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How to plot the graph with different amplitudes at different frequencies?

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TEOH CHEE JIN
TEOH CHEE JIN el 28 de Dic. de 2022
Respondida: Sayan el 15 de Sept. de 2023
For the amplitude of D(w), I have used the command "simplify" and obtained that there are two different amplitudes at different frequency. At w=40pi, the absolute amplitude is 50pi, whereas at w=4960pi and w=5040pi, the amplitudes are 25pi, what should I do to plot the graph showing different amplitudes at different frequency?
%create symbolic functions x, a, b, c, d, e, f_c with independent variable t
syms x(t) a(t) h(t) b(t) c(t) d(t) e f_c(t) f_c1(t) f_c2(t) t tau
%create symbolic functions A, B, C, D, and E with independent variable w
syms A(w) B(w) C(w) D(w) E(w) w
x(t) = cos(100*pi*t);
a(t) = x(0.4*t);
h(t) = dirac(t-0.02);
b(t) = int(a(tau)*h(t-tau), 'tau', -inf, inf);
f_c(t) = 10*cos(2500*pi*t);
f_c1(t) = f_c(t);
f_c2(t) = f_c(t);
c(t) = b(t)*f_c1(t);
d(t) = c(t)*f_c2(t);
% f_c1(t)
% f_c2(t)
% figure
%
% subplot (3,1,1)
% fplot(a(t))
% xlim([-0.05 0.05]),ylim([-1.5 1.5])
% title ('Time domain of signal a(t)')
% xlabel('Time, t')
% ylabel('Amplitude, a(t)')
% grid on
%
% A(w) = fourier(a(t), w);
% w = -60*pi:0.1*pi:60*pi;
% subsA = A(w);
% % Replace Inf value with suitable value
% idx = subsA == Inf;
% subsA(idx) = pi; % choose suitable value from the expression of fourier transform.
% subplot (3,1,2)
% plot(w,real(subsA));
% ylim([0 4])
% title ('Frequency domain of signal A(\omega)')
% ylabel("\Re(A(\omega))");
% xlabel("\omega");
% xticks(-40*pi:20*pi:40*pi);
% xticklabels({'-40\pi', '-20\pi', '0', '20\pi', '40\pi'})
%
% subplot (3,1,3)
% plot(w, angle(A(w)))
% title ('Phase spectrum of signal A(\omega)')
% ylabel("\angle(A(\omega))");
% xlabel("\omega");
%
%
% figure
%
% subplot(3,1,1)
% fplot(b(t))
% xlim([-0.05 0.05]),ylim([-1.5 1.5])
% title ('Time domain of signal b(t)')
% xlabel('Time, t')
% ylabel('Amplitude, b(t)')
% grid on
%
% syms B w
% B(w) = fourier(b(t), w);
% w = -60*pi:0.1*pi:60*pi;
% subsB = B(w);
% % Replace Inf value with suitable value
% idx = abs(subsB) == Inf;
% subsB(idx) = pi; % choose suitable value from the expression of fourier transform.
% subplot (3,1,2)
% plot(w,real(subsB));
% ylim([0 4])
% title ('Frequency domain of signal B(\omega)')
% ylabel("\Re(B(\omega))");
% xlabel("\omega");
% xticks(-60*pi:20*pi:60*pi);
% xticklabels({'-40\pi', '-20\pi', '0', '20\pi', '40\pi'})
%
% subplot (3,1,3)
% plot(w, angle(B(w)))
% ylim([-4 4])
% title ('Phase spectrum of signal B(\omega)')
% ylabel("\angle(B(\omega))");
% xlabel("\omega");
% figure
%
% subplot(3,1,1)
% fplot(c(t))
% xlim([-0.05 0.05]),ylim([-12 12])
% title ('Time domain of signal c(t)')
% xlabel('Time, t')
% ylabel('Amplitude, c(t)')
% grid on
%
% syms C w
% C(w) = fourier(c(t), w);
% simplify(C(w))
% w = -2800*pi:20*pi:2800*pi;
% subsC = C(w);
% % Replace Inf value with suitable value
% idx = abs(subsC) == Inf;
% subsC(idx) = 5*pi; % choose suitable value from the expression of fourier transform.
% subplot (3,1,2)
% plot(w,real(subsC));
%
% ylim([0 20])
% title ('Frequency domain of signal C(\omega)')
% ylabel("\Re(C(\omega))");
% xlabel("\omega");
% xticks([-2540*pi -2460*pi 0 2460*pi 2540*pi]);
% xticklabels({'-2540\pi', '-2460\pi', '0', '2460\pi', '2540\pi'})
%
% subplot (3,1,3)
% plot(w, angle(C(w)))
% ylim([-4 4])
% title ('Phase spectrum of signal C(\omega)')
% ylabel("\angle(C(\omega))");
% xlabel("\omega");
figure
subplot(3,1,1)
fplot(d(t))
xlim([-0.1 0.1]),ylim([-110 110])
title ('Time domain of signal d(t)')
xlabel('Time, t')
ylabel('Amplitude, d(t)')
grid on
syms D w
D(w) = fourier(d(t), w);
w = -5040*pi:10*pi:5040*pi;
simplify(D(w))
ans = 
subsD = D(w);
% Replace Inf value with suitable value
idx = abs(subsD) == Inf;
subsD(idx) = 25*pi; % choose suitable value from the expression of fourier transform.
subplot (3,1,2)
plot(w,real(subsD));
ylim([0 160])
title ('Frequency domain of signal D(\omega)')
ylabel("\Re(D(\omega))");
xlabel("\omega");
subplot (3,1,3)
plot(w, angle(D(w)))
ylim([-4 4])
title ('Phase spectrum of signal D(\omega)')
ylabel("\angle(D(\omega))");
xlabel("\omega");
This is the calculation that I have done to verify the answer.

Respuestas (1)

Sayan
Sayan el 15 de Sept. de 2023
I understand from your query that it is required to plot the co-efficient of the "dirac" function of the fourier transform. However, in the code, all the coefficients of the "dirac" function are assigned to 25*pi. This can be found in the below code snippet.
subsD = D(w);
% Replace Inf value with suitable value
idx = abs(subsD) == Inf;
subsD(idx) = 25*pi;%%here the "idx" contains the indices of dirac functions and all the values of "subsD" corresponding to "idx" is assigned to 25*pi
The possible fix is to find the indices of the correlating angular frequency and use them to assign the correct value. This is shown in the below code snippet.
%create a function "ap" to return the indices of the corresponding
%frequency
function n=ap(f)
n=int32(((5040*pi+f)/(10*pi))+1);%as the value of "w" are in arithmetic progrssion indices are obtained using the formula
end
subsD = D(w);
%manually assign the values to the "dirac" of the corresponding frequency
subsD(ap(-5040*pi))=25*pi;
subsD(ap(5040*pi))=25*pi;
subsD(ap(-4960*pi))=25*pi;
subsD(ap(4960*pi))=25*pi;
subsD(ap(-40*pi))=50*pi;
subsD(ap(40*pi))=50*pi;
%now the amplitude of subsD can be plotted
plot(w,subsD);
Further information on "dirac" function can be found in the following MATLAB documentation:
Hope this helps in resolving the query.

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