Chapter 2:

Challenge: we now give you a problem to test your knowledge of this chapter’s 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 of block diagram parameters shown on the front endpapers for use in subsequent chapter’s 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, θi(t) Voltage to preamp, Vi(t)

Preamp Voltage from potentiometers, Ve(t) = Vi(t) – Vo(t)

Voltage to power amp, Vp(t)

Power amp Voltage from preamp, Vp(t) Voltage to motor, ea(t)

Motor Voltage from power amp, ea(t) Angular rotation to load, θo(t)

Output potentiometer Angular rotation from load, θo(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:

Vi(s)θi (s )

=10π

=3.183

Preamp:

Vp(s )Vi(s)

=K

Power Amp:

Ea(s )Vp(s)

= 150s+150

Motor:

Jm = Ja + JL(50250

¿¿2 = 0.05 + 5(50250

¿¿2 = 0.25

Dm = Da + DL(50250

¿¿2=¿0.01 +3(50250

¿¿2 = 0.13

We have that:

Kt/Ra = 1/5

KtKb/Ra = 1/5

So:

= 0.8

s (s+1.32)

And we also have that: θo(s)Ea(s )

=15θm(s)Ea(s)

= 0.16s(s+1.32)

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 chapter’s 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 state-space representation for these two systems.

Power amplifier:

The transfer function of the power amplifier is given by:

G(s) = Ea(s )Vp(s)

= K 1(s+a)

= 150s+150

When we take the inverse Laplace transform we get the following expression:

deadt

+ 150ea = 150vp(t)

When we rearrange the above equation we get:

deadt

=−150ea+150vp (t)

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) = (RaJmKt

¿ dx 2dt

+(DmRaKt

+Kb) dθmdt

Defining the state variables x1 and x2 as

x1 = θm

x2= dθmdt

Solving for dx2d t

yields

dx2dt

= -1Jm (Dm+ KtKb

Ra )x 2+( KtRaJm )ea(t )

And the state equations are written as follow:

dx1dt

=x 2

dx1dt

=−1Jm

(Dm + KtKbRa

¿x2 + (KtRaJm

¿ea (t)

Using the gear ration, which is 1/5, the output equation is:

y=0.2x 1

In vector – matrix form,

dxdt

=[0 1

0−1jm

(Dm+ KtKbRa )] x+[ 0

KtRaJm ]ea(t )

But we already know that:

Jm=0.25 ; Dm=0.13 ; KtRaJm

=0.8∧1Jm (Dm+ KtKb

Ra )=1.32So our final state and output equations are:

dxdt

=[0 10 −1.32] x+[ 00.8]ea (t)

y= [0.2 0 ] x

Chapter 4:

Challenge: you are now given a problem to test your knowledge of this chapter’s 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 amplifier, 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 equations.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) θ0(s) ᴪ0(s)

Vp(s) ᴪ0(s)

150(s+150)

0.16s (s+1.32)

s

24(s+150)(s+1.32)

Using the transfer function above, we can predict the step response being:

Wo(t) = A + Be−150 t + Ce−1.32 t

b. To find the damping ratio and the natural frequency we need to expand the denominator of the transfer function found in (a), and the resulting transfer function is:

G(s) = 24

s2+151.32 s+198

Thus, 2ᶘwn = 151.32, wn = 14.07, and ᶘ = 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) =24

s (s+150)(s+1.32)

Expanding into partial fractions we get:

Wo(s) = 0.12121s

+0.0010761s+150

−0.12229s+1.32

Transforming the above equation into time domain, we get:

Wo(s) = 0.12121 + 0.0010761e−150 t – 0.12229e−1.32 t

d. First we need to convert the transfer function into the state-space representation:

Wo(s)/Vp(s) = 24

s2+151.32 s+198

ẅo + 151.32ẇo + 198wo = 24Vp(t)

then we define x1 = wo x2 = ẇo

thus, the state – space equation are: ẋ1 = x2ẋ2 = -198x1 – 151.32x2 + 24Vp(t)

writing the above in vector-matrix form we get:

Ẋ = [ 0 1−198 −151.32]X + [ 024]Vp(t)

Y = [1 0 ]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