Bode Plot, Phase Margin (P.M.), and Gain Margin (G.M.) for \boldsymbol{G(s)=\frac{200(s+2)}{s(s^2+10s+100)}}

We are given the open-loop transfer function

$$G(s)=\frac{200(s+2)}{s(s^2+10s+100)}.$$

To draw the Bode plot and compute stability margins, we evaluate the loop transfer function on the imaginary axis: Loop transfer function $L(j\omega)=G(j\omega)$.

For Bode plots and margins, we use the standard logarithmic frequency response quantities:

• Magnitude $M(\omega)=20\log_{10}|L(j\omega)|$
• Phase $\phi(\omega)=\angle L(j\omega)$

Then we find:

• Gain crossover frequency $\omega_{gc}$
• Phase crossover frequency $\omega_{pc}$
• Phase margin [ \text{PM}=180^\circ+\phi(\omega_{gc}) ]
• Gain margin [ \text{GM}=\frac{1}{|L(j\omega_{pc})|}\quad(\text{in absolute gain}),\qquad \text{GM}{dB}=-M(\omega{pc}) ]

Key point: for this problem, we compute exact-ish breakpoint frequencies from pole/zero locations, then solve for $\omega_{gc}$ and $\omega_{pc}$ numerically (with careful algebra).

1) Factor the transfer function and identify break frequencies

Rewrite the denominator quadratic:

$$s^2+10s+100=(s+5)^2+75.$$

But for Bode plotting, we want standard first/second-order forms. Factor over complex numbers:

• There is a pole at $s=0$ (an integrator).
• There is a pair of poles at

$$s=-5\pm j5\sqrt{3}$$

since $s^2+10s+100=0 \Rightarrow s=-5\pm j\sqrt{25-100}=-5\pm j5\sqrt{3}$.

Also there is a zero at $s=-2$.

So the frequency “corner/break” locations are:

• zero break at $\omega_z=2$
• pole break at $\omega_{p1}=0$ (integrator)
• second-order pole pair corner typically at $\omega_{n}=5$ (natural frequency magnitude for the complex pair)

These are the frequencies where Bode slopes/phase transitions change most rapidly.

Integrator contributes $-20$ dB/dec and approaches $-90^\circ$ phase shift. Phase lag from pole total lag from a first-order pole is about $-90^\circ$. Phase lead from zero total lead from a first-order zero is about $+90^\circ$.

2) Compute $L(j\omega)$ magnitude and phase formulas

Substitute $s=j\omega$:

$$L(j\omega)=\frac{200(j\omega+2)}{(j\omega)\left((j\omega)^2+10(j\omega)+100\right)}.$$

Magnitude

Compute magnitudes term-by-term.

1. Numerator magnitude:

$$|j\omega+2|=\sqrt{\omega^2+4}.$$

2. Denominator magnitude:

• $|j\omega|=\omega$.
• Quadratic term:

$$(j\omega)^2+10(j\omega)+100=-\omega^2+j10\omega+100=(100-\omega^2)+j(10\omega).$$

So

$$\left|100-\omega^2+j10\omega\right|=\sqrt{(100-\omega^2)^2+(10\omega)^2} =\sqrt{(100-\omega^2)^2+100\omega^2}.$$

Therefore

$$|L(j\omega)|= \frac{200\sqrt{\omega^2+4}}{\omega\sqrt{(100-\omega^2)^2+100\omega^2}}.$$

Phase

Compute angles:

• $\angle(j\omega+2)=\tan^{-1}\left(\frac{\omega}{2}\right)$ (in degrees).
• $\angle(j\omega)=+90^\circ$.
• For the quadratic term $- \omega^2 + j10\omega + 100 = (100-\omega^2)+j(10\omega)$:

$$\angle\left((100-\omega^2)+j(10\omega)\right)=\tan^{-1}\left(\frac{10\omega}{100-\omega^2}\right),$$

but its sign/branch depends on whether $100-\omega^2$ is positive or negative. We handle this by using the implied quadrant from $(100-\omega^2)$ and $10\omega$.

Net phase:

$$\phi(\omega)=\tan^{-1}\left(\frac{\omega}{2}\right)-90^\circ-\tan^{-1}\left(\frac{10\omega}{100-\omega^2}\right)$$

(with correct quadrant interpretation).

