Error using stem X must be same length as Y.

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TUAN SHARIFAH NUR FATIEHA
评论： TUAN SHARIFAH NUR FATIEHA 2023-4-13
fs = 100000; %sampling frequency
Ts = 1/fs;
Tstart = 22.7; % change to 22.6 for 2.4s time window
N = 2500001;%number of sample for fitting
T = N*Ts;
Electrical_Output_Torque = out.ScopeData12(:,2);
currenta=out.ScopeData7(:,2);
currentb=out.ScopeData7(:,3);
currentc=out.ScopeData7(:,4);
t1 = (1:length(currenta))*Ts;
figure(1)
plot(t1,currenta,'r')
title('Current Measurement in Time Domain');
xlabel('Time [s]');
ylabel('Current [A]');
grid on
currenta_sel = currenta(t1>=Tstart&t1<Tstart+T);
currenta_sel(1) = 0;
t1_sel = t1(t1>=Tstart&t1<Tstart+T);
figure(2)%selected time in current measurement (1.6s:4s)
plot(t1_sel,currenta_sel,'r')
title('Selected Time for the Current Measurement');
xlabel('Time [s]');
ylabel('Current [A]');
grid on
%%
ca_hanning = (currenta_sel .*hann(length(currenta_sel)))';
ca_hanning_fft = fftshift(fft(ca_hanning/length(ca_hanning)))';%2
caA_abs = abs(ca_hanning_fft);
caA_abs(1)= 0;
f1= (-fs/2):1/(T-Tstart):(fs/2);
figure(3);
stem(f1,caA_abs, 'r')
set(gca,'yscal','log')
grid on
title('4.6s, Hanning');
xlabel(' \it f (Hz)');
ylabel('|S(j\omega)| [A]');
xlim([0 100])
CaA_abs(1:2:end-1,:)= [];
f1(:,1:2:end-1)= [];
%%
figure(4);
stem(f1,CaA_abs, 'r')
set(gca,'yscal','log')
grid on
title('4kW motor- Hanning window');
xlabel(' \it f (Hz)');
ylabel('|S(j\omega)| (A)');
xlim([0 100])
  2 个评论
KSSV
KSSV 2023-4-10
To use stem your x and y data should be of same size/ dimensions. In your case they are not. Check why the dimensions are not equal? You need check your variable f1.

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采纳的回答

Walter Roberson
Walter Roberson 2023-4-10
currenta=out.ScopeData7(:,2);
We are not given any information about what out is or size the various ScopeData* are. We can speculate that they are probably values output by Simulink . It is most common for Simulink scope blocks to be configured to remember only the last 10000 points. It is also common in Simulink for different parts of a model to be run at different intervals, so scope blocks in different parts of a model are not expected to have synchronized times (and so are not expected to have the same number of samples.)
currenta_sel = currenta(t1>=Tstart&t1<Tstart+T);
currenta_sel(1) = 0;
t1_sel = t1(t1>=Tstart&t1<Tstart+T);
You select a subset of the scope data anyhow, and you construct time vectors of similar length.
ca_hanning = (currenta_sel .*hann(length(currenta_sel)))';
ca_hanning_fft = fftshift(fft(ca_hanning/length(ca_hanning)))';%2
caA_abs = abs(ca_hanning_fft);
so caA_abs has size according to what was selected out of the current data.
f1= (-fs/2):1/(T-Tstart):(fs/2);
f1 is going to have size according to some fixed values. The length of f1 is not dependent on what size caA_abs turned out to be -- at least not in any direct way.
Because of cumulative floating point round-off error, you should not expect the colon operator to return the number of entries that it would if you were working algebraically. It is common for a colon operator to end up with one fewer entries than would be expected algebraically.
stem(f1,caA_abs, 'r')
We have no direct reason to expect that the sizes are going to match, and we have round-off-error reason to expect they might e different even if the rest of the calculations were correct.
You should probably be taking something like
f1 = linspace(-fs/2, fs/2, size(caA_abs,1));

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