = .8x(n)-.44x(n-1)+.36x(n-2)+.02x(n-3)

clc;clearall;

close all;

b=[0.8 -0.44 0.36 0.02];a=[1 0.7 -0.45 -0.6];

n=0:41;

x=(n==0);

y=(n>=0);imp=filter(b,a,x);

st=filter(b,a,y);

subplot(2,2,1);

stem(n,imp);title('impulse response');

subplot(2,2,2);stem(n,st);

title('step response');

[h,n]=impz(b,a,41);

subplot(2,2,3);stem(n,h);

title('impulse response');

[w,n]=stepz(b,a,41);subplot(2,2,4);

stem(n,w);

title('step response');


14/64


15/64

ANALOG FILTER DESIGN

Program 1

Determine the transfer function of an analog 4th order maximally flat LPF with cut

off frequency of 1rad/second.Plot the gain response also.

clear all;

close all;

w=0:0.01:20;

[z,p,k]=buttap(4);

[b,a]=zp2tf(z,p,k)

h=freqs(b,a,w);

mag=20*log10(abs(h));

plot(w,mag);

grid on;

ylabel('magnitude in db');

xlabel('angular frequency');

OUTPUT

b =

0 0 0 0 1

a =

1.0000 2.6131 3.4142 2.6131 1.0000


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17/64

Program 2

Determine the lowest order Low pass chebyshev filter with 0.25dB cutoff frequency

at 1.5Khz and a minimum attenuation of 25dB at 6Khz without using inbuilt

commands. Verify the results using cheb1ord.

clear all;

close all;

wp=input('enter pass band edge');

ws=input('enter stopband edge');

rp=input('enter passband ripple in db');

rs=input('enter stopband ripple in db');

ds=10^(2*rs/20)

dp=10^(2*rp/20)

x=sqrt(ds-1);

y=sqrt(dp-1);

N=acosh(x/y)/acosh(ws/wp)

[n,wn]=cheb1ord((2500*pi),(12000*pi),.25,25,'s')

OUTPUT

enter pass band edge3000*pienter stopband edge12000*pi

enter passband ripple in db.25

enter stopband ripple in db25

ds =

316.2278

dp =

1.0593

N =

2.4148

n =

3

wn =

7.8540e+003


18/64

Program 3

Design a butterworth LPF with .25 db cut off frequency at 1.5KHz and

minimum attenuation of 25db at 6khz.plot the gain response and verify that the

filter design meets given specification.

clc;

clear all;

close all;

w=0:100:50000;

[n,wn]=buttord(3000*pi,12000*pi,.25,25,'s');

[b,a]=bu tter(n,wn,'s');

h=freqs(b,a,w);

mag=20*log10(abs(h));

plot(w/(2*pi),mag);

grid on;

xlabel(' frequency');

ylabel('magnitude');

title('Magnitude response of butterworth LPF');

OUTPUT


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Program 3

Design a chebychev hpf with 1 db cut off frequency at 700hz and

minimum attenuation of 32db at 500hz.plot the gain response and verify that the

filter design meets given specification.The maximum input frequency is 1000hz.

clc;

clear all;

fp=input('enter passband edge frequency');fs=input('enter stopband edge frequency');


rs=input('enter ripple in stopbandin db');

wp=2*pi*fp;ws=2*pi*fs;

w=0:50:40000;[n,wp] = cheb1ord(wp,ws,rp,rs,'s')

[b,a] = cheby1(n,rp,wp,'high','s')

h=freqs(b,a,w);m=20*log10(abs(h));

plot(w/(2*pi),m);

grid on

title('gain response ');xlabel('frequency in hertz');

ylabel('absolute magnitude in db');

enter passband edge frequency700

enter stopband edge frequency500

enter passband ripple in db1enter ripple in stopbandin db32

n =

6

wp =

4.3982e+003

b =

0.8913 0 0 0 0 0 0


20/64

a =

1.0e+023 *

0.0000 0.0000 0.0000 0.0000 0.0000 0.0002 1.0505


21/64

Program 4

Determine the transfer function of a Low pass type chebyshev1 analog filter with .