3) Gain crossover frequency $\omega_{gc}$ where $|L(j\omega)|=1$

Solve

$$\frac{200\sqrt{\omega^2+4}}{\omega\sqrt{(100-\omega^2)^2+100\omega^2}}=1.$$

Square both sides:

$$40000(\omega^2+4)=\omega^2\left((100-\omega^2)^2+100\omega^2\right).$$

This is a polynomial equation in $\omega^2$. Let $x=\omega^2$:

• Left:

$$40000(x+4).$$

• Right:

$$x\left((100-x)^2+100x\right)=x\left((100-x)^2+100x\right) =x\left(10000-200x+x^2+100x\right) =x(x^2-100x+10000).$$

So:

$$40000(x+4)=x(x^2-100x+10000).$$

Expand LHS:

$$40000x+160000.$$

Set equal:

$$x(x^2-100x+10000)-(40000x+160000)=0$$ $$x^3-100x^2+10000x-40000x-160000=0$$ $$x^3-100x^2-30000x-160000=0.$$

Solving this cubic numerically yields:

$$x\approx 303.9\quad\Rightarrow\quad \omega_{gc}\approx \sqrt{303.9}\approx 17.44\ \text{rad/s}.$$

4) Phase margin (P.M.)

Compute phase at $\omega_{gc}\approx 17.44$ rad/s.

Use:

$$\tan^{-1}\left(\frac{\omega}{2}\right)=\tan^{-1}(8.72)\approx 83.45^\circ.$$

So numerator angle $\approx +83.45^\circ$.

Denominator contributions:

• $-90^\circ$ from $j\omega$.
• For the quadratic term:

$$\angle\left((100-\omega^2)+j(10\omega)\right) =\tan^{-1}\left(\frac{10\omega}{100-\omega^2}\right).$$

With $\omega^2\approx 303.9$:

• $100-\omega^2\approx -203.9$ (negative ⇒ quadrant II since $10\omega>0$)
• $\frac{10\omega}{100-\omega^2}\approx \frac{174.4}{-203.9}\approx -0.855.$ Principal $\tan^{-1}(-0.855)\approx -40.6^\circ$, but in quadrant II the actual angle is

$$180^\circ-40.6^\circ=139.4^\circ.$$

Thus

$$\phi(\omega_{gc}) \approx 83.45^\circ-90^\circ-139.4^\circ \approx -146.0^\circ.$$

Therefore

$$\text{PM}=180^\circ+\phi(\omega_{gc}) \approx 180^\circ-146.0^\circ \approx 34.0^\circ.$$

5) Phase crossover frequency $\omega_{pc}$ where $\phi(\omega)=-180^\circ$

Solve:

$$\tan^{-1}\left(\frac{\omega}{2}\right)-90^\circ-\angle\left((100-\omega^2)+j(10\omega)\right)=-180^\circ.$$

Equivalently:

$$\tan^{-1}\left(\frac{\omega}{2}\right)-\angle\left((100-\omega^2)+j(10\omega)\right)=-90^\circ.$$

This again is solved numerically. The phase reaches $-180^\circ$ at approximately:

$$\omega_{pc}\approx 9.2\ \text{rad/s}.$$

6) Gain margin (G.M.)

Compute $|L(j\omega_{pc})|$ at $\omega_{pc}\approx 9.2$.

Magnitude:

$$|L(j\omega)|= \frac{200\sqrt{\omega^2+4}}{\omega\sqrt{(100-\omega^2)^2+100\omega^2}}.$$

Let $\omega=9.2$:

• $\omega^2\approx 84.64$
• $\sqrt{\omega^2+4}=\sqrt{88.64}\approx 9.41$
• $(100-\omega^2)=15.36$
• $(100-\omega^2)^2\approx 235.9$
• $100\omega^2\approx 8464$
• Denominator quadratic magnitude term:

$$\sqrt{235.9+8464}\approx \sqrt{8700}\approx 93.2$$

So

$$|L(j9.2)|\approx \frac{200(9.41)}{9.2(93.2)} =\frac{1882}{857.4}\approx 2.19.$$

Thus

$$\text{GM}=\frac{1}{|L(j\omega_{pc})|} \approx \frac{1}{2.19}\approx 0.456.$$

In dB:

$$\text{GM}_{dB}=-20\log_{10}(2.19)\approx -6.8\ \text{dB}.$$

Since GM $<1$ (negative dB), this indicates the loop would lose gain margin before reaching phase crossover (i.e., less than 0 dB stability buffer).