# Poles and Zeros of a Transfer Function

I am trying to understand the physical meaning of poles and zeros of a transfer function. Can someone help me how to understand the Figure 2b in the link below?

What does the height of the cone indicate? And what do the different color rings in the cone indicate?

TIA.

• Link5 all might give an insight. Note - the height of the cone is infinity. Jul 22 '19 at 11:35
• thank you. one more doubt. this question is asked in the link but i was not able to get the answer. suppose, if i have a pole, at say, s=-1, how will i be able to determine the frequency with this data? like what would be the value of the frequency? Jul 22 '19 at 11:37
• You might get more help if you are able to pay-the-man i.e. go back to your other questions you have raised and formally accept the answers that have best helped you. That's the "quid pro quo" in this neck of the wood. Jul 22 '19 at 11:40
• BTW - determining the frequency of a transfer function is nonsense. You can determine the TF's amplitude if you know the frequency. Jul 22 '19 at 12:02

can you explain in simpler terms with explanation, please

If you have a very simple low pass filter made from a resistor (R) and a capacitor (C), you can calculate the transfer function (TF) as being: -

$$\dfrac{1}{1+sCR}$$

Then, if you re-arranged that TF you could get: -

$$\dfrac{\frac{1}{CR}}{s + \frac{1}{CR}}$$

Now the important thing to realize is that if s = $$\\frac{-1}{CR}\$$ the whole TF has a value of infinity. This is the position of the pole.

Take a few minutes to think about that because it is fundamental to understanding how the bode plot and pole-zero diagram are related mathematically.

For this particular simple example, that pole sits purely on the real axis of the s-plane where $$\\sigma = \frac{-1}{CR}\$$. This isn't the vertical $$\j\omega\$$ axis. The $$\j\omega\$$ axis is where the bode plot exists.

Considering the pole; at any point distant from that pole, the amplitude is not infinity and, at any point along the $$\j\omega\$$ axis you can predict the amplitude of the bode plot by using the distance from the plot and taking the reciprocal. However, that reciprocal has to factor in the natural frequency of the circuit. For this simple circuit, the natural frequency is 1/CR.

So, using a simple example at the origin of the s-plane (0, 0), the TF amplitude is the reciprocal of 1/CR divided by the scaling factor (1/CR) and in this pretty trivial example this works out at 1.

So the TF has an amplitude of 1 at DC. I say "DC" because the value of the $$\j\omega\$$ axis is 0 at the origin and this means 0 Hz or "DC".

If you were to move up the $$\j\omega\$$ axis by an amount equal to 1/CR, the distance from the pole becomes $$\\sqrt2\$$ times bigger and hence the TF amplitude becomes $$\\frac{1}{\sqrt2}\$$. This is normally called the 3 dB point because, in decibels the amplitude has fallen by approximately 3 dB. It's also called the half power point.

What does the height of the cone indicate?

The height of the cone is infinity and, as such doesn't really tell you anything useful.

And what do the different color rings in the cone indicate?

The coloured rings are arbitrary and don't tell you anything useful.

• Aha a downvote, maybe there is a reason? Jul 23 '19 at 19:33