25 db cutoff frequency at 1.5Khz and minimum attenuation of 25 db at 6Khz.Plot

gain response.

clc;clear all;

close all;

wp=input('enter pass band frequency');

ws=input('enter stop band frequency');rp=input('pass band ripples');

rs=input('stop band ripples');

w=0:100:65000;

[n,wp]=cheb1ord(wp,ws,rp,rs,'s')[b,a]=cheby1(n,rp,wp,'s')

h=freqs(b,a,w);mag=20*log10(abs(h));

plot(w/(2*pi),mag);


ylabel('magnitude');title('Magnitude response of chebyshev 1 LPF');

grid on;

OUTPUT

enter pass band frequency 3000*pienter stop band frequency 12000*pi

pass band ripples .25

stop band ripples 25

n =

3

wp =

9.4248e+003

b =

1.0e+011 *


22/64

0 0 0 8.5980

a =

1.0e+011 *

0.0000 0.0000 0.0017 8.5980


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24/64

n1 =

4

wn =

121.7901


25/64

Program 6

Design an analog chebychev type1 filter with following specifications. Pass band

edges at 4KHz and 7KHz.Stop band edges at 3KHz and 8KHz.Passband ripple of

2dB and stopband attenution of 50dB.

clear all;

close all;

w=0:1000:20000*pi;

wp1=input('enter pass band edge1');


ws1=input('enter stopband edge1');

ws2=input('enter stopband edge2w0');



w0=sqrt(wp2*wp1);

wp=1;

a=ws1*ws2;

b=wp1*wp2;

if(a


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27/64

Program 7

Design an analog chebychev type1 band stop filter with following specifications.Pass

band edges at 100Hz and 600Hz.Stop band edges at 200Hz and 400Hz.Passband

rippleof 3dB and stopband attenution of 20dB.

clear all;

close all;

w=0:100:2000*pi;







w0=sqrt(ws2*ws1);

ws=1;

a=ws1*ws2;

b=wp1*wp2;

if(a>b)

wp11=w0^2/wp2;

wp=(ws2-ws1)/(wp2-wp11);

else

wp22=w0^2/wp1;

wp=(ws2-ws1)/(wp22-wp1);

end

bw=ws2-ws1;

[n,wp]=cheb1ord(wp,ws,rp,rs,'s');

[b,a]=cheby1(n,rp,wp,'s');

[bt,at]=lp2bs(b,a,w0,bw);

h=freqs(bt,at,w);

m1=abs(h);m=20*log10(m1);

plot(w/(2*pi),m);

xlabel('angular frequncy ');

ylabel('magnitude in decibels');

title('magnitude respinse');

grid on;

enter pass band edge1200*pi

enter pass band edge21200*pienter stopband edge1400*pi

enter stopband edge2w0800*pienter passband ripple in db3enter stopband ripple in db20


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29/64

Expt 2 IIRDIGITAL FILTER DESIGN

Program 1

Using bilinear transformation method design digital low pass butterworth filter

with a maximum input frequency of 5KHz. at a sampling rate of with followingspecifications. Pass band edge at 2500Hz Stop band edges at 3000Hz .Pass band

ripple of 1dB Minimum stop band attenuation of 10dB.Plot the gain response.

clear all;

close all;

fp=input('enter pass band edge frequency in hz');

fs=input('enter stopband edge frequency in hz');



fm=input('enter the maxinput frequency');

fsa=2*fm;wp=2*pi*fp/fsa;

ws=2*pi*fs/fsa;

wp1=wp/pi;

ws1=ws/pi;

w=0:0.01:pi;

[n,wp]=cheb1ord(wp/pi,ws/pi,rp,rs);

[b,a]=cheby1(n,rp,wp);

h=freqz(b,a,w);

m=20*log10(abs(h));

plot(w/pi,m);

grid on;

enter pass band edge frequency in hz2500enter stopband edge frequency in hz3000

enter passband ripple in db1

enter stopband ripple in db10enter the maxinput frequency5000


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31/64

Program 2

Using bilinear transformation method design digital high pass butterworth filter at

a sampling rate of 1.5MHz with following specifications. Pass band edge at 600 KHz

