This preview shows page 1. Sign up to view the full content.
Unformatted text preview: MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Aeronautics and Astronautics
16.060: Principles of Automatic Control
Fall 2003 PROBLEM SET 4
Solutions
Problem 1
Step responses A and D clearly exhibit a “long tail” due to a combination of a slow, realvalued p ole and
a nearby zero, so they must correspond to transfer functions (3) and (4). To tell which is which, we need
to ﬁnd the sign of the residue for the realvalued p ole in the step responses for (3) and (4). In (3), there
will b e a p ositive vector from the zero combined with a negative vector from the step input p ole at s = 0,
so the residue will b e negative. That implies that the mode due to the slow p ole will b e subtracted from
the steadystate value, and so the output will approach its steadystate value from b elow. Therefore, step
response D corresponds to transfer function (3). Similarly, the residue for the realvalued p ole in the step
response for (4) will b e p ositive, so the output will approach steadystate from above, so step response A
must correspond to transfer function (4).
The other two step responses are easy: as the zero moves towards the right, p eak time decreases and
P.O. increases. Therefore B corresponds to (2), and C corresponds to (1).
Problem 2
1.
⎤
⎥⎤
⎥
1 −1 0
10
=
A = −1 � A =
2 −1
−1 −2 1
⎤
⎥⎤
⎥
1 8 −3
4 −1.5
−1
B  = 2 � B =
=
−1 0.5
2 −2 1
−1 C  = 0 � C is singular, so C −1 does not exist. 2.
�
�
�
�2 1 �
�
� − (4) � 2 1
A = (1) �
� 0 −1 �
� 1 −1
= (1)(−2) − (4)(−3)
= 10 �
�
�
�
� + (0) � 2 2
�
�1 0 �
�
�
� Since A �= 0, the matrix must have full rank, so A has rank = 3.
�
�
�
�
�
�4
�
� −1 0 �
�
0�
� −1 4
�
�
B  = (1) �
� −8 18 � − (2) � 5 18 � + (9) � 5 −8
= (1)(72) − (2)(−18) + (9)(−12)
=0 1 �
�
�
� Since B  = 0, the matrix must not have full rank. By inspection, the ﬁrst two rows of B are linearly
independent, so the third row must b e a linear combination of the ﬁrst two rows (in fact, the third
row is equal to 2 times the ﬁrst row minus 3 times the second row). So B has rank = 2.
A is nonsingular, but B is singular b ecause B  = 0. A would b e a valid transformation matrix for
a 3element state vector, but B would not, b ecause of its singularity.
3. (a)
�
�
� � − 1 −3 �
�=0
det(�I − A) = 0 � �
�0
�−2 �
� (� − 1)(� − 2) − (−3)(0) = 0
� �2 − 3� + 2 = 0 � �1 = 1, �2 = 2 � = �1 = 1 � (�1 I − A)v1 = 0
�
⎤
⎥
⎤⎥
0 −3
0
�
v1 =
�
0 −1
0
⎤⎥
1
� v1 =
�
0
� = �2 = 2 � (�2 I − A)v2 = 0
�
⎤
⎥
⎤⎥
1 −3
0
�
v2 =
�
00
0
⎤⎥
3
� v2 =
�
1
Similarly for B :
det(�I − B ) = 0 � �2 − 7� + 12 = 0 � �1 = 3, �2 = 4
⎤
⎥
⎤⎥
⎤⎥
−2 2
0
1
� = �1 = 3 �
� v1 =
v=
�
�
−1 1 1
0
1
⎤
⎥
⎤⎥
⎤⎥
0
2
−1 2
� v2 =
�
v=
�
� = �2 = 4 �
1
−1 2 2
0
⎤⎥
1
(b) i. B � =
x
so the vector � has b een rotated from an angle of 63 � to an angle of 79� , and its
x
5 �
�
length has b een scaled from 5 to 26. ⎤⎥
⎤⎥
1 3
ii. If � = v1 =
v
�
, then B� =
v
= 3� . So the vector � has not b een rotated, but its
v
v
1
3
length has b een scaled by a factor of 3, which is equal to the eigenvalue associated with
that eigenvector.
