EE C128 / ME C134 Final Exam Fall 2014
December 19, 2014
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EE C128 / ME C134 Fall 2014 Final Exam
Name:
1. Laplace transform, controllable canonical form
(a) Derive the following Laplace property for convolution integrals
L {g1 (t)} = G1 (s)
Z
L
L {g2 (t)} = G2 (s)
t
g1 (τ )g2 (t − τ )dτ = G1 (s)G2 (s)
0
(b) Find the solution to the following ODE with the given initial conditions:
d2 x dx
−
+ 2x =
dt2
dt
Z
t
δ(τ ) sin (t − τ )dτ
0
ẋ(0) = −2, x(0) = 1
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2. Bode Plotting and Nyquist Stability
Consider the following transfer function:
G(s) =
100(s + 4)
(s2 + 12s + 20)(s + 2)
(a) Sketch a Bode plot of the system (magnitude and phase). Label all slopes and points on the
graph.
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(b) Using the Bode plots you created in Part (a), calculate the phase margin and gain margin for
G(s).
(c) Draw a Nyquist plot of the system and use the Nyquist stability criterion to determine if the
closed loop system under unity feedback is stable.
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(d) Assume we have a closed loop system below where G(s) is given in Part (a) and C(s) = K.
For what values of K is the system stable?
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3. Solution in time domain, stability
(a) For the following system, explicitly determine the time-domain solution x(t)
"
ẋ =
4 1
0 10
#
"
x+
y=
h
1
0
#
"
u,
1 0
i
x(0) =
1
1
#
x
where u(t) is a unit step function.
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(b) Determine the transfer function G(s) for the zero initial state response, given the system in
Part (a).
(c) BONUS: Does the degree of your state space model and transfer function match? Why or
why not?
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4. Electrical Circuit and Root Locus
(a) For the above circuit derive the transfer function C(s) =
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Vout
.
Vin
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(b) Assume the system below, G(s), is in unity negative feedback. Determine the value of K such
1
. Also determine the percent overshoot and
that the steady state error to a step response is 11
settling time of the feedback system at this K.
G(s) =
K
(s + 5)(s + 15)
(c) Now assume that the system is described by the figure below, where C(s) and G(s) are obtained from parts (a) and (b) respectively. Draw the root locus given R1 = 125 MΩ, C1 = 15 µF,
R2 = 625 MΩ and C2 = 0.1 µF. Watch your signs!
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(d) Label on the root locus a suitable region if the goal is to achieve a settling time (Ts ) ≤ 0.4 sec
and percent overshoot (%OS) ≤ 20%.
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5. Controller design using State-feedback
Consider the mechanical system shown above. Here, V denotes the voltage applied to the motor
(control input) and x(t) is the position of the mass. You may assume the back emf from the motor
is negligible (EM F = 0) and the torque supplied by the motor is equal to T = IKm − Jm θ̈
where,
Km : Constant relating T and I
Jm : Inertia of the motor
(a) Show that G(s) is the transfer function from V to x. To do this, you MUST derive the governing
equations for the mechanical/electrical system.
N2 Km r
X(s)
=
G(s) =
V (s)
N1 R + Ls
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"
J1 + J2 + Jm
N2
N1
2 !
#−1
2
2
2
s + r (M s + fv s + k)
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h
iT
as the state vector and x as the output, derive a state space model
(b) Choosing x̄ = x ẋ
(matrices A, B, C and D) for the above system. Use the following parameters:
R = 1, Km = 0.1, L = 0, N2 /N1 = 10, r = 1, J1 = J2 = 1, Jm = 0, M = 1, k = 1, fv = 1.
Note your input to the system should be V .
(c) Explicitly write the observability and controllability matrices. Is the system controllable? Is it
observable? (Use parameters from Part (b))
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(d) Determine the eigenvalues of matrix A you found in Part (b).
