1 ECE-320: Linear Control Systems
Homework 5
Due: Monday April 15 by 5 PM 1) For the following two circuits,
show that the state variable descriptions are given by
[
]
1 1
( ) ( ) ( )
( ) ( ) 0 [0] ( )
1
1 1
( ) ( ) ( )
b
L L L
in B in
c c c
a a
R
t L L t L t
v t y t R v t
t t
i i
t R C
i d
v v v
d
C R
t
C
− −
= + = +
−
1
( ) ( ) ( ) ( )
( ) ( ) ( ) 0 ( )
( ) 1 ( ) ( )
0 0
a b
L a b L a b L b
a b
b
in in
c c a b c a b
R R
t L R R L t R R t R
L R
i i i
d
v t v t R R v t y t R R v t R R v t
dt
C
− −
+ +
= + = − +
+ +
2 2) For the plant
2 2 ( ) 1 2 1 p n n K G s
s ζ s
ω ω
=
+ +
a) If the plant input isu t( ) and the output isx t( ), show that we can represent this system with the differential equation
2 2
( ) 2 n ( ) n ( ) n ( )
x t + ζω x t +ω x t =Kω u t
b) Assuming we use states q t1( )=x t( )and q t2( )=x t( ), and the output is x t( ), show that we can write the state variable description of the system as
[
]
[ ]
1 1 2 2 2 2 1 20 1 0
( ) ( )
( ) 2
( ) ( )
( )
( ) 1 0 0 ( )
( )
n n n
q t q t
d
u t K
q t q t
dt
q t
y t u t
q t
ω ζω ω
= + − − = + or ( ) ( ) ( ) ( ) ( ) ( ) q t =Aq t +Bu t y t =Cq t +Du t Determine the A, B, C and D matrices.
c) Assume we use state variable feedback of the form u t( )=G r tpf ( )−kq t( ), where r t( )is the new input to the system, Gpf is a prefilter (for controlling the steady state error), and kis the state variable
feedback gain vector. Show that the state variable model for the closed loop system is
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
pf
pf
q t A Bk q t BG r t
y t C Dk q t DG r t
= − + = − + or ( ) ( ) ( ) ( ) ( ) ( ) q t Aq t Br t
y t Cq t Dr t
= +
= +
d) Show that the transfer function (matrix) for the closed loop system between input and output is given by
1
( )
( ) ( )( ( ))
( ) pf pf
Y s
G s C Dk sI A Bk BG DG
R s
−
= = − − − +
and ifD is zero this simplifies to
1
( )
( ) ( ( ))
( ) pf
Y s
G s C sI A Bk BG
R s
−
= = − −
e) Assume r t( )=u t( ) and D=0. Show that, in order for lim ( ) 1
t→∞y t = , we must have
1
1
( )
pf
G
C A Bk − B
− =
−
3 If matrix Pis given as
a b P
c d
=
then
1 1 d b
P
c a
ad bc
− = −
−
−
and the determinant of Pis given by ad−bc. This determinant will also give us the characteristic polynomial of the system.
3) For each of the systems below:
• determine the transfer function when there is state variable feedback
• determine if k1and k2exist (k =
[
k1 k2]
) to allow us to place the closed loop poles anywhere. That is, can we make the denominator look like s2+a s1 +a0 for any a1 and any a0. If this is true, the system is said to be controllable.a) Show that for
1 0 0
1 1 1
q= q+ u
[
0 1]
[0]y= q+ u
the closed loop transfer function with state variable feedback is
2
( 1) ( )
( 1)( 1 )
pf
s G
G s
s s k
− =
− − +
b) Show that for
0 1 0
0 1 1
q= q+ u
[
0 1]
[0]y= q+ u
the closed loop transfer function with state variable feedback is 2
2 1
( )
( 1)
pf
sG G s
s k s k
=
+ − +
c) Show that for
0 1 0
1 1 1
q= q+ u
[
1 0]
[0]y= q+ u
the closed loop transfer function with state variable feedback is 2
2 1
( )
( 1) ( 1)
pf
G G s
s k s k
=
4 4) For one of the rectilinear systems in lab, I found the following state variable representations:
0 1 0
174.8205 2.6469 5050 q= q+ u
− −
[
1 0]
[0]y= q+ u
The closed loop transfer function with state variable feedback is
2
2 1
5050 ( )
(2.6469 5050 ) (174.8205 5050 )
pf o
G G s
s k s k
=
+ + + +
a) Design I
• Determine feedback gains k1 and k2 so the resulting system has real poles and a bandwidth of 2 Hz. Set the second pole twice as far from the origin as the first pole.
