EECS C128/ ME C134
Final
Wed. Dec. 15, 2010
0810-1100 am
Name:
SID:
• Closed book. Two pages of formula sheets. No calculators.
• There are 8 problems worth 100 points total.
Problem Points
1
12
2
16
3
10
4
14
5
16
6
10
7
10
8
12
Total
100
Score
In the real world, unethical actions by engineers can cost money, careers, and lives. The
penalty for unethical actions on this exam will be a grade of ‘F’ and a letter will be written
for your file and to the Office of Student Conduct.
tan−1 12 = 26.6◦
1
tan−1 √
= 18.4◦
3
tan−1 3 = 60◦
sin 30◦ =
1
2
tan−1 1 = 45◦
tan−1 41 = 14◦
tan−1 √13 = 30◦
cos 60◦ =
√
3
2
20 log10√1 = 0dB
20 log10 2 = 3dB
20 log10 5 = 20db − 6dB = 14dB
1/e ≈ 0.37
1/e3 ≈ 0.05
20 log10 2 = 6dB
20 log10√21 = −6dB
20 log10 10 = 10 dB
2
1/e
√ ≈ 0.14
10 ≈ 3.16
1
Problem 1 (12 pts)
reference
input
r(t)
control
input
error
Σ
+
Controller
e(t)
u(t)
D(s)
−
output
plant
G(s)
y(t)
grid 8 pixels
You are given the open loop plant:
G(s) =
800
(s + 30)(s2 + 2s + 4)
The Bode plots for the plant G(s) and the plant with 3 different compensators are given
below.
Bode Plot G(s)
Bode Plot D2(s)G(S)
20
Magnitude (dB)
0
−20
−40
0
−20
−40
−60
−60
−80
0
−80
0
−45
−45
Phase (deg)
Phase (deg)
Magnitude (dB)
20
−90
−135
−180
−225
−270
−1
10
−90
−135
−180
−225
0
10
1
10
Frequency (rad/sec)
−270
−1
10
2
10
0
10
2
10
Bode Plot D4(s)G(s)
20
20
0
0
Magnitude (dB)
Magnitude (dB)
Bode Plot D3(s)G(s)
1
10
Frequency (rad/sec)
−20
−40
−20
−40
−60
−60
−80
−90
−80
Phase (deg)
Phase (deg)
0
−180
−270
−360
−1
10
0
10
1
10
Frequency (rad/sec)
−90
−180
−270
−1
10
2
10
0
10
[4 pts] a) For each Bode plot, estimate the phase and gain margin:
(i) G(s): phase margin
gain margin
dB
(ii) D2 (s)G(s): phase margin
gain margin
dB
(iii) D3 (s)G(s): phase margin
gain margin
dB
(iv) D4 (s)G(s): phase margin
gain margin
dB
2
1
10
Frequency (rad/sec)
2
10
Problem 1, cont.
[4 pts] b) For each open-loop Bode plot on the previous page, choose the best corresponding closed-loop root locus (write in letter W,X,Y, or Z). Note: the root locus is zoomed in
and does not show the open-loop pole at s = −30. (Hint: the root locus shows open-loop
DG
pole locations for D(s)G(s), and closed-loop poles for 1+DG
).
(i) G(s): root locus
(ii) D2 (s)G(s): root locus
(iii) D3 (s)G(s): root locus
(iv) D4 (s)G(s): root locus
Root locus X
5
Imaginary Axis
Imaginary Axis
Root Locus W
0
−5
−5
5
0
−5
−4
−3
−2
−1
Real Axis
0
1
−5
−4
Root Locus Y
−3
−2
−1
Real Axis
0
1
Root Locus Z
5
Imaginary Axis
Imaginary Axis
15
0
−5
−20
10
5
0
−5
−10
−15
−10
Real Axis
−5
−15
0
3
−10
−5
Real Axis
0
Problem 1, cont.
[4 pts] c) For each open-loop Bode plot on page 2, choose the best corresponding closedloop step response (A-D)
(i) G(s): step response
(ii) D2 (s)G(s): step response
(iii) D3 (s)G(s): step response
(iv) D4 (s)G(s): step response
Step A
Step B
1
1
Amplitude
1.5
Amplitude
1.5
0.5
0
0
0.5
2
4
6
8
0
0
10
2
4
Time (sec)
6
8
10
Time (sec)
Step C
Step D
1
1
Amplitude
1.5
Amplitude
1.5
0.5
0
0
0.5
1
2
3
0
0
4
Time (sec)
1
2
Time (sec)
4
3
4
Problem 2 (16 pts)
reference
input
r(t)
control
input
error
Σ
+
e(t)
−
Controller
u(t)
D(s)
output
plant
G(s)
y(t)
grid 8 pixels
1
You are given the open loop plant G(s) = s2 +2s+5
. The system is to be controlled using
proportional plus integral “PI” control, that is D(s) = kp + ksI .
[8 pts] a) Sketch the positive root locus as kI varies for fixed kp = 5, noting:
(i) approximate asymptote intersection point s =
(ii) approximate angle of departure for the poles:
[8 pts] b) Sketch the positive root locus as kp varies for fixed kI = 1, noting:
(i) approximate angle of departure for the poles:
Given: the roots of s3 + 2s2 + 5s + 1 ≈ (s + .2168)(s + 0.89 + 1.95j)(s + 0.89 − 1.95j)
5
Problem 3 (10 pts)
You are given the following plant
0
0
1
u(t),
x+
ẋ = Ax + Bu =
1
−6 −5
y = [1 1] x
s+1
1
= s+2
= 1 − s+2
, where
We want to add a lead compensator to the system such that VU (s)
(s)
v(t) is the input to the compensator, and u(t) is the original input to the plant.
[6 pts] a) Determine the new state space equations (that is, fill in the values) for the
combined system of plant plus lead compensator by adding a new state variable z:
 
