Presented By:

Junaid Iqbal ( 2010-EE-42 ) (3rd Year Electrical Engineering) University College of Engineering & Technology BZU Multan

Antenna Azimuth Position Control

System

“ System Concept ”

“ Textbook Schematic ”

“ Textbook Schematic Parameters ”

“ Block Diagram ”

“ Parameters ”

“ Modeling in the frequency domain ”

Subsystem Input Output

Input potentiometer Angular rotation from user, θi(t)

Voltage to preamp, vi (t)

Preamp Voltage from potentiometers,ve(t) = vi (t) – v0(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, θ0 (t)

Output potentiometer Angular rotation from load, θ0(t)

Voltage to preamp, v0(t)

Input Potentiometer; Output Potentiometer:

Vi (s)/θi (s) = 10/10π = 1/π

Pre-amplifier:

Vp (s)/Ve (s) = K

Power Amplifier:

Ea (s)/Vp (s) = K1/(s+a) = 100/(s+100)

Motor and load: Transfer function of motor from armature voltage to

armature displacement

θm(s)/Ea(s) = Km / s(s+am) = 2.083/s(s+1.71)

Transfer function relating load displacement to armature voltage

θ0(s)/Ea(s) = gear ratio* [θm(s)/Ea(s)]

= N1/N2* [2.083/s(s+1.71)]

Given that N1=25 and N2=250. So,

θ0(s)/Ea(s) = 0.1* [2.083/s(s+1.71)]

= 0.2083/s(s+1.71)

“ State-Space Representation ”

We fill find state-space representation of each dynamic subsystem in the azimuth position control system.

“Dynamic” means that system don’t reach its steady state immediately and it is described by the differential equation of 1st order or higher order.

Potentiometers and pre-amplifiers have pure gains and are non-dynamic.

Power amplifiers and motor–load subsystem are dynamic.

Power Amplifier:

G(s) = Ea(s) / Vp(s) = 100 / (s+100)

(s+100)Ea(s) = 100Vp(s)

sEa(s) + 100Ea(s) = 100Vp(s)Taking inverse Laplace Transform we get differential equation as:

dea /dt + 100ea = 100vp(t)So, state equation becomes:

dea /dt = -100ea + 100vp(t)Since output is ea(t) so the output equation is:

y = ea

Motor and Load:

Now find state space representation for load and motor

Ea(t)=Ia(t)Ra + Kbdθm/dt

Torque Tm(t)=K 1i(t)=Jmd2θm /dt2 + Dmdθm /dt

Now further solving we get output equation y=0.1x1

“ Antenna Control: Open Loop Response ”

Poles are used in determining the transient response of the system.

The open loop function we will deal with consist of power amplifier and motor with no load.

Differentiating the angular position of the motor and

load out put by multiplying ‘s’ , we obtain the output angular velocity ω0 .

Using the Transfer function we can predict the nature of the step response.

The step response consists of the steady-state response generated by the input.

Transient response is the sum of two exponentials generated by each pole of the Transfer Function.

The damping ratio and natural frequency of the open loop can be found by expanding the denominator.

First convert the transfer function into state space representation.

Cross multiplying and taking inverse Laplace transform with zero initial conditions.

Finally output equation y = x1

“ Closed loop transfer function using block diagram reduction ”

“ Signal-Flow Graph Representation ”

The original block diagram representation is given below:

In order to obtain the signal-flow graph of each subsystem, we use the following state equations of power amplifier and motor and load which were derived previously.

For Power amplifier, the state equation is:

For motor and load, the state equation is:

So, the signal-flow graph of above block diagram representation is:

“ Closed-Loop Transfer Function using Mason’s

Rule ” Consider the following Signal-Flow Graph representation of

antenna azimuth position control system.

