Assignment about Fundamental Frequency
VerifiedAdded on  2022/09/29
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Question a...............................................................................................................................1
Question b...............................................................................................................................2
Question c...............................................................................................................................3
Question d...............................................................................................................................4
Question e...............................................................................................................................5
clear;
clc;
close all;
T=1e-3;
A=2;
Question a
Fs=1e6;
t=-T:1/Fs:T;
d_c=2*A/pi; %Dc component of the signal
a_c=0;%Ac component calculation
for n=2:200
a_c=a_c+((-1).^(1+n/2))/(n.^2-1)*cos(2*pi*n*t/T);
end
s_t=d_c+4*A./pi.*a_c;
figure;
plot(t,abs(s_t));
grid on;
xlabel('Time in seconds');
ylabel('Amplitude');
amp=max(abs(s_t));% Amplitude calculation
per=T;% Period calculation
fund=1/T;% Fundamental frequency is obtained when n=1
L=length(s_t);
poww=(norm(s_t).^2)/L; % power calculation in watts
[R, SPURPOW, SPURFREQ] = sfdr(abs(s_t));
powed=SPURPOW; % power calculation in dBW
fprintf('Amplitude: %f volts\n Period: %f seconds\n Fundamental frequency: %f Hz\
Question b...............................................................................................................................2
Question c...............................................................................................................................3
Question d...............................................................................................................................4
Question e...............................................................................................................................5
clear;
clc;
close all;
T=1e-3;
A=2;
Question a
Fs=1e6;
t=-T:1/Fs:T;
d_c=2*A/pi; %Dc component of the signal
a_c=0;%Ac component calculation
for n=2:200
a_c=a_c+((-1).^(1+n/2))/(n.^2-1)*cos(2*pi*n*t/T);
end
s_t=d_c+4*A./pi.*a_c;
figure;
plot(t,abs(s_t));
grid on;
xlabel('Time in seconds');
ylabel('Amplitude');
amp=max(abs(s_t));% Amplitude calculation
per=T;% Period calculation
fund=1/T;% Fundamental frequency is obtained when n=1
L=length(s_t);
poww=(norm(s_t).^2)/L; % power calculation in watts
[R, SPURPOW, SPURFREQ] = sfdr(abs(s_t));
powed=SPURPOW; % power calculation in dBW
fprintf('Amplitude: %f volts\n Period: %f seconds\n Fundamental frequency: %f Hz\
n',amp,per,fund);
fprintf('Total power in watts: %f watts\n Total power in dBW: %f dBW\n',poww,powed);
Amplitude: 2.015035 volts
Period: 0.001000 seconds
Fundamental frequency: 1000.000000 Hz
Total power in watts: 2.059740 watts
Total power in dBW: -29.002412 dBW
Question b
k=[0 2 4 6];
figure;
grid on;
xlabel('Time in seconds');
ylabel('Amplitude');
hold on;
for i=1:4
fprintf('Total power in watts: %f watts\n Total power in dBW: %f dBW\n',poww,powed);
Amplitude: 2.015035 volts
Period: 0.001000 seconds
Fundamental frequency: 1000.000000 Hz
Total power in watts: 2.059740 watts
Total power in dBW: -29.002412 dBW
Question b
k=[0 2 4 6];
figure;
grid on;
xlabel('Time in seconds');
ylabel('Amplitude');
hold on;
for i=1:4
n=k(i);
s_e=d_c+((-1).^(1+n/2))/(n.^2-1)*cos(2*pi*n*t/T);
plot(t,abs(s_e));
end
legend('n=0','n=2','n=4','n=6')
hold off;
Question c
figure;
hold on;
k=0:10;
for i=1:11;
n=k(i);
s_t=d_c+((-1).^(1+n/2))/(n.^2-1)*cos(2*pi*n*t/T);
subplot(3,4,i);
stem(abs(s_t));
grid on;
s_e=d_c+((-1).^(1+n/2))/(n.^2-1)*cos(2*pi*n*t/T);
plot(t,abs(s_e));
end
legend('n=0','n=2','n=4','n=6')
hold off;
Question c
