6.3 Application to Classic Control 369
The PI controller used here is one of various possible controllers. Consider a simpler and cheaper
controller such as a proportional controller withHc(s)=K. Would you be able to obtain the same
results? Try it. n
6.3.1 Stability and Stabilization
A very important question related to the performance of systems is: How do we know that a given
causal system has finite zero-input, zero-state, or steady-state responses? This is the stability prob-
lem of great interest in control. Thus, if the system is represented by a linear differential equation
with constant coefficients the stability of the system determines that the zero-input, the zero-state,
as well as the steady-state responses may exist. The stability of the system is also required when con-
sidering the frequency response in the Fourier analysis. It is important to understand that only the
Laplace transform allows us to characterize stable as well as unstable systems; the Fourier transform
does not.
Two possible ways to look at the stability of a causal LTI system are:
n When there is no input so that the response of the system depends on initial energy in the system.
This is related to the zero-input response of the system.
n When there is a bounded input and no initial condition. This is related to the zero-state response
of the system.
Relating the zero-input response of a causal LTI system to stability leads toasymptoticstability. An LTI
system is said to be asymptotically stable if the zero-input response (due only to initial conditions in
the system) goes to zero astincreases—that is,
yzi(t)→ 0 t→∞ (6.5)
for all possible initial conditions.
The second interpretation leads to thebounded-input bounded-output(BIBO) stability, which we
defined in Chapter 2. A causal LTI system is BIBO stable if its response to a bounded input is also
bounded. The condition we found in Chapter 2 for a causal LTI system to be BIBO stable was that
the impulse response of the system be absolutely integrable—that is
∫∞
0
|h(t)|dt<∞ (6.6)
Such a condition is difficult to test, and we will see in this section that it is equivalent to the poles
of the transfer function being in the open left-hands-plane, a condition that can be more easily
visualized and for which algebraic tests exist.
Consider a system being represented by the differential equation
y(t)+
∑N
k= 1
ak
dky(t)
dtk
=b 0 x(t)+
∑M
`= 1
b`
d`x(t)
dt`
M<N
