inverse laplace transform from laplace transfer without using " ilaplace" instruction
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finding the inverse laplace transform from laplace transfer function by matlab not using " ilaplace" instruction
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Walter Roberson
2023 年 11 月 19 日
If you were to convert the symbolic laplace form to a tf() transfer function from the Control System Toolbox, you might be able to express the inverse laplace, at least numerically.
But I suspect that the point of the assignment is to have you go through the algorithm of inverse laplace. For example working with Fourier transform can sometimes be useful in this kind of task
Walter Roberson
2023 年 11 月 19 日
編集済み: Walter Roberson
2023 年 11 月 20 日
https://math.libretexts.org/Courses/Monroe_Community_College/MTH_225_Differential_Equations/8%3A_Laplace_Transforms/8.2%3A_The_Inverse_Laplace_Transform is one tutorial that shows partial fraction expansion
回答 (1 件)
Sam Chak
2023 年 11 月 20 日
Hi @Abdallah
My integral skills get a little rusty due to over-relying on the Laplace transform look-up table. Anyhow, let's try this. The inverse Laplace transform definition or formula is given by

Say, a transfer function of a 1st-order system is given by

Suppose that the initial condition is assumed to be zero and the Laplace transform of the unit-step function is
. The output response in s-domain becomes


Decomposing Y(s) into partial fractions gives

Taking the inverse Laplace transform of the partial fractions without using the ilaplace() command or the look-up table can be mathematically challenging:



Substituting 


Applying Euler's formula, we obtain




Well, up to this point, both
and
are odd functions; thus, the integral over a symmetric interval equals zero, because lower half-area will cancel out the upper half-area. So, we just need to evaluate the integral of the real part.



syms u t T
assume(u, "real");
assume(t > 0);
T = sym('1');
expr1 = (T^2)*1/T*cos(u*t)/((T^2)*u^2 + 1);
f1 = 2*int(expr1, u, [0, Inf])
expr2 = (T^2)*u*sin(u*t)/((T^2)*u^2 + 1);
f2 = 2*int(expr2, u, [0, Inf])
f = 1/(2*sym('pi'))*(f1 + f2)
Therefore, the inverse Laplace transform gives

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