Chapter 2:
Challenge: we now give you a problem problem to test your knowledge of this chapters objectives: objectives: Referring to the antenna azimuth position control system schematic shown on the front endpapers, evaluate the transfer function of each subsystem. Use configuration 2. Record your results in the table table of block diagram parameters shown on the front endpapers for use in subsequent chapters case study challenges. The table below shows individual subsystems for which we must find the transfer functions:
Subsystem
Input
Output
Input potentiometer
Angular rotation from user, i(t)
Voltage to preamp, Vi(t)
Preamp
Voltage from potentiometers,
Voltage to power amp, Vp(t)
Ve(t) = Vi(t) Vo(t) Power amp
Voltage from preamp, Vp(t)
Voltage to motor, ea(t)
Motor
Voltage from power amp, ea(t)
Angular rotation to load, o(t)
Output potentiometer
Angular rotation from load, o(t)
Voltage to preamp, Vo(t)
Table 1: Subsystems of the antenna azimuth position control system
The table that is shown below represent the schematic parameter of the antenna azimuth position control system for configuration 2: Parameter
Configuration 2
V
10
n
1
K
-
K1
150
a
150
Ra
5
Ja
0.05
Da
0.01
Kb
1
Kt
1
N1
50
N2
250
N3
250
JL
5
DL
3
Now we proceed and find each transfer function. Potentiometer:
Preamp:
Power Amp:
Motor:
= 0.05 + 5( = 0.25 ( 0.01 +3( = 0.13
Jm = Ja + JL( Dm
=
Da
+ DL
We have that: Kt/Ra = 1/5 KtKb/Ra = 1/5 So:
=
And we also have that:
The results are summarized in the following block diagram and the table of block diagram parameters: Parameter
Configuration 2
Kpot
3.183
K
-
K1
150
a
150
Km
0.8
am
1.32
Kg
0.2
Chapter 3:
Challenge: You are now given a problem to test your knowledge of this chapters objectives. Referring to the antenna azimuth position control system shown on the front endpapers, find the state-space representation of each dynamic subsystem. Use configuration 2. Since only the power amplifier and the motor and load are dynamic systems, we will only find the statespace representation for these two systems. Power amplifier: The transfer function of the power amplifier is given by: G(s) =
= =
When we take the inverse Laplace transform we get the following expression:
+ 150ea = 150vp(t) When we rearrange the above equation we get:
And since the output of the power amplifier is ea(t), then the output equation is: y(t) = ea(t)
Motor and Load: ea(t) = ( Defining
the state variables x1 and x2 as x1 = m x2=
yields = - Solving for
And the state equations are written as follow:
( m + D
x2 + (
Using the gear ration, which is 1/5, the output equation is:
In vector matrix form,
But we already know that:
So our final state and output equations are:
Chapter 4:
Challenge: you are now given a problem to test your knowledge of this chapters objectives: Referring to the antenna azimuth position control system shown on the front endpapers, configuration 2. Assume an open-loop system (feedback path disconnect) and do the following: a.
Predict the open-loop angular velocity response of the power am plifier, motor, and load to a step voltage at the input to the power amplifier.
b. Find the damping ratio and natural frequency of the open-loop system. c.
Derive
the open-loop angular velocity response of the power amplifier, motor, and load to a
step-voltage input using transfer functions. d. Obtain the open-loop state and output e quations. e. Use MATLAB to obtain a plot of the open-loop angular velocity response to a step-voltage input. Solution: a.
Power amp
motor and load
convert to Angular velocity
Ea(s)
Vp(s)
0(s)
0(s) s
0(s)
Using the transfer function above, we can predict the step response being: Wo(t) = A + B
+ C
b. To find the damping ratio and the natural frequency we nee d to expand the denominator of the transfer function found in (a), and the resulting transfer function is:
G(s) =
Thus, 2wn = 151.32, wn = 14.07, and = 5.38
c.
For us to get the angular velocity response to a step input, we will multiply our transfer function by a step input, 1/s, and we get:
Wo(s)
=
Expanding into partial fractions we get:
Wo(s) =
Transforming the above equation into time domain, we get: Wo(s) = 0.12121 + 0.0010761
0.12229
d. First we need to convert the transfer function into the state-space representation:
Wo(s)/Vp(s) =
o + 151.32o + 198wo = 24Vp(t) then we define x1 = wo x2 = o thus, the state space equation are: 1 = x2 2 = -198x1 151.32x2 + 24Vp(t)
writing the above in vector-matrix form we get:
=
Y=
X +
X
e. Now we run the MATLAB. Program: numg = 24; den = poly([-150 - 1.32]); G = tf(numg, den) Step(G) Response from the computer: Ans = Transfer function: 24 ------------------------S^2 + 151.3s + 198
Vp(t)
