Fourier analysis of an spwm signal

Question

Mosharof Hossain 2023-9-7

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2018121-fourier-analysis-of-an-spwm-signal

评论： Paul 2023-9-8

spwm_one_cycle.mat

I have one cycle of an spwm signal as a time series data. How to do fourier analysis of the periodic signal and plot the spectrum?

The signal is as shown below.

I want to find the harmonics in this periodic signal and filter it to get back the original sine wave.

The time series spwm is attached above.

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Answer 1

Star Strider 2023-9-7

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2018121-fourier-analysis-of-an-spwm-signal#answer_1304531

在 MATLAB Online 中打开

spwm_one_cycle.mat

Try this —

LD = load('spwm_one_cycle.mat')

LD = struct with fields:

ts1: [1×1 timeseries]

ts1 = LD.ts1

timeseries Common Properties: Name: 'unnamed' Time: [224x1 double] TimeInfo: tsdata.timemetadata Data: [224x1 double] DataInfo: tsdata.datametadata

Time = ts1.Time;

Data = ts1.Data;

figure

plot(Time, Data)

xlabel('Time (seconds)')

ylabel('Amplitude')

[FTData,Freqs] = FFT1(Data,Time);

figure

subplot(2,1,1)

plot(Freqs, abs(FTData)*2)

grid

ylabel('Magnitude')

subplot(2,1,2)

plot(Freqs, unwrap(angle(FTData)))

grid

xlabel('Frequency (Hz)')

ylabel('Phase (rad)')

sgtitle('Fourier Transform Of SPWM Signal')

function [FTs1,Fv] = FFT1(s,t)

s = s(:);

t = t(:);

L = numel(t);

Fs = 1/mean(diff(t));

Fn = Fs/2;

NFFT = 2^nextpow2(L);

FTs = fft((s - mean(s)).*hann(L), NFFT)/sum(hann(L));

Fv = linspace(0, 1, NFFT/2+1)*Fn;

Iv = 1:numel(Fv);

FTs1 = FTs(Iv);

end

.

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Star Strider 2023-9-7

As always, my pleasure!

Paul 2023-9-8

在 MATLAB Online 中打开

Is this line

Fs = 1/mean(diff(t));

justified given the wide variability in the delta times in the data?

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Answer 2

Paul 2023-9-7

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2018121-fourier-analysis-of-an-spwm-signal#answer_1304901

编辑：Paul 2023-9-8

在 MATLAB Online 中打开

spwm_one_cycle.mat

Hi Mosharof,

Load the data

load spwm_one_cycle.mat

Plot the differences between time samples. We see that the data is not equally spaced in time, so using discrete-time methods, like fft, is not appropriate without further massaging the data into equally spaced samples that also catch all the transitions. Maybe that could be done by using a much smaller sampling period from the original source (which looks like it might be a Simulink model?).

plot(diff(ts1.Time),'-o')

However, the signal is just the sum of rectangular pulses, and the rectangular pulse has a closed-form expression for its Continuous Time Fourier Transform (CTFT)

Here's one way we can construct the signal for analysis (this would be easier if the original source could save off the switch times and the associated signal levels at the switches).

Verify that the data has unique values:

unique(ts1.Data)
ans = 3×1
    -1
     0
     1

Use diff to find when the signal level switches

index = [1; find(diff(ts1.Data)~=0)];
syms t w real
syms y(t) Y(w)
y(t) = 0;
Y(w) = 0;

The pattern is that the level changes only between 0 and 1, or 0 and -1. Construct the signal as the sum of rectangularPulse and build up its CTFT along the way.

for ii = 2:2:numel(index)-1
    if ts1.Time(index(ii)) < 10
        a = 1;
    else
        a = -1;
    end
    s(t) = a*rectangularPulse(ts1.Time(index(ii)),ts1.Time(index(ii+1)),t);
    y(t) = y(t) + s(t);
    Y(w) = Y(w) + fourier(s(t),t,w);
end

Verify that y(t) matches the original data

figure

subplot(211)

plot(ts1.Time,ts1.Data);

ylim([-1.1 1.1])

subplot(212)

fplot(y(t),[0 20],'MeshDensity',200)

ylim([-1.1 1.1])

fplot was taking too long to make this plot

%figure
%fplot(abs(Y(w)))

Convert to an anonymous function

Yfunc = matlabFunction(Y(w));

and evaluate over some frequency range

wval = -20:.01:20;

figure

subplot(211)

plot(wval,abs(Yfunc(wval))),grid

subplot(212)

plot(wval,180/pi*angle(Yfunc(wval))),grid

xlabel('rad/sec')

The Question stated that the data is actually one period of a periodic singal, which would be represented as a Continuous Fourier Series (CFS). The CFS coefficients are obtained by sampling the CTFT of a single period at multiples of the fundamental frequency (2*pi/T rad/sec, T = 20 sec) and scaling by 1/T. Here's a plot of the magnitude of the CFS coefficients over roughly the same frequency range used to plot the CTFT of the single period.

T = 20;

n = -60:60;

figure

stem(2*pi*n/T,(1/T)*abs(Yfunc(2*pi*n/T)))

xlabel('rad/sec')