y=a1(:,1);
chan1= y-mean(a1);
fs=128;
d=1/fs;
t=[0:length(chan1)-1]*d;
fs=fft(chan1,128);
pp=fs.*conj(fs)/128;
ff=(2:9)/128/d;
figure(4);
plot(ff,pp(2:9));
title('Power Spectrum');
xlabel('Frequency(Hz)');
ylabel('Power Spectrum Density(µv)');
grid on;
Peak.jpg
How to find peak value on y-axis without using Data Cursor tools. Thanks in advance.

 Réponse acceptée

Star Strider
Star Strider le 15 Jan 2020

0 votes

If you have R2017b or later, you can use the islocalmax function.
If you have the Signal Processing Toolbox, you can use the findpeaks function.

4 commentaires

Reff
Reff le 15 Jan 2020
I tried to use findpeaks on y-axis by findpeaks(pp(2:9)), unfortunately the graph shift -1 at x-axis and -1000 on y-axis. Am I missing something?
If you are using:
[pks,locs] = findpeaks(pp(2:9))
you will need to add 1 to the ‘locs’ that findpeaks returns to get the actual indices.
Example —
x = 1:9;
pp = [50 100 1800 300 2500 1500 3000 3500 5800];
[pks,locs] = findpeaks(pp(2:9));
figure
plot(x, pp)
hold on
plot((locs+1), pks, 'r*')
hold off
Note that my code uses ‘(locs+1)’ in the plot call to refer to them correctly.
Reff
Reff le 16 Jan 2020
Hi Strider,
It works well. Really aprreciate it. Thanks.
Star Strider
Star Strider le 16 Jan 2020
As always, my pleasure!

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