Chapter 3

Feedback Systems

3.1 Stability

Consider the feedback system

+ + + + − + r  x_{1}  C  d x_{2}  P  y n x_{3} F

Definition: This feedback system is \underline{BIBO stable} (or just \underline{stable}, for short) if x_{1}, x_{2}, x_{3} are bounded whenever r, d, n are bounded.

Note that all signals in this system are bounded if x_{1}, x_{2}, x_{3}, r, d, n are bounded; for example, y = x_{3} − n, so it’s bounded.

Example:

\frac{1}{s^{2} − 1} \frac{s − 1}{s + 1}  + +  ++ − + r  d  y n

TF from r to y is stable,

TF from d to y is not stable, Hence feedback system is \underline{not} stable.

We want to develop a test for stability. Note that, by definition, the feedback system is stable iff all 9 transfer functions from r, d, n to x_{1}, x_{2}, x_{3} are proper and stable. Derivation of TFs:

\{ \begin{matrix}x̂_{1} & = & r̂ − F x̂_{3} \\ x̂_{2} & = & d̂+ Cx̂_{1} \\ x̂_{3} & = & n̂ + P x̂_{2} \end{matrix}

⇒ [ \begin{matrix}1 & 0 & F \\ −C & 1 & 0 \\ 0 & −P & 1 \end{matrix} ] [ x̂_{1} ; x̂_{2} ; x̂_{3} ] = [ r̂ ; d̂ ; n̂ ]

⇒ [ x̂_{1} ; x̂_{2} ; x̂_{3} ] = [ \begin{matrix}1 & 0 & F \\ −C & 1 & 0 \\ 0 & −P & 1 \end{matrix} ]^{−1} [ r̂ ; d̂ ; n̂ ]

= \frac{1}{1+P CF} [ \begin{matrix}1 & −P F & −F \\ C & 1 & −F C \\ P C & P & 1 \end{matrix} ] [ r̂ ; d̂ ; n̂ ]

Theorem 1: Assume P, C, F are proper and P CF is strictly proper. Then the 9 TFs are all proper.

(Proof is easy.)

Definition: Write P = \frac{N_{P}}{D_{P}} , C = \frac{N_{C}}{D_{C}}, F = \frac{N_{F}}{D_{F}} as ratios of coprime polynomials (i.e. N_{P}, D_{P} have no

common factors, etc.). The \underline{characteristic polynomial} (ch.p.) of the system is

D_{P} D_{C}D_{F} + N_{P} N_{C}N_{F} .

Theorem 2: Assume P, C, F are proper and P CF is strictly proper. Then the feedback system is stable if and only if the ch.p. has no roots in Re s ≥ 0.

Example: (continued)

ch.p. = (s + 1)(s^{2} − 1) + s − 1
= (s − 1)(s^{2} + 2s + 2); root at s = 1

Therefore feedback system is not stable (as we saw before).

Proof of theorem For simplicity assume F = 1 (the general case is similar but a bit messier). Then

[ x̂_{1} ; x̂_{2} ; x̂_{3} ] = \frac{1}{1 + P C} [ \begin{matrix}1 & −P & −1 \\ C & 1 & −C \\ P C & P & 1 \end{matrix} ] [ r̂ ; d̂ ; n̂ ]

= \frac{1}{D_{P} D_{C} + N_{P} N_{C}} [ \begin{matrix}D_{P} D_{C} & −N_{P} D_{C} & −D_{P} D_{C} \\ D_{P} N_{C} & D_{P} D_{C} & −D_{P} N_{C} \\ N_{P} N_{C} & N_{P} D_{C} & D_{P} D_{C} \end{matrix} ] [ r̂ ; d̂ ; n̂ ]  (3.1)

Note that ch.p. = D_{P} D_{C} + N_{P} N_{C}. Thus the feedback system is stable if ch.p. has no roots in Re s ≥ 0.

