Frequency Response Techniques

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

The open-loop poles are $+1$ and $-1$, so P = 1 (one RHP pole). No poles lie on the $j\omega$-axis, so the Nyquist contour needs no indentation.

Frequency response

$$G(j\omega) = \frac{2 + j\omega}{-(1+\omega^2)} = \frac{-2}{1+\omega^2} - j\frac{\omega}{1+\omega^2}$$

• For $\omega > 0$ the imaginary part is always negative, so the plot lies in the third quadrant. It runs from $-2$ (at ω = 0) to the origin (at ω = ∞), arriving along $-90^\circ$.
• The part for $\omega < 0$ is the mirror image about the real axis.
• The large semicircle ($s = Re^{j\theta}$, $R \to \infty$) maps to the origin, because $G \to 1/s \to 0$.

Complete Nyquist plot

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        ..                              ..
--------*---------------@----------------*--------
        **                              **
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            ******              ******   |
                 ****************        |
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       -2              -1                0
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
w = 0 at -2; w -> +-inf at origin

Stability by Nyquist criterion

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 $-\infty$ to $+\infty$, the plot runs from the origin through the upper half to $-2$, then through the lower half back to the origin. This is a closed curve traced counter-clockwise.
• The point $-1$ lies inside it, since the plot crosses the real axis at $-2$, to the left of $-1$.
• So $N = -1$ (one counter-clockwise encirclement).

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

The closed-loop system is stable. Check: $1 + G = 0 \Rightarrow s^2 + s + 1 = 0$, whose roots $-0.5 \pm j0.866$ are in the LHP. ✓

Gain margin

The plot cuts the negative real axis at ω = 0 (phase crossover frequency $\omega_{pc} = 0$), where $|G| = 2$.

$$GM = \frac{1}{|G(j\omega_{pc})|} = \frac{1}{2} = 0.5 \;\Rightarrow\; GM = 20\log_{10}0.5 = -6.02\ \text{dB}$$

Because the open loop is unstable, a negative GM does not mean instability here. It means the loop gain can be reduced by a factor of 2 before the system becomes unstable. (With gain $K$: $s^2 + Ks + 2K - 1 = 0$ is stable only for $K > 0.5$.)

Answer: P = 1, N = −1 (one CCW encirclement), Z = 0, so the closed loop is stable. GM = 0.5 (−6.02 dB).

$$G(s) = \frac{10}{s(s+1)^2}, \qquad G(j\omega) = \frac{10}{\omega(1+\omega^2)}\angle\left(-90^\circ - 2\tan^{-1}\omega\right)$$

Key points of the polar plot

• Low-frequency asymptote: $\text{Re}\,G \to -K\sum T = -10(1+1) = -20$. The plot starts at $\infty\angle -90^\circ$ along the line Re = −20.
• Real-axis crossing: $\angle G = -180^\circ \Rightarrow 2\tan^{-1}\omega = 90^\circ \Rightarrow \omega = 1$, where $|G| = 10/(1 \times 2) = 5$. The plot cuts the negative real axis at −5.
• High frequency: the plot reaches the origin along $-270^\circ$ (from the second quadrant, tangent to the +j axis).

                  .                       |
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                   ...                    |
                     ..                   |
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                        ...               |
                          ...             |
                            ....     **** |
-------------------------------*******--@**-------
                            ****     .... |
                          ***             |
                        ***               |
                      ***                 |
                     **                   |
                   ***                    |
                  **                      |
                  *                       |
      -20                       -5        0
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
crosses real axis at -5 (w = 1); asymptote Re = -20

(iv) Phase crossover frequency (pcf)

$\omega_{pc} = 1$ rad/s.

(i) Gain margin

$$GM = \frac{1}{|G(j\omega_{pc})|} = \frac{1}{5} = 0.2 = 20\log 0.2 = -13.98\ \text{dB}$$

(iii) Gain crossover frequency (gcf)

$$|G| = 1 \Rightarrow \omega(1 + \omega^2) = 10 \Rightarrow \omega^3 + \omega - 10 = 0 \Rightarrow \omega_{gc} = 2\ \text{rad/s}$$

(Check: $8 + 2 = 10$.)

