I am facing a confusion on understanding system open-loop transfer function stability from its Nyquist plot.

According to the formula, for open loop transfer function stability: $$Z=N+P=0$$ where $N$ is the number of encirclements of $(0,0)$ by the Nyquist plot in clockwise direction.

Now how can I verify this formula for the open-loop transfer function: $$H(s)=-\frac{1}{s^3(s+1)}$$ The Nyquist plot is attached in the image.

enter image description here

Here $N=1$. Now how can I find $Z$ and $P$ and verify the formula?

  • 4
    $\begingroup$ Why would you apply a Nyquist plot to see whether an open-loop transfer function is stable? You can easily see that just by looking at the transfer function. The advantage of the Nyquist plot is that one can determine stability of a closed-loop transfer function without calculating the zeros and poles of it, and just using the open-loop information. $\endgroup$ – Tendero Feb 6 '18 at 15:03
  • $\begingroup$ Yeah, I know why we use Nyquist plot; but the above statement about O.L.T.F stability made me confused that's why I have asked it here $\endgroup$ – Suresh Feb 7 '18 at 4:12

As mentioned by Tendero in a comment, you should be interested in the stability of the closed-loop transfer function


where I've assumed unity feedback. Since the stability of $(1)$ is determined by the zeros of the denominator $1+H(s)$, we're interested in the encirclement of the point $-1+j0$ (and not in the encirclement of the origin). In your example, the number of clockwise encirclements of the point $-1+j0$ happens to equal the number of encirclements of the origin, i.e., we have $N=1$. Since the given $H(s)$ has no poles in the right half-plane, the number of zeros in the right half-plane equals the number of clockwise encirclements, i.e., $Z=N=1$, which means that $Q(s)$ has one pole in the right half-plane, and, consequently, the closed-loop transfer function is unstable.

This is also easily verified by computing the poles of $Q(s)$, which are the roots of the polynomial $s^4+s^3-1$:

ans =

  -1.38028 + 0.00000i
  -0.21945 + 0.91447i
  -0.21945 - 0.91447i
   0.81917 + 0.00000i

This shows that one of the four roots is in the right half-plane, as predicted by the Nyquist plot.


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