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Control and PLC Essay

(a) We have to find out transfer function of permanent magnet DC motor. We have two types of control for DC motors, that is, field control and armature control.  A permanent magnet DC motor can only be armature controlled as there is no control on field. We can take the Laplace transform of the given equations as given below assuming zero initial conditions.

This equation relates torque developed by the motor to the armature current. We take its Laplace transform and get

This equation relates back emf to the angular velocity. Its Laplace transform will be as given below.

This equation gives the relation between applied voltage, back emf, armature resistance, armature inductance and armature current. We get this equation by writing KVL across the armature circuit. Its Laplace transform will be as given below.

Or,

This equation gives the relation between the torque developed by the motor, load torque and angular position, moment of inertia and fiction torque. Actually, we can get angular velocity also from this equation. Its Laplace transform will be as given below.

Or,

Angular velocity and angular position are related by the equation

Its Laplace transform gives

We see that  is Laplace transform of angular velocity.

From the above Laplace transforms, we can develop the

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following block diagram.

                                                          Fig.1 Block diagram

(b) The following table gives various symbols and their units.

Sl.no.
Symbol
Unit
1
Ea
Volts
2
Eb
Volts
3
Ra
Ohms
4
La
Henrys
5
Ia
Amperes
6
Km
Newtom-meter/amp
7
T
Newton-meter
8

Newton-meter/rad/sec
9

radians
10
Kb
Volts/rad/sec
(c) From the block diagram, we get the following forward path transfer function.

(d) Feedback path transfer function is given by

(e) From the block diagram, we get

We have to find out closed loop velocity transfer function. This will relate angular velocity to applied voltage. We know that

Hence,

                                                           as Kb = Km in MKS system.

This is the required closed loop transfer function.

We see the denominator will have maximum power of s as 2. Hence, this will be second order system.

(f) We find out the step response of the above transfer function if applied voltage is 240 volts.

We write the following Matlab program. As the applied voltage is 240 volts, we will have product of 240 and 0.12 in numerator.

I=1.8;

B=0.6;

K=0.12;

R=0.6;

L=0.012;

num=240*K;

den=[ (I*L) ((I*R)+(L*B)) ((B*R)+(K^2))];

step(num,den,0:0.1:60)

We get the following step response by running the above program on Matalb.

                                                        Fig.2  Step response

The above plot is for 60 seconds but we can get for any time by putting up the

required time in program.

 

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