Conversely, assume the feedback system is stable. Suppose, for a proof by contradiction, that ch.p. has a root s_{0} in Re s ≥ 0. Then each of the 9 numerator polynomials in (3.1) must also have a root at s_{0} (because all 9 TFs must be stable). Thus

(D_{P} D_{C} + N_{P} N_{C})(s_{0}) = 0 (ch.p.)
(D_{P} D_{C})(s_{0}) = 0 ((1,1)-element in (1))  (3.2)
(N_{P} N_{C})(s_{0}) = 0 ((3,1)-element in (1)).  (3.3)

Then (3.2) implies that either D_{P} (s_{0}) = 0 or D_{C}(s_{0}) = 0 (or both).

Suppose D_{P} (s_{0}) = 0. Then N_{P} (s_{0}) ≠ 0 (by coprimeness), so (3.3) implies that N_{C}(s_{0}) = 0. But also

(N_{P} D_{C})(s_{0}) = 0 ((1,2)-element in (3.1))

so D_{C}(s_{0}) = 0. This implies that N_{C}, D_{C} are not coprime, which is a contradiction.

Similarly, the condition D_{C}(s_{0}) = 0 leads to a contradiction. ■ The final result is another test for feedback stability.

Theorem 3: Assume P, C, F are proper and P CF is strictly proper. Then the feedback system is stable if and only if

1 + P CF ≠ 0

in Re s ≥ 0, and the product P CF has no pole-zero cancellations in Re s ≥ 0.

Example: (continued)

P (s) = \frac{1}{s^{2} − 1}, C(s) = \frac{s − 1}{s + 1}, F (s) = 1

Thus P CF has a pole-zero cancellation at s = 1; hence the feedback system is not stable.

(Proof of Theorem is left as an exercise.)

Summary

Theorem 1 gives a sufficient condition for all closed-loop TFs to be proper (the open-loop TFs P, C, F should be proper and at least one should be strictly proper). Theorem 2 gives a test for stability (the ch.p. must have no roots in Re s ≥ 0). Finally, Theorem 3 gives an equivalent test for stability (the TF 1 + P CF must have no zeros in Re s ≥ 0 and there must be no pole-zero cancellations in Re s ≥ 0 in forming the product).

There is an algebraic test, called the Routh-Hurwitz criterion, for a polynomial to have all its roots in Re s < 0. To check feedback stability we apply this criterion to the ch.p.

Let the polynomial be

a_{n}s^{n} + a_{n−1}s^{n−1} + … + a_{0}, a_{i} real numbers.

We apply an algorithm to generate an array with n + 1 rows; the array looks like

n  a_{n} a_{n−2} a_{n−4} …
n − 1 a_{n−1} a_{n−3} a_{n−5} …
n − 2 b_{n−2} b_{n−4} …
n − 3 c_{n−3} c_{n−5} …
⋮  ⋮

4 CHAPTER 3. FEEDBACK SYSTEMS Note that the first two rows are written down immediately from the polynomial. The 3rd row is defined as follows:

b_{n−2} := −\frac{1}{a_{n−1}} | \begin{matrix}a_{n} & a_{n−2} \\ a_{n−1} & a_{n−3} \end{matrix} |
b_{n−4} := −\frac{1}{a_{n−1}} | \begin{matrix}a_{n} & a_{n−4} \\ a_{n−1} & a_{n−5} \end{matrix} |

etc.

The fourth row is generated in the same way, but using rows 2 and 3. Theorem 4: The roots of the polynomial all lie in Re s < 0 ⇔ all elements in the first column of the array are nonzero and have the same sign. A necessary condition is that a_{n}, ..., a_{0} all be nonzero and have the same sign. Example 1: s^{2} + a_{1}s + a_{0}

\begin{matrix}2 & 1 & a_{0} \\ 1 & a_{1} & 0 \\ 0 & a_{0} & 0 \end{matrix}

all roots in Re s < 0 ⇔ a_{1} > 0 & a_{0} > 0 Example 2: s^{3} + a_{2}s^{2} + a_{1}s + a_{0}

