Antenna Azimuth Position Control System with answers PDF

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Antenna Azimuth Position Control System Layout

Schematic

Block Diagram: Desired azimuth angle

θi (s)

Potention meter

Preamplifier

Power amplifier

V i (s ) +

Motor and load

Azimuth angle V e (s) Vap(s) E (s)

Gears

θm (s)

θo (s)

_ Potentionmeter

Transfer Functions: 1. Potentionmeter: 5 turns towards either the positive 10 volts or negative 10 volts yields a voltage change of 10 volts. V i (s) 10 1 = = θ i (s) 10 π π 2. Preamplifier: V p (s) =K V e (s) 3. Power amplifier: E a (s ) 100 = V p (s) s +100 4. Motor and Load: K t /( Ra J m ) θm (s) = E a (s) K K 1 Dm + t b s s+ Ra Jm

[ (

)]

DC Motor schematic:

θm (s) =? E a (s)

Armature rotating in a magnetic field, its voltage is proportional to speed: v b ( t )=K b

d θ m (t) dt

->

V b ( s ) =K b s θm( s)

Use impedance method to get the equation for armature circuit:

Torque developed by the motor is proportional to the armature current:

Typical equivalent mechanical loading of a motor:

DC motor driving a rotational mechanical load:

Schematic Parameters: Parameter

Configuration 1

Configuration 2

Configuration 3

V

10

10

10

n

10

1

1

K

-

-

-

K1

100

150

100

a

100

150

100

Ra

8

5

5

Ja

0.02

0.05

0.05

Da

0.01

0.01

0.01

Kb

0.5

1

1

Kt

0.5

1

1

N1

25

50

50

N2

250

250

250

N3

250

250

250

JL

1

5

5

DL

1

3

3

For configuration 1 parameters, the block diagrams are as following:

Desired azimuth angle

θi (s)

Potention meter

Preamplifier

V i (s1) +

Power amplifier

Motor and load

100 s+100

K

π _

Gears

Azimuth angle

θm (s) 2.083 V e (s)θ (s) o s (s +1.71

Eap(s) V (s)0.1

1 π Potentionmeter

Using simplification of block diagrams (please refer to week 7 classroom exercise solutions on BB), we can get the following transfer functions: Desired azimuth angle

θi (s)

6.63 K s (s +1.71)( s+100)

Azimuth angle

θo (s)

First sketch the root-locus and find the preamplifier gain required for 25% overshoot. This we can do using Matlab.

>> s=tf('s') >> G=1/(s*(s+1.71)*(s+100)) >> rlocus(G) >> axis([-4 4 0 0]) z=-log(0.25)/sqrt(pi^2+(log(0.25)^2)) z= 0.4037 >> sgrid(z,0)

RootLocus 10 0. 404 8

I magi nar yAxi s( seconds-1)

6 4 2 0 2 4 6 8 0. 404 10 4

3

2

1

0

1

1) Real Axi s( seconds

Now find the K to give 25% overshoot: [k,poles]=rlocfind(G) Select a point in the graphics window selected_point = -0.8294 + 1.8478i k= 411.2239 poles = 1.0e+02 * -1.0004 -0.0083 + 0.0185i -0.0083 - 0.0185i

2

3

4

Therefore 6.63K=411, and K=62. Do a simulation using the command, >> sys=feedback(k*G, 1) sys = 411.2 ------------------------------s^3 + 101.7 s^2 + 171 s + 411.2 Continuous-time transfer function. >> step(sys)

We can see the step response is in the following figure and 25% overshoot is met.

St epResponse 1. 4

1. 2

Ampl i t ude

1

0. 8

0. 6

0. 4

0. 2

0 0

1

2

3

4 Ti me( seconds)

Lag-Lead Compensation:

5

6

7

8

In the above example we obtained 25% overshoot by adjusting the gain K, and the corresponding settling time is determined. If we want to improve the settling time and steady-state error performance further we have to use a cascade compensation. Next, we are going to design cascade compensation to meet the following requirements: (1) 25% overshoot, (2) 2-second settling time, (3). Kv=20.

Steps: I.

Where should the new closed-loop poles be:

The original dominant closed-loop poles are at -0.83 +/- 1.85i . The desired closed-loop poles are at ? :

II.

Design lead compensator to improve transient response, using angle criterion. How much angle needs to compensate?

Choose a pair of zero and pole to meet the angle criterion.

III.

Design Lag compensator to improve the steady-state response. Current Kv with lead compensation?

New zero and pole location for lag compensator?

IV: Determine the new gain K using new root locus or gain criterion....