for some constants c 1 , c 2 ≥ 0 that are independent of R.
A necessary and sufficient condition for the feedback system (3.6.1) to be
internally stable is that each element of H( s) has stable poles, i.e., poles in
the open left half s-plane [231].
The concept of internal stability may be confusing to some readers be-
cause of the fact that in most undergraduate books the stability of the feed-
back system is checked by examining the roots of the characteristic equation
1 + F CG 0 = 0
The following example is used to illustrate such confusions.
Example 3.6.1 Consider
1
s − 2
G 0( s) =
, C( s) =
, F ( s) = 1
s − 2
s + 5
136
CHAPTER 3. STABILITY
for which the characteristic equation
( s − 2)
1
1 + F CG 0 = 1 +
= 1 +
= 0
(3.6.3)
( s + 5)( s − 2)
s + 5
has a single root at s = − 6, indicating stability. On the other hand, the transfer
matrix H( s) calculated from this example yields
e
s + 5 −( s + 5) /( s − 2) −( s + 5) −( s + 5)
y∗
u 0
1 s − 2
s + 5
−( s − 2) −( s − 2) du
y = s + 6 1
( s + 5) /( s − 2)
s + 5
1
d (3.6.4)
yn
1
( s + 5) /( s − 2)
s + 5
s + 5
dn
indicating that a bounded du will produce unbounded e, y, yn, i.e., the feedback
system is not internally stable. We should note that in calculating (3.6.3) and
(3.6.4), we assume that s+ α g( s) and g( s) are the same transfer functions for any
s+ α
constant α. In (3.6.3) the exact cancellation of the pole at s = 2 of G 0( s) by
the zero at s = 2 of C( s) led to the wrong stability result, whereas in (3.6.4)
such cancellations have no effect on internal stability. This example indicates that
internal stability is more complete than the usual stability derived from the roots
of 1 + F CG 0 = 0. If, however, G 0 , C, F are expressed as the ratio of coprime
polynomials, i.e., G 0( s) = n 0( s) , C( s) = nc( s) , F ( s) = nf( s) then a necessary and p 0( s)
pc( s)
pf ( s)
sufficient condition for internal stability [231] is that the roots of the characteristic
equation
p 0 pcpf + n 0 ncnf = 0
(3.6.5)
are in the open left-half s-plane. For the example under consideration, n 0 = 1 , p 0 =
s − 2 , nc = s − 2 , pc = s + 5 , nf = 1 , pf = 1 we have
( s − 2)( s + 5) + ( s − 2) = ( s − 2)( s + 6) = 0
which has one unstable root indicating that the feedback is not internally stable.
3.6.3
Sensitivity and Complementary Sensitivity Functions
Although internal stability guarantees that all signals in the feedback system
are bounded for any bounded external inputs, performance requirements put
restrictions on the size of some of the signal bounds. For example, one of the
main objectives of a feedback controller is to keep the error between the plant
output y and the reference signal y∗ small in the presence of external inputs,
such as reference inputs, bounded disturbances, and noise. Let us consider
3.6. STABILITY OF LTI FEEDBACK SYSTEMS
137
the case where in the feedback system of Figure 3.2, F ( s) = 1 , du = 0. Using
(3.6.1), we can derive the following relationships between the plant output
y and external inputs y∗, d, dn:
y = T 0 y∗ + S 0 d − T 0 dn
(3.6.6)
where
1
CG
S
0
0 =
,
T
1 + CG
0 =
0
1 + CG 0
are referred to as the sensitivity function and complementary sensitivity
function, respectively. It follows that S 0 , T 0 satisfy
S 0 + T 0 = 1
(3.6.7)
It is clear from (3.6.6) that for good tracking and output disturbance rejec-
tion, the loop gain L 0 = CG 0 has to be chosen large so that S 0 ≈ 0 and
T 0 ≈ 1. On the other hand, the suppression of the effect of the measure-
ment noise dn on y requires L 0 to be small so that T 0 ≈ 0, which from
(3.6.7) implies that S 0 ≈ 1. This illustrates one of the basic trade-offs in
feedback design, which is good reference tracking and disturbance rejection
( |L 0 |
1 , S 0 ≈ 0 , T 0 ≈ 1) has to be traded off against suppression of
measurement noise ( |L 0 |
1 , S 0 ≈ 1 , T 0 ≈ 0). In a wide class of control
problems, y∗, d are usually low frequency signals, and dn is dominant only at
high frequencies. In this case, C( s) can be designed so that at low frequen-
cies the loop gain L 0 is large, i.e., S 0 ≈ 0 , T 0 ≈ 1, and at high frequencies L 0
is small, i.e., S 0 ≈ 1 , T 0 ≈ 0. Another reason for requiring the loop gain L 0
to be small at high frequencies is the presence of dynamic plant uncertainties
whose effect is discussed in later chapters.
