Graphing Fourier magnitude and phase spectra using fft, error: Array indices must be positive integers or logical values.

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Luke McDevitt
Luke McDevitt 2022-10-21
评论： Paul 2022-10-22
I am writing a script to get the frequency spectra of a sawtooth function with a period of 2pi
The problem I am facing is I cannot find a way to use the frequency domain in my code. I tried fft() and fourier() but have been met with errors either way. any tips appreciated!
Goal:
Plotting magitude and phase of Cn and Dn with respect to frequency (see attached images for what I mean)
fft=[];ann=[];cnplot=[];cnt=[];jnt=[];
T = 2*pi;
A=1;
funct1 = @(t) t/T;
funct = @(t) ones(size(funct1));
w0=2*pi*(1/T);
a0 = (1/(T))*integral(funct1,0,T);
for n=1:1:20 % Change the range from 3, 10, 100
for t=1:1:360*5
anfunct= @(t) funct1(t).*cos(n*t*w0);
bnfunct= @(t) funct1(t).*sin(n*t*w0);
an = (1/(T/2))*integral(anfunct,0,T);
bn = (1/(T/2))*integral(bnfunct,0,T);
cn= (an*an+bn*bn)^(1/2);
theta=atan((-bn)/an);
dn= (.5*cn)*theta;
ft(t)=an*cos(n*w0*deg2rad(t))+bn*sin(n*w0*deg2rad(t));
x(t)=-cn*cos(n*w0*deg2rad(t)+theta);
d(t)=dn*exp(1i*n*w0*t);
k(w)=fft(x(t)); %error here: Array indices must be positive integers or logical values.
end
fft=[fft;ft];
cnt=[cnt;x];
jnt=[jnt;d];
end
F_A = sum(fft)+a0;
C_A = sum(cnt)+a0;
D_A=sum(jnt)+a0;
tt=deg2rad(1:1:t);
subplot(3,1,1)
plot(tt,F_A)
title("Signal X(t) derived from trigonomtric form")
xlabel("t")
ylabel("f(t)")
subplot(3,1,2)
plot(tt,C_A)
title("Signal X(t) derived from compact trigonomtric form")
xlabel("t")
ylabel("f(t)")
ww=(1:1:w);
nn=(1:1:n);
%any code below here does not work
subplot(3,1,3);
plot(w,k(w))
title("amplitude spectra")
xlabel("w")
ylabel("k(w)")

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回答（1 个）

Paul
Paul 2022-10-22
Ran in:
Hi Luke,
At the line in question, we see that x(t) is non-integer numeric value, which can't be used to index into the array fft. Also, at that point in the code, w is not defined so that will be a problem as well.
It looks like fft is intended to become a 2D array; are you sure you mean to index into it with a single number?
Finally, consider renaming the variable fft so as not to shadow the built-in function fft.
fft=[];ann=[];cnplot=[];cnt=[];jnt=[];
T = 2*pi;
A=1;
funct1 = @(t) t/T;
funct = @(t) ones(size(funct1));
w0=2*pi*(1/T);
a0 = (1/(T))*integral(funct1,0,T);
for n=1:1:20 % Change the range from 3, 10, 100
for t=1:1:360*5
anfunct= @(t) funct1(t).*cos(n*t*w0);
bnfunct= @(t) funct1(t).*sin(n*t*w0);
an = (1/(T/2))*integral(anfunct,0,T);
bn = (1/(T/2))*integral(bnfunct,0,T);
cn= (an*an+bn*bn)^(1/2);
theta=atan((-bn)/an);
dn= (.5*cn)*theta;
ft(t)=an*cos(n*w0*deg2rad(t))+bn*sin(n*w0*deg2rad(t));
x(t)=-cn*cos(n*w0*deg2rad(t)+theta);
d(t)=dn*exp(1i*n*w0*t);
x(t)
k(w)=fft(x(t)); %error here: Array indices must be positive integers or logical values.
end
fft=[fft;ft];
cnt=[cnt;x];
jnt=[jnt;d];
end
ans = -0.0056
Array indices must be positive integers or logical values.
  2 个评论
Luke McDevitt
Luke McDevitt 2022-10-22
Would a solution be to do something like finding the nearest integer of abs(1000*x(t)) then plotting that in order to find the general shape of the amplitude graph?
Paul
Paul 2022-10-22
I did not look closely, if at all, at what the code is actually trying to compute, so I'm afraid I can't comment on that.

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