Driving MiP: MAE143A Signals & Systems final exam solutions, S1 2014

1a. Taking u(t) = constant and assuming small perturbations from pure motion along the x1 axis, dropping the primes for

notational clarity, we have

dr1/dt = r u, dr2/dt = r uα dα/dt = r u∆ /(L/2), δ = ℓα+ r2, γ =−2δ/d2, ω = γr u, u∆ = (L/2)ω/r.

1b. dr1/dt = constant, and thus r1(t) increases linearly with t in this approximation. Also, d(t) = g1(t)− r1(t) = ℓ.

1c. Defining f1 = r u and f2 = 2r u/ℓ2, we have

dr2/dt = r uα

dα/dt = r u∆ /(L/2) = . . .= ω =−(2δ/d2)r u =−(2ru/ℓ2)(ℓα+ r2)⇒

dr2/dt = f1 α

dα/dt =− f2 [ℓα+ r2]

Taking the Laplace transform, and multiplying the second equation by s and substituting the first, we have

sR2(s) = f1 α(s)

sα(s) =− f2 [ℓα(s)+R2(s)]⇒ [s2 + f2 ℓs+ f2 f1]α(s) = 0 ⇒ p± = [− f2 ℓ±

√

f 22 ℓ

2 − 4 f2 f1]/2 =−σ± iωd .

1d. Substituting, we get f1 = 0.1 and f2 = 20 and thus [s2+2s+2]α(s) = 0, with roots p± =−1±1i, with (moderate) damping

ζ= sin[atan(1/1)] = 0.707, implying about a 5% overshoot to a step input (if an input is put on the system). Taking α(0) = 0.05

and r2(0) = 0 gives α′(0) =−0.1. Following (17.51) in NR [derive it from scratch if you forgot it] gives

α(s) =c1s+ c0

(s− p+)(s− p−)=

d+

s− p++

d−

s− p−,

α(t) = d+ep+t + d−ep−t

= e−σt [(d++ d−)cosωd t + i(d+− d−)sinωd t]

where

d+ =c1 p++ c0

p+− p−

d− =c1 p−+ c0

p−− p+

and

c1 = α(0)

c0 = α′(0)+ 2α(0)

from which it follows that c1 = 0.05, c0 = 0, and d± = 0.025± 0.025i. Thus, α(t) = e−t [0.05cos t − 0.05sin t]. A plot of this

result is given below (solid).

1e. Repeating questions 1c and 1d with the second equation in (4) replaced with u∆ = K(L/2)ω/r gives

dr2/dt = r uα

dα/dt = r u∆ /(L/2) = Kω =−(2Kru/ℓ2)(ℓα+ r2)⇒

dr2/dt = f1 α

dα/dt =− f2 [ℓα+ r2]

with f1 = r u and f2 = 2Kr u/ℓ2. The formulae resulting upon taking the Laplace transform follow as before.

Substituting with K = 2, we get f1 = 0.1 and f2 = 40 and thus [s2 + 4s+ 4]α(s) = 0. Both roots of the polynomial in s are at

p =−2, with (strong) damping ζ = 1, implying reduced transients, and reduced overshoot to a step input (if an input is put on

the system). Taking α(0) = 0.05 and r2(0) = 0 gives α′(0) =−0.2.

The solution in this case is slightly different than (17.51) in NR, as we have repeated roots. In this case, we have

α(s) =c1s+ c0

(s− p)2=

d1

s− p+

d2

(s− p)2

α(t) = d1 ept + d2 t ept

where

d1 = c1

d2 = c0 + pc1

and

c1 = α(0)

c0 = α′(0)+ 4α(0)

From which is follows that c1 = 0.05, c0 = −0.1, d1 = 0.05, d2 = −0.2 and thus α(t) = 0.05e−2 t − 0.2 t e−2 t . A plot of this

result is give below (dashed); note that the response settles almost a factor of 2 faster than the case considered previously.

