6.3 Application to Classic Control 371
Ha(s)
x(t) y(t)
G(s) K
x(t) y(t)
+ G(s)
−
(a)(b)
FIGURE 6.8
Stabilization of an unstable plantG(s)using (a) an all-pass filter and (b) a proportional controller of gainK.
To get rid of the pole ats=2 and to replace it with a new pole ats=−2, we let the all-pass filter be
Ha(s)=
s− 2
s+ 2
To see that this filter has a constant magnitude response consider
Ha(s)Ha(−s)=
(s− 2 )(−s− 2 )
(s+ 2 )(−s+ 2 )
=
(s− 2 )(s+ 2 )
(s+ 2 )(s− 2 )
= 1
If we lets=j, the above gives the magnitude-squared function
Ha(j)H(−j)=Ha(j)H∗(j)
=|H(j)|^2
which is unity for all values of frequency. The cascade of the unstable system with the all-pass
system gives a stable system
H(s)=G(s)Ha(s)=
1
s+ 2
with the same magnitude response asG(s). This is an open-loop stabilization and it depends on
the all-pass system having a zero exactly at 2 so that it cancels the pole causing the instability. Any
small change on the zero and the overall system would not be stabilized. Another problem with
the cascading of an all-pass filter to stabilize a filter is that it does not work when the pole causing
the unstability is at the origin, as we cannot obtain an all-pass filter able to cancel that pole.
Consider then a negative-feedback system (Figure 6.8(b)). Suppose we use a proportional
controller with a gainK, then the overall system transfer function is
H(s)=
KG(s)
1 +KG(s)
=
K
s+(K− 2 )
and if the gainKis chosen so thatK− 2 >0 orK>2, the feedback system will be stable. n
6.3.2 Transient Analysis of First- and Second-Order Control Systems
Although the input to a control system is not known a-priori, there are many applications where the
system is frequently subjected to a certain type of input and thus one can select a test signal. For
instance, if a system is subjected to intense and sudden inputs, then an impulse signal might be the
