MAE 143BLinear Control Prof. M. Krstic

MIDTERM May 7, 2013

• One sheet of hand-written notes (two pages). Write answers only in the blue book.• Present your reasoning and calculations clearly. Inconsistent etchings will not be graded.• Total points: 25. Time: (1 hour 15 minutes).

Problem 1: Block Diagram Reduction (4 points) Find the transfer function Y (s)/R(s).

Solution:

=⇒

=⇒

Problem 2: Block Diagram Reduction (4 points) Find the transfer function Y (s)/R(s).

Note: Do not use Mason’s rule. Please use standard block diagram modification rules.

Solution:

=⇒

=⇒

=⇒

=⇒

Problem 3: Sensitivity (4 points) Control of the speed of an aircraft is represented by the modelbelow. Find the sensitivity of the closed-loop transfer function T (s) to the parameter a.

Solution: The closed-loop transfer function T (s) is given by

T (s) =

5(s+6)s(s+a)(s+3)

1 + 5(s+6)s(s+a)(s+3)

=5(s+ 6)

s(s+ a)(s+ 3) + 5(s+ 6).

The sensitivity of T to a is

STa =

a

T

dT

da.

Because

dT

da= − 5s(s+ 3)(s+ 6)

[s(s+ a)(s+ 3) + 5(s+ 6)]2,

we have

STa = − as(s+ 3)

s(s+ a)(s+ 3) + 5(s+ 6).

Problem 4: Time Domain Specifications An automatic insulin injection system for blood-sugarlevel control in diabetic persons is shown in the following figure below.

(a) (3 points) Determine K so that overshoot is 7%.

(b) (1 points) What is the settling time for K determined in a)?

Solution:

a) The closed-loop transfer function is given by

Tcl(s) =K 1

s(s+3)

1 +K 1s(s+3)

=K

s2 + 3s+K.

For this second-order system, we have

ω2n = K, 2ζωn = 3 ⇒ K =

√ωn =

√1.5

ζ.

The overshoot Mp solely depends on ζ:

Mp = e−πζ√1−ζ2 × 100%.

It follows from the above equation that

ζ =

√(lnMp)2

π2 + (lnMp)2.

In order for the overshoot to be 7%, we insert Mp = 0.07 into the above equation and willobtain

ζ = 0.6461 ⇒ K = 5.39.

b) We have

Ts =4.6

ζωn

, ζωn = 1.5 ⇒ Ts = 3.067.

Problem 5: Routh’s criterion, tracking, and disturbance rejection. Consider the feedback systembelow.

(a) (3 points) Determine conditions on KP , KD, ω so that the closed-loop system is asymptoticallystable.

(b) Specialize the conditions from point (a) to the following two separate cases:

(b1) (1 points) ω = 0

(b2) (1 points) KP = KD

(c) (2 points) Let d(t) = 0 and r(t) = 1(t). For ω 6= 0, do there exist values of KP , KD such thaty(∞) = r(∞) = 1?

(d) (2 points) Let r(t) = 0, d(t) = sin(ωt), and ω 6= 0, namely, let the user’s command be zero butlet the disturbance be a persistent sinusoid. If KP , KD are selected to satisfy the stabilizingconditions from point (a), what is y(∞)?

Solution:

a) The characteristic equation is 1 + C(s)P (s) = 0. It can be written as

(s2 + ω2)(s2 + s+ 1) + (KP +KDs)s = 0,

which is equivalent to

s4 + s3 + (ω2 +KD + 1)s2 + (ω2 +KP )s+ ω2 = 0.

The Routh array is as follows:

According to the Routh’s Criterion, to guarantee the asymptotical stability, we need

KD −KP + 1 > 0,(KD −KP + 1)(ω2 +KP )− ω2

KD −KP + 1> 0, ω2 > 0.

Hence, the conditions are

KD −KP + 1 > 0, (KD −KP + 1)(ω2 +KP )− ω2 > 0, ω2 > 0.

b1) If ω = 0, the system will be marginally stable from above. Although the pole-zerocancelation should be treated with caution, if we do it, an s will be canceled in the numeratorand denominator of C(s). Accordingly, the Routh array is

Thus, the conditions will be

KP > 0, KD > KP − 1.

b2) If KP = KD, the conditions are

KP = KD > 0, ω 6= 0.

c) The closed-loop transfer function from R(s) to Y (s) is given by

Try(s) =C(s)P (s)

1 + C(s)P (s)

=(KP +KDs)s

(s2 + ω2)(s2 + s+ 1) + (KP +KDs)s.

Since r(t) = 1(t), R(s) = 1/s. According to the Finite Value Theorem, we have

y(∞) = lims→0

sY (s) = lims→0

sTry(s)R(s) = lims→0

Try(s).

We observer that, if ω 6= 0, then y(∞) = 0, no matter what values KP and KD take. Thus,there does not exist values of KP and KD such that y(∞) = r(∞) = 1.

d) The closed-loop transfer function from D(s) to Y (s) is given by

Tdy(s) =P (s)

1 + C(s)P (s)

=s2 + ω2

(s2 + ω2)(s2 + s+ 1) + (KP +KDs)s.

When r(t) = 0, the system is excited only by d(t) = sin(ωt). We have D(s) = ωs2+ω2 . Then

by the Finite Value Theorem, we have

y(∞) = lims→0

sY (s) = lims→0

sTdy(s)D(s)

= lims→0

ss2 + ω2

(s2 + ω2)(s2 + s+ 1) + (KP +KDs)s

ω

s2 + ω2

= 0.