Stop band edges at 210 KHz .Pass band ripple of 0.4dBMinimum stop bandattenuation of 45dB.What are the specifications of analog prototype LPF .Show all

TFs. Plot the gain response of prototype analog LPF, analog HPF and desired

digital HPF

clc;

clear all;

close all;

ft=input('entere sampling frequency');

T=1/ft;

fp=input('enter passband edge frequency');

fs=input('enter stopband edge frequency');


rs=input('enter ripple in stopbandindb');w1=0:0.1:10;

w=0:0.01:pi;

w2=0:1000:10000000;

wp=2*pi*fp/ft

ws=2*pi*fs/ft

omp=2/T*(tan(wp/2))

oms=2/T*(tan(ws/2))

oms1=omp/oms

omp1=1

[n,omp1] = cheb1ord(omp1,oms1,rp,rs,'s')

[b,a] = cheby1(n,rp,omp1,'s');

h1=freqs(b,a,w1);

m1=20*log10(abs(h1));figure;

plot(w1,m1);

grid on;

xlabel('angular frequency ');


title('magnitude response of lpf');

[bt,at]=lp2hp(b,a,omp);

h2=freqs(bt,at,w2);

m2=20*log10(abs(h2));

figure;

plot(w2,m2);

grid on;

xlabel('angular frequency ');

ylabel('magnitude in decibels');title('magnitude response of analog hpf');

[nr,dr]=bilinear(bt,at,ft);

h=freqz(nr,dr,w);

m=20*log10(abs(h))

figure;

plot((w*ft)/(2*pi),m);

xlabel(' frequency ');



32/64

title('magnitude response of digital hpf');

grid on;

OUTPUTentere sampling frequency1500000

enter passband edge frequency600000

enter stopband edge frequency210000enter passband ripple in db.4

enter ripple in stopbandindb45

wp =

2.5133

ws =

0.8796

omp =

9.2331e+006

oms =

1.4117e+006

oms1 =

6.5404

omp1 =

1

n =

3

omp1 =

1


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34/64

Program 3

Design a digital band pass filter with following specifications

Pass band edges at .45 pi and .65 pi

Stopband edges at.3pi and .75pi

Pass band ripple -1dBStopband attenuation-40dB

clear all;

close all;

w=0:0.01:pi;

fs=input('enter the sampling frequency');





rp=input('enter passband ripple in db');rs=input('enter stopband ripple in db');

T=1/fs;

wp1new=(2/T)*tan(wp1*pi/2);


ws1new=(2/T)*tan(ws1*pi/2);


w0=sqrt(wp2new*wp1new);

wp=1;

a=ws1new*ws2new;

b=wp1new*wp2new;

if(a


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enter stopband edge1.3

enter stopband edge2w0.75

enter passband ripple in db1enter stopband ripple in db40

>>


36/64

Program 4

Design a digital band stop filter with following specifications

Pass band edges at .45 pi and .65 pi

Stopband edges at.3pi and .75pi

Pass band ripple -1dBStopband attenuation-40dB

clear all;

close all;

w=0:0.1:pi;






rp=input('enter passband ripple in db');rs=input('enter stopband ripple in db');

T=1/fs;





w0=sqrt(ws2new*ws1new);

ws=1;

a=ws1new*ws2new;

b=wp1new*wp2new;

if(a>b)

wp11=w0^2/wp2new;

wp=(ws2new-ws1new)/(wp2new-wp11);else

wp22=w0^2/wp1new;

wp=(ws2new-ws1new)/(wp22-wp1new);

end

bw=ws2new-ws1new;

[n,wp]=cheb1ord(wp,ws,rp,rs,'s');

[b,a]=cheby1(n,rp,wp,'s');

[bt,at]=lp2bs(b,a,w0,bw);

[nr,dr]=bilinear(bt,at,fs);

h=freqz(nr,dr,w);

m1=abs(h);

m=20*log10(m1);

plot(w/pi,m);

xlabel('angular frequncy ');ylabel('magnitude in decibels');

title('magnitude respinse');

grid on;

