Signals and Systems - Electrical Engineering

220 C H A P T E R 3: The Laplace Transform

after replacingX(s)= 1 /s. We find thatB 1 = 1 /2,B 2 =1, andB 3 =− 1 /2, so that the complete

response is

y(t)=[0.5+e−t−0.5e−^2 t]u(t) (3.36)

Again, we can check that this solution satisfies the initial conditiony( 0 )anddy( 0 )/dt(this

is particularly interesting to see, try it!). The steady-state response is 0.5 and the transient

[e−t−0.5e−^2 t]u(t).

According to Equation (3.36), the complete solutiony(t)is composed of the zero-state response,

due to the input only, and the response due to the initial conditions only or the zero-input

response. Thus, the system considers two different inputs: One that isx(t)=u(t)and the other

the initial conditions.

If we are able to find the transfer functionH(s)=Y(s)/X(s)its inverse Laplace transform would be

h(t). However that is not possible when the initial conditions are nonzero. As shown above, in the

case of nonzero initial conditions, we get that the Laplace transform is

Y(s)=

X(s)

A(s)

+

I(s)

A(s)

where in this caseA(s)=(s+ 1 )(s+ 2 )andI(s)=s+3, and thus we cannot find the ratio

Y(s)/X(s). If we make the second term zero (i.e.,I(s)=0), we then have thatY(s)/X(s)=H(s)=

1 /A(s)andh(t)=e−tu(t)−e−^2 tu(t). n

nExample 3.24

Consider an analog averager represented by

y(t)=

1

T

∫t

t−T

x(τ)dτ (3.37)

wherex(t)is the input andy(t)is the output. The derivative ofy(t)gives the first-order differential

equation

dy(t)

dt

=

1

T

[x(t)−x(t−T)]

with a finite difference for the input. Let us find the impulse response of this analog averager.

Solution

The impulse response of the averager is found by lettingx(t)=δ(t)and the initial condition be

zero. Computing the Laplace transform of the two sides of the differential equation, we obtain

sY(s)=

1

T

[1−e−sT]X(s)