# FFT plot in frequency domain, error help

4 views (last 30 days)
Nina Perf on 25 Oct 2021
Commented: Nina Perf on 27 Oct 2021
Hi,
I did a myfft function in order to plot the (PSD,f) of a signal: 10696x1 double
[f, ~, ~, psd, ~] = myfft(Data, fs);
function [f, amplitude, phase, PSD, power] = myfft(signal, samplingRate)
if ~isempty(signal)
Fs = samplingRate;
T = 1/Fs;
L = length(signal);
t = (0:L-1)*T;
Y = fft(signal);
P2 = abs(Y);
P1 = P2(1:floor(L/2)+1,:);
P1(2:end-1,:) = 2*P1(2:end-1,:);
f = Fs*(0:(L/2))/L;
amplitude = P1;
phase = unwrap(angle(Y));
phase = phase(1:floor(L/2)+1,:);
% Power Spectrum
PSD = Y.*conj(Y);
PSD = (1/(Fs*L)) .* PSD(1:floor(L/2)+1,:);
PSD(2:end-1,:) = 2*PSD(2:end-1,:);
power = sum(PSD);
else
f = [];
amplitude = [];
phase = [];
power = NaN;
PSD = [];
end
I get this error:
% Error using *
% Incorrect dimensions for matrix multiplication. Check that the number of columns in the first matrix matches the
% number of rows in the second matrix. To perform elementwise multiplication, use '.*'.
% Error in myfft (line 10)
% t = (0:L-1)*T; % signal duration (time vector)
% Error in s (line 19)
% [f, ~, ~, psd, ~] = myfft(Data, fs); ;

Dave B on 25 Oct 2021
What is size(Data) and size(fs)? Your function doesn't error for me when I give it a scalar fs and a vector signal:
size(y)
ans = 1×2
73113 1
size(Fs)
ans = 1×2
1 1
[f, ~, ~, psd, ~] = myfft(y, Fs);
plot(f,psd) function [f, amplitude, phase, PSD, power] = myfft(signal, samplingRate)
if ~isempty(signal)
Fs = samplingRate;
T = 1/Fs;
L = length(signal);
t = (0:L-1)*T;
Y = fft(signal);
P2 = abs(Y);
P1 = P2(1:floor(L/2)+1,:);
P1(2:end-1,:) = 2*P1(2:end-1,:);
f = Fs*(0:(L/2))/L;
amplitude = P1;
phase = unwrap(angle(Y));
phase = phase(1:floor(L/2)+1,:);
% Power Spectrum
PSD = Y.*conj(Y);
PSD = (1/(Fs*L)) .* PSD(1:floor(L/2)+1,:);
PSD(2:end-1,:) = 2*PSD(2:end-1,:);
power = sum(PSD);
else
f = [];
amplitude = [];
phase = [];
power = NaN;
PSD = [];
end
end
Nina Perf on 27 Oct 2021
Thanks! That worked

R2021b

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