Matlab Code for Circular convolution using DFT

1. ECE324: DIGITAL SIGNAL PROCESSING LABORATORY
Practical No.:-05
Roll No.: B-54 Registration No.:11205816 Name:Shyamveer Singh
Program Codes: (Function files)
Circular Convolution:
Marks Obtained
Job Execution (Out of 40):_________
Online Submission (Out of 10):________
Aim: To compute the convolution linear and curricular both using DFT and IDFT techniques.
Mathematical Expressions Required:
Inputs (Should be allocated by the Instructor, Individually):
1. Linear Convolution
x=[1 2 3 4]
h=[1 2 3 4]
2. Circular Convolution
X=[1 2 3 4]
H=[1 2 3 4]
3. Circular Convolution
X=[1 2 3 4]
H=[1 2 3 4]

2. function[y]=shyamcconv(x,h)
m=length(x)
l=length(h)
n=m+l-1
for t=1:n
y(t)=0;
for k=max(1,t-(m-1)):min(t,m)
y(t)=y(t)+x(k).*h(mod(t-k+1,n))
end
end
z=y
y=cconv(x,h)
LINEAR CONVOLUTION USING DFT AND IDFT:
function[y]=shyamlinconv(x,h)
Nx=length(x)
Nh=length(h)
Ny=Nx+Nh-1
X=[x,zeros(1,Ny)]
H=[h,zeros(1,Ny)]
p1=fft(X)
p2=fft(H)
p=p1.*p2;
y=ifft(p)
z=conv(x,h)
end
CIRCULAR CONVOLUTION USING DFT OR IDFT
function[y]=shyamcirconv(x,h)
n=input('valu of n')
X=[x,zeros(1,n)]
H=[h,zeros(1,n)]
p1=fft(X)
p2=fft(H)
p=p1.*p2;
y=ifft(p)
z=cconv(x,h)
end
Outputs/ Graphs/ Plots:
1. Linear Convolution :
>> x=[1 2 3 4]
x =
1 2 3 4
>> h=[1 2 3 4]
h =

3. 1 2 3 4
>> shyamlinconv(x,h)
Nx =
4
Nh =
4
Ny =
7
X =
1 2 3 4 0 0 0 0 0 0 0
H =
1 2 3 4 0 0 0 0 0 0 0
p1 =
Columns 1 through 3
10.0000 3.3595 - 7.7695i -3.9717 - 2.9596i
Columns 4 through 6
-0.5014 + 2.5041i 2.6283 - 0.7046i -1.0147 - 1.9645i
Columns 7 through 9
-1.0147 + 1.9645i 2.6283 + 0.7046i -0.5014 - 2.5041i
Columns 10 through 11
-3.9717 + 2.9596i 3.3595 + 7.7695i

4. p2 =
Columns 1 through 3
10.0000 3.3595 - 7.7695i -3.9717 - 2.9596i
Columns 4 through 6
-0.5014 + 2.5041i 2.6283 - 0.7046i -1.0147 - 1.9645i
Columns 7 through 9
-1.0147 + 1.9645i 2.6283 + 0.7046i -0.5014 - 2.5041i
Columns 10 through 11
-3.9717 + 2.9596i 3.3595 + 7.7695i
y =
Columns 1 through 6
1.0000 4.0000 10.0000 20.0000 25.0000 24.0000
Columns 7 through 11
16.0000 0.0000 0 -0.0000 -0.0000
z =
1 4 10 20 25 24 16
ans =
Columns 1 through 6
1.0000 4.0000 10.0000 20.0000 25.0000 24.0000
Columns 7 through 11
16.0000 0.0000 0 -0.0000 -0.0000
2. Circular Convolution Using DFT IDFT
>> shyamcirconv(x,h)
valu of n 4

5. n =
4
X =
1 2 3 4 0 0 0 0
H =
1 2 3 4 0 0 0 0
p1 =
Columns 1 through 3
10.0000 -0.4142 - 7.2426i -2.0000 + 2.0000i
Columns 4 through 6
2.4142 - 1.2426i -2.0000 2.4142 + 1.2426i
Columns 7 through 8
-2.0000 - 2.0000i -0.4142 + 7.2426i
p2 =
Columns 1 through 3
10.0000 -0.4142 - 7.2426i -2.0000 + 2.0000i
Columns 4 through 6
2.4142 - 1.2426i -2.0000 2.4142 + 1.2426i
Columns 7 through 8
-2.0000 - 2.0000i -0.4142 + 7.2426i
y =
Columns 1 through 6

6. 1.0000 4.0000 10.0000 20.0000 25.0000 24.0000
Columns 7 through 8
16.0000 -0.0000
z =
Columns 1 through 6
1.0000 4.0000 10.0000 20.0000 25.0000 24.0000
Column 7
16.0000
ans =
Columns 1 through 6
1.0000 4.0000 10.0000 20.0000 25.0000 24.0000
Columns 7 through 8
16.0000 -0.0000
2. Circular Convolution
>> shyamcconv(x,h)
m =
4
l =
4
n =
7
y =

7. 1
y =
1 2
y =
1 4
y =
1 4 3
y =
1 4 7
y =
1 4 10
y =
1 4 10 4
y =
1 4 10 10
y =
1 4 10 16
y =
1 4 10 20

8. y =
1 4 10 20 8
y =
1 4 10 20 17
y =
1 4 10 20 25
y =
1 4 10 20 25 12
y =
1 4 10 20 25 24
y =
1 4 10 20 25 24 16
z =
1 4 10 20 25 24 16
y =
Columns 1 through 6
1.0000 4.0000 10.0000 20.0000 25.0000 24.0000
Column 7
16.0000
ans =

9. Columns 1 through 6
1.0000 4.0000 10.0000 20.0000 25.0000 24.0000
Column 7
16.0000
Analysis/ Learning outcomes: After performing this experiment we know about the various
MATLAB commands and how to pad zeros in any signal and how to match the dimensions of
matrix.