figure;
hold on;
k=0:10;
for i=1:11;
n=k(i);
s_t=d_c+((-1).^(1+n/2))/(n.^2-1)*cos(2*pi*n*t/T);
subplot(3,4,i);
stem(abs(s_t));
grid on;
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xlabel('frequency');
ylabel('Power spectral density');
end
hold off;
Question d
fcut=7000*pi*2;
sot=lowpass(abs(s_t),fcut/(2*pi*Fs));
figure;
plot(t,sot);
grid on;
xlabel('Time in seconds');
ylabel('Amplitude');
L=length(sot);
poww1=(norm(sot).^2)/L; % power calculation in watts
powrat=poww1/poww;
f=-1/T:(200/T-7000);
ylabel('Power spectral density');
end
hold off;
Question d
fcut=7000*pi*2;
sot=lowpass(abs(s_t),fcut/(2*pi*Fs));
figure;
plot(t,sot);
grid on;
xlabel('Time in seconds');
ylabel('Amplitude');
L=length(sot);
poww1=(norm(sot).^2)/L; % power calculation in watts
powrat=poww1/poww;
f=-1/T:(200/T-7000);
highest_freq=max(f);
hh=highest_freq/fund;
fprintf('The highets frequency of So(t) is as multiple of Fo: %f \n',hh);
fprintf('The highets frequency of So(t) in Hz is: %f \n',highest_freq);
fprintf('The total power of So(t) is: %f \n',poww1);
fprintf('The total power of So(t) as a ratio of power of s(t): %f \n',powrat);
% Comment
% The signal resembles the original signal s(t).
The highets frequency of So(t) is as multiple of Fo: 193.000000
The highets frequency of So(t) in Hz is: 193000.000000
The total power of So(t) is: 1.617594
The total power of So(t) as a ratio of power of s(t): 0.785339
Question e
figure;
sfdr(sot);
hh=highest_freq/fund;
fprintf('The highets frequency of So(t) is as multiple of Fo: %f \n',hh);
fprintf('The highets frequency of So(t) in Hz is: %f \n',highest_freq);
fprintf('The total power of So(t) is: %f \n',poww1);
fprintf('The total power of So(t) as a ratio of power of s(t): %f \n',powrat);
% Comment
% The signal resembles the original signal s(t).
The highets frequency of So(t) is as multiple of Fo: 193.000000
The highets frequency of So(t) in Hz is: 193000.000000
The total power of So(t) is: 1.617594
The total power of So(t) as a ratio of power of s(t): 0.785339
Question e
figure;
sfdr(sot);
r=sfdr(sot);
hhh=1/T;
a=(200*hhh-7000);
f=-hhh:1e4:a;
pww=[];
for j=1:length(f);
ff=f(j);
for n=2:200;
ss_e=d_c+((-1).^(1+n/2))/(n.^2-1)*cos(2*pi*n*t*ff);
end
L=length(ss_e);
pw=(norm(ss_e).^2)/L; % power calculation in watts
pww=[pww pw];
end
pwwr=pww./poww;
pww_db=10*log10(pww);
q=[pww;pwwr;pww_db];
fprintf('power in watts Normalised power Normalised watts in dB')
q'
fprintf('Dynamic power ratio is: %f\n',r);
power in watts Normalised power Normalised watts in dB
ans =
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
hhh=1/T;
a=(200*hhh-7000);
f=-hhh:1e4:a;
pww=[];
for j=1:length(f);
ff=f(j);
for n=2:200;
ss_e=d_c+((-1).^(1+n/2))/(n.^2-1)*cos(2*pi*n*t*ff);
end
L=length(ss_e);
pw=(norm(ss_e).^2)/L; % power calculation in watts
pww=[pww pw];
end
pwwr=pww./poww;
pww_db=10*log10(pww);
q=[pww;pwwr;pww_db];
fprintf('power in watts Normalised power Normalised watts in dB')
q'
fprintf('Dynamic power ratio is: %f\n',r);
power in watts Normalised power Normalised watts in dB
ans =
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
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1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
Dynamic power ratio is: 257.678527
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
Dynamic power ratio is: 257.678527
References
[1] M. J. Roberts, Signals and systems Analysis Using Transform methods and MATLAB,
New York: McGraw-Hill, 2012.
[2] D. Silage, Digital communication Systems using Matlab and Simulink, Bookstand
Publishiing, 2009.
[1] M. J. Roberts, Signals and systems Analysis Using Transform methods and MATLAB,
New York: McGraw-Hill, 2012.
[2] D. Silage, Digital communication Systems using Matlab and Simulink, Bookstand
Publishiing, 2009.
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