# Question: inverse Laplace of transfer function has hyperbolic sines and cosines in the solution --why?.

## Question: inverse Laplace of transfer function has hyperbolic sines and cosines in the solution --why?.

Maple 2020

Hello,

I use Syrup to solve for the transfer function of a circuit, and in particular, the voltage at a given node.   If I use the "trans" option, the solution generates the expected result, in the expected format.  Syrup solves the differential equations and the resulting time domain equation is as expected.

If I then use the 'AC'  option, I get the result in the s-domain.  I then solve for the step response, and take the inverse Laplace transform.  The solution is given in terms of sinh and cosh.

If I plot the voltage in either solution as a function of time, the plots are identical.  So, the sinh/cosh solution must be somehow simplifiable so that it "looks" like the time domain solution I get from the transient analysis.

 (1)

 (2)

 (3)

Solution doing Transient Analysis

 > sol1 := dsolve(deqs):

Reassign rest to include the solution.

 > rest := eval(rest,sol1):

Compute the voltage across the capacitor Cm.

 > tmp1:=simplify(eval(v[Cm](t), sol1));
 (4)
 >
 (5)
 > solve(tmp1=2.5*(1-exp(-1)),t)
 p10 := plot(tmp1, t = 0 .. 0.00050000, color = 'blue', linestyle = 'dash')

Solution from AC analysis:

 (7)
 (8)

Impulse Response

 (9)

 (10)

Step Response

 (11)

 (12)

What is going on?  Where are the sinh and cosh terms in the solution coming from and why wont it simplify so that it looks like the other solution if they are identical?  Is there something special I need to do when using the invlaplace function?

Thanks!