I am trying to store all the values ​​of (f) and all the corresponding peaks of the resulting wave in two columns and then plot f vs. peaks, but there is a problem.

clc
clear all

tspan = [0 5];
Ic = 0;  
A=1;
results = [];

% Simulate using ode45 in a loop
for f = 2:6 
    
    signal = @(t, y) A*2*pi*f*cos(2*pi*f*t);
    
    [t, y] = ode45(signal, tspan, Ic);
    
    pks = findpeaks(y(:,1),'minpeakHeight',0.99);
    
    results = [results; f; pks]; 
end
plot(results(:,1),results(:,2),'bo-')
xlabel('f (Hz)')
ylabel('peaks (cm)')

Answer 1

[Walter Roberson]

findpeaks() is returning different numbers of peaks for each frequency.

clc
clear all

tspan = [0 5];
Ic = 0;  
A=1;
results = {};

% Simulate using ode45 in a loop
for f = 2:6 
    
    signal = @(t, y) A*2*pi*f*cos(2*pi*f*t);
    
    [t, y] = ode45(signal, tspan, Ic);
    
    pks = findpeaks(y(:,1),'minpeakHeight',0.99);
    
    results{end+1} = {f, pks};
end
celldisp(results)

 
results{1}{1} =
 
     2

 
 
results{1}{2} =
 
    0.9985
    0.9988
    0.9992
    0.9941
    0.9976

 
 
results{2}{1} =
 
     3

 
 
results{2}{2} =
 
    0.9985
    0.9988
    0.9992
    0.9941
    0.9976
    0.9951
    0.9970
    0.9984

 
 
results{3}{1} =
 
     4

 
 
results{3}{2} =
 
    0.9985
    0.9988
    0.9992
    0.9941
    0.9976
    0.9951
    0.9970
    0.9984
    0.9987
    0.9992
    0.9941

 
 
results{4}{1} =
 
     5

 
 
results{4}{2} =
 
    0.9985
    0.9988
    0.9992
    0.9941
    0.9976
    0.9951
    0.9970
    0.9984
    0.9987
    0.9992
    0.9941
    0.9975
    0.9953
    0.9971
    0.9984

 
 
results{5}{1} =
 
     6

 
 
results{5}{2} =
 
    0.9985
    0.9988
    0.9992
    0.9941
    0.9976
    0.9951
    0.9970
    0.9984
    0.9987
    0.9992
    0.9941
    0.9975
    0.9953
    0.9971
    0.9984
    0.9987
    0.9992

 

plot(t,y)

[mohammed]

@Walter Roberson I am sorry, the diagram must be between f and pks (plot (f, pks)), where for each frequency there are several peaks.

[Walter Roberson]

clc
clear all

tspan = [0 5];
Ic = 0;  
A=1;
results = {};

% Simulate using ode45 in a loop
for f = 2:6 
    
    signal = @(t, y) A*2*pi*f*cos(2*pi*f*t);
    
    [t, y] = ode45(signal, tspan, Ic);
    
    pks = findpeaks(y(:,1),'minpeakHeight',0.99);
    
    results{end+1} = {f, pks};
end

for K = 1 : length(results)
    plot(results{K}{1}, results{K}{2}, 'bo-');
    hold on
end
hold off

[mohammed]

@Walter Roberson Tank you

Good job, but I wished f and pks were collected in two columns (two vectors), I think it required to be extracted from the cells into two columns.

[Walter Roberson]

The different f have different number of pks . You cannot extract the pks into a numeric array -- not unless you pad unused columns with NaN or inf.

tspan = [0 5];
Ic = 0;  
A=1;
f_vector = [];
pks_array = [];

% Simulate using ode45 in a loop
f_list = 2:6;
for fidx = 1:length(f_list) 
    f = f_list(fidx);
    signal = @(t, y) A*2*pi*f*cos(2*pi*f*t);
    
    [t, y] = ode45(signal, tspan, Ic);
    
    pks = findpeaks(y(:,1),'minpeakHeight',0.99);

    f_vector(fidx) = f;
    pks_array(1:fidx-1,end+1:length(pks)) = nan;
    pks_array(fidx,1:length(pks)) = pks;
    pks_array(fidx,length(pks)+1:end) = nan;
end
plot(f_vector, pks_array.', 'bo')

pks_array

pks_array = 5×17

0.9985 0.9988 0.9992 0.9941 0.9976 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN 0.9985 0.9988 0.9992 0.9941 0.9976 0.9951 0.9970 0.9984 NaN NaN NaN NaN NaN NaN NaN NaN NaN 0.9985 0.9988 0.9992 0.9941 0.9976 0.9951 0.9970 0.9984 0.9987 0.9992 0.9941 NaN NaN NaN NaN NaN NaN 0.9985 0.9988 0.9992 0.9941 0.9976 0.9951 0.9970 0.9984 0.9987 0.9992 0.9941 0.9975 0.9953 0.9971 0.9984 NaN NaN 0.9985 0.9988 0.9992 0.9941 0.9976 0.9951 0.9970 0.9984 0.9987 0.9992 0.9941 0.9975 0.9953 0.9971 0.9984 0.9987 0.9992

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