Problem in finding inverse Laplace

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chess
chess 2015-6-23
评论： chess 2015-6-23
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?

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回答（1 个）

Azzi Abdelmalek
Azzi Abdelmalek 2015-6-23
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 个评论
chess
chess 2015-6-23
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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