Question

Suppose a causal CT LTI system has bilateral Laplace transform H(s) 2s - 2 $2 + (10/3)s + 1 (8)

(a) Write the linear constant coefficient differential equation (LCCDE) relating a general input x(t) to its corresponding output y(t) of the system corresponding to this transfer function in equation (8).
(b) Suppose the input x(t) = e-tu(t). Find the output y(t). In part (c), the output signal can be expressed as y(t) = - e-(1/3)t u(t) + e-tu(t) e-3tu(t), - 019 Where a, b, and care positive integers. What are they? a = b = C=

Answer 1

Given :

`$H(S) =(2S-2)/(S^2+\left((10)/(3)\right) S+1)$`

Transfer function,
`$H(S) =(Y(S))/(K(S))= (2S-2)/(S^2+\left((10)/(3)\right) S+1)$`

`$Y(S) \left(S^2+(10)/(3)S+1\right) = (2S-2) * (S)$`

`$S^2Y(S) + (10)/(3)(SY(S)) + Y(S) = 2(S * (S)) - 2 * (S)$`

Apply Inverse Laplace Transforms,

`$(d^2y(t))/(dt^2) + (10)/(3) (dy(t))/(dt) + y(t)=2 (dx(t))/(dt) - 2x(t)$`

The above equation represents the differential equation of transfer function.

Given :
`$x(t)=e^(-t) u(t) \Rightarrow X(S) = (1)/(S+1)$`

We have :
`$H(S) =(Y(S))/(K(S))= (2S-2)/(S^2+\left((10)/(3)\right) S+1)$`

`$Y(S) = X(S) * (6S-6)/(3S^2+10 S + 3) = (6S-6)/((S+1)(3S+1)(S+3))$`

`$Y(S) = (A)/(S+1)+(B)/(3S+1) + (C)/(S+3)`

`$A = Lt_(S \to -1) (S+1)Y(S)=(6S-6)/((3S+1)(S+3)) = (-6-6)/((-3+1)(-1+3)) = 3$`

`$B = Lt_(S \to -1/3) (3S+1)Y(S)=(6S-6)/((S+1)(S+3)) = (-6/3-6)/((1/3+1)(-1/3+3)) = (-9)/(2)$`

`$C = Lt_(S \to -3) (S+3)Y(S)=(6S-6)/((S+1)(3S+1)) = (-18-6)/((-3+1)(-9+1)) = (-3)/(2)$`

So,

`$Y(S) = (3)/(S+1) - (9/2)/(3S+1) - (3/2)/(S+3)$`

`$=(3)/(S+1) - (3/2)/(S+1/3) - (3/2)/(S+3)$`

Applying Inverse Laplace Transform,

`$y(t) = 3e^(-t)u(t)-(3)/(2)e^(-t/3)u(t) - (3)/(2)e^(-3t) u(t)$`

`$=(-3)/(2)e^{-(1)/(3)t}u(t) + (3)/(1)e^(-t)u(t)-(3)/(2)e^(-3t) u(t)$`

where, a = 2

b = 1

c= 2