Signals and Systems - Electrical Engineering

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

n If the system is BIBO stable and causal, then the region of convergence includes thejaxis so that the

frequency responseH(j)exists, and all the poles ofH(s)are in the open left-hands-plane (thejaxis is

not included).

3.5 Analysis of LTI Systems

Dynamic linear time-invariant systems are typically represented by differential equations. Using the

derivative property of the one-sided Laplace transform (allowing the inclusion of initial conditions)

and the inverse transformation, differential equations are changed into easier-to-solve algebraic equa-

tions. The convolution integral is not only a valid alternate representation for systems represented by

differential equations, but for other systems. The Laplace transform provides a very efficient com-

putational method for the convolution integral. More important, the convolution property of the

Laplace transform introduces the concept oftransfer function, a very efficient representation of LTI

systems whether they are represented by differential equations or not. In Chapter 6, we will present

applications of the material in this section to classic control theory.

3.5.1 LTI Systems Represented by Ordinary Differential Equations

Two ways to characterize the response of a causal and stable LTI system are:

n Zero-stateandzero-inputresponses, which have to do with the effect of the input and the initial

conditions of the system.

n Transientandsteady-stateresponses, which have to do with close and faraway behavior of the

response.

The complete responsey(t)of a system represented by anNth-order linear differential equation with constant

coefficients,

y(N)(t)+

N∑− 1

k= 0

aky(k)(t)=

∑M

`= 0

b`x(`)(t) N>M (3.29)

wherex(t)is the input andy(t)is the output of the system, and initial conditions

{y(k)(t), 0≤k≤N− 1 } (3.30)

is obtained by inverting the Laplace transform

Y(s)=

B(s)

A(s)

X(s)+

1

A(s)

I(s) (3.31)

whereY(s)=L[y(t)],X(s)=L[x(t)], and

A(s)=

∑N

k= 0

aksk aN= 1

B(s)=

∑M

`= 0

b`s`