Signals and Systems - Electrical Engineering

3.5 Analysis of LTI Systems 215

I(s)=

∑N

k= 1

ak





k∑− 1

m= 0

sk−m−^1 y(m)( 0 )





That is,I(s)depends on the initial conditions.

The notationy(k)(t)andx()(t)indicates thekth and theth derivatives ofy(t)and ofx(t), respectively
(it is to be understood thaty(^0 )(t)=y(t)and likewisex(^0 )(t)=x(t)in this notation). The assump-
tionN>Mavoids the presence ofδ(t)and its derivatives in the solution, which are realistically not
possible. To obtain the complete responsey(t)we compute the Laplace transform of Equation (3.29):

[N

∑

k= 0

aksk

]

︸ ︷︷ ︸

A(s)

Y(s)=

[M

∑

`= 0

b`s`

]

︸ ︷︷ ︸

B(s)

X(s)+

∑N

k= 1

ak





k∑− 1

m= 0

s(k−^1 )−my(m)( 0 )





︸ ︷︷ ︸

I(s)

which can be written as

A(s)Y(s)=B(s)X(s)+I(s) (3.32)

by definingA(s),B(s), andI(s)as indicated above. Solving forY(s)in Equation (3.32), we have

Y(s)=

B(s)

A(s)

X(s)+

1

A(s)

I(s)

and finding its inverse we obtain the complete responsey(t).

Letting

H(s)=

B(s)

A(s)

and H 1 (s)=

1

A(s)

thecomplete responsey(t)=L−^1 [Y(s)]of the system is obtained by the inverse Laplace transform of

Y(s)=H(s)X(s)+H 1 (s)I(s) (3.33)

which gives

y(t)=yzs(t)+yzi(t) (3.34)

where

zero-state response: yzs(t)=L−^1 [H(s)X(s)]

zero-input response: yzi(t)=L−^1 [H 1 (s)I(s)]

In terms of convolution integrals,

y(t)=

∫t

0

x(τ)h(t−τ)dτ+

∫t

0

i(τ)h 1 (t−τ)dτ (3.35)