Question

The signal y_b(t) = (V + v(t))w(t), where v(t) = V cos (2πt/T) and w(t) = u(t + T/2 ) − u(t − T/2), is called a Hanning window function. Using a computer, calculate the percent of the signal’s total energy that is located in its side lobes. A side lobe is any part of a function that is not a central lobe, and a central lobe is the nonzero portion of a function between its first negative and positive zero crossings.

Answer 1

MATLAB:

clc;clear all;close all;
V= 0.5; T=20
t=-T/2:0.01:T/2
v=V*cos(2*pi*t/T)
w=(t>=-T/2 &t<=T/2)
yb=(V+(v.*w))
subplot(211)
plot(t,yb);xlabel('t');ylabel('yb(t)')
title('hanning window')

W=-pi:pi/500:pi
%Fourier transform oof yb(t)
k=0
for W=-pi:pi/1000:pi
k=k+1
Yb(k)=trapz(t,yb.*exp(-j*W*t));
end
W=-pi:pi/1000:pi
subplot(212)
plot(W/pi,(abs(Yb)),'r')
title('Fourier transfrom of yb(t)')
xlabel('w/pi');ylabel('|Yb(jw)|')

E=trapz(W,(abs(Yb).^2))
%Energy presen in side lobes
%The non zero portion of the side lobe is there between w=0.2pi and 0.5pi
Yside=Yb.*((W>=0.2*pi &W<=0.5*pi)+(W>=-0.5*pi &W<=-0.2*pi))
Eside=trapz(W,(abs(Yside).^2))
energypercentage=Eside/E

Plots:

hanning window 1 0.8 0.6 0.4 0.2 0 10 0 -5 -10 Fourier transfrom of yb (t) 12 10 8 6 4 2 0 0.5 0 -0.5 -1 w/pi |Yb(jw)] (1)q

Command window:

>> E
E = 47.124
>> Eside
Eside = 0.024033
>> energypercentage
energypercentage = 0.00050999