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
      
      
 on 19 Nov 2023
				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
      
      
 on 19 Nov 2023
				
      Edited: Walter Roberson
      
      
 on 20 Nov 2023
  
			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
Answers (1)
  Sam Chak
      
      
 on 20 Nov 2023
        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
. 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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