ME 132
Homework # 1
Issued: January 22, 2016
Due: January 29, 2016
Suggested Readings:
1 My lecture notes, Introduction, posted on bspace.
2 Chapter 1, Astrom & Murray. This is a great book, and is posted on b-courses.
Problems to be turned in:
1 Structur
ME 132
Solutions # 1
1 Structure of Control Systems
Shown below is a block diagram for the temperature control system in a refrigerator. A relay
is an on-o switch, and is often the most elementary actuation in control systems. Not shown
are humidity eects
ME 132 Fall 2009 Solutions to Homework 1
1. There are several examples from each eld that are valid descriptions of feedback systems. As long as you showed the dependence of the control signal on some measured variable, your example should be sucient. 2.
ME 132
Issued: Monday, November 23, 2009 1 Observers Consider the plant y= 1 u+n s2
Homework # 12
Due: Monday, November 30, 2009
Here n represents measurement noise. (a) Design an observer to estimate the state of the plant using measurements of the outpu
ME 132
Solutions # 5
1 Complex Arithmetic
(a) Notice that for all real ,
2 + a2
|j a|
j a
=1
=
=
j + a
|j + a|
2 + a2
(b) In all four cases the magnitude plots are identical and are shown in the top panel. The
phase plots are shown in the bottom panel.
University of California at Berkeley Department of Mechanical Engineering ME 132: Dynamic Systems and Feedback Instructor: Kameshwar Poolla Office: 5105 or 5141 Etcheverry Hall Email: poolla@berkeley.edu Office Hours: Th 1-3 pm Teaching Assistants: GSI 1:
ME 132
Issued: October 09, 2009
Homework # 7
Due: October 17, 2009
1 Block diagrams Consider the block diagram shown below. Each block is a gain with the letter in the block denoting the numerical value of the gain. Find the matrix M (which is 1 3) such t
ME 132
Solutions # 11
1 Equilibrium points, Linearization
(a) If x1 = y , x2 = y , then we have the following state-space description:
x1 = x2
x2 = cos(x2 ) x1 2 + u.
(b) We are given that u = 0. To nd equilibrium points, we solve 0 = f (x, u) as
0 = x2
0
ME 132
Solutions # 10
1 PID Controller Design
(a) The characteristic equation of the ODE describing the system is s 1. The root of the
equation is s1 = 1 > 0 and thus the system is unstable.
(b) We start by taking the time derivative of the ODE y (t) = y
ME 132
1 The state-variable equations for the system are x1 = x2 x2 = g y = x1 c u2 m x2 1
Solutions # 13
(a) The control u needed to maintain y = is found from evaluating the system at equilibrium. This gives the equilibrium x1 x2 , u, y = 0 , mg , c
The
ME 132
Issued: December 02, 2009
Homework # 13
Due: December 09, 2009
1 Consider the magnetically-suspended ball treated in class. The dierential equation relating the control input u to the height of the ball y is y=g with the following constants: m 0.15
University of California at Berkeley Department of Mechanical Engineering ME132 Dynamic Systems and Feedback Homework Set #6 Assigned: April 3 (Th) Due: April 15 (Tu) [1] 18.1-1 [2] 21.3-1 Spring 2008
ME 132
Solutions # 4
1 Programming in matlab
First set x1 = y and x2 = y.
Then, the given second order differential equation can be
written as
x1 = x2 ,
x2 = 4x1
and we can now write our differential equation solver (shown below). Incidentally, the
analy
G(s) H (s) C1 (s) C2 (s)
R Y (R Y ) N E (N E )
N = 0
G( s) =
s2
1
,
+ 2s + 2
H ( s) = 1 ,
K
R Y
C1 (s) = K,
C 2 ( s) = 0 .
R Y
R Y
K
T
1
z1 z2
mincfw_Re(z1 ),Re(z2 )
T=
N = 0
ME 132
Homework # 12
Issued: Monday, November 21, 2011
Due: Monday, November 28, 2011
1 Observers
Consider the plant
1
y = 2 u+n
s
Here n represents measurement noise.
(a) Design an observer to estimate the state of the plant using measurements of the
o
ME 132
Solutions # 12
1 Observers
(a) Since the noise is unknown and zero-mean, the best model we can use is y =
A state-space realization of this model is
0
0
1
u
x +
x =
1
0 0
1
s2
u.
y = [ 1 0 ] x
The observable canonical form of this realization is
ME 132
Solutions # 2
1 Model Properties
Many of you had difficulty with this problem, particularly part (c).
(a) This model is memoryless, nonlinear, time-invariant, causal.
(b) This model is memoryless, nonlinear (because of the affine term b), time-inva
ME 132
Solutions # 13
1 Equilibrium points, Linearization
(a) A state variable realization of this system can be obtained by setting
x1 = y
x2 = y
which will give the following state variable realization:
x 1 = f1 (x1 , x2 , u) = x2
x 2 = f2 (x1 , x2 , u)
ME 132
Solutions # 1
1 Structure of Control Systems
Shown below is a block diagram for the temperature control system in a refrigerator. A relay is an on-off
switch, and is often the most elementary actuation in control systems. Not shown are humidity eff
ME 132
Solutions # 7
1 PI position control for a DC Motor
First observe that the motor dynamics in transfer function notation are given by
1.9
1600
=
u
TL
s(s + 0.97)
s(s + 0.97)
Next, our PI controller has the form
sKP + KI
(r )
u=
s
Combining these equa
UNIVERSITY OF CALIFORNIA AT BERKELEY
Department of Mechanical Engineering
ME132 Dynamic Systems and Feedback
Midterm Examination I
Spring 2007
Closed Book and Closed Notes. Two 8.5 11 pages of handwritten notes allowed.
Your Name:
Please answer all questi
ME 132
Solutions # 11
1 State-space realizations
The controllable canonical form can be calculated from the methodology presented in
class/previous homework solutions.
(a)
x 1
x 2
x 3
x 4
x 5
y
=
0
0
0
0
3
1
1 0
0
0 0
0 1
0
0 0
0 0
1
0 0
0 0
0
1
ME 132
Solutions # 8
1 Poles and Zeros
The plant on the left (L) has a zero in the right-half complex plane, which the plant on
the right (R) does not. Otherwise, both plants have identical poles. The typical unit-step
responses for these plants are shown
ME 132
Solutions # 9
1 Matrix Exponentials
(a) Since v1 and v2 are linearly independent, we can write the matrix A as
A = T T 1
where T = [ v1 v2 ] and
1
A = [ v1 v2 ]
0
1
1 2
=
0
0 1
1 2
=
0 2
= diagcfw_1 , 2 . For this problem,
0
1
[ v1 v2 ]
2
1 2
0
0
ME 132
Solutions # 10
1 Linear Algebra
(a) Note that e(K+L)t = eKt eLt only if AB = BA (this can be verified from the exponential
expansion). In this case, we can write the A matrix as
0
0
+
A=
0
0
=K +L
and note that KL = LK. Therefore,
eAt = eKt e