x˙1
 x˙2  = 
ż

 
x1
  x2  + 
z


 v(t)
y=[

x1
]  x2  + [ ]v(t)
z
(1)

[4 pts] b) Prove that the compensated system is observable (hint: consider new state and
output matrices Ā and C̄).
6
Problem 4 (14 pts)
Given the following model of the inverted pendulum
0
0 1
u(t)
x+
ẋ = Ax + Bu =
1
4 0
y = [1 0]x
A full order observer is given x̂˙ = (A − T C)x̂ + Bu + T Cx
[3 pts] a) For combined plant and full state estimator with state feedback u = −F x̂ and
observer gain T , derive the combined state equations for [x e]′ using e = x̂ − x. (Please
leave equations in terms of A, B, C, T rather than numerical values.)
[3 pts] b) Using a), show that the eigenvalues for the plant and the observer can be set
independently (separation principle). (Hint: use property of determinants.)
[3 pts] c) Find feedback gains F such that the controller has closed loop poles at -2 and
-4.
F = [f1 f2 ] = [
]
[3 pts] d) Find observer gains T such that the observer has closed loop poles at -10 and
-6.
T = [t1 t2 ]′ = [
]’
[2 pts] e) What would the effect be on system performance if the observer poles were
slower than the controller poles?
7
Problem 5 (16 pts)
You are given the following
1
−1 −8
u(t),
x+
ẋ = Ax + Bu =
1
12 −29
y = [3 − 1] x
[2 pts] a) Determine if the system is controllable and observable.
[4 pts] b) Find the transformation P and Ā such that Ā = P −1 AP is in modal canonical (diagonal) form.
P =
Ā =
[2 pts] c) Find B̄, C̄ such that x̄˙ = Āx̄ + B̄u and y = C̄x̄.
B̄ =
C̄ = [
]
[4 pts] d) Find eĀt and eAt :
Āt
e =
At
e
=
[4 pts] e) Draw a block diagram representation for the system in modal form with input
u and output y. Note any modes which are not observable or not controllable.
8
Problem 6. (10 pts)
You are given a continuous time plant described by the following state equation.
1
−1 −3
x
x(t = 0) =
ẋ = Ax =
1
0 −2
Every T seconds the state of the system is measured with an A/D converter, that is
x[n] = x(t = nT ).
[2 pts] a) For the zero input response, the sampled state can be represented by:
x[n + 1] = Gx[n]. What are the elements of G(T )? (Leave in terms of an exact expression.)






