How to get the fundamental frequency magnitude ?

Hi,

I have an array of a pseudo-sin signal.

I want to get the fundamental magnitude of it knowing the fundamental frequency.

My approach is (about 4-5 command lines)...,

a) fft the signal; b) search for the magnitude of the fundamental frequency; c) show the magnitude.

There is no command to get the fundamental magnitude easily ?

Thanks.

Andre

Answer 1

[Star Strider]

If you have the Signal Processing Toolbox, use the findpeaks function.

[Andre]

Hi,

my code goes like this:

y=ySignal(m,:);             
ydft = fft(y,NFFT);
freq=Fs/2*linspace(0,1,NFFT/2+1);
data = 2*abs(ydft(1:NFFT/2+1)/size(y,2));
index60=find(freq(:)==60);
YAAPreFaltaMagnitude60=abs(data(index60));

For example, the fundamental frequency of ySignal is on ~60Hz.

The problem is that, I can´t get the correct Amplitude of ySignal.

Thanks.

Andre

[Star Strider]

You need to attach (use the ‘paperclip’ icon) the ‘m’ row for Signal, so we can run your code to see what the problem is.

[Andre]

Hi,

its attached as a csv file.

Work like this, (1:400, 6)

for col 6, first 400 rows.

Almost 60Hz fundamental frequency signal.

Thank you.

[Andre]

Forgot the attached file.

[Star Strider]

What is the sampling frequency?

[Andre]

Sorry,

Fs=15360; NFFT=512;

[Star Strider]

Thanks!

[Star Strider]

This is what I used to find the frequency and amplitude of the first 100 ms ( 1537 samples ) of the signal. The fprintf statement at the end writes the maximum amplitude and its frequency to the Command Window. This is the only way I know of to do what you want:

filedat = which('Andre_signal.csv');
x = csvread(filedat);
ad = x(:,6);
lendat = size(x,1);
Fs=15360; NFFT=512;
Ts = 1/Fs;

Tv = linspace(0,length(ad)/Fs,length(ad));
Pls1 = find(Tv > 0.1, 1, 'first');              % Initial ‘pulse’ about 100 ms long
Fv = [0:length(ad)/2-1]*Fs/length(ad);

ad2 = ad(1:Pls1);
Tv2 = linspace(0,length(ad2)/Fs,length(ad2));   % Time vector
Fv2 = [0:length(ad2)/2-1]*Fs/length(ad2);       % Frequency vector
fftad2 = fft(ad2);
afftad2 = abs(fftad2(1:fix(length(ad2)/2)));

figure(1)
plot(Tv2, ad2)
grid

maxix2 = find(afftad2 == max(afftad2));         % Find index of maximum fft amplitude
FVmax2 = [2*afftad2(maxix2)/length(ad2)    Fv2(maxix2)]

figure(2)
plot(Fv2, 2*afftad2/length(ad2))
axis([0  350    ylim])
grid

fprintf(1,'\n\tMaximum amplitude = %13.5E at %.2f Hz\n\n', FVmax2)

[Andre]

Thank you. Gotta 2 questions:

1) Why can´t I: ad = x( 1:400 ,6); ?

2) What is Pls1 ?

Andre

[Star Strider]

My pleasure!

1. You probably can. I used the entire column in one part of my code because I didn’t get sufficiently detailed FFT results with only 400 samples ( 26 ms ) of a 60 Hz signal. That was only about 1.56 cycles, not enough for a reliable analysis, and gave a peak frequency of about 75 Hz, obviously wrong in the context of a longer and more reliable signal, and an inaccurate amplitude.
2. Your signal had three components, an initial 100 ms = 1537 sample segment (that I used as Pls1), a second smaller-amplitude segment with the same frequency about as long, and a third segment that seemed to repeat the first 100 ms. (There also appeared to be switching transients. They appeared in the analysis of the entire signal, but I did not include them in Pls1. I ignored all but the first 100 ms in my analysis that I posted here.) That was a longer version of the first 400 samples you wanted to use, and gave better results. I called it ‘Pulse#1’ for lack of a better term, and abbreviated it to Pls1. The FFT of Pls1 gave good results. The FFT of the first 400 samples did not.

[Andre]

Thank you very much.

Andre

[Star Strider]

Again, my pleasure!