second largest peak value

I'm getting the amp of second largest peak as correct but the time is wrong ?

How to obtain the second highest value X,Y from this and plot as already Y is correct ,X seems to be in wrong region?

s=load ('30827.mat');

signal(1,1:length(s.ans))=s.ans;

load t.mat;

[PkAmp, PkTime] = findpeaks(abs(hilbert(signal(1,:))),t);

[highestPeak, indexOfHighestPeak] = max(PkAmp);

[a(1),b2]=max(abs(hilbert(signal(1,:))));

[a2(1),b6]=max(PkAmp(PkAmp<max(PkAmp))); % to get second largest peak value

% error is the time value calcluate and plotted wrong??

Highesttime(1,:) = t(b2);

Highest2time(1,:) = t(b6);

figure;

plot(t,(signal(1,:)),t,abs(hilbert(signal(1,:))),t(b2),a(1),'ro')

hold on

plot(PkTime, PkAmp, '^r', 'MarkerFaceColor','r')

xlim([0 0.0006]);ylim([-inf inf])

grid on

hold on,plot(t(b6),a2(1),'ko')

hold off

Answer 1

[Alex Sune]

To obtain the coordinates of the second peak you just need:

[PkAmp, PkTime] = findpeaks(abs(hilbert(signal(1,:))),t); 
[~,idx] = sort(PkAmp,'descend');
PkAmp(idx(2)) %Amplitude of the second peak
PkTime(idx(2)) %Time of the second peak

[Ramesh Bala]

Danke Schon Alex,

Lemme have a check

[Ramesh Bala]

Alright Alex ! I tried to use the method but it shows not compatible error ??

close all

s=load ('30827.mat');

signal(1,1:length(s.ans))=s.ans;

s= load ('30802.mat');

signal(2,1:length(s.ans))=s.ans;

load t.mat;

for k = 1:size(signal,1)

[PkAmp, PkTime] = findpeaks(abs(hilbert(signal(k,:))),t);

[a(k),b2]=max(abs(hilbert(signal(k,:))));

[a2(k),b6] = sort (PkAmp, 'descend' );

M(k) =PkAmp (b6(2)); % Amplitude of the second peak

Z(k) =PkTime (b6(2)); % Time of the second peak

figure;

plot(t,(signal(k,:)),t,abs(hilbert(signal(1,:))),t(b2),a(k),'ro')

hold on

plot(PkTime, PkAmp, '^r', 'MarkerFaceColor','r')

xlim([0 0.0006]);ylim([-inf inf])

grid on

hold on,plot(Z(k),M(k),'ko')

hold off

end

[Ramesh Bala]

Now it works well with some changes !!!But one question is how do i get the changing index values of b2 and b6 ,as it gives only the end values ( reason the X*Y matrix are not equal )

now it gives a single scalar value ,but I want all the index values b2 = [ xyx xyz] and b6 as such ??

close all

s = load ('30827.mat');

signal (1.1: length (s.ans)) = s.ans;

s = load ('30802.mat');

signal (2.1: length (s.ans)) = s.ans;

load t.mat;

for k = 1: size (signal, 1)

[PkAmp, PkTime] = findpeaks(abs(hilbert(signal(k,:))),t);

[a(k),b2]=max(abs(hilbert(signal(k,:))));

[a2, b6] = sort (PkAmp, 'descend' );

M(k)=PkAmp (b6(2)); % Amplitude of the second peak

Z(k)=PkTime (b6 (2)); % Time of the second peak

figure;

plot(t,(signal(k,:)),t,abs(hilbert(signal(k,:))),t(b2),a(k),'ro')

hold on

plot(PkTime, PkAmp, '^r', 'MarkerFaceColor','r')

xlim([0 0.0006]);ylim([-inf inf])

grid on;hold on

plot(Z(k),M(k),'ko')

hold off

end

[Alex Sune]