3.6.4
Internal Model Principle
In many control problems, the reference input or setpoint y∗ can be modeled
as
Qr( s) y∗ = 0
(3.6.8)
where Qr( s) is a known polynomial, and s = d is the differential operator.
dt
For example, when y∗ =constant, Qr( s) = s. When y∗ = t, Qr( s) = s 2 and 138
CHAPTER 3. STABILITY
when y∗ = Asinω 0 t for some constants A and ω 0, then Qr( s) = s 2 + ω 20, etc.
Similarly, a deterministic disturbance d can be modeled as
Qd( s) d = 0
(3.6.9)
for some known Qd( s), in cases where sufficient information about d is avail-
able. For example, if d is a sinusoidal signal with unknown amplitude and
phase but with known frequency ωd then it can be modeled by (3.6.9) with
Qd( s) = s 2 + ω 2 d.
The idea behind the internal model principle is that by including the
factor
1
in the compensator C( s), we can null the effect of y∗, d on
Qr( s) Qd( s)
the tracking error e = y∗ − y. To see how this works, consider the feedback
system in Figure 3.2 with F ( s) = 1 , du = dn = 0 and with the reference
input y∗ and disturbance d satisfying (3.6.8) and (3.6.9) respectively for some
known polynomials Qr( s) , Qd( s). Let us now replace C( s) in Figure 3.2 with
¯
C( s)
C( s) =
,
Q( s) = Q
Q( s)
r( s) Qd( s)
(3.6.10)
where C( s) in (3.6.10) is now chosen so that the poles of each element of
H( s) in (3.6.1) with C( s) replaced by ¯
C( s) are stable. From (3.6.6) with
du = dn = 0 and C replaced by C/Q, we have
1
1
1
e = y∗ − y =
y∗ −
d =
Q( y∗ − d)
1 + CG 0
1 + CG 0
Q + CG
Q
Q
0
Because Q = QrQd nulls d, y∗, i.e., Q( s) d = 0 , Q( s) y∗ = 0, it follows that 1
e =
[0]
Q + CG 0
which together with the stability of 1 /( Q + CG 0) guaranteed by the choice
of C( s) imply that e( t) = y∗( t) − y( t) tends to zero exponentially fast.
The property of exact tracking guaranteed by the internal model principle
can also be derived from the values of the sensitivity and complementary
sensitivity functions S 0 , T 0 at the frequencies of y∗, d. For example, if y∗ =
sin ω 0 t and d = sin ω 1 t, i.e., Q( s) = ( s 2 + ω 20)( s 2 + ω 21) we have Q
CG
S
0
0 =
, T
Q + CG
0 =
0
Q + CG 0
3.7. PROBLEMS
139
and S 0( jω 0) = S 0( jω 1) = 0, T 0( jω 0) = T 0( jω 1) = 1.
A special case of the internal model principle is integral control where
Q( s) = s which is widely used in industrial control to null the effect of
constant set points and disturbances on the tracking error.
3.7
Problems
3.1
(a) Sketch the unit disk defined by the set x x ∈ R 2 , |x|p ≤ 1 for (i)
p = 1, (ii) p = 2, (iii) p = 3, and (iv) p = ∞.
(b) Calculate the L 2 norm of the vector function
x( t) = [ e− 2 t, e−t]
by (i) using the | · | 2-norm in R 2 and (ii) using the | · |∞ norm in R 2.
3.2 Let y = G( s) u and g( t) be the impulse response of G( s). Consider the induced norm of the operator T defined as T
y ∞
∞ = sup u
, where T : u ∈
∞=1
u ∞
L∞ → y ∈ L∞. Show that T ∞ = g 1.
