Nyquist stability of (4s+1)/[s²(s+1)(2s+1)]

$$G(s)H(s) = \frac{4s+1}{s^2(s+1)(2s+1)}$$

Open-loop stability

Open-loop poles: $0, 0, -1, -0.5$. None lie in the RHP, so P = 0. However, the double pole at the origin makes the open-loop system itself not stable: its impulse response contains a term $t$ that grows without bound. In Nyquist terms we take P = 0 and indent the contour around the origin.

Mapping the Nyquist contour

$$\angle G(j\omega) = -180^\circ + \tan^{-1}4\omega - \tan^{-1}\omega - \tan^{-1}2\omega$$

1. ω = 0⁺ → ∞: the plot starts at infinity just below the negative real axis (angle slightly above −180°). It cuts the negative real axis where

$$\tan^{-1}4\omega = \tan^{-1}\omega + \tan^{-1}2\omega \Rightarrow 4\omega = \frac{3\omega}{1 - 2\omega^2} \Rightarrow \omega_{pc} = \frac{1}{\sqrt8} = 0.354\ \text{rad/s}$$ $$|G(j0.354)| = \frac{\sqrt{1+2}}{0.125\sqrt{1.125}\sqrt{1.5}} = 10.67$$

So the crossing is at −10.67. The plot then enters the second quadrant and reaches the origin along −270°. 2. ω = −∞ → 0⁻: the mirror image. 3. Infinite semicircle: maps to the origin. 4. Small semicircle around the double pole: $G \approx 1/s^2 = (1/\varepsilon^2)e^{-j2\theta}$. As θ goes from −90° to +90°, $G$ traces a full 360° clockwise circle of infinite radius, from +180° through 0° to −180°.

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                           ***************    |
...                 ********             **** |
  .......     *******                       **|
--------*******---------------------------@--**---
  *******     .......                       ..|
***                 ........             .... |
                           ...............    |
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        -10.67                                0
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
crosses at -10.67 (w = 0.354)

(An infinite clockwise circle joins the ω = 0⁻ and ω = 0⁺ ends.)

Closed-loop 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 $Z = 0$.

The point −1 lies between the crossing at −10.67 and the origin. Following the plot together with the infinite 360° clockwise arc, the point (−1, j0) is encircled twice clockwise: N = 2.

$$Z = P + N = 0 + 2 = 2$$

The closed-loop system is unstable, with 2 poles in the RHP.

Check (Routh): $2s^4 + 3s^3 + s^2 + 4s + 1 = 0$

There are two sign changes, so there are 2 RHP poles (they are at $0.26 \pm j1.02$). ✓

Answer: Open loop: no RHP poles (P = 0), but the repeated pole at the origin makes it unstable in the BIBO sense. Closed loop: N = 2, Z = 2, so it is unstable.