Problem 3
1. Deﬁne x1 = y , x2 = y and x3 = y . So we have the state equations:
˙,
¨
x1 = x2
˙
x2 = x3
˙
x3 = −2x1 − 3x2 + 2x3 + u
˙ 2 And the output equation:
y = x1
The statespace model is then:
� 0
˙
� = �0
x
−2
�
y= 10 ⎡
�⎡
10
0
0 1 ⎣ � + �0⎣ u
x
−3 2
1
�
��
0 �+ 0 u
x 2. Split up the system into p oles and zeros as follows: U (s) � X1 (s�
) 1
s2 +4s−1 3s − 2 Y (s) � From the ﬁrst block, we have:
x1 + 4x1 − x1 = u
¨
˙
Let x1 b e the ﬁrst state variable, and x2 = x1 b e the second state variable. Then the state equation
˙
is:
⎥
⎤⎥
˙= 0 1 �+ 0 u
�
x
x
1 −4
1
Looking at the second block, we have:
y = 3x˙1 − 2x1 = 3x2 − 2x1
And so the output equation is: Problem 4 �
�
��
y = −2 3 � + 0 u
x Use the formula:
G(s) = C (sI − A)−1 B + D
(a)
⎤ ⎥
s
1
sI − A =
0 s+2
�
⎥ ⎦1
⎤
−1
1
s + 2 −1
+2)
= s s(s1
� (sI − A)−1 =
0
s
0
s(s + 2)
s+2
�⎤ ⎥ ⎦
�
⎤
⎥ ⎦1
−1
− s(s1
1 0 s s(s+2) 0
+2)
=
� G(s) =
1
1
01 0
1
s+2
s+2
3 (b) sI − A = ⎤ s −1
2 s+3 ⎥ ⎤
⎥
⎤
⎥
1
1
s+3 1
s+3 1
� (sI − A)
=
=
(s + 1)(s + 2) −2 s
s(s + 3) + 2 −2 s
⎤
⎥⎤
⎥⎤
⎥
1
10
s+3 1 0 1
� G(s) =
−2 s 1 0
(s + 1)(s + 2) 1 −1
⎤
⎥⎤
⎥
1
10
1 s+3
=
(s + 1)(s + 2) 1 −1 s −2
⎦
�
−1 = 1
(s+1)(s+2)
s−1
− (s+1)(s+2) s+3
(s+1)(s+2)
s+5
(s+1)(s+2) Problem 5
Deﬁne d1 as the p osition of m1 , d2 as the p osition of m2 , and d3 as the p osition of m3 , where d1 , d2 ,
and d3 are measured from the left of the diagram (the same way as y is measured). Then write down the
massspringdamper equation for each mass:
¨
m1 d1 = F − c1 (d˙1 − d˙2 ) − k1 (d1 − d2 ) ¨
m2 d2 = −c2 (d˙2 − d˙3 ) − k2 (d2 − d3 ) + c1 (d˙1 − d˙2 ) + k1 (d1 − d2 )
¨
m3 d3 = c2 (d˙2 − d˙3 ) + k2 (d2 − d3 )
Now deﬁne our states as: x1 = d1 , x2 = d2 , x3 = d3 , x4 = d˙1 , x5 = d˙2 , x6 = d˙3 . Also deﬁne the input
as u = F . So the above equations can b e rewritten as: x˙4 =
x˙5 =
x˙6 = 1
(−k1 x1 + k1 x2 − c1 x4 + c1 x5 + u) m1 1
(k1 x1 − (k1 + k2 )x2 + k2 x3 + c1 x4 − (c1 + c2 )x5 + c2 x6 )
m2
1
(k2 x2 − k2 x3 + c2 x5 − c2 x6 ) m3 In addition, from the way we deﬁned our states, we have:
x1 = x4
˙
x2 = x5
˙
x3 = x6
˙ And ﬁnally, the output equation is: y = x3 So now we can write the statespace model: 4 � 0
0
0 0
0
0 0
0
0
0 1
0
0
c
− m11 �
�
�
�
˙
� = � − k1
x
k1
� m1
m1
� k1
k2
c1
k1 +k2
� m2 − m2
m2
m2
k
k2
− m23
0
0
m3
�
�
��
y = 0 0 1 0 0 0 �+ 0 u
x Problem 6 + R1 0
1
0
c1
m1
c1 +c2
− m2
c2
m3 0
0
1
0
c2
m2
c
− m23 ⎡ �⎡
0
⎢
�0⎢
⎢
�⎢
⎢
�0⎢
⎢
x �⎢
� + � ⎢u
⎢
⎢
�1⎢
⎢
�0⎣
⎣
0 + R2
+ e_i C v i e_o L   Deﬁne our states to b e x1 = v , the voltage across the capacitor, and x 2 = i, the current through the
inductor. The input is u = ei and the output is y = eo . Applying KCL at the node b etween R1 and R2
gives:
dv
v − ei
+C
+i=0
R1
dt � dv
1
1
1
=−
ei
v− i+
dt
R1 C
C
R 1 C Applying KVL around the loop containing C , R 2 , and L, gives: di
=0
dt
Finally, the ouptut voltage eo is given by: v − R2 i − L � di
1
R2
= v−
i
dt
L
L eo = v − R2 i So the statespace model is: ⎤1⎥
1⎥
−C
� + R1 C u
x
1
0
− R2
L
L
�
�
��
x
y = 1 −R2 � + 0 u ˙
�=
x ⎤ 1
− R1 C 5 ...
View
Full
Document
This note was uploaded on 11/06/2011 for the course AERO 100 taught by Professor Willcox during the Fall '03 term at MIT.
 Fall '03
 willcox
 Aeronautics, Astronautics

Click to edit the document details