3.3 Take u( t) = f ( t) where f ( t), shown in the figure, is a sequence of pulses centered at n with width 1 and amplitude n, where n = 1 , 2 , . . . , ∞.
n 3
f(n)
k
1/27
1/8
2
1
1
n
1
2
3
k
(a) Show that u ∈ L 1 but u ∈ L 2 and u ∈ L∞.
(b) If y = G( s) u where G( s) = 1 and u = f , show that y ∈ L
s+1
1
L∞ and
|y( t) | → 0 as t → ∞.
3.4 Consider the system depicted in the following figure:
140
CHAPTER 3. STABILITY
u 1 ✲
+ ❧ e 1 ✲
y 1
✲
Σ
H 1
✻
−
❄
+
y 2
✛ e 2
✛
❧ u
H
2
2
Σ +
Let H 1 , H 2: L∞e → L∞e satisfy
( H 1 e 1) t ≤ γ 1 e 1 t + β 1
t
( H 2 e 2) t ≤ γ 2 e 2 t +
e−α( t−τ) γ( τ ) e 2 τ dτ + β 2
0
for all t ≥ 0, where γ 1 ≥ 0 , γ 2 ≥ 0, α > 0, β 1 , β 2 are constants, γ( t) is a nonnegative continuous function and ( ·) t , ( ·) denote the L∞e, L∞ norm
respectively. Let u 1 ≤ c, u 2 ≤ c for some constant c ≥ 0, and γ 1 γ 2 < 1
(small gain). Show that
(a) If γ ∈ L 2, then e 1 , e 2 , y 1 , y 2 ∈ L∞.
(b) If γ ∈ S( µ), then e 1 , e 2 , y 1 , y 2 ∈ L∞ for any µ ∈ [0 , µ∗), where µ∗ =
α 2(1 −γ 1 γ 2)2 .
2 c 0 γ 21
3.5 Consider the system described by the following equation:
˙ x = A( t) x + f ( t, x) + u
(3.7.1)
where f ( t, x) satisfies
|f ( t, x) | ≤ γ( t) |x| + γ 0( t)
for all x ∈ Rn, t ≥ 0 and f ( t, 0) = 0, where γ( t) ≥ 0 , γ 0( t) are continuous functions. If the equilibrium ye = 0 of ˙ y = A( t) y is u.a.s and γ ∈ S( µ), show that the following statements hold for some µ∗ > 0 and any µ ∈ [0 , µ∗):
(a) u ∈ L∞ and γ 0 ∈ S( ν) for any ν ≥ 0 implies x ∈ L∞
(b) u ≡ 0 , γ 0 ≡ 0 implies that the equilibrium xe = 0 of (3.7.1) is u.a.s in
the large
(c) u, γ 0 ∈ L 2 implies that x ∈ L∞ and lim t→∞ x( t) = 0
(d) If u ≡ γ 0 ≡ 0 then the solution x( t; t 0 , x 0) of (3.7.1) satisfies
|x( t; t 0 , x 0) | ≤ Ke−β( t−t 0) |x( t 0) | for t ≥ t 0 ≥ 0
where β = α−c 0 µK > 0 for some constant c 0 and K, α > 0 are constants
in the bound for the state transition matrix Φ( t, τ ) of ˙ y = A( t) y, i.e.,
Φ( t, τ ) ≤ Ke−α( t−τ) ,
f or t ≥ τ ≥ t 0
3.7. PROBLEMS
141
3.6 Consider the LTI system
˙ x = ( A + B) x
where
> 0 is a scalar. Calculate ∗ > 0 such that for all
∈ [0 , ∗), the
equilibrium state xe = 0 is e.s. in the large when
(a)
− 1
10
1 0
A =
,
B =
0
− 2
2 5
(b)
0
10
5 − 8
A =
,
B =
0 − 1
0
2
using (i) the Euclidean norm and (ii) the infinity norm.
3.7 Consider the system given by the following block diagram:
r
✲
+ ♥ u
y
Σ
✲ G( s)(1 + ∆( s))
✲
✻
−
✛
F ( s)
where F ( s) is designed such that the closed-loop system is internally stable
when ∆