Question

Please Solve this Question anyway either by using MATLAB or any other way .....

QUESTION :

Generate a square wave with peak as last 2 digits of your roll number.

Find the overshoot value for this square wave.

How many harmonics are required to generate this square wave with 9% overshoot.

Find the overshoot when 101 harmonics are used.

Find the overshoot when 401 harmonics are used.

Exploit the effect of odd harmonics in generating a square wave.

Draw necessary conclusions about Gibbs phenomenon and its effects on Fourier series.

Hint : Use Gibbs Phenomenon  

Answer 1

When 101 harmonics are used:

MATLAB:

clc;
clear all;
close all;
T=4;t=0:0.01:T;A=10;
x=A.*(t>=0 &t<=2)

%To caclculate fourier series coefficients of x(t)
w0=2*pi/T;N=50
k=0;
for n=-N:1:N
k=k+1;
a(k)=(trapz(t,x.*exp(-j*n*w0*t)));
end
a=a/T;n=-N:1:N;

%output xN(t)
k=0;
w0=2*pi/T;
for t=0:0.01:T
k=k+1;
xN(k)=sum(a.*(exp(j*n*w0*t)));
end
t=0:0.01:T;
plot(t,xN,'r',t,x,'b')
title('Fourier approx of squre wave for N=101')
xlabel('t')
ylabel('xN(t)')

stepinfo(real (xN))

Fourier approx of squre wave for N=101 12 4 2 -2 2 3 4

Command window:

RiseTime: 0
SettlingTime: 400.9515
SettlingMin: -0.8645
SettlingMax: 5.0000
Overshoot: 117.2936
Undershoot: 17.2901
Peak: 10.8647
PeakTime: 198

% overshoot:

overshoot -100%=117.29-100%=17.29%

___________________________________________________________________________

When 401 harmonics are used:

MATLAB:

clc;
clear all;
close all;
T=4;t=0:0.01:T;A=10;
x=A.*(t>=0 &t<=2)

%To caclculate fourier series coefficients of x(t)
w0=2*pi/T;N=200
k=0;
for n=-N:1:N
k=k+1;
a(k)=(trapz(t,x.*exp(-j*n*w0*t)));
end
a=a/T;n=-N:1:N;

%output xN(t)
k=0;
w0=2*pi/T;
for t=0:0.01:T
k=k+1;
xN(k)=sum(a.*(exp(j*n*w0*t)));
end
t=0:0.01:T;
plot(t,xN,'r',t,x,'b')
title('Fourier approx of squre wave for N=401')
xlabel('t')
ylabel('xN(t)')

stepinfo(real (xN))

Command window:

RiseTime: 0
SettlingTime: 400.9799
SettlingMin: -0.0250
SettlingMax: 5.0000
Overshoot: 100.5000
Undershoot: 0.5000
Peak: 10.0250
PeakTime: 201

% overshoot:

Overshoot -100%=100.5-100%=0.5%

Fourier approx of squre wave for N-401 12 10 6 4 2 4 -2

_________________________________________________________________________

To achieve 9% overshoot

clc;
clear all;
close all;
T=4;t=0:0.01:T;A=10;
x=A.*(t>=0 &t<=2)

%To caclculate fourier series coefficients of x(t)
w0=2*pi/T;
k=0;
for n=-180:1:181
k=k+1;
a(k)=(trapz(t,x.*exp(-j*n*w0*t)));
end
a=a/T;n=-180:1:181

%output xN(t)
k=0;
w0=2*pi/T;
for t=0:0.01:T
k=k+1;
xN(k)=sum(a.*(exp(j*n*w0*t)));
end
t=0:0.01:T;
plot(t,xN,'r',t,x,'b')
title('Fourier approx of squre wave for N=362')
xlabel('t')
ylabel('xN(t)')

stepinfo(real (xN))

Fourier approx of squre wave for N-362 12 10 6 4 2 4 -2

Command window:

RiseTime: 2.8422e-14
SettlingTime: 400.9781
SettlingMin: -0.4610
SettlingMax: 5.0000
Overshoot: 109.2182
Undershoot: 9.2199
Peak: 10.4609
PeakTime: 200

% overshoot:

Overshoot -100%=109.21-100%=9.21%

Observation

For 362 harmonics , the overshoot in % is 9.21%

For 401 harmonics , the overshoot in % is 0.5%

For 101 harmonics , the overshoot in % is 17.29%