# Transfer function based question

1. Mar 30, 2015

### Dhruv

1. The problem statement, all variables and given/known data
In order to increase number of poles in system we need to include
A) zero at origin
B) zero at infinity
C) pole at origin
D) pole at infinity

2. Relevant equations

3. The attempt at a solution
Aren't the options linked ? I mean zero at infinity means a pole at origin and ques is already giving pole at origin as an option (c) while pole at infinity is itself a zero at origin I guess.

2. Mar 30, 2015

### Hesch

I assume that the question concerns number of poles/zeroes in the controller of the system ? ( The number of poles/zeroes in the whole system will be affected by that).

But which domain are you referring to ?

Laplace domain (analog control) ?

z-domain (digital control) ?

Last edited: Mar 30, 2015
3. Mar 30, 2015

### donpacino

If it were z-domain he would say, zero at orgin or zero at mag of 1, etc

4. Mar 30, 2015

### donpacino

in what way is a zero at infinity a pole at the origin?

(s/inf+1) != 1/s

5. Mar 30, 2015

### rude man

Noned of the above.,
I can add as many poles as I want to without adding any poles or zeros at the origin.

6. Mar 30, 2015

### Dhruv

An
any value of s that make transfer function go to zero is its zero correct? Then if I have a pole at origin and if I keep s = infinity then my transfer function value will will become zero which means there is a zero at infinity.

7. Mar 30, 2015

### Dhruv

Ho
how to do that sir ?

8. Mar 30, 2015

### rude man

Give me 10 resistors and 10 capacitors. I can make a transfer function with 10 real and distinct poles with no zeros!
F(s) = 1/Π(s+ai), i = 1 to 10.

9. Mar 30, 2015

### Hesch

I don't understand the question: What is a "system"? Is it a motor to be position controlled? How do you increase the number of poles (not magnetic poles) in a motor? Or is the "system" a motor + a control loop?

The last mentioned makes sense as we can put as many poles into the controller (well, as many real poles, and as many conjugate polepairs) as we want (rude man). That is not a problem. Just solder a number of op-amps, resistors, capacitors into a PCB. That's it.

When we turn on power a problem may arise: Can we make the system stable? Say we have 19 poles at s = 0 and 1 zero at s = -1. Now loop amplification slowly is increased from 0: The rootcurves will immediately explode, also into the right half of the s-plane (which is the unstable area), and 8 of the roots will never return to stable area.

Have I completely misunderstood, what a "system" eventually might be?

Last edited: Mar 30, 2015
10. Mar 31, 2015

### Dhruv

But what will be the answer to this question ?

11. Mar 31, 2015

### milesyoung

If you have some transfer function:
$$G(s) = \frac{N(s)}{D(s)}$$
where $N(s)$ and $D(s)$ are polynomials of degree $n$ and $m$, respectively.

If $G(s) \rightarrow 0$ for $s \rightarrow \infty$, which it does for $m > n$, then $G(s)$ is said to have a zero at infinity of multiplicity $m - n$.

If $G(s)$ is proper, then I think the idea is that you can't add a pole without adding a zero at infinity.

The problem statement is somewhat brief.