# POLE ZERO and its significance

Discussion in 'Homework Help' started by syed_husain, Aug 11, 2010.

1. ### syed_husain Thread Starter Active Member

Aug 24, 2009
61
5
hello everybody,

i have a problem understanding the significance of pole-zero in transfer function especially the pole. i understand the mathematical concept of it but in the sense of electrical circuit how we interpret pole and zero?

thnx

2. ### Papabravo Expert

Feb 24, 2006
11,064
2,152
In the vicinity of a zero the output of the system is very much less than the input. Zeros occur in the numerator of the transfer function.

In the vicinity of a pole the output of the system is very much greater than the input. Poles occur in the denominator of the transfer function.

The location of the pole or the zero in the s-plane also has a qualitative effect on the output of the system.

A solitary pole on the negative real axis corresponds to a decaying exponential in the time domain. If the pole is on the positive real axis it corresponds to an exponential that increases without bound.

Poles located off the real axis always occur in conjugate pairs. In the left half-plane they correspond to decaying exponentials, on the jω-axis they correspond to oscillations, and in the right half plane the systems destroy themselves.

"The positive real roots must disappear or the system will"

3. ### Ghar Active Member

Mar 8, 2010
655
73
I don't really agree with that.

A pole is the root of the denominator but it's a complex variable whereas signals are real.
When you're at a pole in a first order system you're at 1/Sqrt(2) of the input, it's not a gain.

Similarly with a zero you won't actually be at zero output.

These things get more complicated with higher order systems.

I consider poles and zeros as the frequencies when the response starts to behave differently, due to different components emerging relative to others.
In a low pass RC circuit at very low frequencies the capacitor is essentially an open, it doesn't do anything. Eventually it starts to conduct, eventually almost shorting out. The pole is the frequency at which the capacitor starts to make a noticeable difference by reducing the output amplitude.

In a damped series RLC circuit at very low frequencies the capacitor is a very high impedance, there is no current, which means you have a zero in the transfer function for current. At some frequency the resistor starts taking over and you have approximately I = V/R. This frequency is a pole.
Eventually the inductor starts reducing current since its impedance is increasing. This marks another pole.

4. ### Papabravo Expert

Feb 24, 2006
11,064
2,152
Of course poles are complex variables. What do you think a "conjugate pair" refers to?? Boston Cream Pie?

5. ### Ghar Active Member

Mar 8, 2010
655
73
Don't give me this nonsense.
The point is that no, the output of the system is not very much greater than the input at a pole. Either you misspoke or that's the Boston Cream Pie kind of pole.

6. ### Papabravo Expert

Feb 24, 2006
11,064
2,152
So tell me. If you multiply the transform of the input by a transfer function evaluated near a pole are you telling me that you get something the same size or smaller?

7. ### Ghar Active Member

Mar 8, 2010
655
73
Rather than mix math here, let's go to something clearer.

First order RC low pass filter.
It has one pole.

Does it ever have a gain greater than 1?

8. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
In my field of study, Control Systems, poles and zeros have significant values and it is often useful to represent them in the s-plane in order to examine the response characteristics of the system at hand. So, system-wise, I would agree with Papabravo. That is because we often write the transfer functions in the pzk (Pole-Zero-Gain) form:
$K\cdot \frac{(s-z1)(s-z2)\cdots (s-zn)}{(s-p1)(s-p2)\cdots (s-pn)}$
This form makes it obvious how the output is nullified on the zeros and destabilized on the poles.
However, in the low pass filter case, isn't the pole minus 1/RC? Even if the bode diagram "breaks" at the frequency of 1/RC, this isn't the actual pole.

I haven't mastered the art of the Bode Plot yet, so I can't be absolute here.

9. ### Papabravo Expert

Feb 24, 2006
11,064
2,152
I was never talking about the gain in the time domain. I was always talking about the frequency domain. The transformed input F(s) times the transfer Function H(s) gives the output function C(s). Clearly the values of C(s) become large in the vicinity of poles that are not canceled by zeros.

10. ### Ghar Active Member

Mar 8, 2010
655
73
The OP clearly asked for significance of poles and zeros in electronic circuits.
You spoke of inputs and outputs of the system yet clearly you were not referring to circuit inputs and outputs.

Again you make it misleading, gain in the frequency domain is equivalent to gain in the time domain for signals. What you're talking about is the transfer function when you manage to set s equal to the pole.
How do you actually excite the transfer function like this in a physical system? How does this ever manifest itself as a gain?

You can say the transfer functions becomes infinity at a pole all you want but what does that even mean? That is not applying an input to the system.

Poles are the roots of the characteristic equation of the differential equation you're solving.

@Georacer:
Of course he's technically correct but re-read the first post - he explicitly wrote "i understand the mathematical concept of it" and the point both of you are making is almost purely mathematical (and hence I still think it's misleading).

Last edited: Aug 22, 2010
11. ### Flow Member

May 30, 2010
37
1
Have you learnt to construct bode diagrams out of more complex transfer functions yet? Or the other way - to construct a transfer function out of a bode plot? That's where your answer lies, at least in the frequency domain.

Analogue to the frequency domain, poles and zeros have an impact in the time domain, e.g. have an impact on responses of the system to certain inputs e.g. impulses, steps, sinusoidals etc. etc. .

12. ### syed_husain Thread Starter Active Member

Aug 24, 2009
61
5

yea i just started to learn to draw bode plot from a transfer func but not the other way round. i will try that later after i master the previous.

btw, cud u be more specific what do u mean by saying that pole zero will effect on the system inputs?

thnx.

13. ### Flow Member

May 30, 2010
37
1
The output of an LTI system depends on the input signal and the system itself. There are two ways to analyze the system; either you look at it in the frequency domain, or in the time domain. Both are different sides of the same coin. With either the system is fully described.

A pole zero will therefore affect the frequencies that will "pass through" it, and it will affect timing characteristics such as the systems impulse/step/sinusoidal/whatever response to such an input.

Is it clear now?