THE LAPLACE TRANSFORM - Signals, Systems, and Transforms - page 370

h ( t ) y ( t )

1 1

0 1 t 0 1 t

Figure 7.12 Signals for Example 7.14.

(a) (b)

Using the real-shifting property, we find the Laplace transform of h(t) to be

1 - e -s
s

H(s) = .

Note that this transfer function is not a rational function. From (7.55), the system output
is then

Y(s)

1 - e -s
s

1

1
s 2 [1 - e -s ].

Y(s) = H(s)X(s) =

s =

From the Laplace transform table and the real-shifting property, we find the system output
to be

y(t) = tu(t) - [t - 1]u(t - 1).

This inverse transform is verified with the MATLAB program
syms F s
F = ( 1 exp ( s) ) / (s^2)
ilaplace (F)

The results of running this program contains the expression Heaviside (t - 1), which is the

MATLAB expression for the unit step function
This response is shown in Figure 7.12(b).

u(t - 1).

■

Transforms with Complex Poles

We next consider a transformed function that has a pair of complex poles. Suppose
that F(s) is n th-order with two complex poles. For convenience, we let the other

poles be real, so that (n - 2)

N(s)
(s - p 1 )(s - p 2 )(s - p 3 ) Á (s - p n ) ,

F(s) =

where is the numerator polynomial. Let and then,
with the order of the numerator less than that of the denominator, the partial-fraction
expansion for

N(s) p 1 = a - jb p 2 = a + jb;

F(s) can be written as

k 1
s - a + jb +

k 2
s - a - jb +

k 3
s - p 3

k n
s - p n .

+ Á + F(s) = (7.57)