\begin{matrix}3 & 1 & a_{1} \\ 2 & a_{2} & a_{0} \\ 1 & b_{1} & 0 \\ 0 & a_{0} & 0 \end{matrix} b_{1} = \frac{a_{2}a_{1} − a_{0}}{a_{2}}

all roots in Re s < 0 ⇔ a_{2} > 0, a_{1} > 0, a_{0} > 0, a_{2}a_{1} > a_{0}. Example 3: s^{4} + s^{3} + s^{2} + s + K

\begin{matrix}4 & 1 & 1 & K \\ 3 & 1 & 1 \\ 2 & 0 \\ 1 \\ 0 \end{matrix}

for every K there is some root in Re s ≥ 0. Additional Information:

A row vanishes identically. Preceding row gives a factor of p(s). Example:

p(s) = s^{5} + s^{4} + 4s^{3} + 24s^{2} + 3s + 63

\begin{matrix}s^{5} : & 1 & 4 & 3 \\ s^{4} : & 1 & 24 & 63 \\ s^{3} : & −20 & −60 & 0 \\ s^{2} : & 21 & 63 & 0 \\ s^{1} : & 0 & 0 & 0 \end{matrix}

21s^{2} + 63 = 0 is a factor s = ±i\sqrt{3}

s^{3} + s^{2}  + s + 21
s^{2} + 3 s^{5} + s^{4} + 4s^{3} + 24s^{2} + 3s + 63
s^{5}  + 3s^{3}
s^{4} + s^{3} + 24s^{2}
s^{4}  + 3s^{2}
s^{3} + 21s^{2}
s^{3}  +3s
21s^{2} +  63

3.2 Steady state performance

Define the trading error e(t) = r(t) − y(t). Then the steady state error = lim_{t→∞} e(t) = e_{∞}, if it exists. Consider a unity feedback system

+ -  G

In terms of unit step, we have

e(s) = \frac{1}{1 + G(s)}r(s)

To get e_{∞}, use the final value theorem of Laplace transform:

Let fϵP S[0, ∞). Suppose f_{∞} exists and is finite. Then f_{∞} = lim ; \begin{matrix}s → 0 \\ s real \end{matrix} sf̂(s)

(Alternatively, suppose f̂(s), strictly proper rational, has no poles in Re s ≥ 0 except possibly for simple pole at s = 0. Then lim_{t→∞} f(t) exists and equals lim_{s→0} sf̂(s).)

Warning: Laplace transform of cos ωt = \frac{s}{s^{2}+ω^{2}} .

Apply to our problem to give (assuming closed-loop system is stable so that y_{∞} exists)

e_{∞} = lim_{s→0} s\frac{1}{1 + G(s)} \frac{1}{s} = \frac{1}{1 + G(0)},

G(0) often referred to as the dc gain of the system.

If G(s) = \frac{q(s)}{sp(s)} can interpret G(0) = ∞ or

lim_{s→0} \frac{1}{1 + G(s)} = \frac{sp(s)}{sp(s) + q(s)} = 0

If G(0) ↑, e_{∞} ↓ as long as system remains stable. If e_{∞} = 0, we say asymptotic tracking is achieved. We can combine the above ideas to give a controller structure that will provide asymptotic tracking of steps.

Example:

r −  C  P  y

Let P (s) = \frac{1}{s^{2}+4s−5}. We would like to design a controller C(s) to achieve asymptotic tracking of step reference inputs. The two ingredients we need are (i) An integrator at 0 in C(s).

(ii) Stability of the closed loop system.

Consider the controller C(s) = \frac{K(s+1)}{s}. K is to be chosen to give closed loop stability. The ch. p. is given by s^{3} + 4s^{2} + (K − 5)s + K. Applying Routh-Hurwitz, we find that the necessary and sufficient condition for stability is K > \frac{20}{3}. Thus for

C(s) = \frac{7(s + 1)}{s}

asymptotic tracking of steps is achieved.