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

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?

Answer 1

[Star Strider]

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:

[Mark Rzewnicki]

Beautiful. Thanks very much!

[Star Strider]

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.