(ii) Phase margin

$$\begin{aligned} \angle G(j2) &= -90^\circ - 2\tan^{-1}2 = -90^\circ - 126.87^\circ = -216.87^\circ \\ PM &= 180^\circ + \angle G(j\omega_{gc}) = -36.87^\circ \end{aligned}$$

(v) Stability

GM < 0 dB and PM < 0. The polar plot crosses the real axis at −5, to the left of −1, so the $(-1, j0)$ point is enclosed. With P = 0, the complete Nyquist plot encircles −1 twice clockwise, so Z = 2. The closed-loop system is unstable, with two RHP poles.

Check with Routh: $s^3 + 2s^2 + s + 10$ gives $2 \times 1 < 10$, so it is unstable. For stability $K$ must be below 2, i.e. the gain must be reduced by a factor of 5.

Answer: GM = −13.98 dB, PM = −36.87°, gcf = 2 rad/s, pcf = 1 rad/s. The system is unstable.

$$G(s)H(s) = \frac{50}{(s+1)(s+2)}, \qquad G(j\omega) = \frac{50}{\sqrt{(1+\omega^2)(4+\omega^2)}}\angle\left(-\tan^{-1}\omega - \tan^{-1}\tfrac{\omega}{2}\right)$$

Open-loop poles

$-1$ and $-2$, so P = 0. There are no poles on the $j\omega$-axis.

Polar (ω: 0 → ∞) data

• The plot starts at 25 on the positive real axis, goes through the 4th and 3rd quadrants, and reaches the origin along $-180^\circ$.
• The phase reaches $-180^\circ$ only at ω = ∞, so the plot never cuts the negative real axis at a finite point.
• The $\omega < 0$ part is its mirror image. The infinite semicircle of the s-plane maps to the origin.

        |      ................
        | ......              .......
       ....                         ....
     ...|                              ....
    ..  |                                 ..
   ..   |                                  ...
   ..   |                                    ..
    ..  |                                     .
-----*@**-------------------------------------*---
    **  |                                     *
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   **   |                                  ***
    **  |                                 **
     ***|                              ****
       ****                         ****
        | ******              *******
        |      ****************
        0                                    25
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
w = 0 at 25; ends at origin along -180 deg

Nyquist 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 closed curve passes between the origin and the far right (25). The point $-1$ lies outside the curve, so $N = 0$.

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

The closed-loop system is stable for this gain (and in fact for any $K > 0$, because the plot never crosses the negative real axis, so GM = ∞).

Phase margin

Gain crossover: $|G| = 1$:

$$\begin{aligned} (1+\omega^2)(4+\omega^2) &= 2500 \\ \omega^4 + 5\omega^2 - 2496 &= 0 \\ \omega^2 &= \frac{-5 + \sqrt{25 + 9984}}{2} = 47.52 \\ \omega_{gc} &= 6.894\ \text{rad/s} \end{aligned}$$ $$\begin{aligned} \angle G(j\omega_{gc}) &= -\tan^{-1}6.894 - \tan^{-1}3.447 = -81.75^\circ - 73.82^\circ = -155.57^\circ \\ PM &= 180^\circ - 155.57^\circ = 24.43^\circ \end{aligned}$$

Answer: N = 0, P = 0, so Z = 0 and the closed loop is stable. PM ≈ 24.4° at $\omega_{gc}$ = 6.89 rad/s (GM = ∞).

$$G(s) = \frac{1}{s(1+2s)(1+s)}, \qquad G(j\omega) = \frac{1}{\omega\sqrt{1+4\omega^2}\sqrt{1+\omega^2}}\angle\left(-90^\circ - \tan^{-1}2\omega - \tan^{-1}\omega\right)$$

Nyquist contour

There is a pole at the origin, so the contour is indented around it by a small semicircle $s = \varepsilon e^{j\theta}$ with θ from −90° to +90°. Open-loop poles in the RHP: P = 0.

Mapping of each section

1. ω = 0⁺ to ∞ (positive $j\omega$-axis):

• Low-frequency asymptote: $\text{Re} \to -(2+1) = -3$. The plot starts at $\infty\angle -90^\circ$ along Re = −3.
• Phase crossover: $\tan^{-1}2\omega + \tan^{-1}\omega = 90^\circ \Rightarrow 2\omega^2 = 1 \Rightarrow \omega_{pc} = 0.707$ rad/s.
• $|G(j0.707)| = \dfrac{1}{0.707 \times \sqrt3 \times \sqrt{1.5}} = 0.667$, so it crosses at −0.667.

