11
\$\begingroup\$

Can someone please explain, provide a link or cite a book where the properties of the zeros for continuous and discrete time systems are explained? I know that the zeros are the frequencies where the numerator of a transfer function becomes zero.

$$ H(s) = \frac{A(s)}{B(s)} $$

But I would like to know what role the location plays in the pole-zero plot? All I can find are pole-zero plots and that basically the poles define the system stability and time response. However, what are the zeros "doing"? What happens if the zeros are in the right or left half plane? Are the zeros describing the damping or also stability?

Here is a link to a pdf of MIT explaining the pole zeros. However, I am missing details about zeros.

\$\endgroup\$
10
  • \$\begingroup\$ I found that too already. I am looking for a pole zero plot where the locations for zeros get explained. But it doesn't give much information about the zeros either. What happens if a zero lies in the right half plane? They describe excess of poles and zeros and what happens if zeros are on/close the imaginary axis and at zero. \$\endgroup\$
    – fjp
    Mar 10, 2016 at 1:15
  • \$\begingroup\$ Are there also unstable zeros just like unstable poles? \$\endgroup\$
    – fjp
    Mar 10, 2016 at 1:16
  • \$\begingroup\$ It might help to look at the sections on drawing Bode plots. These give an intuitive explanation of the effect of the zeros in a continuous time system. \$\endgroup\$
    – Chris Hansen
    Mar 10, 2016 at 1:48
  • 1
    \$\begingroup\$ there are zeros that can be located in the same region as unstable poles (that is in the right-half s-plane or outside the unit circle in the z-plane). but when zeros are out there, it doesn't cause the system to be unstable. it does cause it to be non-minimum-phase, though. so both zeros and poles have to be in the left half s-plane or inside the unit circle in the z-plane for the system to be both stable and minimum phase. and a minimum-phase system can be inverted (which causes swapping of poles and zeros) and be stable. not so with a non-minimum-phase system. \$\endgroup\$
    – robert bristow-johnson
    Mar 13, 2016 at 3:51
  • 1
    \$\begingroup\$ @Alvaro , i just now say your 10-week-old question. You can have a state-variable system where the input-output transfer function looks stable (no poles in the right half s-plane) but internally is unstable because a pole that exists in the right half-plane was canceled by a zero. you can have a 3rd-order system with two stable poles and one unstable pole that is canceled by a zero. there are 3 states in this system. put it into a black box and it might appear stable at first, but internally some state inside is going to hell. \$\endgroup\$
    – robert bristow-johnson
    May 10, 2019 at 22:35

4 Answers 4

Reset to default
6
\$\begingroup\$

1)zeros with positive real part give a negative phase contribution, reducing the phase margin (which is bad) thus limits the performance of the system.

2)Time delay in the system can also be approximated as a zero with positive real part (see first order Pade approximation 1), similar effect as previous point.

3)Blocking property of zeros, If you have a transfer function with a zero in the right hand plane, and an input tuned to that zero, then the output is at 0 for any time t. Example: enter image description here Proof for blocking property of zeros:3

\$\endgroup\$
1
  • 1
    \$\begingroup\$ I like this answer because it lists multiple effects of a zero rather than focussing in on one (such as pole cancellation). To me, the blocking property (the effect on the amplitude ratio at certain frequencies) is important and easy to understand yet others haven't mentioned it. \$\endgroup\$
    – Bill
    Jul 3, 2020 at 17:09
6
\$\begingroup\$

There are zeros that can be located in the same region as unstable poles (that is in the right-half \$s\$-plane or outside the unit circle in the \$z\$-plane). But when zeros are out there, it doesn't cause the system to be unstable. It does cause it to be non-minimum-phase, though.

enter image description here

So both zeros and poles have to be in the left half \$s\$-plane or inside the unit circle in the \$z\$-plane for the system to be both stable and minimum phase. And a minimum-phase system can be inverted (which causes swapping of poles and zeros) and will continue to be stable. That is not the case with a non-minimum-phase system. If one inverts a non-minimum-phase system, the result will have poles in the unstable region and will be unstable.

