Nyquist plot of (s+10)/[(s−3)(s+3)] and gain margin

$$G(s)H(s) = \frac{s+10}{(s-3)(s+3)} = \frac{s+10}{s^2 - 9}$$

Open-loop poles: $+3$ and $-3$, so P = 1.

Nyquist contour

The contour encloses the whole right half of the s-plane:

• the $j\omega$-axis from $-j\infty$ to $+j\infty$;
• a semicircle of infinite radius on the right, traversed clockwise.

No poles lie on the $j\omega$-axis, so no indentation is needed. The RHP pole at $s = +3$ lies inside the contour.

     jw
      ^  +j inf
      |\
      | \     R -> inf
      |  \
------+---x----> σ
      |  /  +3 (inside)
      | /
      |/
       -j inf

Mapping

$$G(j\omega) = \frac{10 + j\omega}{-(\omega^2 + 9)} = \frac{-10}{\omega^2 + 9} - j\frac{\omega}{\omega^2 + 9}$$

• For ω > 0 the plot lies in the third quadrant, from $-1.111$ to the origin.
• The ω < 0 part is the mirror image.
• The infinite arc maps to the origin.

                                         |
                                         |
                                         |
                                         |
                                         |
                                         |
               .......................   |
           .....                     ....|
-----------*--@-------------------------**--------
           *****                     ****|
               ***********************   |
                                         |
                                         |
                                         |
                                         |
                                         |
                                         |
         -1.11                           0
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
w = 0 at -1.111; w -> +-inf at origin

(i) Stability

Nyquist criterion: $Z = P + N$, where $P$ = number of open-loop poles in the RHP, $N$ = net clockwise encirclements of $(-1, j0)$ by the Nyquist plot of $G(s)H(s)$, and $Z$ = number of closed-loop poles in the RHP. The closed loop is stable only if $Z = 0$.

As ω goes from −∞ to +∞, the closed curve runs from the origin through the upper half to $-1.111$, then through the lower half back to the origin. This is counter-clockwise, and $-1$ lies inside because $-1.111 < -1$. So N = −1.

$$Z = P + N = 1 - 1 = 0$$

The closed-loop system is stable. Check: $s^2 - 9 + s + 10 = s^2 + s + 1$, with roots in the LHP. ✓

(ii) Gain margin

Phase crossover is at ω = 0, where $|G| = 10/9 = 1.111$.

$$GM = \frac{1}{1.111} = 0.9 \;\Rightarrow\; 20\log0.9 = -0.92\ \text{dB}$$

Since the open loop is unstable, this means the gain may be reduced only by a factor of 0.9 before instability. With a gain $K$, stability needs $K > 0.9$.

Answer: P = 1, N = −1, Z = 0, so the system is stable. GM = 0.9 (−0.92 dB).