# Homework Help: Analysis of a servo motor

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1. Dec 15, 2017

### Granger

1. The problem statement, all variables and given/known data
2. Relevant equations

I'm studying a mathematical behaviour of a servo motor and I need some help to understand it.

The output signal is \$\beta (t)\$, representing the angle rotated by the axis at instant t, in relation to the equilibrium position.
On the servomotor there are two torques:

- the first torque is produced by the electric current \$i(t) \$ that does through the motor. It is proportional to the electric current being given by \$Ki(t)\$.
- the other torque is given by friction and is given by \$-2 \beta'(t)\$ where \$\beta'(t)\$ is the derivative of \$\beta (t)\$, meaning it represents the angular velocity.

The feedback control system is given by:

[![enter image description here][1]][1]

Where Motor represents the differential equation
\$4\beta''(t)=100i(t) - 2 \beta'(t) \$

And \$\alpha(t)\$ is the input angle that give the desired position (angle) for the servomotor.

B is unknown but we will assume A = -1 so that if the input is constant we'll have the output equal to the input (unitary static gain).

The transfer function of the system is:

\$H(s)= \frac{100}{4s^2+(2-100B)s-100A}\$

By this expression we conclude that our system is a second order system with no zeros.
By controlling the value of B we will control where the poles are in our system and by that the type of system we get. More specifically:

- If \$B<-0.38\$ we have two real poles (and an overdamped system)
- If \$B=-0.38\$ we have o double pole (critically damped system)
- If \$-0.38<B<0.02\$ we have two complex conjugated poles (underdamped system).

So until this I don't have questions and everything makes sense to me.
The next affirmation is what leaves me doubtful:

- By controlling the value of B we are controlling the friction of the system.

3. The attempt at a solution
No more explanation is given to this and I'm kinda worried if my interpretation of this fact is correct or not. Here it goes:

- We have a 2nd order system without zeros. By controlling B we will control the type of poles our system will have. For low values (in module) of B we have an underdamped system. For B=-0.38 the system is critically damped. For B<-0.38 (meaning, large absolute values of B) the system will be overdamped. The system damping is related to the existence of friction in the motor. For low absolute values we will have little friction and so the system will oscillate a lot until it reaches its final value. As we increase B, we will increase friction which will oppose to the natural oscillatory tendency of the system, making it monotonous until it reaches its final values. We have to be careful however because as we increase the friction the system will also take more time to reach the value (it has to "beat" the friction).

Is this interpretation correct? I'm not sure if this is the correct relationship between friction and the type of system I get? I'm also curious if there is any more implication or consequence of this relationship. Can anyone clarify me please?

Thanks!

2. Dec 15, 2017

### donpacino

your image is messed up, we cannot see it.
That being said, I can guess from your transfer function what it is.

Your interpretation is very close, but there are a few things wrong or slightly off.

First of all its generally better to talking about your damping ratio (zeta) than your beta values. This will scale the damping relative to your dc offset.

so it becomes

- If zeta>1 we have two real poles (and an overdamped system)
- If zeta=1 we have o double pole (critically damped system)
- If zeta<1 we have two complex conjugated poles (underdamped system).

The above statement will be true for any 2nd order system with no dominant zeros. With beta, as your transfer function changes from controllers, the beta value required for the system to be critically damped with change.

The second thing, what you are calling friction... If in your paragraph you replace 'friction' with 'viscous friction, motor moment of inertia, electromotive force, and torque constant" then it is correct. There are many more items at play than friction. take a look at the below example of how the motor characteristics are defined. The important bits are equation 7 (fill in numbers for the variables, look familiar???) and the physical parameters defining the variables.

http://ctms.engin.umich.edu/CTMS/index.php?example=MotorSpeed&section=SystemModeling

The field is called control SYSTEMS. Everything the motor touches (the system) will effect its performance. Increasing friction would cause a change in the characteristics, as well as changing magnetic fields, changing the motor build, etc. Overall though I'd say you have a decent grasp, you just need to refine a few thoughts you have on the subject.

Last edited: Dec 15, 2017