2. ω = −∞ to 0⁻: mirror image of the above.
3. Infinite semicircle: $G \to 0$, so it maps to the origin.
4. Small semicircle at origin: $G \approx 1/s = (1/\varepsilon)e^{-j\theta}$. As θ goes from −90° to +90°, $G$ sweeps an infinite-radius arc clockwise from +90° through 0° to −90° (right half of the G-plane), joining the ω = 0⁻ end to the ω = 0⁺ end.

                ..                     |
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-----------------------------@**********----------
                          *****        |
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                      **               |
                    ***                |
                   **                  |
                  **                   |
                 **                    |
                **                     |
         -3                    -0.67   0
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
crosses at -0.667 (w = 0.707); asymptote Re = -3

(The infinite clockwise arc on the right closes the curve.)

Gain margin

$$GM = \frac{1}{0.667} = 1.5 = 20\log1.5 = 3.52\ \text{dB}$$

The phase margin (for reference) is $11.4^\circ$ at $\omega_{gc} = 0.572$ rad/s.

Stability by Nyquist criterion

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 plot crosses the negative real axis at −0.667, to the right of −1, so the point $(-1, j0)$ is not encircled: N = 0.

$$Z = P + N = 0$$

The closed-loop system is stable. Check: $2s^3 + 3s^2 + s + 1$ has Routh first column $2, 3, 1/3, 1$, all positive. ✓

Answer: GM = 1.5 (3.52 dB) at $\omega_{pc}$ = 0.707 rad/s. N = 0, P = 0, Z = 0, so the system is stable.

A Bode magnitude asymptote is read segment by segment. The initial slope gives the type and gain, and each change of slope gives a corner frequency.

Reading the plot

1. Initial slope −20 dB/dec: one pole at the origin (type 1), i.e. a factor $K/s$.
2. Gain K: the −20 dB/dec line (or its extension) cuts 0 dB at $\omega = K$. It crosses 0 dB at ω = 1, so K = 1. Check at ω = 0.1: $20\log(1/0.1) = 20$ dB, which matches the plot.
3. At ω = 2 the slope changes from −20 to −40 dB/dec: the change is −20 dB/dec, so there is a simple pole with corner frequency 2 rad/s, i.e. a factor $1/(1 + s/2)$.
4. There are no further changes, so there are no more poles or zeros.

Transfer function

$$G(s) = \frac{1}{s\left(1 + \dfrac{s}{2}\right)} = \frac{2}{s(s+2)}$$

Check

Answer: $G(s) = \dfrac{2}{s(s+2)}$, i.e. $\dfrac{1}{s(1+0.5s)}$.

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

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.

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-----------*--@-------------------------**--------
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               ***********************   |
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         -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).

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

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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***                 ........             .... |
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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.

$$G(s)H(s) = \frac{s+2}{s(s+1)(s+3)} = \frac{\tfrac23\left(1 + \tfrac{s}{2}\right)}{s(1+s)\left(1 + \tfrac{s}{3}\right)}$$

Open-loop poles: $0, -1, -3$, so P = 0. There is one pole at the origin, so the contour is indented there.

1. Magnitude and phase

$$|G(j\omega)| = \frac{\sqrt{\omega^2 + 4}}{\omega\sqrt{\omega^2 + 1}\sqrt{\omega^2 + 9}}, \qquad \angle G = -90^\circ + \tan^{-1}\frac{\omega}{2} - \tan^{-1}\omega - \tan^{-1}\frac{\omega}{3}$$

2. Real and imaginary parts

$$G(j\omega) = \frac{(2 + j\omega)\left[-4\omega^2 - j(3\omega - \omega^3)\right]}{16\omega^4 + (3\omega - \omega^3)^2}$$ $$\text{Re} = \frac{-5\omega^2 - \omega^4}{16\omega^4 + (3\omega - \omega^3)^2}, \qquad \text{Im} = \frac{-6\omega - 2\omega^3}{16\omega^4 + (3\omega - \omega^3)^2}$$

• Im < 0 for every ω > 0, so the plot never crosses the real axis at a finite frequency. The phase approaches −180° only as ω → ∞.
• Low-frequency asymptote: as ω → 0, Re → $-5/9 = -0.556$. (Equivalently $K\sum T = \tfrac23(1 + \tfrac13 - \tfrac12) = 0.556$.)

