Signals and Systems - Electrical Engineering

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

nExample 3.7

Compute the Laplace transform of the ramp functionr(t)=tu(t)and use it to find the Laplace of

a triangular pulse3(t)=r(t+ 1 )− 2 r(t)+r(t− 1 ).

Solution

Notice that although the ramp is an ever-increasing function oft, we still can obtain its Laplace

transform

R(s)=

∫∞

0

te−stdt=

e−st

s^2

(−st− 1 )

∣

∣∞t= 0 =^1

s^2

where we letσ >0 for the integral to exist. Thus,R(s)= 1 /s^2 with region of convergence{(σ,):

σ >0,−∞<  <∞}. The above integration can be avoided by noticing that if we find the

derivative with respect tosof the Laplace transform ofu(t), or

d U(s)

ds

=

∫∞

0

de−st

ds

dt

=

∫∞

0

(−t)e−stdt

=−R(s)

where we assumed the derivative and the integral can be interchanged. We then have

R(s)=−

d U(s)

ds

=

1

s^2

The Laplace transform of3(t)can then be shown to be (try it!)

3(s)=

1

s^2

[es− 2 +e−s]

The zeros of3(s)are the values ofsthat makees− 2 +e−s=0, or multiplying bye−s,

1 − 2 e−s+e−^2 s=( 1 −e−s)^2 = 0

which is equivalent toe−s= 1 =ej^2 πk, for integerk, or double zeros at

sk=j 2 πk k=0,±1,±2,...

In particular, whenk=0 there are two zeros at 0, which cancel the two poles at 0 resulting from

the denominators^2. Thus,3(s)has an infinite number of zeros but no poles given this pole-zero

cancellation (see Figure 3.9). Therefore,3(s)has the wholes-plane as its region of convergence,

and can be calculated ats=j. n