Signals and Systems - Electrical Engineering

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

Assume the above equation represents a system with inputx(t)and outputy(t). Find the impulse

responseh(t)and the unit-step responses(t)of the system.

Solution

If the initial conditions are zero, computing the two- or one-sided Laplace transform of the two

sides of this equation, after lettingY(s)=L[y(t)] andX(s)=L[x(t)], and using the derivative

property of Laplace, we get

Y(s)[s^2 + 3 s+2]=X(s)

To find the impulse response of this system (i.e., the system responsey(t)=h(t)), we letx(t)=δ(t)

and the initial condition be zero. SinceX(s)=1, thenY(s)=H(s)=L[h(t)] is

H(s)=

1

s^2 + 3 s+ 2

=

1

(s+ 1 )(s+ 2 )

=

A

s+ 1

+

B

s+ 2

We obtain valuesA=1 andB=−1, and the inverse Laplace transform is then

h(t)=

[

e−t−e−^2 t

]

u(t)

which is completely transient.

In a similar form we obtain the unit-step responses(t), by lettingx(t)=u(t)and the initial

conditions be zero. CallingY(s)=S(s)=L[s(t)], sinceX(s)= 1 /s, we obtain

S(s)=

H(s)

s

=

1

s(s^2 + 3 s+ 2 )

=

A

s

+

B

s+ 1

+

C

s+ 2

It is found thatA= 1 /2,B=−1, andC= 1 /2, so that

s(t)=0.5u(t)−e−tu(t)+0.5e−^2 tu(t)

The steady state ofs(t)is 0.5 as the two exponentials go to zero. Interestingly, the relationsS(s)=

H(s)indicates that by computing the derivative ofs(t)we obtainh(t). Indeed,

ds(t)

dt

=0.5δ(t)+e−tu(t)−e−tδ(t)−e−^2 tu(t)+0.5e−^2 tδ(t)

=[0.5− 1 +0.5]δ(t)+[e−t−e−^2 t]u(t)

=[e−t−e−^2 t]u(t)=h(t) n

Remarks

n Because the existence of the steady-state response depends on the poles of Y(s)it is possible for an unstable

causal system (recall that for such a system BIBO stability requires all the poles of the system transfer

function be in the open, left-hand s-plane) to have a steady-state response. It all depends on the input.

Consider, for instance, an unstable system with H(s)= 1 /(s(s+ 1 )), being unstable due to the pole at