---
title: "Polar plot of 1/[s(1+s)(1+2s)]"
subject: "Control Systems"
chapter: 6
key: "semester_3/control_systems:polar-1-s-1s-12s"
source: https://www.bctengineer.com/q/semester_3/control_systems/polar-1-s-1s-12s
site: BCT Engineering
---

# Polar plot of 1/[s(1+s)(1+2s)]

Control Systems · Chapter 6: Frequency Response Techniques

- Asked in: 2076 Chaitra, 8 marks
- Video: [polar plot in control system](https://www.youtube.com/watch?v=xWw7z5QzYgE) — Smart Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/polar-1-s-1s-12s
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6

## Question

Sketch the polar plot of the system whose open loop transfer function is given by G(s)H(s) = 1/[s(1+s)(1+2s)]. Also comment on stability.

## Short answer

The polar plot starts from an asymptote Re = -3 at ω→0+, crosses the negative real axis at −0.667 (ω = 0.707 rad/s), and ends at the origin tangent to the positive imaginary axis. The closed-loop system is stable with gain margin 1.5 (3.5 dB) and phase margin ~11°.

## Answer

#### Frequency response

$$
G(j\omega)H(j\omega)=\frac{1}{j\omega(1+j\omega)(1+j2\omega)}
$$

$$
|GH|=\frac{1}{\omega\sqrt{1+\omega^2}\sqrt{1+4\omega^2}},\qquad \angle GH=-90^\circ-\tan^{-1}\omega-\tan^{-1}2\omega
$$

Rationalised form:

$$
GH=\frac{-3\omega^2 - j\omega(1-2\omega^2)}{9\omega^4+\omega^2(1-2\omega^2)^2}
$$

| $\omega$ (rad/s) | $0^+$ | 0.2 | 0.4 | 0.5 | 0.707 | 1 | 2 | $\infty$ |
|---|---|---|---|---|---|---|---|---|
| $|GH|$ | $\infty$ | 4.55 | 1.81 | 1.26 | 0.667 | 0.316 | 0.054 | 0 |
| $\angle GH$ | $-90^\circ$ | $-123^\circ$ | $-150^\circ$ | $-162^\circ$ | $-180^\circ$ | $-198^\circ$ | $-229^\circ$ | $-270^\circ$ |

#### Key points

- **Start ($\omega\to0^+$):** magnitude $\infty$ at $-90^\circ$. The real part tends to $-(T_1+T_2)=-(1+2)=-3$, so the plot comes from $-j\infty$ along the asymptote $\text{Re}=-3$.
- **Negative real-axis crossing:** imaginary part zero when $1-2\omega^2=0\Rightarrow\omega_{pc}=\dfrac{1}{\sqrt2}=0.707$ rad/s.
$$
\begin{aligned}
GH(j0.707)&=\frac{-3(0.5)}{9(0.25)+0}=-\frac{1.5}{2.25}=-0.667
\end{aligned}
$$
- **End ($\omega\to\infty$):** magnitude 0 at $-270^\circ$; the plot reaches the origin tangent to the positive imaginary axis.

```text
              Im
              |
     -0.667   |
 ----x--o-----+--------- Re
    -1  |\_/  |   (ends at origin
        | w=0.707   from 2nd quad.)
     ___|     |
    /   |     |
   |  Re=-3 asymptote
   |   w -> 0+
   v  -j inf
```

#### Comment on stability

- Open-loop poles: $0,-1,-0.5$, so $P=0$.
- The polar plot cuts the negative real axis at $-0.667$, which is **to the right of $-1$**; with its mirror image and the infinite clockwise arc for the pole at the origin, the point $-1+j0$ is not encircled: $N=0$, $Z=0$.

The closed-loop system is **stable**.

Relative stability:

$$
GM=\frac{1}{0.667}=1.5\ (3.52\ \text{dB}),\qquad PM=180^\circ-168.6^\circ=11.4^\circ\ \text{at}\ \omega_{gc}=0.572\ \text{rad/s}
$$

The margins are small, so the response will be quite oscillatory; the gain can be raised by a factor of 1.5 before instability (Routh: $2s^3+3s^2+s+K$ needs $K<1.5$).

**Answer: the plot crosses the negative real axis at −0.667 (ω = 0.707 rad/s); the closed-loop system is stable with GM = 1.5 (3.5 dB) and PM ≈ 11°.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For G(s)H(s) = 1/[s(1+s)(1+2s)], where does the polar plot begin as w -> 0+?
   A. At magnitude 0 and angle -270 degrees
   B. At magnitude infinity, angle -90 degrees, along the asymptote Re = -3
   C. At magnitude 1 and angle 0 degrees
   D. At magnitude infinity, angle -180 degrees, along the negative real axis
   Answer: B. The pole at the origin gives infinite magnitude at -90 degrees, and the real part tends to -(T1+T2) = -3. (Key point: Start of polar plot)

2. How is the frequency where the plot crosses the negative real axis found, and what is it?
   A. Set the real part to zero; w = 0.5 rad/s
   B. Set the magnitude to 1; w = 0.572 rad/s
   C. Set the imaginary part to zero, 1 - 2w^2 = 0; w = 0.707 rad/s
   D. Set the phase to -90 degrees; w = 1 rad/s
   Answer: C. The phase is -180 degrees when the imaginary part is zero, giving 1 - 2w^2 = 0 and w = 1/sqrt(2) = 0.707 rad/s. (Key point: Negative real-axis crossing)

3. The plot crosses the negative real axis at -0.667. What does this tell us about closed-loop stability (P = 0)?
   A. Unstable, because the crossing is on the negative real axis
   B. Stable, because the crossing lies to the right of -1 so -1+j0 is not encircled
   C. Unstable, because the plot starts at infinity
   D. Marginally stable, because the crossing is at -0.667 not 0
   Answer: B. Crossing between -1 and 0 means -1+j0 is not encircled, so N = 0 and Z = P + N = 0: stable. (Key point: Stability from crossing point)

4. With the phase-crossover magnitude 0.667, what is the gain margin?
   A. 1.5 (about 3.5 dB)
   B. 0.667 (about -3.5 dB)
   C. 1 (0 dB)
   D. 11.4 dB
   Answer: A. GM = 1/|GH| at phase crossover = 1/0.667 = 1.5, which is 3.52 dB; gain can rise 1.5 times before instability. (Key point: Gain margin)

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