MEEN 411
Spring 2010
Solutions to Homework set - 2
Sol (1):
(c) L( s ) =
1
.
s ( s + 1)(1 + 0.02 s )
The transfer function is converted to the following form by substituting s = j ,
1
L ( j ) =
.
j ( j + 1)(1 + 0.02 j )
From the above relation, the corner
MEEN 411
Spring 2010
Solutions to Homework set - 2
Sol (1):
(c) L( s ) =
1
.
s ( s + 1)(1 + 0.02 s )
The transfer function is converted to the following form by substituting s = j ,
1
L ( j ) =
.
j ( j + 1)(1 + 0.02 j )
From the above relation, the corner
MEEN 411
Spring 2010
Solutions to Homework Set - 3
Sol (1):
(a)
The closed loop transfer function is:
1
2
1
Ts + s
=2
T=
1 Ts + s + 1
1+ 2
Ts + s
The error is given by:
Ts 2 + s
R
Ts 2 + s + 1
E = R Y = R TR = (1 T) R =
(i)
For the step input: R(s) =
Solution Set # 4
1) Problem 6.57.
a) The plant transfer function is given as
G ( s) =
1
s
s
s
s + 1
+ 0 .5
+ 1
2
100
20 100
2
.
The lead compensator is
s
+ 1
20
.
D( s) = K1
s
+ 1
100
Therefore the loop transfer function is G(s)*D(s), given by
G (
MEEN 411
Spring 2010
Solutions to Homework Set - 5
Sol (1):
We have,
4 0 0
0
1 4 0 0
A=
5
7 1 15
0 3 3
0
For the system to be stable, the poles of the system should lie in the left half plane. The
poles of the system are nothing but the eigenvalues of
7.14 Solution
&
&
We are given x = Fx + Gu. Steady-state means that x = 0 and a step input (or unit step)
means u = 1(t). Thus, assuming that F is invertible, we have,
0 = Fx +G
1
4 1 0 1 / 6
x = -F G =
=
2 1 1 2 / 3
-1
This can be veried in MATLAB w
7. 34
Controllability
det[ B
A 2 B L A n 1 B] 0
AB
C
CA
Observability CA 2 0
M
CA n 1
Stability
Recfw_eig ( A) < 0
So, in general, there is no connection between these three properties. However, for a minimal
realization (controllable and observable)
33 For the compensation
.r +
5+15'
use Euler's forward retangular method to determine the difference equations for a digital
implementation with a sample rate of 80 Hz. Repeat the calculations using the hackward
retangular method (see Problem 3.2) and
MEEN 411
Spring 2010
Solutions to Homework set - 1
Sol (1):
The differential equations of motion for the system defined are given as
.
.
J + B = K T ia ,
L
.
di a
+ Ri a = v a K e .
dt
Taking the Laplace transforms of the first equation, we get
Js 2 ( s )
MEEN 411
Spring 2010
Solutions to Homework set - 1
Sol (1):
The differential equations of motion for the system defined are given as
.
.
J + B = K T ia ,
L
.
di a
+ Ri a = v a K e .
dt
Taking the Laplace transforms of the first equation, we get
Js 2 ( s )
MEEN 411
Spring 2010
Solutions to Homework Set - 3
Sol (1):
(a)
The closed loop transfer function is:
1
2
1
Ts + s
=2
T=
1 Ts + s + 1
1+ 2
Ts + s
The error is given by:
Ts 2 + s
R
Ts 2 + s + 1
E = R Y = R TR = (1 T) R =
(i)
For the step input: R(s) =
Solution Set # 4
1) Problem 6.57.
a) The plant transfer function is given as
G ( s) =
1
s
s
s
s + 1
+ 0 .5
+ 1
2
100
20 100
2
.
The lead compensator is
s
+ 1
20
.
D( s) = K1
s
+ 1
100
Therefore the loop transfer function is G(s)*D(s), given by
G (
MEEN 411
Spring 2010
Solutions to Homework Set - 5
Sol (1):
We have,
4 0 0
0
1 4 0 0
A=
5
7 1 15
0 3 3
0
For the system to be stable, the poles of the system should lie in the left half plane. The
poles of the system are nothing but the eigenvalues of
7.14 Solution
&
&
We are given x = Fx + Gu. Steady-state means that x = 0 and a step input (or unit step)
means u = 1(t). Thus, assuming that F is invertible, we have,
0 = Fx +G
1
4 1 0 1 / 6
x = -F G =
=
2 1 1 2 / 3
-1
This can be veried in MATLAB w
7. 34
Controllability
det[ B
A 2 B L A n 1 B] 0
AB
C
CA
Observability CA 2 0
M
CA n 1
Stability
Recfw_eig ( A) < 0
So, in general, there is no connection between these three properties. However, for a minimal
realization (controllable and observable)