1st Step: Find the forward-path gain:

Here k = no. of forward paths = 1. So, Tk = T1

T1 = (1/π)(K)(100)(1/s)(2.083)(1/s)(1/s)(0.1) = 6.63K/s3

2nd Step: Identify the Closed-Loop Gains

There are 3 closed loops whose gains are as follows:Power amplifier loop gain = GL1(s) = -100/s

Motor loop gain = GL2(s) = -1.71/s

Entire system loop = GL3(s) = (K)(100)(1/s)(2.083)(1/s)(1/s)(0.1)(-1/π) = -6.63K/s3

3rd Step: Find the non-touching

loop gain:Only GL1(s) and GL2(s) are non-touching loops.

So, non-touching loop gain becomes:GL1(s)GL2(s)=171/s2

4 Step: Find the values of ∆ and ∆k :

Since,∆ = 1 - sum of loop gains + sum of non-touching loop

gains taken two at a timeSo, ∆ = 1 – [GL1(s) + GL2(s) + GL3(s)] + [GL1(s)GL2(s)]

∆ = 1 + 100/s + 1.71 /s + 6.63K/s + 171/s2 Now, ∆K= ∆1= ∆ - sum of loop gains that touch the k th

forward path gains. So, ∆1 = 1

Now, put the above values in Mason’s Rule given as:Closed-Loop transfer function = T(s) = C(s)/R(s) = T1∆1/∆

T(s) = 6.63K/(s3 + 101.71s2 + 171s + 6.63K)

“ Finding the values of Tp, %OS, Ts ”

Consider again the block diagram representation of antenna azimuth position control system.

Replacing the power amplifier gain with unity and letting the pre-amplifier gain K equal to 1000, we find G(s) and closed loop transfer function T(s) as:

G(s) = 66.3/s(s+1.71) and T(s) = 66.3/s2 + 1.71s + 66.3

For Tp:

For %OS:

For Ts:

From above equation of T(s) we get ωn= 8.14 and damping ratio = 0.105. Putting these values in following equations we get values of Tp, %OS, Ts .

Hence we get, Tp = 0.388s, %OS = 71.77% and

Ts = 4.68s.

We found above that closed loop transfer function T(s) = C(s) / R(s) = 66.3/s2 + 1.71s + 66.3

From above equation we get, C(s) = R(s)* [66.3/s2 + 1.71s + 66.3] To find the step response we take input signal as 1/s So, the output becomes: C(s) = 66.3/s(s2 + 1.71s + 66.3) Expanding the above equation by partial fractions we

get, C(s) = 1/s – (s+0.855)+0.106(8.097)/[(s+0.855)2+(8.097)2] Taking the inverse Laplace Transform we get, c(t) = 1 – e - 0.855r (cos8.097t + 0.106sin8.097t)

“ Finding the value of K that gives a specific value of

%OS ” Consider again the following block diagram:

Given that power amplifier gain= unity. So,G(s)=0.0663K/s(s+1.71) and T(s)=0.0663K/s2+1.71s+0.0663K

We have to find the value of K for 10% Overshoot.

From the following equation, a 10% Overshoot yields a damping ratio of 0.591.

Using the denominator of T(s) we get ωn= (0.0663K)1/2 and a=1.71

Thus damping ratio = a / 2ωn= 1.71/2*(0.0663K)1/2 = 0.591

from which K=31.6

“ Finding the Stability of antenna azimuth position

control system ” The antenna azimuth position control system is

represented by block diagram as:

Its Closed-Loop transfer function was found as:

The ‘K’ in the above expression is the pure gain of a non-dynamic system which is pre-amplifier.

As pre-amplifier is non-dynamic which means that it reaches the steady state instantaneously.

So the gain ‘K’ of pre-amplifier determines the stability of the closed-loop system.

When the loop gain is changed then there is a possibility that the location of poles may also change and go into right half plane.

So, proper gain selection is essential for the stability of closed loop systems.

To determine the value of ‘K’ we make the Routh table as follows:

There will be no sign changes in the first column if 0<K<2623. System will be stable on the value of ‘K’ in this range.

Thanks!