Nyquist's Stability Criterion for linear-time-invariant systems makes use of Cauchy's argument principle to determine if any zeros in the characteristic equation are in the right-half plane (positive real roots), given a closed loop transfer function of the form:

$$G_c(s) = \frac{G_f(s)}{1+G_o(s)}$$


$G_f(s)$ is the forward transfer function

$G_o(s)$ is the open loop transfer function

$G_c(s)$ is the closed loop transfer function

$1+G_o(s)$ is the Characteristic Equation

Thus if there are any zeros in the right half plane for the characteristic equation, then there will be poles in the right half plane for the closed loop transfer function and therefore unstable.

Cauchy's argument principle states that if we take any analytic closed contour in the complex plane and plot the resulting function $G(s)$ with $s$ being the complex values along our contour of choice, the number of times the resulting plot encircles the origin in the same direction is equal to the number of zeros minus the number of poles enclosed by the contour. Such as the example image below from user Archibald on Wikipedia: If the five red dots were zeros and four blue dots were poles, the following contour would result in $5-4 = 1$ encirclement counter-clockwise.

enter image description here

(image credit: Wikipedia)

More specifically, for every zero the contour encloses, the result is an encirclement of the origin in the same direction (if the countour is taken clockwise, the encirclement of the origin will be clockwise), and for every pole the contour encloses, the result will be an encirclement of the origin in the opposite direction. Thus if there are the same number of poles and zeros, the net encirclement will be zero.

That said, the argument principle can be used as partial input to determine existence of zeros in the right half plane, by making the contour be the entire right half plane as depicted in the image below. We use a radius $R$ to enclose more and more of the right half plane (RHP), extending R to infinity, and in the limit, the entire RHP is enclosed. Thus by evaluating $G(j\omega)$ with $\omega$ from $-\infty$ to $+\infty$ we are enclosing a contour around the RHP and can therefore use the resulting plot as input toward determining how many zeros the RHP may contain (based on the resulting number encirclements of the origin).

enter image description here

Further, instead of taking the contour around the poles of the characteristic equation $1+G_o(j\omega)$ itself, since $G_o(j\omega)$ is often readily available in measurement (frequency response of the open loop gain), Nyquist identified that we can take the contour around the RHP of $G_o(s)$ (which is $G_o(j\omega)$) and the number of resulting encirclements $N$ around $-1$ in the same direction is equal to the number of zeros minus the number of poles in the RHP for the characteristic equation.

This is because the roots (zeros) of the characteristic equation are found from solving:


Which is equivalent to: $$G_o(s) = -1$$

So the origin simply shifts and instead of encircling the origin as Cauchy's argument principle states, the result encircles $-1$. So Nyqist's Criterion involves determining number of encirclements of $-1$ after evaluating $G_o(j\omega) \forall \omega\in(-\infty,+\infty)$.

The Question:

That said, without elaborating further on the complete process outlined by Nyquist to determine stability (although I gave most of it), here is the question:

We encircle the RHP by taking $j\omega$ from $-\infty$ to $+\infty$ and have convinced ourselves using a semi-circle with radius $R$ extending into the RHP with the limit of $R$ going to infinity and thus have correctly completed a clock-wise contour around the RHP in search of zeros using Cauchy's principle. However that seems somewhat arbitrary to say that positive infinity connects back to minus infinity by route of the right half plane. We could have just have easily convinced ourselves that with the exact same sweep along the $j\omega$ axis we are incorrectly taking a counter-clockwise contour around the LHP (by extending a radius into the left half plane and similarly taking the limit as R goes to infitiny- with no change to our sweep!); in which case we would not have encircled ANY of the zeros that we were looking to find. What is the intuitive explanation for this apparent signal processing mystery?

The Question (stated another way by Tendero)

To draw a Nyquist plot, one could write down a table with different values of $\omega$ and their corresponding values of $|G_o (j\omega)|$ and $\angle{G_o (j\omega)}$, and then draw the phasors with those magnitudes and phases in a complex plane, to finally join all the dots with a continuous line, getting the Nyquist plot. So the curve could be drawn using just the frequency response of $G_o(s)$ (i.e. $G_o(j\omega)$). Therefore, there appears to be no relationship between the Nyquist plot and the way in which we close the contour that includes the $j\omega$-axis: we just said that the curve depends only in the frequency response! How can this be explained?