(e) We will now control the system
" using
# a state feedback controller as shown in the diagram below
h
i
r1
where K = k1 , k2 and r =
.
r2
Write the dynamics of the closed-loop system as ẋ = Ãx + B̃r. That is, find both à and B̃ in
terms of the system parameters given in Part (b) and the elements of the controller gain matrix
K.
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6. State-feedback and observer design
Consider the following system:
"
ẋ = Ax + Bu,
y = Cx with A =
7 1
−2 4
#
"
, B=
0
1
#
, C=
h
0 1
i
(a) Compute the eigenvalues and eigenvectors for A.
(b) Use state-feedback of the form of u = −Kx. Determine the gain K = [k1 k2 ] such that the
poles of the closed loop system are located at s1,2 = −2 ± 5j.
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(c) Unfortunately for this system we are unable to measure all the states. In order to do state
feedback we must use a Luenberger observer of the form:
x̂˙ = Ax̂ + Bu + L(y − ŷ)
ŷ = C x̂
and the system is controlled using state feedback, given by
u = −K x̂
Determine the error dynamics of the system, ė, where e = x̂ − x. The result must be in terms
of e only.
(d) Determine the observer matrix L = [l1 l2 ]T such that the error dynamics have poles at
s1,2 = −2 ± 5j.
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(e) Comment on the performance of the state observer given the previously placed poles. What
are we interested in when designing an observer and how could we improve the observer?
(f) Complete the following block diagram of the system described in part(c).
Plant
Controller
y
u
Observer
x̂
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7. Linear Quadratic Regulator
Consider the LTI system
ẋ = Ax + Bu
x = [x1 , x2 ]T
where
"
#
1 1
A=
,
0 −1
" #
1
B=
0
We would like to solve an LQR problem for the system. That is, we want to find the optimal
control u∗ (t) that minimizes the cost functional
Z
∞
J=
(x21 (t) + u2 (t)) dt
t=0
(a) Solve the Algebraic Riccati Equation for the infinite horizon LQR. Hint: the solution of the
Algebraic Riccati equation is a positive semi-definite matrix. A (2 × 2) matrix P is positive
semi-definite everywhere when:
"
#
p11 p12
2
p11 ≥ 0
p22 ≥ 0
p12 ≤ p11 p22
for P =
p12 p22
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(b) Determine the optimal feedback matrix K1 such that the optimal control is u∗ (t) = −K1 x(t).
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8. Linear Quadratic Regulator
Consider the system where the dynamics are scalar:
ẋ = ax + bu
x is a scalar
We want to create a finite horizon optimal controller given the cost function:
Z
tf
J=
(qx2 (t) + ru2 (t))dt
t=0
(a) Write the Ricatti Equation for this system as well as the terminal condition.
(b) Find P for the static case where tf = ∞. Your answer should be in terms of q, r, a, and b. Note
the P should be positive semi-definite everywhere.
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(c) For this scalar system, given any tf , the Riccati equation can be analytically solved:
P (τ ) =
(aP (tf ) + q) sinh(βτ ) + βP (tf ) cosh(βτ )
2
b P (tf )
−
a
sinh(βτ ) + β cosh(βτ )
r
r
b2 q
and sinh(.) and cosh(.) are the hyperbolic trigonometric
r
functions. Taking the limit as tf → ∞, the solution becomes:
where τ = tf − t, β =
a2 +
P (τ ) =
q
−a + β
Show that this is equivalent to your solution from Part (b).
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f (t)
δ(t)
F (s)
1
u(t)
1
s
1
s2
n!
sn+1
ω
s2 +ω 2
s
s2 +ω 2
ω
(s+α)2 +ω 2
s+α
(s+α)2 +ω 2
tu(t)
tn u(t)
sin(ωt)u(t)
cos(ωt)u(t)
e−αt sin(ωt)u(t)
e−αt cos(ωt)u(t)
Table 1: Laplace transforms of common functions
eθ − e−θ
sinh(θ)
2
eθ + e−θ
cosh(θ)
2
sinh(θ) 1 − e−2θ
tanh(θ)
=
cosh θ
1 + e−2θ
Table 2: Trigonometric functions
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