3. Table (ω > 0)

4. Mapping of the full contour

• ω = 0⁺ → +∞: from $\infty\angle-90^\circ$ (along Re = −0.556) down to the origin, arriving at −180° (third quadrant).
• ω = −∞ → 0⁻: mirror image in the second quadrant.
• Infinite semicircle: maps to the origin.
• Small indentation at s = 0: $G \approx \tfrac23/s$ maps to an infinite arc, clockwise through the right half-plane (from +90° to −90°).

                       ..              |
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                           ...         |
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                               .....   |
-------@---------------------------*****----------
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                          **           |
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                        *              |
                       **              |
                  -0.556               0
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
asymptote Re = -0.556; no negative real crossing

5. 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 plot never cuts the negative real axis, so $(-1, j0)$ is not encircled: N = 0, and $Z = 0 + 0 = 0$.

The closed-loop system is stable. Also, GM = ∞ and PM = 64.8° (at $\omega_{gc}$ = 0.588 rad/s).

Check: $s^3 + 4s^2 + 4s + 2 = 0$ gives Routh first column $1, 4, 3.5, 2$, all positive. ✓

Time-constant form

$$G(s) = \frac{512(s+3)}{s(s^2 + 16s + 256)} = \frac{6\left(1 + \dfrac{s}{3}\right)}{s\left(1 + \dfrac{s}{16} + \dfrac{s^2}{256}\right)}$$

Quadratic factor: $\omega_n = 16$ rad/s and $2\zeta\omega_n = 16$, so $\zeta = 0.5$.

Factors and corner frequencies

Magnitude asymptotes

With ζ = 0.5 the resonant correction at $\omega_n$ is $-20\log(2\zeta) = 0$ dB, so the asymptotes are close to the true curve.

Phase

$$\phi(\omega) = -90^\circ + \tan^{-1}\frac{\omega}{3} - \tan^{-1}\frac{\omega/16}{1 - (\omega/16)^2}$$

Bode plot

   40 |*
      |*******
      |      ******
   10 |            **************
      |-------------------------****-----------------
      |                            ****
  -20 |                               ***
      |                                  ***
      |                                    ****
  -50 |                                       ****
      |                                          ****
      |                                             *
  -80 |
      +----------------------------------------------
       0.1        1          10          100        1000
Magnitude asymptotes (dB) vs w (rad/s)

    0 |
      |
      |
  -45 |                  ****
      |              *****  ***
      |        *******        **
  -90 |*********               **
      |                         *
      |                         **
 -135 |                          **
      |                           ***
      |                             ******
 -180 |----------------------------------************
      +----------------------------------------------
       0.1        1          10          100        1000
Phase (deg); -180 line shown

Margins and stability

• Gain crossover: the exact $|G| = 1$ at $\omega_{gc} = 24.4$ rad/s (asymptotic estimate 22.6 rad/s). There, φ = −137.9°, so

$$PM = 180^\circ - 137.9^\circ = 42.1^\circ$$

• Phase crossover: the phase tends to −180° only as ω → ∞, so it never actually crosses −180°. Hence $\omega_{pc} = \infty$ and GM = ∞.

Both margins are positive, so the closed-loop system is stable, with PM ≈ 42° (reasonably damped). Check: $s^3 + 16s^2 + 768s + 1536$ gives $16 \times 768 > 1536$. ✓

Answer: GM = ∞, PM ≈ 42.1° (at 24.4 rad/s). The system is stable.

$$G(s) = \frac{K}{s(s^2 + 2s + 2)}$$

Poles: $0,\ -1 \pm j1$. P = 0. The pole at the origin needs an indentation.

Frequency response

$$G(j\omega) = \frac{K}{-2\omega^2 + j\omega(2 - \omega^2)}$$ $$\text{Re} = \frac{-2K}{4\omega^2 + (2 - \omega^2)^2}, \qquad \text{Im} = \frac{-K(2 - \omega^2)}{\omega\left[4\omega^2 + (2 - \omega^2)^2\right]}$$

• Low-frequency asymptote: Re → −K/2.
• Real-axis crossing: Im = 0 when $2 - \omega^2 = 0$, so $\omega_{pc} = \sqrt2 = 1.414$ rad/s, where $G = K/(-4)$. The crossing is at −K/4.
• High frequency: reaches the origin along −270° (from the second quadrant).

