Use L34 and L2 to find the inverse transform of G(p)H(p)whenand G(p)=1/(p+a)andH(p)=1/(p+b); your result should be L7 .

Short Answer

Expert verified

The inverse transforms of is GpHpis1b-ae-at-e-bτ.

Step by step solution

01

Given information.

The given expressions are andGpHpis1p+aandHpis1p+b .

02

Inverse transform.

The piecewise-continuous and exponentially-restricted real function f(t) is the inverse Laplace transform of a function F(s), and it has the property:

L{f}(s)=L{f}(s)=F(s)

where L is the Laplace transform.

03

Find the inverse transform of G(p)H(p). .

Calculate Laplace inverse.

L-1GpHp=g*h=0tgt-τhτdτ

Calculate Laplace inverse of Gp.

role="math" localid="1659264162995" L-1Gp=L-11p+agt=e-atgt-τ=e-at-τ

Calculate Laplace inverse of H(p).

L-1Hp=L-11p+bht=e-btht-τ=e-bτ

Calculate Laplace inverse of G(p)H(p) .

L-11p+a1p+b=1b-ae-at-e-bτ

Thus, the Laplace inverse of is GpHpis1b-ae-at-e-bτ..

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