pade-laplace: stuck with pade approximation!
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hi..im try to use pade-laplace method in matlab to model multi exponential,however im stuck at part with pade approximation. here my code,
for t=1:100
syms A1 A2 L1 L2
y(t)=A1*exp(-L1*t)+A2*exp(-L2*t);
end
lap=laplace(y)
i dont know how to use pade approximation with this function for the next process. can someone guide me? thanks.
3 Kommentare
Walter Roberson
am 10 Aug. 2011
What are "L1" and "L2", and do they relate to "l1" and "l2" ?
Amir Hamzah UTeM
am 10 Aug. 2011
Amir Hamzah UTeM
am 10 Aug. 2011
Antworten (1)
Paulo Silva
am 11 Aug. 2011
Why do you define the symbolic variables in every loop? there's no need for it, just define the symbolic variables once before the for loop.
Now for the time it takes you probably know the reason, the symbolic operations and mostly the laplace is causing the slowness.
Try the pade function from the Control System Toolbox
This version of your code should be faster
syms A1 A2 L1 L2
t=1:100;
y(t)=A1*exp(-L1*(t-1))+A2*exp(-L2*(t-1)); %y(t-1)
y1(t)=A1*exp(-L1*t)+A2*exp(-L2*t); %y(t)
lap=laplace(y); % get L(s-1)
lap1=laplace(y1); % get L(s)
pad=lap./lap1; % take too long to process L(s-1)/L(s)
6 Kommentare
Amir Hamzah UTeM
am 11 Aug. 2011
Amir Hamzah UTeM
am 11 Aug. 2011
Paulo Silva
am 11 Aug. 2011
Looking at your code more carefully I think that you are making some errors, first the pade approximation is for time delays on the frequency domain in the form exp(-s*T), your expression is in the time domain and when converted to the frequency domain gives exponential values but they are constant not in the form exp(-s*T) so they are not delays.
Amir Hamzah UTeM
am 11 Aug. 2011
Paulo Silva
am 11 Aug. 2011
here's a clue
s=sym('s')
il=ilaplace(exp(-s*0.1))
laplace(il)
Paulo Silva
am 11 Aug. 2011
http://www.mathworks.com/help/toolbox/control/ug/bstzkhr.html
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