Complete Nyquist plot

1. ω = 0⁺ → ∞: from $\infty\angle-90^\circ$ (along Re = −K/2), crossing at −K/4, to the origin.
2. ω = −∞ → 0⁻: mirror image.
3. Infinite semicircle: maps to the origin.
4. Indentation at s = 0: maps to an infinite arc clockwise from +90° through 0° to −90°.

                       .                 |
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----@------------------------*************--------
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                       **                |
                       *                 |
                       *                 |
                       *                 |
                       *                 |
   -1                -0.5     -.25
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
drawn for K = 1: crosses at -0.25 (w = 1.414);
asymptote Re = -0.5. For gain K, scale every point by K.

Maximum K for 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 $Z = 0$.

With P = 0, stability needs N = 0, i.e. the point −1 must lie to the left of the crossing:

$$\frac{K}{4} < 1 \;\Rightarrow\; K < 4$$

If $K > 4$ the crossing moves left of −1, giving N = 2 and Z = 2 (unstable). At $K = 4$ the plot passes through −1: marginal stability with oscillation at $\omega = \sqrt2$ rad/s.

Check (Routh): $s^3 + 2s^2 + 2s + K$ is stable if $2 \times 2 > K$, i.e. $K < 4$. ✓

Answer: K_max = 4. The system is stable for 0 < K < 4 and oscillates at 1.414 rad/s when K = 4.

$$G(s) = \frac{2.2}{s(s+1)(s^2 + 2s + 2)}$$

Poles: $0, -1, -1 \pm j1$, so P = 0. The pole at the origin is indented.

Frequency response

$$\angle G = -90^\circ - \tan^{-1}\omega - \tan^{-1}\frac{2\omega}{2 - \omega^2}, \qquad |G| = \frac{2.2}{\omega\sqrt{1 + \omega^2}\sqrt{(2 - \omega^2)^2 + 4\omega^2}}$$

• Low-frequency asymptote: $G = \dfrac{1.1}{s(1+s)(1 + s + 0.5s^2)}$, so Re → $-1.1(1 + 1) = -2.2$.
• Phase crossover: $s^4 + 3s^3 + 4s^2 + 2s$ at $s = j\omega$ has imaginary part $-3\omega^3 + 2\omega = 0$, so $\omega_{pc} = \sqrt{2/3} = 0.816$ rad/s. The real part is $\omega^4 - 4\omega^2 = 0.444 - 2.667 = -2.222$, so $G = 2.2/(-2.222) = -0.990$.
• The plot cuts the negative real axis at −0.99, then passes through the second quadrant (crossing the +j axis at ω = 2, where ∠G = −270°) and reaches the origin along −360°.

              ..                          |
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                      .....     **********|
--------------------------**@****--------**-------
                      *****     ..........|
                    ***                   |
                  ***                     |
                 **                       |
                **                        |
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               *                          |
              **                          |
         -2.2             -0.99           0
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
crosses at -0.99 (w = 0.816), just right of -1

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 crossing at −0.99 lies just to the right of −1, so the point $(-1, j0)$ is not encircled: N = 0 and $Z = 0 + 0 = 0$.

The closed-loop system is stable, but only barely:

$$GM = \frac{1}{0.99} = 1.0101 = 0.087\ \text{dB}, \qquad PM \approx 0.5^\circ \;(\omega_{gc} = 0.811\ \text{rad/s})$$

The closed-loop poles are at $-0.0026 \pm j0.814$ and $-1.497 \pm j1.039$. The dominant pair is almost on the $j\omega$-axis, so the step response will be a very lightly damped oscillation at about 0.81 rad/s. The gain limit (Routh) is $K < 2.222$, so a 1% gain increase would make the system unstable.

Answer: N = 0, Z = 0, so the system is stable (marginally: GM ≈ 0.09 dB, PM ≈ 0.5°).

