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Consider a unit feedback system $$ X(s) = \frac{G(s)}{1+G(s)} $$

where the open loop transfer function of the system is $$ G(s) = \frac{s-1}{s(s+1)} $$

Open loop Bode & Nyquist plots: http://www.wolframalpha.com/input/?i=nyquist+plot+%28s-1%29%2F%28s%28s%2B1%29%29

It's easy to see from the Bode plot that the system is stable (phase does not cross $ 180^\circ $).

According to the textbook I'm using (Modern Control Systems, Dorf, 12th ed), the system is stable iff no poles of $ X(s) $ are on the right half s-plane. However this case seems to be a contradiction, because $$ X(s) = \frac{s-1}{s^2+2s-1} $$ has a pole $ -1+\sqrt{2} $, which is located on the right half s-plane.

Could anyone please point out what/where I missed?

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The closed-loop system $X(s)$ is unstable. You can see this by noting the pole it has in the right half plane, or by applying the Nyquist Stability Criterion to the open-loop system $G(s)$.

The Nyquist Stability Criterion (I grabbed this definition from the book by Ogata) states that the system is stable if

$$Z = N + P$$

where:

$Z =$ Number of zeros of 1+G(s) in the right half plane.

$N =$ Number of clockwise encirclements of the -1 point by G(s) in the complex plane.

$P =$ number of poles of G(s) in the right half plane.

The transfer function $1+G(s)$, which is the characteristic equation of $X(s)$, has a zero in the right half plane, but all its poles are stable. Hence, $N=1$ and $P=0$, and the Nyquist plot of $G(s)$ must encircle the -1 point once, in a clockwise direction, for the closed-loop system to be stable. It does not, and $X(s)$ is unstable.

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Your open loop gain will be one at a frequency of one radian/sec. At this frequency you will have phase shift of 90 degrees from the pole at zero, 45 degrees from the left half plane pole, and another 45 degrees from the right half plane zero. Total of 180 degrees.

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  • $\begingroup$ But $ G(i)=1+0i $ with $ 0^\circ $ phase shift... $\endgroup$ – smilekzs Jun 21 '13 at 0:01
  • $\begingroup$ No. You would get that result if you swapped the pole and zero. $\endgroup$ – Bill Kleinhans Jun 21 '13 at 20:13
  • $\begingroup$ Please, http://www.wolframalpha.com/input/?i=%28i-1%29%2F%28i*%28i%2B1%29%29 $\endgroup$ – smilekzs Jun 22 '13 at 8:46

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