• Use the final value Theorem to determine the prefilter gain so the steady state error for a step input is zero.
• Estimate the settling time for the step response based on the system bandwidth.
• Use Matlab to plot the Bode plot of the closed loop system (to verify the bandwidth) and plot the step response (to verify the estimated settling time).
You should get numbers like 0.028, 0.0069, 0.06, and 0.32. When you look at the Bode plot, the bandwidth will not be 2 Hz, primarily because the second pole is too close to the first pole.
b) Design II
• Determine feedback gains k1 and k2 so the resulting (ideal second order) system has a percent
overshoot of 15% and a settling time of 0.5 seconds.
• Use the final value Theorem to determine the prefilter gain so the steady state error for a step input is zero.
• Use Matlab to plot the step response (to verify the percent overshoot and settling time).
• Determine (analytically) the steady state output if the input to the closed loop system is r(t) = 1 cos(4t)cm and determine the time delay between the input and the output.
You should get numbers like 0.0128, 0.0026, 0.047, and a time delay of 0.07 seconds. Your step response should be fairly close to the requirements.
Prelab for Lab 4, to be turned in with your homework
5) One of the things that will be coming up in lab more and more is the limitation of the amplitude of the control signal, or the control effort. This is also a problem for most practical systems. In this problem we will do some simple analysis to better understand why Matlab's sisotool won't give us a good
5 a) For the system below,
show that U(s) and R(s) are related by
( ) ( )
( ) ( )
1 ( ) ( )
c pf
p c
G s G s
U s R s
G s G s =
+
b) For many types of controllers, the maximum value of the control signal is just after the step is applied, at t=0+. Although most of the time we are concerned with steady state values and use the final value Theorem in the s-plane, in this case we want to use the initial value Theorem, which can be written as
0
lim ( ) lim ( )
s
t→ +u t = →∞sU s
If the system input is a step of amplitude A, show that
( ) ( ) (0 ) lim
1 ( ) ( )
c pf
s
p c
AG s G s u
G s G s
+ →∞
= +
This result shows very clearly that the initial control signal is directly proportional to the amplitude of the input signal, which is pretty intuitive.
c) Now let's assume
( ) ( )
( ) ( ) ( )
( ) ( )
( ) p ( ) pf
c
c p pf
c p pf
N s N s
N s
G G G
D s D s D
s s s
s
= = =
If we want to look at the initial value for a unit step, we need to look at
( ) ( ) 1 ( ) ( )
(0 ) lim lim
1 ( ) ( ) 1 ( ) ( )
c pf c pf
s s
c p c p
sG s G s G s G s
u
G s G s s G s G s
+
→∞ →∞
= =
+ +
Let's also then define
( ) ( )
( )
1 ( ) ( )
c pf
c p
G s G s
U s
G s G s =
+
so that
(0 ) lim ( )
s
u + U s
→∞
=
+ -
( )
6 Show that
( ) ( )
( ) ( )
( ) ( )
( ) ( )
pf
p c
pf pf
c p
N s
U s
N s
D s
D s D s
N s D s
=
+
and
degU =degNpf −max deg Dc−degNc+degDpf, degNp−degDp+degDpf
where degYis the degree of polynomial Y.
d) Since we are going to take the limit as s→ ∞, we need the degree of U s( ) to be less than or equal to zero for a step input to have a finite (0 )u + . Why?
For our 1 dof systems in lab, we have degNp =0 and degDp =2. Use this for the remainder of this problem
e) If the prefilter is a constant, show that in order to have a finite (0 )u + we must have
degDc≥degNc
f) If the numerator of the prefilter is a constant, then in order to have a finite (0 )u + we must have
degDc−degNc+degDpf ≥0 or − +2 degDpf ≥0
g) For P, I, D, PI, PD, PID, and lead controllers, determine if u(0 )+ is finite if the prefilter is a constant.
