Question

1. 25pts) The input signal z(t) is given to the LII system with its impulse reponuoh(t) where r(t) = sin(2t), h(t) - t). Caculate the corresponding output response y(t).

Answer 1

Given.

          

(t) = sin(2)

         
Laplace

transform
     
X(S) = 52 +4

        

h(t) = e2u(t) = H(S) = 27.

        

Output y(t) = r(t) * h(t)

                       

=Y(S) = X(S) * H(S)

                                       

7+8 x ++zS -

                                       

2 (52 + 4) (S+2)

Now let us do partial fractions,

2 A (S2 + 4) (S + 2) 5+2 BS + C 52 + 4

AS2 + 4) + (BS+C)(8 +2) (S2 + 4)( +2) (S2 + 4) (S + 2)

2 = ASP + 4A + B S2 + 2BS + CS +20

By equating corresponding coefficients, we get

A+B=0
     $\rightarrow (1)$

2B+C =0=C= -2B
   
(2)

4A + 2C = 2
     
→ (3

Substitute (2) in (3), we get

4A – 4B = 2
   
D)

Multiply equation(1) with '4' :

4A + 4B=0
   
→ (6)

By solving (a) and (b), we get

A = -and B=-

Substitute this in equation(2), then

C=-2B = -21-1-1

Now substitute these values in output equation ,

* + E FSH" = (){

(9) = 4(5+2) 4(52 + 4) * 2(32 + 4)

Now apply inverse laplace transform,

(77 , 2S)) + (777,28}_ [2 +S)1 = (s)x.

y(t) = 7e-2 – cos(2t) + =sin(2t)

:: y(t) = -(e-2 – cos(2t) + sin(2t)