# Nyquist Criterion: Encirclement around origin

I am studying the Nyquist stability criterion and I am stuck on the following problem.

The open-loop transfer function with unity feedback is $$\\small\mathrm{ G(s)=\large\frac{1}{s-3}}\$$.

Following is the Nyquist plot for G(s): Here, the origin is not encircled at all by the plot, so N=0. But G(s) has a pole in RHP (at s=3) and no zero in RHP.

Assuming counter-clockwise encirclement to be positive, N=P-Z.. but N=0, P=1 and Z=0.

So it seems the equation for N does not hold up here. What am I missing?

I can see that point (-1,0) is also not encircled at all. So for (-1,0), N=0, P=1 and Z=1 (at s=2). So N=P-Z works here and the closed loop system is unstable with one RHP pole at s=2.

My problem is with the validity of N=P-Z for the origin as per the argument principle. What fundamental thing am I missing here?

You have an unstable open loop pole (open loop pole on rhs of s-plane) and so you need a counter-clockwise encirclement of the -1,j0 point to cancel it out to stabilize the closed loop system.

Your open loop transfer function of 1/(s-3) gives a closed loop transfer function of 1/(s-2). There is a closed loop pole on rhs of s-plane and so the closed loop system will be unstable.

If you increase the open loop gain above a value of 3 you should get that anti-clockwise encirclement and a conditionally stable system.

The open loop transfer function will then become 4/(s-3) (If, for example, k is set to 4) and the closed loop transfer function will become 4/(s+1) moving the pole to the lhs of s plane.

System stability is all about whether or not there are any closed loop poles on the rhs of the s-plane. For a system to be stable there must be no closed loop poles on the rhs of the s-plane.

Nyquist says that the number of closed loop poles on the rhs of the s plane (unstable closed loop poles) is equal to the number of open loop poles on the rhs of the s-plane (unstable open loop poles) plus the number of clockwise encirclements to the left of the -1,j0 point.

So, for a system to be stable, there must be no clockwise encirclements to the left of the -1,j0 point but a further requirement for a stable system is that for every unstable open loop pole there must be a counter-clockwise encirclement to the left of the -1,j0 point.

Cauchy's Argument Principle is the predecessor to Nyquist Stability Criteria. Both were developed as methods designed to give an indication of the number of unstable closed loop poles, that is the number of closed loop poles on the rhs of the s-plane. Where a lot of confusion can be created is that the poles of the closed loop transfer function are the zeros of the characteristic equation (the characteristic equation is the denominator of the closed loop transfer function, 1 + Go(s), where Go(s) is the loop transfer function) and as such are sometimes referred to as zeros!

Cauchy's Argument Principle is about defining some enclosed contour and then creating a plot using the s values which define the contour (s=jw). If we define the contour as surrounding the rhs of the s-plane then we can plot 1 + Go(jw) as w varies from -infinity to + infinity. If you did this I would expect you to get no encirclements of the origin. This is because you have one unstable open loop pole and one unstable closed loop pole and N = P-Z. (assuming anti-clockwise encirclements to be positive). N = 1 - 1 = 0. Z in the equation refers to the zeros of the characteristic equation which are the unstable closed loop poles.

So, you are making two errors, firstly if you are using Cauchy's Argument principle to look at encirclements of the origin then you should be plotting 1 + Go(s) instead of Go(s) and secondly your Z value is equal to 1 because you have an unstable closed loop pole sometimes referred to as a zero because it is a root of the characteristic equation, solved by equating 1 + Go(s) to zero.

Some time later Nyquist came along and said that if we plot the loop gain, Go(s) as w varies from -infinity to +infinity and look at encirclements of the -1,j0 point then we can obtain the same information as using Cauchy's argument Principle, plotting 1 + Go(s) and looking at encirclements of the origin.

Nyquist realised that this would make life easier because obtaining experimental Gain and Phase information of Go(jw) is much easier than for 1+Go(jw).

You have plotted Go(jw) (so you must use Nyquist not Cauchy's) and you have no encirclements of the -1,j0 point because you have one unstable closed loop pole and one unstable open loop pole. N = P - Z = 1 - 1 = 0.

• I had made one calculation mistake in the closed loop pole location (s=4 instead of s=2). I have edited that part in the original post. I understand that the closed loop system is unstable. Sep 16, 2022 at 4:07
• Continuing my previous comment: Here, my question is about the open loop transfer function only. If the open loop TF has one pole in RHP (at s=3) and no zero in RHP, the plot should encircle the origin once in CCW direction as per Cauchy's argument principle. But I don't see any encirclement of the origin. And that makes me wonder how N=P-Z holds up here when P=1, Z=0, N=0. Sep 16, 2022 at 4:27
• I think that, for the argument principle, the number of zeros minus the number of poles within the contour is equal to the number of encirclements of the origin but it is about the number of each within the contour. That open loop pole is not within the contour and therefore the number of zeros within the contour minus the number of poles within the contour is equal to 0 - 0 = 0 and therefore no encirclements of the origin.
– user173271
Sep 16, 2022 at 8:16
• "That open loop pole is not within the contour ". This plot is based on the Nyquist contour which includes the entire RHP and the jw axis. So the open loop pole at s=3+j0 is included in the contour right? Sep 16, 2022 at 9:40