  • $\begingroup$ I don't know if I understand your question. Why would you take a contour around the LHP if you are looking for zeros in the RHP? $\endgroup$
    – Tendero
    Commented Mar 19, 2017 at 23:13
  • $\begingroup$ I'm not saying I would, I am saying if you consider the logic in how we describe that we are taking a contour around the right half plane when we are sweeping $j\omega$ from $-\infty$ to $+\infty$, you could apply that same logic to show that with the same sweep in the same direction, we are taking a contour around the left half plane, except in that case the contour is going counter-clockwise around the region. So I did not change anything except the description (using the same appraach of applying a radius with a limit)- Yet there is no paradox! $\endgroup$ Commented Mar 19, 2017 at 23:16
  • $\begingroup$ I will try to make my final question a little clearer. $\endgroup$ Commented Mar 19, 2017 at 23:18
  • $\begingroup$ Let's assume that somebody comes to me and shows me a Nyquist plot. Is your question related to how I could be sure that that plot was made via mapping a RHP contour and not a LHP one? Or am I still on the wrong track? $\endgroup$
    – Tendero
    Commented Mar 19, 2017 at 23:32
  • $\begingroup$ Off track- sorry I'm not being clearer. I don't want to explain too much since this is a "puzzle" but review the graph above showing the semi circle around the RHP, and the route we take from positive infinity down to minus infinity via an "infinity path" around the right half plane. That part of the path could just as well been around the left half plane WITHOUT CHANGING ANYTHING except convincing ourselves that we went around the right half plane. $\endgroup$ Commented Mar 19, 2017 at 23:37

3 Answers 3


Let's assume a rational open-loop transfer function $G_o(s)$ of order $N$ with no zeros or poles on the imaginary axis. Let $Z_l$ and $P_l$ denote the numbers of zeros and poles in the left half-plane, respectively, and $Z_r$ and $P_r$ are the numbers of zeros and poles in the right-half plane, respectively. We have $$Z_l+Z_r=P_l+P_r=N$$ and, consequently, $$Z_r-P_r=P_l-Z_l$$ If we use a closed contour moving clockwise from $-jR$ to $jR$ and then on a semi-circle of radius $R$ in the right half-plane back to the starting point ($-jR$), and if we let $R\rightarrow\infty$, then we encircle the complete right-half plane, and the number of clockwise encirclements of the point $-1$ in the $G_o(s)$-plane equals $Z_r-P_r$. However, if we were to move on a semi-circle across the left half-plane (on a counter-clockwise contour), the number of encirclements would equal $P_l-Z_l$, which, as shown above, equals $Z_r-P_r$. So no matter how we close the contour, the result and its meaning is always the same: $Z_r-P_r=P_l-Z_l$. Also note that the value of $G_o(s)$ doesn't change when moving on a semi-circle (of infinite radius); it always equals the limit $\lim_{|s|\rightarrow\infty}G_o(s)$, which is equal to the value of $G_o(j\omega)$ for $\omega\rightarrow\infty$; that's why we only need the values of $G_o(s)$ on the imaginary axis, and the way we close the contour (via the right or left half-plane) doesn't make any difference.