OUTPUTenter the sampling frequency.5


37/64

enter pass band edge1.3

enter pass band edge2.75

enter stopband edge1.45enter stopband edge2w0.65

enter passband ripple in db1

enter stopband ripple in db40


38/64

Program 5

Design digital low pass butterworth filter with following specifications. Pass band

edge at .25pi. Stop band edges at .55pi.Pass band ripple of 0.5dBMinimum stop

band attenuation of 15dB. Plot the gain response

clear all;

close all;wp=input('enter pass band edge');

ws=input('enter stopband edge');



w=0:0.01:pi;

[n,wn]=buttord(wp,ws,rp,rs);

[b,a]=butter(n,wn);

h=freqz(b,a,w);

m=20*log10(abs(h));

plot(w/pi,m);

grid on;

enter pass band edge .25enter stopband edge.55

enter passband ripple in db.5enter stopband ripple in db15


39/64

Program 6

Design digital high pass butterworth filter with following specifications. Pass band

edge at 700Hz. Stop band edges at 500Hz.Pass band ripple of 1dB.Minimum stop

band attenuation of 32dB.The maximum input frequency is 1KHz. Plot the gain

responseclear all;

close all;

fp=input('enter pass band edge frequency in hz');

fs=input('enter stopband edge frequency in hz');



fm=input('enter the maxinput frequency');

fsa=2*fm;

wp=2*pi*fp/fsa;

ws=2*pi*fs/fsa;

wp1=wp/pi;

ws1=ws/pi;

w=0:0.01:pi;

[n,wn]=cheb1ord(wp1,ws1,rp,rs);

[b,a]=cheby1(n,rp,wn,'high');h=freqz(b,a,w);

m=20*log10(abs(h));

plot(w/pi,m);

grid on;

enter pass band edge frequency in hz 700

enter stopband edge frequency in hz500

enter passband ripple in db1enter stopband ripple in db32

enter the maxinput frequency1000


40/64

I


41/64

Program 7

Using impulse invariant transformation method design digital high pass

butterworth filter at a sampling rate 0.5 with following specifications. Pass band

edge at .25*pi.Stop band edges at .55*pi .Pass band ripple of 0.5dB.Minimum stop

band attenuation of 15dB. Plot the gain response

clc;

clear all;

close all;

w=0:0.01:pi;

fs=input('enter sampling frequency');

T=1/fs;

omp=input('enter passband edge frequency');

oms=input('enter stopband edge frequency');

rp=input('enter ripple in passband in db');

rs=input('enter ripple in stopbandin db');

wp=omp/pi;

ws=oms/pi;

[n,wn] = buttord(wp,ws,rp,rs,'s')

[b,a] =butter(n,wn,'s');[nr,dr]=impinvar(b,a,fs);

h=freqz(nr,dr,w);

m=20*log10(abs(h))

plot(w/pi,m);

grid on

title('magnitude response');

xlabel('normalised frequency in radians');


enter sampling frequency0.5enter passband edge frequency0.25*pi

enter stopband edge frequency0.55*pi

enter ripple in passband in db0.5enter ripple in stopbandin db15


42/64


43/64

Expt 3 FIR DIGITAL FILTER DESIGN

Program 1

Design an FIR LPF for cut off frequency of.5*pi. Let the order of the filter be

49.Use rectangular and hann windows.

clear all;

close all;

wp=input('enter cut off frequency');

N=input('enter the length of filter');

t=(N-1)/2;

e=0.001;

n=0:N-1;

hd=(sin((n-t+e)*wp))./((n-t+e)*pi);

wr=boxcar(N);

hn=hd.*wr';

w=0:0.01:pi;

h=freqz(hn,1,w);

figure;

plot(w,(20*log10(abs(h))));

hold on;

plot(w,unwrap(angle(h)),'r');

grid on;


ylabel('magnitude and phase ');

title('magnitude and phase response of lpf using rectangular window');

hd=(sin((n-t+e)*wp))./((n-t+e)*pi);

wr=hann(N);

hn=hd.*wr';

w=0:0.01:pi;

h=freqz(hn,1,w);

figure;plot(w,(20*log10(abs(h))));

hold on;


grid on;


ylabel('magnitude ');

title('magnitude response of lpf using hann window');

enter cut off frequency.5*pi

enter the length of filter7


44/64


45/64

Program 2

Design an FIR HPF for cut off frequency of.5*pi.Let the order of the filter be 49.Use

hamming and blackmann windows.