G=

[2 pts] b) If you were given G and x[0], how would you find x[n]?
x[n] =
[2 pts] c) For what conditions will the sampled system x[n + 1] = Gx[n] be stable?
[4 pts] d) You are given a continuous time system ẋ(t) = Ax(t) + Bu(t) and its zero-order
hold equivalent system x[n + 1] = Gx[n] + Hu[n], with sample rate T . The continuous time
system is controlled using full state feedback u = −kx, that is
ẋ = Ax + Bu = Ax + B(−k)x = [A − Bk]x,
with k such that the system is asymptotically stable.
Will the zero-order-hold equivalent sampled system with the same state feedback, that
is x[n + 1] = (G − Hk)x[n], necessarily be asymptotically stable? Why or why not?
9
Problem 7 (10 pts)
The simplified dynamics of a magnetically suspended steel ball are given by:
mÿ = mg − c
u2
.
y2
where m is mass of ball, g is gravity, y is ball position, and u is the control input.
[2 pts] a) using the states x1 = y and x2 = ẏ write down a nonlinear state space description
of this system.
x˙1 =
x˙2 =
[2 pts] b) What equilibrium control input must be applied to suspend the ball at position
y = yo ?
ue =
[2 pts] c) Write the linearized state space equations for state and input variables representing perturbations away from the equilibrium of part b).
x1
x˙1
+
u(t)
(2)
=
x2
x˙2
[2 pts] d) Is the linearized model stable? What can you conclude about the stability of
the nonlinear system close to the equilibrium point xe ?
[2 pts] e) Briefly describe, using a block diagram, how to implement a controller to regulate steel ball position at y = yo , specifying u(t). You have access to only output y.
10
Problem 8. (12 pts)
True/False questions. +1 point for correct answer, -1 point for incorrect answer, 0 for
blank. Write “T” or “F” in blank. (Minimum points on problem 8 is zero).
[7 pts] a) In lab 3, we used a PD controller on the Quanser carts to track a reference
position input. Which of the following are true about PD control?
(i)
PD control allows us to have both a shorter rise time and a smaller overshoot for a
step input compared to just P control.
(ii)
PD control is useful for eliminating steady state error.
(iii)
A PD controller can be approximated by a lag compensator.
(iv)
A PD controller can always be converted into an equivalent state-feedback controller.
(v)
A PD controller amplifies noise.
(vi)
The derivative term of a PD controller is used to increase the damping.
(vii)
The best way of estimating velocity for PD control is to use numerical differentiation.
[5 pts] b) In Lab 5a, state feedback controller for the inverted pendulum, we linearized the
system around the inverted position, and get the state space model as ẋ = Ax + Bu, y = Cx.
The linearized model is controllable and observable. The objective of that lab is to regulate
the pendulum at the inverted position (θ = 0). In the simulation, the controller is able to
regulate the system with zero steady state error. However, in actual hardware response, we
observe that the pendulum usually continues to oscillate about the equilibrium point. Please
check the correctness of the following statements:
(i)
The continuous oscillation behavior is mostly because it is impossible to initialize
the pendulum perfectly vertical, so the desired θ = 0 is slightly tilted.
(ii)
The difference between simulation and actual hardware response is partially due to
the linearization error of the system model at actual regulation position.
(iii)
Using a full-order observer to estimate the system states and state feedback will
eliminate the continuous oscillation behavior.
(iv)
The LQR control can eliminate this continuous oscillation behavior.
R
R
(v) Assume we want to implement the integral control as u = vdt = K(r − x)dt or
u̇ = v = K(r − x), where r is the reference input. The extended system with [x; u] as
the system states and v as the system input will become uncontrollable.
11