This document describes a student's laboratory experiment on designing different types of FIR filters, including low pass, high pass, band pass, and band stop filters. It provides the mathematical expressions, input parameters, program codes, and output responses for each filter. The student learned about the design and graphical working of each filter type through this experiment.

1. ECE324: DIGITAL SIGNAL PROCESSING LABORATORY
Practical No.: 6
Roll No.:B-54 Registration No.: 11205816 Name: Shyamveer Singh
Aim: Design a FIR filter
1. Low pass filter
Mathematical Expressions Required:
Wc = 2π(fc / fs)
Hd(w) =
Inputs :
wc=pi/2
m=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

2. Aim: Design a FIR filter
1. 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 and
helps me to understand its working graphically.
Aim: Design a FIR filter
2. High pass filter
Mathematical Expressions Required:
Wc = 2π(fc / fs)
Hd(w) =

3. Inputs:
Wc=pi/2
m=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

4. Aim: Design a FIR filter
3. Band pass filter
Mathematical Expressions Required:
Wc = 2π(fc / fs)
Hd(w) =
Inputs:
Wc1=pi/4 ,wc2=pi/6
m=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 and
helps me to understand its working graphically

5. 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 filter
4. Band stop filter
Mathematical Expressions Required:
Wc = 2π(fc / fs)
Hd(w) =
Inputs:
Wc1=pi/4 , wc2=pi/6
m=7
Analysis/ Learning outcomes
This experiment give me a better understanding of band pass filter and
helps me to understand its working graphically

6. 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 and
helps me to understand its working graphically