Asked by Lisa Justin
on 3 Dec 2019 at 12:52

Hi,

I need to extract only the peaks of three signals so I have one plot without changing the amplitude of the individual plots.

fs = 2e3;

t = (0:1/fs:1-1/fs)';

A= sin(2*pi*19*t);

B= chirp(t-0.6,61,t(end),603,'quadratic');

C=cos(2*pi*19*t);

peakSig= A+B+C;

Answer by Image Analyst
on 3 Dec 2019 at 14:50

Accepted Answer

Lisa, try using max() function to compute the max.

fs = 2e3;

t = (0:1/fs:1-1/fs)';

A= sin(2*pi*19*t);

B= chirp(t-0.6,61,t(end),603,'quadratic');

C=cos(2*pi*19*t);

plot(t, A, 'r-', 'LineWidth', 2);

hold on;

plot(t, B, 'g-', 'LineWidth', 2);

plot(t, C, 'b-', 'LineWidth', 2);

% Get the highest of any signal:

peakSig = max([A, B, C], [], 2);

% Get the lowest of any signal:

valleySig = min([A, B, C], [], 2);

plot(t, peakSig, 'k-', 'LineWidth', 3);

plot(t, valleySig, 'k-', 'LineWidth', 3);

grid on;

% Zoom in so we can actually see what's happening:

xlim([0, 0.04]);

legend('A', 'B', 'C', 'peakSig', 'valleySig');

If you have the valley signal in there also, you'll get this:

KALYAN ACHARJYA
2019 年 12 月 3 日 15:07

Great Sir!

Lisa Justin
about 11 hours ago

Thanks Image Analyst. Just what I wanted.

Great.

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Answer by KALYAN ACHARJYA
on 3 Dec 2019 at 13:58

fs = 2e3;

t = (0:1/fs:1-1/fs)';

A= sin(2*pi*19*t);

B= chirp(t-0.6,61,t(end),603,'quadratic');

C=cos(2*pi*19*t);

peakSig= max(A)+max(B)+max(C) % You can find the maximum of A/B/C using max function

Or:

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Answer by ABHILASH SINGH
on 3 Dec 2019 at 14:02

Edited by ABHILASH SINGH
on 3 Dec 2019 at 14:03

I think you are looking for this.

fs = 2e3;

t = (0:1/fs:1-1/fs)';

A= sin(2*pi*19*t);

B= chirp(t-0.6,61,t(end),603,'quadratic');

C=cos(2*pi*19*t);

peakSig= A+B+C;

plot(t,A,t,B,t,C)

figure

plot(t,envelope(peakSig))

Let me know if you need something different. This is the peakSig vs t

This is the peak of the three signals together

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