1f. Combining (2), (3), and (4) with the extra factor of K = 2 results in the full feedback rule (along segment k) of

u∆(t) =−K Lu

d2[(g1− r1)sin α− (g2− r2)cosα] where c(t) = pk−1 −[pk−1−r(t)]′sksk, g(t) = c(t)+ ℓsk, d = ‖g−r‖.

Simplifing this rule based on the assumptions stated gives the simplified feedback rule

u∆(t) =−K Lu

ℓ2(ℓα+ r2).

0 1 2 3 4 5 6 7−0.03

−0.02

−0.01

0

0.01

0.02

0.03

0.04

0.05

10−4

10−3

10−2

10−1

100

101

102

103

104

102

103

104

10−4

10−3

10−2

10−1

100

101

102

103

104

100

150

200

250

2a. Denoting the six intermediate variables V1,V2, I1, IR3, I2, IC3

as, respectively, the voltage between C1 and R1, the voltage

between C2 and R2, and the currents through the four horizontal branches of the circuit, the relevant current/voltage relationship

across the six passive components are:

I1 =C1d(V1 −Vin)

dt, V−−V1 = R1 I1, V−−Vin = IR3

R3 I2 =C2d(V2 −V−)

dt, Vout−V2 = R2 I2, IC3

=C3d(Vout−V−)

dt.

The current flowing into the input of the op amp is negligible; thus, Kirchoff’s current law gives I1 + IR3= I2 + IC3

. We also

see that there is stabilizing feedback into the op amp (into its negative terminal), and may thus simplify the analysis by setting

V− = V+ = 0. We thus have seven equations, and may combine them to eliminate the six intermediate variables mentioned

previously to arrive at a single equation relating Vout to Vin.

2b. Taking the Laplace transform of these seven equations (replacing d/dt with s everywhere) and writing in matrix form gives

A =

−sC1 0 1 0 0 0 0

−1 0 −R1 0 0 0 0

0 0 0 −R3 0 0 0

0 −sC2 0 0 1 0 0

0 −1 0 0 −R2 0 1

0 0 0 0 0 1 −sC3

0 0 1 1 −1 −1 0

, x =

V1

V2

I1

IR3

I2

IC3

Vout

, b =

−sC1 Vin

0

Vin

0

0

0

0

Solving in Matlab and looking at the last component of the solution gives immediately

Vout(s)

Vout(s)=−

1

R3 (C2 +C3)·

[C2 R2 s+ 1] [C1 (R1 +R3)s+ 1]

s [C1 R1 s+ 1] [C2C3 R2 s/(C2 +C3)+ 1]=−K ·

(s/z1 + 1)(s/z2 + 1)

s(s/p1 + 1)(s/p2 + 1).

2c. Combining the seven equations written out in question 2a to eliminate the six intermediate variables is doable by hand in

about 30 minutes (be patient and methodical!), and gives the same solution as found above. Comparing to the desired transfer

function gives the K,z1,z2, p1, p2 in terms of R1,R2,R3,C1,C2,C3 immediately:

K =1

R3 (C2 +C3), z1 =

1

C2 R2, z2 =

1

C1 (R1 +R3), p1 =

1

C1 R1, p2 =

C2 +C3

C2 C3 R2.

Assuming a given value for R3 (e.g., R3 = 1 kΩ), combining the equations above for z2 and p1 gives simple expressions for C1

and R1, and combining the equations for K, z1 and p2 gives simple expressions for C3, C2, and R2:

C1 =p1 − z2

R3 p1 z2, R1 =

R3 z2

p1 − z2, C3 =

z1

K p2 R3, C2 =

p2 − z1

K R3 p2, R2 =

K R3 p2

z1 (p2 − z1).

2d. The Bode plot of the transfer function given, taking K = 1, z1 = 0.01, z2 = 1, and p1 = p2 = 100, is shown above right.