Problem in finding inverse Laplace

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chess
chess el 23 de Jun. de 2015
Comentada: chess el 23 de Jun. de 2015
I have a problem with finding inverse laplace of function with Matlab. I obtained following transfer function :
I tried methods like
syms s t;
num=[0 0 0 1.658e24 -1.163e14 6.076e15];
den=[1 3.334e09 1.005e15 1.675e24 5.025e27 1.675e33];
numsym=poly2sym(num);
densym=poly2sym(den);
transfer=numsym./densym;
ilaplace(transfer,s,t)
but did not get any result. Also I tried to use partial fraction expansion and I wrote transfer function respect to poles and zeros but when I added all the pieces I did not get the original transfer function.
syms s t;
num=[0 0 0 1.658e24 -1.163e14 6.076e15];
den=[1 3.334e09 1.005e15 1.675e24 5.025e27 1.675e33];
numsym=poly2sym(num);
densym=poly2sym(den);
transfer=numsym./densym;
[z,p,r]=residue(num,den);
transfer1=(z(1)/(s-p(1)))+(z(2)/(s-p(2)))+(z(3)/(s-p(3)))+(z(4)/(s-p(4)))+(z(5)/(s-p(5)));
transfer1=vpa(simplifyFraction((transfer1),'Expand',true),2)
The transfer1 variable after simplification is order 4 in numerator while the original transfer function is order 2. I would be very thankful if anyone help me to find the transfer function of following expression by any method?

Respuestas (1)

Azzi Abdelmalek
Azzi Abdelmalek el 23 de Jun. de 2015
num=[0 0 0 1.658e24 -1.163e14 6.076e15];
den=[1 3.334e09 1.005e15 1.675e24 5.025e27 1.675e33];
numsym=poly2sym(num,'s');
densym=poly2sym(den,'s');
transfer=numsym./densym;
H=ilaplace(transfer)
  1 comentario
chess
chess el 23 de Jun. de 2015
That returns 1658000000000000130023424*sum((r3^2*exp(r3*t))/(5*r3^4 + 13336000000*r3^3 + 3015000000000000*r3 which is not what I want

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