Time-constant form

$$G(s)H(s) = \frac{50(s+10)}{(s+1)(s+100)} = \frac{5\left(1 + \dfrac{s}{10}\right)}{(1+s)\left(1 + \dfrac{s}{100}\right)}$$

It is a type-0 system with $K = 5$, so $20\log5 = 13.98$ dB.

Corner frequencies

Magnitude asymptotes

The asymptotic gain crossover is where $13.98 - 20\log\omega = 0$, i.e. ω = 5 rad/s. The exact value is 5.64 rad/s.

Phase

$$\phi = \tan^{-1}\frac{\omega}{10} - \tan^{-1}\omega - \tan^{-1}\frac{\omega}{100}$$

The phase dips to about −55° near ω ≈ 4 and tends to −90° at high frequency. It never reaches −180°.

Bode plot

   15 |*************
      |            ****
      |               ****
    0 |------------------***-------------------------
      |                     ***************
      |                                   ***
  -15 |                                     ****
      |                                        ****
      |                                           ***
  -30 |
      |
      |
  -45 |
      +----------------------------------------------
       0.1        1          10          100        1000
Magnitude asymptotes (dB) vs w (rad/s)

    0 |***
      |  ******
      |        ****             *******
  -45 |           ****     ******     ****
      |              *******             ****
      |                                      ******
  -90 |                                           ***
      |
      |
 -135 |
      |
      |
 -180 |----------------------------------------------
      +----------------------------------------------
       0.1        1          10          100        1000
Phase (deg); never reaches -180

Stability from the plot

• Phase margin: at $\omega_{gc} = 5.64$ rad/s, φ = −53.7°, so $PM = 180^\circ - 53.7^\circ = 126.3^\circ$.
• Gain margin: the phase never crosses −180°, so $\omega_{pc} = \infty$ and GM = ∞.

Both margins are positive and large, so the closed-loop system is stable (and stays stable for any increase in gain). Check: the closed loop $s^2 + 151s + 600 = 0$ has both roots in the LHP. ✓

Answer: PM ≈ 126°, GM = ∞. The system is stable.

Nyquist contour

The Nyquist contour is a closed path in the s-plane that encloses the entire right half-plane. It consists of:

1. the whole $j\omega$-axis from $-j\infty$ to $+j\infty$;
2. a semicircle of infinite radius ($s = Re^{j\theta}$, $R \to \infty$, θ from +90° to −90°), traversed clockwise;
3. small semicircular indentations (radius ε → 0) around any open-loop poles that lie on the $j\omega$-axis, so the path avoids them.

Mapping this contour through $G(s)H(s)$ gives the Nyquist plot. By the argument principle, the encirclements of $(-1, j0)$ count the RHP zeros of $1 + G(s)H(s)$, i.e. the RHP closed-loop poles.

Mapping $G(s) = \dfrac{200}{(s+1)(s+2)(s+3)}$

Open-loop poles: $-1, -2, -3$, so P = 0. No indentation is needed.

$$(j\omega+1)(j\omega+2)(j\omega+3) = (6 - 6\omega^2) + j(11\omega - \omega^3)$$

         |........................
      .....                      ......
    ...  |                            ....
  ...    |                               ...
 ..      |                                 ...
 .       |                                   ..
 ..      |                                    ..
  ..     |                                     .
---*****@*-------------------------------------*--
  **     |                                     *
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 *       |                                   **
 **      |                                 ***
  ***    |                               ***
    ***  |                            ****
      *****                      ******
         |************************
   -3.33                                     33.3
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
w = 0 at 33.3; crosses at -3.33 (w = 3.317)

Nyquist criterion

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 plot crosses the negative real axis at −3.33, to the left of −1. The closed curve (ω from −∞ to +∞) goes round the point $(-1, j0)$ twice clockwise, so N = 2.

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

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

• Gain margin: $GM = 1/3.33 = 0.3 = -10.5$ dB (negative).
• The gain would have to be reduced below $200/3.33 = 60$ for stability.
• Check (Routh): $s^3 + 6s^2 + 11s + 206$ gives $6 \times 11 = 66 < 206$, so it is unstable. ✓

Time-constant form

$$G(s)H(s) = \frac{60}{s(s+2)(s+6)} = \frac{5}{s\left(1 + \dfrac{s}{2}\right)\left(1 + \dfrac{s}{6}\right)}$$

$K = 5$ ($20\log5 = 13.98$ dB), type 1. Corner frequencies: 2 and 6 rad/s.