clc;

clear all;

close all;wp=input('enter cut off frequency');


t=(N-1)/2;

e=0.001;

n=0:N-1;

hd=(sin((n-t+e)*pi)-sin((n-t+e)*wp))./((n-t+e)*pi);

wr=hamming(N);

hn=hd.*wr';

w=0:0.01:pi;

h=freqz(hn,1,w);

figure;


hold on;


grid on;



title('magnitude response of lpf using hamming window');


wr=blackman(N);

hn=hd.*wr';

w=0:0.01:pi;

h=freqz(hn,1,w);

figure;


hold on;

plot(w,unwrap(angle(h)),'r');grid on;



title('magnitude response of lpf using blackman window');




46/64


47/64

Program 3

Design an FIR BPF for cut off frequency of.5*pi.Let the order of the filter be 49.Use

bartlett window.

clc;

clear all;close all;

wp1=input('enter cut off frequency1');



t=(N-1)/2;

e=0.001;

n=0:N-1;

hd=(sin((n-t+e)*wp2)-sin((n-t+e)*wp1))./((n-t+e)*pi);

wr=bartlett(N);

hn=hd.*wr';

w=0:0.01:pi;

h=freqz(hn,1,w);


hold on;


grid on;



title('magnitude response of bpf using bartlett window');

enter cut off frequency1.25*pi

enter cut off frequency2.75*pi



48/64


49/64

Program 4

Design an FIR BSF for cut off frequency of.5*pi.Let the order of the filter be 49.Use

hann window.

clc;

clear all;

close all;




t=(N-1)/2;

e=0.001;

n=0:N-1;

w=0:0.01:pi;

hd=(sin((n-t+e)*pi)-sin((n-t+e)*wp2)+sin((n-t+e)*wp1))./((n-t+e)*pi);

wr=hann(N);

hn=hd.*wr';

h=freqz(hn,1,w);figure;


hold on;


grid on;



title('magnitude response of bsf using hann window');

enter cut off frequency1.25*pienter cut off frequency2.75*pi



50/64


51/64

Program 5

Design a high pass FIR filter using Dolph Chebychev window Let the order of the

filter be 49.Let the cut off frequency is 0.5*pi

clc;clear all;

close all;

wp=input('enter cut off frequency');


t=(N-1)/2;

e=0.001;

n=0:N-1;


wr=chebwin(N);

hn=hd.*wr';

w=0:0.01:pi;

h=freqz(hn,1,w);


hold on;


grid on;



title('magnitude response of hpf using dolph window');




52/64


53/64

Program 6

Design a FIR bandpass filter using Kaiser window. Let the order of the filter be 49.

Let the cut off frequencies are 1500Hz and 3000Hz.Let the sampling frequency is

8KHz.

clc;clear all;

fp1=input('enter cut off frequency1');

fp2=input('enter cut off frequency2');

N=input('enter the order of filter');


wp1=2*pi*fp1/fs;

wp2=2*pi*fp2/fs;

hn=fir1(N,[(wp1/pi) (wp2/pi)],'bandpass',kaiser(N+1));

w=0:0.01:pi;

h=freqz(hn,1,w);

figure;

plot((w*fs)/(2*pi),(20*log10(abs(h))));

hold on;

plot((w*fs)/(2*pi),unwrap(angle(h)),'r');

grid on;


ylabel('magnitude and phase ');

title('magnitude and phase response of bpf using kaiser window');

enter cut off frequency11500

enter cut off frequency23000

enter the order of filter50enter the sampling frequency8000


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Program 7

Design a FIR bandpass filter using Kaiser window. Let the order of the filter be 49.