I don't understand what do you want exactly, but to extract the peaks you want you don't need to use the max function:

for k = 1: size (signal, 1)
    [PkAmp, PkTime] = findpeaks(abs(hilbert(signal(k,:))),t); 
     
    [~,idx] = sort (PkAmp, 'descend' );
    b(k) = idx(1);  % Index of the largest peak
    a(k) = PkAmp (b(k)); % Amplitude of the largest peak
    time(k) = PkTime (b(k)); % Time of the largest peak
    
    b2(k) = idx(2);  % Index of the second largest peak
    a2(k) = PkAmp(b2(k)); % Amplitude of the second largest peak
    time2(k) = PkTime (b2(k)); % Time of the second largest peak
    
    %Some plots
end

[Jan]

Use maxk(Signal, 2) instead of sorting the complete vector.

[Ramesh Bala]

Thank you it works well.

I would like to know if there is a way to collect all the peak values in a matrix ?

for k = 1: size (signal, 1)

[PkAmp, PkTime] = findpeaks(abs(hilbert(signal(k,:))),t);

b2(k) = idx(2); % Index of the second largest peak

a2(k) = PkAmp(b2(k)); % Amplitude of the second largest peak

time2(k) = PkTime (b2(k)); % Time of the second largest peak

%this helps me to idenitfy largest peaks ,but how can I collect all the %peak values of a figure into a matrix??

end

[Alex Sune]

In fact, you have all the peaks in PkAmp variable and the time they occur in Pktime.

[Ramesh Bala]

Yeah you're right .But I can't group as it says " 1-by-85 and the size of the right side is 1-by-87."

for the obtained values

[Ramesh Bala]

[PkAmp(k,:), PkTime(k,:)] = findpeaks(abs(hilbert(signal(k,:))),t);

[Ramesh Bala]

The above one doesn't work as the matrix is changing its size.so,how to group such changing matrix

[Alex Sune]

Using cell array:

[PkAmp{k}, PkTime{k}] = findpeaks(abs(hilbert(signal(k,:))),t); 

[Ramesh Bala]

Dear Alex

I have one more silly doubt in removing the index,I'm trying to find now the the largest time instead of Amp.So,I have used 'ascend' function which helps me to plot the the required time.

clc

clear all

close all

format long;

filenames={"30827"}

S(:,1) = [25 25 21.4 17.92 28.5 32.07 30 35 20 15];

S(:,2)= [10 15 8.5 12.07 8.5 12.07 5 5 5 5 ];

for i=1:numel(filenames)

s(i)=load (filenames{i});

signal(i,1:length(s(i).ans))=s(i).ans; % loading the signals from the scanning pts

end

load t.mat; %loading the time values

for k=1:1:length(S)

[a(k),b2]=max(abs(hilbert(signal(k,:)))); % first peak amp & index

%finding the peaks

set(0,'DefaultFigureWindowStyle','docked')

[PkAmp, PkTime] = findpeaks(abs(hilbert(signal(k,:))),t);

[~,idx] = sort (PkTime, 'ascend' );

b6(k) = idx(4); % Index of the fourth largest peak

a6(k) = PkAmp(b6(k)); % Amplitude of the fourth largest peak

time4(k)= PkTime (b6(k)); % Time of the fourth largest peak

figure;

plot(t,(signal(k,:)),t,abs(hilbert(signal(k,:))),t(b2),a(k),'ro')

hold on;grid on

plot(PkTime, PkAmp, '^r', 'MarkerFaceColor','r')

%plotting 4th largest

hold on,plot(time4(k), a6(k),'ko')

hold off

end

As per the script it plots the 4th largest PKTime peak .But how to remove other PkTime based on highest "a" highest PKAmp and then recontinue to plot the graph by taking highest(PKAMP) as PKTIME 1?

[Ramesh Bala]

Answer 2

[Steven Lord]

Use the islocalmax function with the 'MaxNumExtrema' option.