Magnitude asymptotes

Phase

$$\phi = -90^\circ - \tan^{-1}\frac{\omega}{2} - \tan^{-1}\frac{\omega}{6}$$

Bode plot

   40 |
      |********
      |       ********
   10 |              ********
      |---------------------****---------------------
      |                        ****
  -20 |                            ***
      |                              ****
      |                                 ***
  -50 |                                   ****
      |                                      ***
      |                                        ****
  -80 |                                           ***
      +----------------------------------------------
       0.1            1              10             100
Magnitude asymptotes (dB) vs w (rad/s)

  -90 |*****
      |    *********
      |            ****
 -135 |               ****
      |                  ***
      |                    ***
 -180 |----------------------***---------------------
      |                        ***
      |                          ***
 -225 |                            ****
      |                               ****
      |                                   ********
 -270 |                                          ****
      +----------------------------------------------
       0.1            1              10             100
Phase (deg); crosses -180 at 3.46 rad/s

Gain margin

Phase crossover:

$$\tan^{-1}\frac{\omega}{2} + \tan^{-1}\frac{\omega}{6} = 90^\circ \;\Rightarrow\; \frac{\omega^2}{12} = 1 \;\Rightarrow\; \omega_{pc} = \sqrt{12} = 3.464\ \text{rad/s}$$ $$|G(j3.464)| = \frac{60}{3.464 \times \sqrt{12 + 4} \times \sqrt{12 + 36}} = \frac{60}{3.464 \times 4 \times 6.928} = 0.625$$ $$GM = \frac{1}{0.625} = 1.6 = 20\log1.6 = 4.08\ \text{dB}$$

(The asymptotic plot gives about −1.6 dB at 3.46 rad/s, i.e. GM ≈ 1.6 dB; the exact value is 4.08 dB.)

Phase margin

Exact gain crossover: $\omega_{gc} = 2.71$ rad/s, where φ = −167.8°, so $PM = 12.2^\circ$.

Stability

GM = +4.08 dB > 0 and PM = +12.2° > 0, and $\omega_{gc} < \omega_{pc}$. The closed-loop system is stable, but with small margins, so the response will be quite oscillatory.

Check: $s^3 + 8s^2 + 12s + 60$ gives $8 \times 12 = 96 > 60$. ✓ The gain could rise by a factor of 1.6 (to 96) before instability.

Answer: GM = 4.08 dB at $\omega_{pc}$ = 3.46 rad/s. PM ≈ 12.2°. The system is stable.

Time-constant form

$$G(s) = \frac{40(s+2)}{s(1+2s)(25s^2 + s + 2)} = \frac{40\left(1 + \dfrac{s}{2}\right)}{s(1+2s)\left(1 + 0.5s + 12.5s^2\right)}$$

Quadratic factor: $\omega_n^2 = 2/25$, so $\omega_n = 0.283$ rad/s; $2\zeta\omega_n = 1/25$, so $\zeta = 0.0707$ (very lightly damped).

Factors

Magnitude asymptotes

Resonant correction at $\omega_n$: $-20\log(2\zeta) = +17$ dB, so the true peak is about 59 dB near 0.28 rad/s.

Phase

$$\phi = -90^\circ - \tan^{-1}2\omega + \tan^{-1}\frac{\omega}{2} - \tan^{-1}\frac{0.5\omega}{1 - 12.5\omega^2}$$

Bode plot

   90 |
      |****
      |   *************
   45 |               *********
      |                       *****
      |                           ***
    0 |-----------------------------****-------------
      |                                ****
      |                                   *****
  -45 |                                       *****
      |                                           ***
      |
  -90 |
      +----------------------------------------------
       0.01           0.1            1              10
Magnitude asymptotes (dB); true peak ~59 dB at 0.28

  -90 |***************
      |              *******
      |                    **
 -150 |                     *
      |---------------------*------------------------
      |                      *
 -210 |                      *
      |                      *
      |                      *
 -270 |                       *                    **
      |                       *****       **********
      |                           ********
 -330 |
      +----------------------------------------------
       0.01           0.1            1              10
Phase (deg); crosses -180 at 0.275 rad/s