Let the passband edge frequency 0.3pi and stopband edge frequency is 0.4pi.

clc;clear all;

close all;

w=0:0.01:pi;

wp=input('enter passband edgefrequency');

ws=input('enter stopband edge frequency');

f=[wp/pi ws/pi];

a=[1 0];

dev = [0.05 0.01];

[n,wn,beta] = kaiserord(f,a,dev);

b=fir1(n,wn,kaiser(n+1,beta));

h=freqz(b,1,w);

figure

plot(w/pi,20*log10(abs(h)));grid on

hold on

a=unwrap(angle(h));

plot(w/pi,a,'--');

title('low pass filter using kaiser window');



a2=[0 1];

[n1,wn1,beta2,high] = kaiserord(f,a2,dev);

b1=fir1(n1,wn1,'high',kaiser(n1+1,beta2));

h1=freqz(b1,1,w);

figure

plot(w/pi,20*log10(abs(h1)));grid on

hold on

a1=unwrap(angle(h1));

plot(w/pi,a1,'--');

title('highpass filter using kaiser window');



enter passband edgefrequency0.3*pienter stopband edge frequency0.4*pi


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Expt 4 Fast Fourier Transform

Program 1

Implement FFT using DITclc;clear all;close all;

N=input('enter the lenth of fft');

x=input('enter the sequence');

X=bitrevorder(x); %taking the bit reversal of input sequencedisplay(X);

j=sqrt(-1);

w=exp(-j*2*pi/N);

M=log2(N);%total no. of stages

display(M);

for m=1:M

for t= 1:(2^(m-1))

k(t)=N*(t-1)/2^m;%calculating exp factor

end;

l=k;

for n=1:2^(M-m)-1

l=[l k ];%replicating exp fator based on exp repeat factor(ERF)

end

display(l);

inc=2^(m-1)

jump=0;jump1=0;

for c=1:N/2

y(1+jump)=X(1+jump)+w^(l(c))*X(1+jump+2^(m-1));%(l(c)) taking

individual elements from array

y(1+jump+2^(m-1))=X(1+jump)-w^(l(c))*X(1+jump+2^(m-1));

inc=inc-1

if(inc==0)

jump1=2^(m-1)+1%calculating the jumping location

jump=jump+jump1

inc=2^(m-1)

else

jump1=1;jump=jump+jump1%calculating the jumping location

end

end

display(y)%output of each stage

X=y;

end

display(X)%fft of input sequence

xx=fft(x);

display(xx)%fft of input sequence

enter the lenth of fft8enter the sequence[1 2 3 4 4 3 2 1]

X =

1 4 3 2 2 3 4 1


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M =

3

l =

0 0 0 0

inc =

1

inc =

0

jump1 =

2

jump =

2

inc =

1

inc =

0

jump1 =

2

jump =


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4

inc =

1

inc =

0

jump1 =

2

jump =

6

inc =

1

inc =

0

jump1 =

2

jump =

8

inc =


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1

y =

5 -3 5 1 5 -1 5 3

l =

0 2 0 2

inc =

2

inc =

1

jump =

1

inc =

0

jump1 =

3

jump =

4

inc =

2


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inc =

1

jump =

5

inc =

0

jump1 =

3

jump =

8

inc =

2

y =

Columns 1 through 5

10.0000 -3.0000 - 1.0000i 0 -3.0000 + 1.0000i 10.0000

Columns 6 through 8

-1.0000 - 3.0000i 0 -1.0000 + 3.0000i

l =

0 1 2 3


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inc =

4

inc =

3

jump =

1

inc =

2

jump =

2

inc =

1

jump =

3

inc =

0

jump1 =

5


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jump =

8

inc =

4

y =

Columns 1 through 5

20.0000 -5.8284 - 2.4142i 0 -0.1716 - 0.4142i 0

Columns 6 through 8

-0.1716 + 0.4142i 0 -5.8284 + 2.4142i

X =

Columns 1 through 5

20.0000 -5.8284 - 2.4142i 0 -0.1716 - 0.4142i 0

Columns 6 through 8

-0.1716 + 0.4142i 0 -5.8284 + 2.4142i

xx =

Columns 1 through 5

20.0000 -5.8284 - 2.4142i 0 -0.1716 - 0.4142i 0

Columns 6 through 8

-0.1716 + 0.4142i 0 -5.8284 + 2.4142i

>>


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clc;

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