ECE324: DIGITAL SIGNAL PROCESSING LABORATORY

Practical No.: 6

Roll No.:B-54 Registration No.: 11205816 Name: Shyamveer Singh

Aim: Design a FIR filter1. Low pass filter

Mathematical Expressions Required:

Wc = 2π(fc / fs)Hd(w) =

Inputs :

wc=pi/2m=7

Program Code:

function[]=lpfshyam(wc,N)

i=(N-1)/2;

wn=wc/pi;

syms w

for n=-i:i

y(n+i+1)=(int(exp(w*n*j),-wc,wc))*1/(2*pi);

end

y=sym2poly(y)

for n=0:N-1

w(n+1)=0.54-0.46*cos(2*pi*n/(N-1));

h(n+1)=y(n+1).*w(n+1);

end

h=sym2poly(h)

fir1(N-1,wn)

end

Aim: Design a FIR filter1. Low pass filter

Output:>> lpfshyam((pi/2),7)

y =

-0.1061 0 0.3183 0.5000 0.3183 0 -0.1061

h =

-0.0085 0 0.2451 0.5000 0.2451 0 -0.0085

ans =

-0.0087 0.0000 0.2518 0.5138 0.2518 0.0000 -0.0087

Analysis/ Learning outcomes

This experiment give me a better understanding of low pass filter andhelps me to understand its working graphically.

Aim: Design a FIR filter2. High pass filter

Mathematical Expressions Required:

Wc = 2π(fc / fs)Hd(w) =

Inputs:

Wc=pi/2m=7

Program Codes:

function[]=hpfshyam(wc,N)

i=(N-1)/2;

wn=wc/pi;

syms w

for n=-i:i

y(n+i+1)=(1/2*pi)*(int(exp(j*w*n),-pi,-wc)+int(exp(j*w*n),wc,pi));

end

y=sym2poly(y)

for n=0:N-1

w(n+1)=0.54-0.46*cos(2*pi*n/(N-1));

h(n+1)=y(n+1).*w(n+1);

end

h=sym2poly(h)

fir1(N-1,wn)

end

Output:>> hpfshyam((pi/2),7)

y =

1.0472 0 -3.1416 4.9348 -3.1416 0 1.0472

h =

0.0838 0 -2.4190 4.9348 -2.4190 0 0.0838

ans =

-0.0087 0.0000 0.2518 0.5138 0.2518 0.0000 -0.0087

Aim: Design a FIR filter3. Band pass filter

Mathematical Expressions Required:

Wc = 2π(fc / fs)Hd(w) =

Inputs:

Wc1=pi/4 ,wc2=pi/6m=7

Program Codes:

function[]=bpfshyam(wc1,wc2,N)

i=(N-1)/2;

wn1=wc1/pi;

wn2=wc2/pi;

syms w

for n=-i:i

y(n+i+1)=(1/2*pi)*(int(exp(j*w*n),-wc2,-wc1)+int(exp(j*w*n),wc1,wc2));

end

y=sym2poly(y)

for n=0:N-1

w(n+1)=0.54-0.46*cos(2*pi*n/(N-1));

h(n+1)=y(n+1).*w(n+1);

end

h=sym2poly(h)

fir1(N-1,wn1)

end

Analysis/ Learning outcomes

This experiment give me a better understanding of High pass filter andhelps me to understand its working graphically

Outputs/ Graphs/ Plots:>> bpfshyam((pi/4),(pi/6),7)

y =

0.3067 -0.2104 -0.6506 -0.8225 -0.6506 -0.2104 0.3067

h =

0.0245 -0.0652 -0.5010 -0.8225 -0.5010 -0.0652 0.0245

Aim: Design a FIR filter4. Band stop filter

Mathematical Expressions Required:

Wc = 2π(fc / fs)Hd(w) =

Inputs:

Wc1=pi/4 , wc2=pi/6m=7

Analysis/ Learning outcomes

This experiment give me a better understanding of band pass filter andhelps me to understand its working graphically

Program Codes:

function[]=bsfshyam(wc1,wc2,N)

i=(N-1)/2;

wn1=wc1/pi;

wn2=wc2/pi;

syms w

for n=-i:i

y(n+i+1)=(1/2*pi)*(int(exp(j*w*n),-pi,-wc2)+int(exp(j*w*n),-wc1,wc1)+int(exp(j*w*n),wc2,pi));

end

y=sym2poly(y)

for n=0:N-1

w(n+1)=0.54-0.46*cos(2*pi*n/(N-1));

h(n+1)=y(n+1).*w(n+1);

end

h=sym2poly(h)

fir1(N-1,wn1)

end

Outputs/ Graphs/ Plots:

>> bsfshyam(pi/4,pi/6,7)

y =

-0.3067 0.2104 0.6506 10.6921 0.6506 0.2104 -0.3067

h =

-0.0245 0.0652 0.5010 10.6921 0.5010 0.0652 -0.0245

Analysis/ Learning outcomes

This experiment give me a better understanding of band stop filter andhelps me to understand its working graphically