\$\endgroup\$
8
  • \$\begingroup\$ Just as a fair warning, recognition may never come, but out of all answers, this is the one that directly addresses OP's question. \$\endgroup\$
    – a concerned citizen
    Nov 30, 2018 at 20:15
  • \$\begingroup\$ @robert I have one question. If a pole is placed in the right side of the s plane and there is a zero that cancels that pole, can it be considered that this zero may affect the stability of the system? – \$\endgroup\$
    – Alvaro
    Feb 28, 2019 at 11:07
  • 1
    \$\begingroup\$ @Alvaro, from the POV of the input-output relationship (pretend the system is in a black box and all you can see is the input and output) the cancellation of the pole makes the system stable. perhaps the way you cancelled the unstable pole was replace it with a wire inside. but it's possible that your system internally is going to hell when there is pole-zero cancellation and it was an unstable pole that was cancelled. suppose you had an unstable 1st-order filter and that followed by a zero that cancels the pole. it might look okay on the outside, but inside it is blowing up. \$\endgroup\$
    – robert bristow-johnson
    Feb 28, 2019 at 20:17
  • 1
    \$\begingroup\$ I'm in the same boat as the OP. I find a lot of info on poles in relation to stability, but I want to know more about the importance of zeros. I think the question might be too broad to get a single answer here. You provide some valuable insights in your answer, but is that the only significance of zeros? It seems like there is more to the story? It sounds like you might know of some good references to follow up on? \$\endgroup\$
    – ThatsRightJack
    May 10, 2019 at 20:35
  • 1
    \$\begingroup\$ @ThatsRightJack , i s'pose Oppenhiem and Schafer. but there might be a good reference on-line. so do you know what All-pass filters are? do you know what pole/zero cancellation is? an non-minimum phase filter can be thought of as a minimum phase filter (having the same magnitude response) cascaded with an all-pass filter that has poles cancelling some of the zeros and introducing zeros in that "unstable" region (which is the right half of the \$s\$-plane or outside the unit circle in the \$z\$-plane. \$\endgroup\$
    – robert bristow-johnson
    May 10, 2019 at 20:42
1
\$\begingroup\$

Zeros are very import for the system behavior. They influence the stability and the transient behavior of the system. The referenced document is a good start.

When dealing with transfer functions it is important to understand that we are usually interested in the stability of a closed loop feedback system. In order for the closed loop system to be stable, the poles have to be located in the left half plane. The zeros have no importance, since the stability of a linear system is solely determined by the position of the poles.

When designing a closed loop system (i.e. a circuit), this is usually done by analyzing the open loop system. Because for the open loop system it is easier to understand how the circuit parameters are going to influence the system behavior.

It can be shown that the position of zeros of the open loop system are important for the stability of the closed loop system. When closing the loop slowly by increasing the feedback while monitoring the poles, it can be seen that the poles are attracted by the zeros. The poles move towards the zeros and if there are zeros in the right half plane, the tendency for the system to become unstable is higher because finally the pole will assume the position of the zero. Such a system would be called a non-minimum phase system, and they are quite common.

\$\endgroup\$
15
  • 1
    \$\begingroup\$ zeroes do not influence the stability or the asymptotic transient behavior of the system unless they precisely cancel a pole. but even if there is pole/zero cancellation, an unstable pole that is canceled by a zero will still result in some internal states going to hell inside the system. \$\endgroup\$
    – robert bristow-johnson
    Mar 13, 2016 at 3:53
  • \$\begingroup\$ No, you are wrong. Please reread my answer and pay attention to the fact that the discussion is about open loop vs. closed loop behavior. \$\endgroup\$
    – Mario
    Mar 13, 2016 at 6:12
  • \$\begingroup\$ i read your answer before making my comment. i am not wrong. your system with the feedback is a different system. systems are unstable if they have poles in the right half-plane (s) or outside the unit circle (z). systems are stable if all of their poles are in the left half-plane or inside the unit circle. that's it. there is no other attribute that determines stability. \$\endgroup\$
    – robert bristow-johnson
    Mar 13, 2016 at 7:51
  • \$\begingroup\$ @robert bristow-johnson No, once again. If the system is to be used in a feedback configuration we have to take care about the zeros. You obviously don't have a background in circuit design so this concept is not familiar to you. \$\endgroup\$
    – Mario
    Mar 13, 2016 at 8:02
  • \$\begingroup\$ I suppose the discrepancy between both positions comes from the fact that open-loop and closed -loop zeros are mixed. \$\endgroup\$
    – LvW
    Mar 13, 2016 at 10:39
1
\$\begingroup\$

all of the answers are correct but one subject is missing: zero in the right-hand side of the s plane can cause undershoot in the time response of the system, and this can be very very dangerous in some cases.

\$\endgroup\$
1
  • \$\begingroup\$ I do think this effect is worth mentioning. Good article on it here: Nonminimum-Phase Zeros \$\endgroup\$
    – Bill
    Jul 3, 2020 at 18:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.