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

14 Ansichten (letzte 30 Tage)
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

Antworten (4)

William Rose
William Rose am 20 Feb. 2024
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 Kommentare
William Rose
William Rose am 20 Feb. 2024
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 am 21 Feb. 2024
@William Rose Thank you for giving me this information
But, i am working on Wavelets So, i need to do it using wavelet transform

Melden Sie sich an, um zu kommentieren.


Alexander
Alexander am 20 Feb. 2024
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.
  3 Kommentare
Alexander
Alexander am 21 Feb. 2024
@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.
A Lumbini
A Lumbini am 21 Feb. 2024
Thank you, your inputs are really helpful
But,i want to do it using wavelets

Melden Sie sich an, um zu kommentieren.


Alexander
Alexander am 21 Feb. 2024
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 am 21 Feb. 2024
@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.
  6 Kommentare
A Lumbini
A Lumbini am 23 Feb. 2024
@William Rose Thank you mam
The information you provided is really beneficial
I will work with CWT

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Time-Frequency Analysis finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by