Inverse Laplace transform - there has to be a better way?

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Mark Rzewnicki
Mark Rzewnicki 2019-11-6
评论： Star Strider 2019-11-7
I have a transfer function and I am applying a step input . My ultimate goal is to find the time response .
Solving by hand, I know that output .
Then by partial fraction expansion I know that .
From the above I can easily take the inverse Laplace transform and see that .
My goal is to obtain with as few keystrokes as possible.
The fastest way I have found is to perform the partial fraction expansion using residue():
num = 9;
denom = [1 9 9 0];
[r,p,k] = residue(num,denom);
which gives the result:
r =
0.1708
-1.1708
1.0000
p =
-7.8541
-1.1459
0
k =
[]
From which I can write:
sym s
F = 0.1708/(s+7.8541) -1.1459/(s+1.1708) + 1/s;
ilaplace(F)
which gives the result:
ans =
(427*exp(-(78541*t)/10000))/2500 - (11459*exp(-(2927*t)/2500))/10000 + 1
which is the answer I want!
But there has to be a better way of doing this! Can someone please advise?

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Star Strider
Star Strider 2019-11-6
Try this:
syms s t
num = 9;
denom = [1 9 9 0];
% [r,p,k] = residue(num,denom);
nums = num;
dens = poly2sym(denom, s); % Create Symbolic Polynomial
F = nums / dens * 1/s % Transfer Function With Step Input
Fpf = partfrac(F) % Partial Fraction Expansion
f = ilaplace(Fpf) % Time Domain Expression
f = rewrite(f, 'exp') % Rewrite As Exponential Terms
f = vpa(f,4) % Convert Fractions To Decimal Equivalents
flatex = latex(f) % LaTeX Expression
producing:
  2 个评论
Star Strider
Star Strider 2019-11-7
As always, my pleasure!
The partfrac function was introduced in R2015a, and seems to be invoked automatically in the last two or three releases. It was not automatic before then, so I specifically included it here.

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