Gain margin

The phase crosses −180° at $\omega_{pc} = 0.275$ rad/s, just below $\omega_n$. Because of the resonance, the magnitude there is very large: $|G| = 871.6$ (58.8 dB).

$$GM = -58.8\ \text{dB} \quad (\text{i.e. } 1/871.6)$$

Phase margin

Gain crossover is at $\omega_{gc} = 1.16$ rad/s (asymptotic estimate 1.12), where φ = −304.5°.

$$PM = 180^\circ - 304.5^\circ = -124.5^\circ$$

Stability

GM is negative (in dB), PM is negative, and $\omega_{pc} < \omega_{gc}$. The closed-loop system is unstable.

Check: $50s^4 + 27s^3 + 5s^2 + 42s + 80 = 0$ gives Routh first column $50, 27, -72.8, \dots$. There are two sign changes, so there are 2 RHP poles. ✓

Answer: GM ≈ −58.8 dB, PM ≈ −124.5°. The system is unstable.

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

Open-loop poles: $0, -1, -0.5$, so P = 0. The contour is indented around $s = 0$.

Frequency response

$$G(j\omega) = \frac{2}{\omega\sqrt{1+\omega^2}\sqrt{1+4\omega^2}}\angle\left(-90^\circ - \tan^{-1}\omega - \tan^{-1}2\omega\right)$$

• Low-frequency asymptote: Re → $-K(T_1 + T_2) = -2(1 + 2) = -6$.
• Phase crossover: $\tan^{-1}\omega + \tan^{-1}2\omega = 90^\circ \Rightarrow 2\omega^2 = 1 \Rightarrow \omega_{pc} = 0.707$ rad/s.

$$|G| = \frac{2}{0.707 \times 1.225 \times 1.732} = 1.333$$

• Indentation at the origin: maps to an infinite arc clockwise from +90° through 0° to −90°.
• Infinite semicircle: maps to the origin.

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--------------------------------*****@******------
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                      ***                  |
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               **                          |
     -6                          -1.33     0
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
crosses at -1.333 (w = 0.707); asymptote Re = -6

Nyquist criterion

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 plot cuts the negative real axis at −1.333, to the left of −1. Together with its mirror image and the infinite clockwise arc, it encircles $(-1, j0)$ twice clockwise: N = 2.

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

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

$$GM = \frac{1}{1.333} = 0.75 = -2.5\ \text{dB}$$

The gain must be reduced below 1.5 for stability. Check: $2s^3 + 3s^2 + s + 2$ gives $3 \times 1 < 2 \times 2$, so it is unstable. ✓

$$G(s)H(s) = \frac{s+5}{(s-2)(s+2)} = \frac{s+5}{s^2 - 4}$$

Open-loop poles: $+2, -2$, so P = 1 (one RHP pole). No poles lie on the $j\omega$-axis.

Frequency response

$$G(j\omega) = \frac{5 + j\omega}{-(\omega^2 + 4)} = \frac{-5}{\omega^2 + 4} - j\frac{\omega}{\omega^2 + 4}$$

• For ω > 0 the plot is in the third quadrant, from −1.25 to the origin (arriving along −90°).
• For ω < 0 it is the mirror image in the second quadrant.
• The infinite semicircle maps to the origin.

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           ....                     .....|
         ...                            ..
---------*-----@-------------------------*--------
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       -1.25                             0
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
w = 0 at -1.25; w -> +-inf at origin

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 curve runs from the origin through the upper half to −1.25 and back through the lower half. This is a counter-clockwise loop, and it contains −1 because −1.25 < −1. So N = −1.

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

The closed-loop system is stable. Check: $s^2 - 4 + s + 5 = s^2 + s + 1 = 0$ has roots $-0.5 \pm j0.866$. ✓

Gain margin

Phase crossover is at ω = 0, where $|G| = 5/4 = 1.25$.

$$GM = \frac{1}{1.25} = 0.8 = 20\log0.8 = -1.94\ \text{dB}$$

For this open-loop unstable system, the negative GM means the gain can be reduced by at most a factor of 0.8 (gain $K > 0.8$ is needed) before the system becomes unstable.

Answer: P = 1, N = −1, Z = 0, so the system is stable. GM = 0.8 (−1.94 dB).