  • $\begingroup$ >! Excellent Matt- A key point that helped me was the number of zeros must always equal the number of poles-- so what is not in the RHP must be in the LHP....so the "opposites" are there and we rotate in the opposite direction, so the result is the same.... $\endgroup$ Commented Mar 20, 2017 at 20:43
  • $\begingroup$ Nice reasoning, that cleared it out for me. One question though: why does the equality $Z_l+Z_r=P_l+P_r$ hold? For example, if $G_o(s) = \frac{1}{(s-2)}$, there are no zeros and there is one pole. I think the equality holds for $1+G_o(s)$, which is the characteristic equation and is in fact what goes inside the complex integral to use Cauchy's argument principle. $\endgroup$
    – Tendero
    Commented Mar 21, 2017 at 14:09
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    $\begingroup$ @Tendero: In your example of $G_o(s)$ there is one pole at $s=2$, but there is also one zero at $s=\infty$. $\endgroup$
    – Matt L.
    Commented Mar 21, 2017 at 16:04
  • $\begingroup$ @MattL When we limit our use Cauchey's Argument in Nyquist's applications to proper G_o(s) only and using it to solve for the roots of $1+G_o(s)$ in the RHP, we will always have the same number of finite poles and zeros as far as I can tell- (his example would be $1+\frac{1}{s-2}$ which is $\frac{s-1}{s-2}$. However I have thought about the case you bring up as I could state that the missing pole must be at infinity in the LHP-- but could not completely make the mathematical connection that forces it to absolutely be there vs RHP besides making Cauchey's Argument work with my mental picture. $\endgroup$ Commented Mar 24, 2017 at 6:16
  • $\begingroup$ @MattL. And in that case, that zero at $s=\infty$ would be part of the RHP zeros or the LHP ones? That's what I don't understand yet. $\endgroup$
    – Tendero
    Commented Mar 26, 2017 at 19:35

After reading Matt's resolution, I understood the puzzle (that was rather tricky by the way). I'll write a more formal demonstration for those who are interested:

Let $\Gamma_L$ be a counter-clockwise contour that embraces the whole LHP. Let $\Gamma_R$ be a clockwise contour that embraces the whole RHP. Finally, let $\Gamma_T$ be a clockwise contour that embraces the whole complex plane. So it is easy to note that: $$\oint\limits_{\Gamma_R}f(s) \ \mathrm{d}s - \oint\limits_{\Gamma_L}f(s) \ \mathrm{d}s=\oint\limits_{\Gamma_T}f(s) \ \mathrm{d}s \ \ \ \ \ \ \forall f(s)$$ If our open-loop transfer functions is $L(s)$, then we want to know the poles of our closed-loop transfer functions, whose denominator is $1+L(s)$. Using Cauchy's argument principle: $$-\oint\limits_{\Gamma_T}\frac{(1+L(s))'}{1+L(s)} \ \mathrm{d}s = Z-P$$ where $Z$ and $P$ are the total number of zeros and poles, respectively, of $1+L(s)$ in the whole complex plane. We can write $1+L(s)$ as follows: $$1+L(s) = 1+\frac{\prod\limits_{m=1}^M (s-z_m)}{\prod\limits_{n=1}^N (s-p_n)} = \frac{\prod\limits_{n=1}^N (s-p_n)+\prod\limits_{m=1}^M (s-z_m)}{\prod\limits_{n=1}^N (s-p_n)}$$ If $L(s)$ is not improper, then $N\geq M$. This means that the grades of the numerator and denominator of $1+L(s)$ are equal, so they have the same number of roots (i.e. $Z=P \implies Z-P=0$). Finally, this leads to: $$\oint\limits_{\Gamma_T}\frac{(1+L(s))'}{1+L(s)} \ \mathrm{d}s = P-Z = 0$$ $$\oint\limits_{\Gamma_R}\frac{(1+L(s))'}{1+L(s)} \ \mathrm{d}s - \oint\limits_{\Gamma_L}\frac{(1+L(s))'}{1+L(s)}\ \mathrm{d}s=0$$ $$\oint\limits_{\Gamma_R}\frac{(1+L(s))'}{1+L(s)} \ \mathrm{d}s = \oint\limits_{\Gamma_L}\frac{(1+L(s))'}{1+L(s)} \ \mathrm{d}s$$ So it is the same to close the contour in either way.

Even though the two integrals lead to the same result, I think that using the LHP one would not be conceptually correct. The result is the same, as has just been proved, but the whole Nyquist plot procedure was made to know whether there are unstable poles, which requires to close the contour via the RHP. The fact that the other contour leads to the same result is by 'pure chance' (kind of).

