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

Question

Mark Rzewnicki 2019-11-6

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https://ww2.mathworks.cn/matlabcentral/answers/489597-inverse-laplace-transform-there-has-to-be-a-better-way

评论： Star Strider 2019-11-7

采纳的回答： Star Strider

在 MATLAB Online 中打开

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:

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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Answer 1

Star Strider 2019-11-6

1
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https://ww2.mathworks.cn/matlabcentral/answers/489597-inverse-laplace-transform-there-has-to-be-a-better-way#answer_400213

在 MATLAB Online 中打开

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:

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Mark Rzewnicki 2019-11-6

Beautiful. Thanks very much!

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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