Question

Convolve the following two series: f=(0,0,1,1,1,1,1,1,1,1,1,1,1,1,0,0) g=(1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0). using the following methods: (a) Using Fast Convolution (Using MATLAB and show codes) (b) Using Fast Convolution with sufficient zeros added to avoid wrap-around effects (c) Using your convolution function. Plot all of the outputs in the time domain (Using MATLAB and show codes), using the same scale to enable better comparison. Comment on, and explain in detail, the differences between the results.

Answer 1

MATLAB CODE

x=[0,0,1,1,1,1,1,1,1,1,1,1,1,1,0,0]; % Input Signal
h=[1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0]
%% Part a
Ly=length(x)+length(h)-1; %
Ly2=pow2(nextpow2(Ly)); % Find smallest power of 2 that is > Ly
X=fft(x, Ly2);       % Fast Fourier transform
H=fft(h, Ly2);   % Fast Fourier transform
Y=X.*H;     %
y=real(ifft(Y, Ly2)); % Inverse fast Fourier transform
disp('Convolution Result of Part a is :');
y=int16(y(1:1:Ly)) % Take just the first N elements

% %Part b
% part b will be same as part a except we will add 0 to the end of the array
% whose dimension is less than other to equalise the dimension
Lx=length(x);
Lh=length(h);
if(Lx>Lh)
Zero_M=zeros(1,Lx-Lh); % Will create an array of zeros to desired length
h=[h Zero_M]; % Appending to h as its length is less than x
else
Zero_M=zeros(1,Lh-Lx); % Will create an array of zeros to desired length
x=[x Zero_M]; % Appending to x as its length is less than h
end
Ly=length(x)+length(h)-1; %
Ly2=pow2(nextpow2(Ly)); % Find smallest power of 2 that is > Ly
X=fft(x, Ly2);       % Fast Fourier transform
H=fft(h, Ly2);   % Fast Fourier transform
Y=X.*H;     %
y=real(ifft(Y, Ly2)); % Inverse fast Fourier transform
disp('Convolution Result of Part b is :');
y=int16(y(1:1:Ly)) % Take just the first N elements
%% Part c
% Using Convolution function
disp('Convolution Result of Part c is : ');
conv(x,h)

x=[0,0,1,1,1,1,1,1,1,1,1,1,1,1,0,0];
h=[1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0]
%% Part a
Ly=length(x)+length(h)-1; %
Ly2=pow2(nextpow2(Ly)); % Find smallest power of 2 that is > Ly
X=fft(x, Ly2);       % Fast Fourier transform
H=fft(h, Ly2);   % Fast Fourier transform
Y=X.*H;     %
y=real(ifft(Y, Ly2)); % Inverse fast Fourier transform
disp('Convolution Result of Part a is :');
y=int16(y(1:1:Ly)) % Take just the first N elements

% %Part b
% part b will be same as part a except we will add 0 to the end of the array
% whose dimension is less than other to equalise the dimension
Lx=length(x);
Lh=length(h);
if(Lx>Lh)
Zero_M=zeros(1,Lx-Lh); % Will create an array of zeros to desired length
h=[h Zero_M]; % Appending to h as its length is less than x
else
Zero_M=zeros(1,Lh-Lx); % Will create an array of zeros to desired length
x=[x Zero_M]; % Appending to x as its length is less than h
end
Ly=length(x)+length(h)-1; %
Ly2=pow2(nextpow2(Ly)); % Find smallest power of 2 that is > Ly
X=fft(x, Ly2);       % Fast Fourier transform
H=fft(h, Ly2);   % Fast Fourier transform
Y=X.*H;     %
y=real(ifft(Y, Ly2)); % Inverse fast Fourier transform
disp('Convolution Result of Part b is :');
y=int16(y(1:1:Ly)) % Take just the first N elements
%% Part c
% Using Convolution function
disp('Convolution Result of Part c is : ');
conv(x,h)

RESULT:

Convolution Result of Part a is :
y =

Columns 1 through 26:

0 0 1 2 3 4 5 6 7 8 8 8 8 8 7 6 5 4 3 2 1 0 0 0 0 0

Columns 27 through 31:

0 0 0 0 0

Convolution Result of Part b is :
y =

Columns 1 through 26:

0 0 1 2 3 4 5 6 7 8 8 8 8 8 7 6 5 4 3 2 1 0 0 0 0 0

Columns 27 through 31:

0 0 0 0 0

Convolution Result of Part c is :
ans =

Columns 1 through 20:

0 0 1 2 3 4 5 6 7 8 8 8 8 8 7 6 5 4 3 2

Columns 21 through 31:

1 0 0 0 0 0 0 0 0 0 0