  • $\begingroup$ Yes very good Tendero, yet there is one more thing I am looking for that you did not mention- I would argue that even though we close the LHP, we can still determine, using Cauchy's Argument Principle and the process outlined by Nyquist, the number of zeros in the RHP--- how come? What I mean is you can proceed with all logical argument that we are encircling the LHP and you will find that it does not matter! And there is no paradox, there is a very good explanation why it still works, with no conflict with what Cauchey and Nyquist have shown... $\endgroup$ Commented Mar 20, 2017 at 19:30
  • $\begingroup$ I have just changed completely my answer thanks to Matt's. It has the same information but presented a little bit more formally. I didn't want to leave my answer as it was because it was not correct and could lead to confusion. $\endgroup$
    – Tendero
    Commented Mar 21, 2017 at 14:35
  • $\begingroup$ Regarding your final comment I would think the other contour is completely valid in that we can still use the procedure to count number of zeros by determining the number of unpaired poles (or other process strictly following Nyquist with N=Z-P just in reverse with the contour also going in the reverse direction). So that said it still works all the same. Thanks for your added contribution, well done! What is also interesting to me, ignoring Nyquist and sticking with Cauchey is what occurs when we use this thought process with improper functions.... $\endgroup$ Commented Mar 21, 2017 at 15:41
  • 1
    $\begingroup$ Even with improper functions we can still make the argument that it has the same number of poles and zeros, it is just some of them will be at infinity! (For a single zero (s-2) for example it has a single pole at s= infinity since that is the value of s that makes the function go to infinity). $\endgroup$ Commented Mar 21, 2017 at 15:42
  • $\begingroup$ But then this is where my knowledge of math and infinity breaks down and I get less formal: I "see" how we can circle a pole in the right half plane clockwise, or similarly circle the companion zero at negative infinity counter clockwise... but can we really say there is a negative infinity? I do note in root locus how the closed loop pole goes to the open loop zero (so it's trajectory gives us a directional clue as to where that zero is). Interesting in the case of zeros at infinity how the pole can go toward negative infinity like the case I described earlier, or in other directions of inf! $\endgroup$ Commented Mar 21, 2017 at 15:46

Using infinity as a $point$ is really an abbreviation for a limiting process. If you draw a sequence of regions comprising an increasing portion of the $j\omega$-axis along with corresponding semicircular arcs lying in the RHP you will find using the Cauchy theorem a value for the number of zeros in each region. If there are no zeros in the RHP, each value will be zero, as is then the limit. If you were to do this with regions in the LHP, you would have different answers.

  • $\begingroup$ close @oldlab! However the question is noting that you get the same answer for the Nyquist plot in either case (whether you say you are encircling the RHP clockwise or the LHP counter-clockwise, which in the limit for both cases both results from sweeping the $j\omega$ axis from minus infinity to positive infinity---- how can that be based on what you said? $\endgroup$ Commented Mar 20, 2017 at 17:35
  • $\begingroup$ Thanks for your comment @DanBoschen. But please reread my post. I am not encircling the RHP or LHP as goes the common way to think about this. The so-called "point at infinity" is an informal artifice only. The more formal mathematics IMHO is a limiting argument. And I do believe that this approach using limits addresses the original question correctly. Using this approach, you will not get the same answers. $\endgroup$
    – oldlab
    Commented Mar 21, 2017 at 19:43
  • $\begingroup$ But the essence of the original question was why do we get the same answers- but you are saying that we don't get the same answer? (Note! I did not down-vote your answer) We know when we sweep from minus infinity to infinity we get the result we get from the Nyquist plot. We have also made the argument with the responses given that such a sweep can be interpreted as a clockwise rotation around the entire RHP or a counter clockwise rotation around the entire LHP and Matt I believe explained best why the same answer would result. Are you saying that you disagree with those conclusions? $\endgroup$ Commented Mar 21, 2017 at 20:11
  • $\begingroup$ (So if you do not get the same answer with a different approach, that approach must not be accurate as we DO get the same answer--- your response is a good one but I think it is not specific to the question at hand- the question again was why DO we get the same answer? $\endgroup$ Commented Mar 21, 2017 at 20:14

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