Bode plot of 4(s+4)/[(s+2)(s²+2s+4)]

Step 1: Time-constant (Bode) form

$$G(s)=\frac{4(s+4)}{(s+2)(s^2+2s+4)} = \frac{4\times4\,(1+s/4)}{2\times4\,(1+s/2)\left(1+\frac{s}{2}+\frac{s^2}{4}\right)} = \frac{2(1+0.25s)}{(1+0.5s)\left(1+0.5s+0.25s^2\right)}$$

• $K=2$: $20\log2 = 6.02$ dB, type-0 (initial slope 0 dB/dec).
• Simple pole: corner $\omega=2$ rad/s.
• Quadratic poles: $\omega_n=2$ rad/s, $2\zeta/\omega_n=0.5 \Rightarrow \zeta=0.5$ (corner 2 rad/s).
• Simple zero: corner $\omega=4$ rad/s.

Step 2: Asymptotic magnitude plot

Corrections at $\omega=2$: quadratic with $\zeta=0.5$ gives 0 dB, the simple pole $-3.01$ dB, the zero $+0.97$ dB, so the actual value there is $6.02-3.01+0.97=3.98$ dB.

Step 3: Phase plot

$$\phi(\omega)=\tan^{-1}\frac{\omega}{4}-\tan^{-1}\frac{\omega}{2}-\tan^{-1}\frac{2\zeta(\omega/\omega_n)}{1-(\omega/\omega_n)^2}$$

 dB 6 |------\  0 dB/dec
    0 |-------\-------------- w
      |      2 \  -60
  -12 |         \__ 4
      |            \__ -40
 deg 0|--__
  -90 |    \__
 -180 |- - - - -\____________ (approaches -180)

Step 4: Margins

• Gain crossover: the asymptote gives $6.02-60\log(\omega/2)=0 \Rightarrow \omega\approx2.52$ rad/s; the exact value is $\omega_{gc}=2.59$ rad/s.

$$PM = 180^\circ + \phi(\omega_{gc}) = 180^\circ - 136.8^\circ = 43.2^\circ$$

• Phase crossover: the net phase goes from $0^\circ$ toward $-180^\circ$ (three poles $-270^\circ$, one zero $+90^\circ$) and reaches $-180^\circ$ only at $\omega=\infty$. So $\omega_{pc}=\infty$ and

$$GM = \infty\ \text{dB}$$

Stability

All open-loop poles and zeros are in the left half plane (minimum phase), GM $=\infty$ and PM $=43.2^\circ>0$, so the closed-loop system is stable, with a reasonable phase margin (moderate overshoot, $\zeta\approx0.43$).

Answer: $\omega_{gc}\approx2.59$ rad/s, PM ≈ 43°, GM = ∞; the closed-loop system is stable.