% A Program For Analysis of Sinusoidal Pulse Width Modulation Of An AC
% Signal.
% By: Yasin A. Shiboul
% Date 15/8/2003
%
% PART I (preparation)
% In this part the screen is cleared, any other functions, figures and
% variables are also cleared. The name of the programm is displayed.
clc
clear all
disp('Sinusoidal Pulse Width Modulation of AC Signal')
disp(' ')
%
% PART II
% In this part the already known variables are entered, the user is
% asked to enter the other variables.
% Vrin is the rms value of the input supply voltage in per unit.
Vrin=1;
% f is the frequency of the input supply voltage.
f=input('The frequency of the input supply voltage, f = ');
% Z is the load impedance in per unit.
Z=1;
% ma is the modulation index
ma=input('the modulation index,ma, (0<ma<1), ma = ');
% phi is load-phase-angle
phi=input('the phase angle of the load in degrees = ');
% Q is the number of pulses per half-period of the supply voltage.
Q=input('The number of pulses per half period = ');
%
% PART III
% Calculating load parameters.
%
phi=phi*pi/180;
% R and L are the load resistance and inductance respectively.
R=Z*cos(phi);
L=(Z*sin(phi))/(2*pi*f);
%
% PART IV
% Calculating the number of pulses per period,N
N=2*Q;
%
%PART V
% Building the Sawtooth signal,Vt, the Input voltage waveform,Vin,
% the Output voltage waveform, Vout, and finding the beginning (alpha)
% and the end (beta)for each of the output pulses.
% In each period of the sawtooth, there is one increasing and decreasing part of
% the sawtooth, thus the period of the input supply is divided into
% into 2N sub-periods, k is used as a counter of these sub-periods.
% for calculation purposes each of these sub-periods is divided into 50 points, i.e., the
% input supply period is divided into 100N points.
% % j is a counter inside the sub-period
% i is the generalized time counter
for k=1:2*N
for j=1:50
% finding the generalized time counter
i=j+(k-1)*50;
% finding the time step
wt(i)=i*pi/(N*50);
% calculating the input supply voltage.
Vin(i)=sqrt(2)*Vrin*sin(wt(i));
ma1(i)=ma*abs(sin(wt(i)));
% calculating the sawtooth waveform
if rem(k,2)==0
Vt(i)=0.02*j;
if abs(Vt(i)-ma*abs(sin(wt(i))))<=0.011
m=j;
beta(fix(k/2)+1)=3.6*((k-1)*50+m)/N;
else
j=j;
end
else
Vt(i)=1-0.02*j;
if abs(Vt(i)-ma*abs(sin(wt(i))))<0.011
l=j;
alpha(fix(k/2)+1)=3.6*((k-1)*50+l)/N;
else
j=j;
end
end
if Vt(i)>ma*abs(sin(wt(i)))
Vout(i)=0;
else
Vout(i)=Vin(i);
end
end
end
beta(1)=[];
% PART VI
% Displaying the beginning (alpha), the end (beta) and the width
% of each of the output pulses.
disp(' ')
disp('......................................................................')
disp('alpha beta width')
[alpha' beta' (beta-alpha)']
% PART VII
% Plotting the input voltage waveform Vin, the triangular carrier
% signal,Vt,% the modulating signal and the output voltage
% waveform, Vout.
a=0;
subplot(3,1,1)
plot(wt,Vin,wt,a)
axis([0,2*pi,-2,2])
title('Generation Of The Output Voltage Pulses ')
ylabel('Vin(pu)');
subplot(3,1,2)
plot(wt,Vt,wt,ma1,wt,a)
axis([0,2*pi,-2,2])
ylabel('Vt, m(pu)');
subplot(3,1,3)
plot(wt,Vout,wt,a)
axis([0,2*pi,-2,2])
ylabel('Vo(pu)');
xlabel('Radian');
% PART VIII
% Analyzing the output voltage waveform
% Finding the rms value of the output voltage
Vo =sqrt(1/(length(Vout))*sum(Vout.^2));
disp('The rms Value of the Output Voltage ')
Vo
% finding the harmonic contents of the output voltage waveform
y=fft(Vout);
y(1)=[];
x=abs(y);
x=(sqrt(2)/(length(Vout)))*x;
disp('The rms Value of the output voltage fundamental component = ')
x(1)
% Findint the THD of the output voltage
THDVo = sqrt(Vo^2 -x(1)^2)/x(1);
% PART IX
% calculating the output current waveform
m=R/(2*pi*f*L);
DT=pi/(N*50);
C(1)=-10;
i=100*N+1:2000*N;
Vout(i)=Vout(i-100*N*fix(i/(100*N))+1);
for i=2:2000*N;
C(i)=C(i-1)*exp(-m*DT)+Vout(i-1)/R*(1-exp(-m*DT));
end
% PART X
% Analyzing the output current waveform
% finding the harmonic contents of the output current waveform
for j4=1:100*N
CO(j4)=C(j4+1900*N);
CO2= fft(CO);
CO2(1)=[];
COX=abs(CO2);
COX=(sqrt(2)/(100*N))*COX;
end
% Finding the RMS value of the output current.
CORMS = sqrt(sum(CO.^2)/(length(CO)));
disp(' The RMS value of the load current is')
CORMS
%Finding the THD for the output current
THDIo = sqrt(CORMS^2-COX(1)^2)/COX(1);
% PART XI
% Finding the supply current waveform
for j2=1900*N+1:2000*N
if Vout(j2)~=0
CS(j2)=C(j2);
else
CS(j2)=0;
end
end
% PART XII
% Analyzing the supply current waveform
%Supply current waveform and its RMS
for j3=1:100*N
CS1(j3)=CS(j3+1900*N);
end
CSRMS= sqrt(sum(CS1.^2)/(length(CS1)));
disp('The RMS value of the supply current is')
CSRMS
% Finding the Fourier analysis
CS2= fft(CS1);
CS2(1)=[];
CSX=abs(CS2);
CSX=(sqrt(2)/(100*N))*CSX;
% Finding the THD of the Supply current
THDIS = sqrt(CSRMS^2-CSX(1)^2)/CSX(1);
% Finding the Displacement Factor of the supply current
phi1 = atan(real(CS2(1))/imag(CS2(1)))-pi/2;
PF=cos(phi1)*CSX(1)/CSRMS;
% PART XIII
% Displaying the calculated parameters.
disp(' Performance parameters are')
THDVo
THDIo
THDIS
PF
a=0;
%PART XIV
% Openning a new figure window for plotting of
% the output voltage,output current, supply current and the harmonic
% contents of these values
figure(2)
subplot(3,2,1)
plot(wt,Vout(1:100*N),wt,a);
title('');
axis([0,2*pi,-1.5,1.5]);
ylabel('Vo(pu)');
%
subplot(3,2,2)
plot(x(1:100))
title('');
axis([0,100,0,0.8]);
ylabel('Von(pu)');
subplot(3,2,3)
plot(wt,C(1900*N+1:2000*N),wt,a);
title('');
axis([0,2*pi,-1.5,1.5]);
ylabel('Io(pu)');
subplot(3,2,4)
plot(COX(1:100))
title('');
axis([0,100,0,0.8]);
ylabel('Ion(pu)');
subplot(3,2,5)
plot(wt,CS(1900*N+1:2000*N),wt,a);
axis([0,2*pi,-1.5,1.5]);
ylabel('Is(pu)');
xlabel('Radian');
subplot(3,2,6)
plot(CSX(1:100))
title('');
axis([0,100,0,0.8]);
ylabel('Isn(pu)');
xlabel('Harmonic Order');
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