Problem: 7.9.9.10
There are a couple of ways to work this problem. One is to nd the
equivalent transfer function for the inner feedback look. The other way is
to put the inner summer past the block with Ks in it. We elect the rst
method. Then
60
(s ; 1)(s
solution 9.10.1.1
The rst step is to put the transfer function in time constant form. So we
have
10(s + 1)
G(s) =
s(s + 10)(s + 50)
10(s + 1)
=
(10)(50)s(1 + s=10)(1 + s=50)
0:02(1 + s=1) :
=
s(1 + s=10)(1 + s=50)
Then the terms to be plotted are
0:02 1 1
Solution: 10.8.1.1
For
GH (s) =
K (s + 4)
(s + 2)(s ; 1)
The angle contributions of each of the poles and zeros can be determined
from the vector diagram of gure 1. For ! = 0, = 2 = 0 while 1 = 180 .
Im(s)
2
-4
1
-2
Re(s)
1
Figure 1: Vector Components of
Solution 11.9.2.5
The compensated plant is shown in Figure 1. In order to meet the speci cations we have added a double integrator and cancelled both plant poles.
Then
100(1 + s=1:5)2
Gp (s) = 3
s (1 + s=60)(1 + s=200)
The crossover frequency is !c = 15 r
Solution 11.9.2.11
R+
+
Gc
G1
G2
C
Kr
Figure 1: Maglev block diagram
For the system of Figure 1 we have
1 and
G (s) =
1
Js
( ) = 1:
G2 s
s
We rst note that the proposed design turns a type 2 system into a type
1. The reason is that if we nd the equivalent
ECE 470
Quiz 0827, 10 points
Use a separate sheet for calculation details and summarize
only the important steps of your analysis below.
The open loop and closed loop controllers discussed in class last time are to be compared
for a system whose transfer