How to find out the amplitude and duration of a sine wave of different amplitudes?

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aswathy
aswathy 2013-8-9
评论： William Rose 2024-2-23
Hi How to find out the amplitude and duration of a sine wave of different amplitudes?
Consider a voltage sag ie, a 50hz signal of amplitude less than the normal one.I have to find out the amplitude and duration at wich this disturbance occurs.
I have read some articles which gives some information that we can extract time and frequency information using wavelets db4 using multi resolution analysis .As i am a beginner i have just a rough idea about the work.Can anyone please direct me to accomplish my works?
Thanks

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回答（4 个）

William Rose
William Rose 2024-2-20
Ran in:
@aswathy, @A Lumbini,
I would not use wavelets. , I would use stft().
Make a signal which is 3 seconds long. The frequency is 50 Hz. The amplitude is 230 V rms (i.e. peak amplitude=325 V). The peak amplitude dips to 275 V for 1 second in the middle of the 3 second long signal.
fs=1000; % sampling rate (Hz)
t=0:1/fs:3-1/fs; % time (s)
f=50; % frequency (Hz)
a1=325; a2=275; % amplitude (V)
x=a1*sin(2*pi*f*t); % signal
x(t>=1 & t<2)=a2*sin(2*pi*f*t(t>=1 & t<2));
Plot the signal versus time.
plot(t,x,'-r');
grid on; xlabel('Time (s)'); ylabel('Voltage (V)');
The plot above shows that the amplitude is lower from t=1 to t=2.
Compute the time-varying Fourier transform, and plot it:
nfft=100; % number of points to use for FFT
[s,f1,t1]=stft(x,fs,Window=ones(1,nfft),OverlapLength=nfft/2,FFTLength=nfft,FrequencyRange="onesided");
% the STFT scales like nfft/2, so we adjust the plotted value accordingly
surf(t1,f1,abs(s)/(nfft/2));
colorbar; ylim([0,100])
xlabel('Time (s)'); ylabel('Frequency (Hz)'); zlabel('Amplitude')
Plot the ampltude of the STFT for frequency=50, as a function of time. This means plotting the amplitude along the top ridge of the 3D plot above.
i50=find(f1==50); % frequency index for 50 Hz
y=abs(s(i50,:))/(nfft/2);
figure; plot(t1,y,'-rx')
grid on; xlabel('Time (s)'); ylabel('Amplitude'); title('Amplitude at 50 Hz')
Now let us find the time period when the signal amplitude is abnormal, and let us find the mean amplitude during that time period. The normal voltage is 325 V (=230V rms x sqrt(2)). Any amplitude that deviates by more than 3 V from the normal value will be considered abnormal.
t_ab=t1(abs(y-325)>3); % time vector for abnormal amplitude
tabs=min(t_ab); tabe=max(t_ab); % start, end times for abnormal amplitude
a_ab=mean(y(abs(y-325)>3)); % mean amplitude during abnormal time period
fprintf('Ampltude is abnormal from t=%.2f to %.2f.\n',tabs,tabe)
Ampltude is abnormal from t=1.00 to 2.00.
fprintf('Mean abnormal amplitude=%.1f.\n',a_ab)
Mean abnormal amplitude=277.4.
The calculated mean abnormal ampltude = 277.4 V, not 275 V, because the first and last abnormal values are halfway between 325 V (normal) and 275 V (abnormal).
  2 个评论
William Rose
William Rose 2024-2-20
@A Lumbini,
There are simpler ways to extract the amplitude, as a function of time, for the signal above, if you do not care about frequency. For example, you could compute the maximum value of the absolute value of the signal, using a moving window that you slide across the signal.
A Lumbini
A Lumbini 2024-2-21
@William Rose Thank you for giving me this information
But, i am working on Wavelets So, i need to do it using wavelet transform

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Alexander
Alexander 2024-2-20
Very easy approach with "movmax":
clear;clf;
fs=1000; % sampling rate [Hz]
ts=1/fs; % sampling rate [Hz]
t=0:0.001:5; % time [s]
tlength = length(t);
fl=50; % frequency [Hz] to look for
tl=1/50; % period time s
nl=round(tl/ts);
a=(1:tlength)/100; % amplitude
y=a.*sin(2*pi*fl*t); % signal
subplot(211)
plot(t,y);grid minor;
yAmp = movmax(y,nl); % look for maxima
subplot(211)
plot(t,y);grid minor;
subplot(212)
plot(t,yAmp);grid minor;
I think it can be done much more elegant by performing a FFT at a single frequency, in this case "fl". But I have to think about it and if I get a solution I will come back later.
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Alexander
Alexander 2024-2-21
@William Rose you are right, but @A Lumbini was investigating the jitter of the amplitude, so I think he can work with this. but I will post another possible solution.

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Alexander
Alexander 2024-2-21
Here is another approach as promissed above:
clear;clf;commandwindow;
fs=1000; % sampling rate [Hz]
ts=1/fs; % sampling rate [Hz]
t=0:ts:5-ts; % time 0-5 s
tlength = length(t);
fl=50; % frequency [Hz] to look for
tl=1/50; % period time s
N=round(tl/ts); % number of elements in period
a=ones(1,tlength)*4.5+rand(1,tlength); % amplitude 5 and +-0.5 random noise
y=a.*sin(2*pi*fl*t); % signal
yRes = reshape(y,N,[]); % Preparing the time slots
tN = (1:N)*ts;
Real = 2*fl/fs*sum(yRes.*sin(2*pi*fl*tN)');
Simag = 2*fl/fs*sum(yRes.*cos(2*pi*fl*tN)');
yARes = sqrt(Simag.^2+Real.^2); % vector of amplitudes
tRes = reshape(t,N,[]); % Preparing the time slots
tRes = mean(tRes); % t-axis according to the amplitudes in the middle of a time slot
plot(tRes,yARes);grid minor;
ylabel('Amplitude');xlabel('Time [sec]');
The advantage is, that the frequency fl is easy to adapt if used in (e.g.) order analysis of an engine runup.
Just an remainder: the equations Real and Simag are handmade (to be honest: I didn't found something appropriate in the net). Hence, check your results carefully.
William Rose
William Rose 2024-2-21
Ran in:
@A Lumbini, If you must use wavelets, do a continuous wavelet transform (cwt). THe Matlab documentation is good.
Make a 50 Hz signal, 3 seconds long, with an amplitude dip in the middle:
fs=1000; % sampling rate (Hz)
t=0:1/fs:3-1/fs; % time (s)
f=50; % frequency (Hz)
a1=325; a2=275; % amplitude (V)
x=a1*sin(2*pi*f*t); % signal
x(t>=1 & t<2)=a2*sin(2*pi*f*t(t>=1 & t<2));
Do the cwt:
[cfs,f] = cwt(x,fs);
Plot the cwt:
surf(t,f,abs(cfs),Edgecolor='none')
xlabel("Time (s)"); ylabel("Frequency (Hz)");
ylim([0 200]); grid on; colorbar
title('C.W.T. of x(t)')
The surface plot, above, shows the dip in 50 Hz amplitude, from t=1 to t=2.
Make a 2D plot of amplitude versus time, for the 50 Hz component of the CWT:
[~,i50]=min(abs(f-50)); % frequency index closest to 50 Hz
y=abs(cfs(i50,:));
figure; plot(t,y,'-rx')
grid on; xlabel('Time (s)'); ylabel('Amplitude'); title('Amplitude at 50 Hz')
The plot above shows the ampltude of the 50 Hz component versus time. It shows the dip in amplitude from t=1 to t=2.
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