How can I find the frequency of this signal?

how can i find the frequency and period of this signal? Thank you

Answer 1

[Star Strider]

The obvious solution is to use the fft (link) function. There is a low-frequency baseline drift and a high-frequency signal in that appear to be spikes (EKG, perhaps).

To find the peaks and the associated frequencies, use the findpeaks function.

See this Answer (link) for an illustration.

[Jose Sandoval]

it is possible apply fft to a data set, and how i can get in a matrix the values of the points that are closed in the circles, it is possible make this for all the peaks in the graph?

[Star Strider]

That seems different from your original Question. It is definitely possible to find the times (or indices) as well as the amplitudes of the positive and negative peaks. For your signal, it would first be necessary to use a highpass (or bandpass) filter to filter out the low-frequency baseline variation. You did not post your data with the sampling times, so I cannot design a specific filter for it. If this is an EKG signal, this filter would work:

Fs =  ...;                                              % Sampling Frequency (Hz)
Fn = Fs/2;                                              % Nyquist Frequency (Hz)
Wp = [1.5   100]/Fn;                                    % Passband Frequencies (Normalised)
Ws = [0.5   102]/Fn;                                    % Stopband Frequencies (Normalised)
Rp = 10;                                                % Passband Ripple (dB)
Rs = 50;                                                % Stopband Ripple (dB)
[n,Ws] = cheb2ord(Wp,Ws,Rp,Rs);                         % Filter Order
[z,p,k] = cheby2(n,Rs,Ws);                              % Filter Design
[sosbp,gbp] = zp2sos(z,p,k);                            % Convert To Second-Order-Section For Stability

figure(3)
freqz(sosbp, 2^17, Fs)

filtered_signal = filtfilt(sosbp, gbp, original_signal); % Filter Signal

You will need to supply the sampling frequency, ‘Fs’. If it is not an EKG signal, you will also have to do a fft on it to determine the low-frequency cutoff to determine the frequency of the baseline variation and the signal frequencies you are interested in. Then design the filter passband appropriately.

To identify the positive peaks, use the findpeaks function on the filtered data:

[pospks,poslocs] = findpeaks(filtered_signal, 'MinPeakHeight',100);

and to find the negative peaks, use findpeaks on the negative of your signal:

[negpks,neglocs] = findpeaks(-filtered_signal, 'MinPeakHeight',300);

These return the amplitudes in the first output and the indices in the second output. To convert the indices to times, if ‘t’ is your vector of sampling times:

postimes = t(poslocs);
negtimes = t(neglocs);

To plot them:

figure(1)
plot(t, filtered_signal)
hold on
plot(postimes, pospeaks, '^r')
plot(negtimes, -negpeaks, 'vg')
hold off
grid

Again, without your data, I cannot provide definitive code, only code that I know works and that you will have to change to be certain it works with your data.

[Jose Sandoval]

Sorry, I thought I loaded my data, these are my data, it is in format .xls (Excel book 97 - 2003), please help me I've been delayed for several days with this, thank you

[Star Strider]

I am not certain exactly what you want from your data. This is the best I can do to identify the positive and negative peaks:

[d,s] = xlsread('data2.xls');
t = d(:,1);
y = d(:,2);

Td = std(diff(t));
Ts = mean(diff(t));                                     % Sampling Interval
Fs = 1/Ts;                                              % Sampling Frequency
Fn = Fs/2;                                              % Nyquist Frequency
L = length(t);
Fy = fft(y-mean(y))/L;                                  % Fourier Transform
Fv = linspace(0, 1, fix(L/2)+1)*Fn;                     % Frequency Vector
Iv = 1:length(Fv);                                      % Index Vector

figure(1)
plot(Fv, abs(Fy(Iv))*2)
grid
axis([0  15    0  20])

Wp = [1.5   25]/Fn;                                     % Passband Frequencies (Normalised)
Ws = [0.5   26]/Fn;                                     % Stopband Frequencies (Normalised)
Rp = 10;                                                % Passband Ripple (dB)
Rs = 50;                                                % Stopband Ripple (dB)
[n,Ws] = cheb2ord(Wp,Ws,Rp,Rs);                         % Filter Order
[z,p,k] = cheby2(n,Rs,Ws);                              % Filter Design
[sosbp,gbp] = zp2sos(z,p,k);                            % Convert To Second-Order-Section For Stability

figure(2)
freqz(sosbp, 2^17, Fs)

y_filtered = filtfilt(sosbp, gbp, y);                  % Filter Signal

figure(3)
plot(t, y_filtered)
grid

[pospks,poslocs] = findpeaks(y_filtered, t, 'MinPeakDistance',4, 'MinPeakHeight',100);
[negpks,neglocs] = findpeaks(-y_filtered, t, 'MinPeakDistance',4, 'MinPeakHeight',100);

postimes = poslocs;
negtimes = neglocs;

figure(4)
plot(t, y_filtered)
hold on
plot(postimes, pospks, '^r')
plot(negtimes, -negpks, 'vg')
hold off
grid

Note that the period is not constant. It decreases with time.

[Jose Sandoval]

thank you very much with postime and negtime, i can obtain the time interval

[Star Strider]

As always, my pleasure.