HELP! Error using plot Vectors must be the same length. Error in Doc (line 94) plot(t, y);

Hello everyone!! I'm trying to plot: x,y,X,Y. But when my matlab try to plot y, the following error appears: Error using plot Vectors must be the same length. Error in doc (lne 94) plot(t, y);

Please fell free to check the code below:

Fs = 5000;

T = 1/Fs;

t = 0:T:1-T;

NFFT = 2^nextpow2(length(t));

x = cos(2*pi*100*t) + cos(2*pi*500*t) + cos(2*pi*1000*t);

[X, f, ~] = fftm(x, Fs, NFFT);

H = double(f <= 600);

Y = X .* H;

y = ifft(Y);

figure;

subplot(2,1,1);

plot(t, x);

title('x(t)');

xlabel('Time(s)');

ylabel('Amp');

subplot(2,1,2);

plot(t, y);

title('y(t)');

xlabel('Time(s)');

ylabel('Amp');

figure;

subplot(2,1,1);

plot(f, abs(X));

title('X(\omega)');

xlabel('Freq(Hz)');

ylabel('Mag');

subplot(2,1,2);

plot(f, abs(Y));

title('Y(\omega)');

xlabel('Freq(Hz)');

ylabel('Mag');

Oh, the following function fftm in line 6, is a function downloaded by me, it's a fft function modified, and you can see it below:

% Modified fft algorithm

%

% This function estimates the fft of a signal x using the Fs sampling

% frequency, NFFT bins, and returns the frequencies values along with the spectrum X.

%

% [X,f] = fftm(x,Fs,NFFT)

%

% Example:

% Fs = 1e3;

% t = 0:0.001:1-0.001;

% x = cos(2*pi*100*t)+sin(2*pi*202.5*t);

% [X,f]=fftm(x,Fs,2000);

% plot(f,abs(X));

% http://www.mathworks.com/help/signal/ug/amplitude-estimation-and-zero-padding.html

% http://www.mathworks.com/help/signal/ug/psd-estimate-using-fft.html

function [xdft,f, psdx] = fftm(x,Fs,NFFT)

L = length(x);

if nargin < 3

NFFT = 2^nextpow2(L);

end

xdft = fft(x,NFFT);

xdft = xdft(1:length(xdft)/2+1);

psdx = (1/(Fs*L)).*abs(xdft).^2;

psdx(2:end-1) = 2*psdx(2:end-1);

psdx = 10*log10(psdx);

xdft = 1/L.*xdft;

xdft(2:end-1) = 2*xdft(2:end-1);

f = 0:Fs/NFFT:Fs/2;

Answer 1

[Paul]

Hi Matheus,

The explanation for the error is that the fftm function is doing a "one-sided" FFT, hence it returns an output with a different number of elements than in the input signal, x. Based on what it appears you're trying to do, you should reconsider using fftm, because the so-called "one-sided" FFT is probably not what should be used for this problem.

Fs = 5000;
T = 1/Fs;
t = 0:T:1-T;
NFFT = 2^nextpow2(length(t));
x = cos(2*pi*100*t) + cos(2*pi*500*t) + cos(2*pi*1000*t);

x has 5000 elements, as does t

[numel(x) numel(t)]

ans = 1x2

5000 5000

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

[X, f, ~] = fftm(x, Fs, NFFT);

X is derived from an 8192 point DFT

NFFT

NFFT = 8192

But it looks like fftm is trying to do some sort of one-sided DFT (I'm not even sure it's doing it correctly),

numel(X)

ans = 4097

Looks like this is trying to remove frequencies greater than 600. But this approach won't really work becasue half of the DFT is now missing

H = double(f <= 600);
Y = X .* H;
y = ifft(Y);

y will have as many elements as X, which is not the same number of elements as in t, hence the error below on the plot command.

figure;
subplot(2,1,1);
plot(t, x);
title('x(t)');
xlabel('Time(s)');
ylabel('Amp');
subplot(2,1,2);

plot(t, y);

Error using plot
Specify the coordinates as vectors or matrices of the same size, or as a vector and a matrix that share the same length in at least one dimension.

title('y(t)');
xlabel('Time(s)');
ylabel('Amp');
figure;
subplot(2,1,1);
plot(f, abs(X));
title('X(\omega)');
xlabel('Freq(Hz)');
ylabel('Mag');
subplot(2,1,2);
plot(f, abs(Y));
title('Y(\omega)');
xlabel('Freq(Hz)');
ylabel('Mag');


% Modified fft algorithm
%
% This function estimates the fft of a signal x using the Fs sampling
% frequency, NFFT bins, and returns the frequencies values along with the spectrum X.
%
% [X,f] = fftm(x,Fs,NFFT)
%
% Example:
% Fs = 1e3;
% t = 0:0.001:1-0.001;
% x = cos(2*pi*100*t)+sin(2*pi*202.5*t);
% [X,f]=fftm(x,Fs,2000);
% plot(f,abs(X));
% http://www.mathworks.com/help/signal/ug/amplitude-estimation-and-zero-padding.html
% http://www.mathworks.com/help/signal/ug/psd-estimate-using-fft.html
function [xdft,f, psdx] = fftm(x,Fs,NFFT)
L = length(x);
if nargin < 3
    NFFT = 2^nextpow2(L);
end
xdft = fft(x,NFFT);
xdft = xdft(1:length(xdft)/2+1);
psdx = (1/(Fs*L)).*abs(xdft).^2;
psdx(2:end-1) = 2*psdx(2:end-1);
psdx = 10*log10(psdx);
xdft = 1/L.*xdft;
xdft(2:end-1) = 2*xdft(2:end-1);
f = 0:Fs/NFFT:Fs/2;

end

[Matheus Victor]

Hi Paul! Many thanks for the reply!! So do you think that I should use fft function instead of fftm? Because this fftm function was given by my professor, but there's no problem if I use a different function, I just need this works 😅.

So, many thanks!

[Paul]

Yes, I suspect that the fft function is probably all that's needed if the goal is to compute the FFT of a signal, set to zero those elements of the FFT corresponding to frequences abs(f) > 600, and then computing the IFFT of the result (the intentional use of abs() is a hint).

If you try it and it still doesn't work, feel free to post back with the updated code showing where you're having a problem.