findpeaks in a for loop
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I would like to use the findpeaks function to calculate the peaks of the different rows of the DT matrix obtained with the for loop.
Could someone explain to me how to do it?
Thanks in advance
% Input parameters
r= 0.006; % radius[m]
k= 5e-7; % thermal diffusivity[m^2/s]
pc= 1e6; % volumetric heat capacity[J/m*3*K]
qs= 62.5; % heat input per unit of lenght per unit of time[J/m*s]
t0= [1 10 100 1000]; % heating duration[s]
Qs= qs./pc; % source strenght per unit time[m*2*K/s]
n=100;
% for loop
for i=1:length(t0)
time = logspace(0,4,n);
idH = find(time < t0(i));
idC = find(time>=t0(i));
tH = time(idH);
tC = time(idC);
t= [tH tC];
DeltaTH = (-Qs./(4.*pi.*k)).*-expint((r.^2)./(4.*k.*tH));
DeltaTC = (Qs./(4.*pi.*k)).*-(expint((r.^2)./(4.*k.*(tC-t0(i))))-expint((r.^2)./(4.*k.*tC)));
DT(i,:) = [DeltaTH DeltaTC];
end
采纳的回答
Star Strider
2019-8-7
Since ‘DT’ appears to be two concatenated row vectors, add this line just after ‘DT’:
[pks{i},locs{i}] = findpeaks(DT(i,:));
so the loop is now:
for i=1:length(t0)
time = logspace(0,4,n);
idH = find(time < t0(i));
idC = find(time>=t0(i));
tH = time(idH);
tC = time(idC);
t= [tH tC];
DeltaTH = (-Qs./(4.*pi.*k)).*-expint((r.^2)./(4.*k.*tH));
DeltaTC = (Qs./(4.*pi.*k)).*-(expint((r.^2)./(4.*k.*(tC-t0(i))))-expint((r.^2)./(4.*k.*tC)));
DT(i,:) = [DeltaTH DeltaTC];
[pks{i},locs{i}] = findpeaks(DT(i,:));
end
Cell arrays are best for this, even though there is only one identified peak and index for each iteration. That could change if you change ‘DT’ or add name-value pairs to your findpeaks call.
4 个评论
Massimiliano Gamba
2019-8-7
Star Strider
2019-8-7
As always, my pleasure!
NIKHIL MC
2023-3-15
can we convert the cell array into matrix?? i have a cell array with each column havig different number of elements
Star Strider
2023-3-15
@NIKHIL MC — Unless all the cell array contents are of equal size in each cell, it is not possible to concatenate them into a matrix.
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