Bode plot of 48(1+s)/[s²(1+3s)(1+0.5s)(2+0.2s)]

Step 1: Time-constant form

$$G(s)=\frac{48(1+s)}{s^2(1+3s)(1+0.5s)(2+0.2s)} = \frac{24(1+s)}{s^2(1+3s)(1+0.5s)(1+0.1s)}$$

• $K=24$, type-2 (double pole at origin): initial slope $-40$ dB/dec.
• Corner frequencies: pole $1/3=0.333$, zero $1$, pole $1/0.5=2$, pole $1/0.1=10$ rad/s.

Step 2: Asymptotic magnitude plot

$$\begin{aligned} \omega=0.1:&\ 20\log(24/0.01) = 67.6\ \text{dB}\\ \omega=0.333:&\ 67.6-40\log(3.33)=46.7\ \text{dB}\\ \omega=1:&\ 46.7-60\log 3 = 18.1\ \text{dB}\\ \omega=2:&\ 18.1-40\log 2 = 6.0\ \text{dB}\\ \omega=10:&\ 6.0-60\log 5 = -35.9\ \text{dB} \end{aligned}$$

Step 3: Phase plot

$$\phi(\omega) = -180^\circ + \tan^{-1}\omega - \tan^{-1}3\omega - \tan^{-1}0.5\omega - \tan^{-1}0.1\omega$$

 dB  67 |\ -40
     46 | \__ -60
     18 |     \__ -40
      6 |        \__ -60
      0 |-----------x-------- w
                 wgc~2.5
deg -180|------------------- w
        |\___
        |    \____ (always
        |         \__ below -180)

(i) Phase crossover frequency

$\tan^{-1}3\omega > \tan^{-1}\omega$ for every $\omega>0$, so the phase is always below $-180^\circ$; it equals $-180^\circ$ only in the limit $\omega\to0$. The phase curve never crosses $-180^\circ$ at a finite non-zero frequency, so $\omega_{pc}$ does not exist (it is at $\omega=0$).

(ii) Gain crossover frequency

From the asymptotic plot, the slope is $-60$ dB/dec after $\omega=2$ (6 dB): $6.0 - 60\log(\omega/2)=0 \Rightarrow \omega_{gc}=2.52$ rad/s. The exact value from $|G(j\omega)|=1$ is $\omega_{gc}=2.34$ rad/s.

(iii) Phase margin

$$\begin{aligned} \phi(2.34) &= -180^\circ + 66.8^\circ - 81.9^\circ - 49.4^\circ - 13.1^\circ = -257.6^\circ\\ PM &= 180^\circ + (-257.6^\circ) = -77.6^\circ \end{aligned}$$

(iv) Gain margin

The phase crossover is effectively at $\omega=0$, where the magnitude is infinite, so GM is negative (−∞ dB). Since the phase stays below $-180^\circ$ at every frequency, no reduction in gain can make the system stable.

(v) Stability

PM is negative and GM is negative, so the closed-loop system is unstable. Check: the characteristic equation $0.3s^5+3.7s^4+7.2s^3+2s^2+48s+48=0$ has roots $0.995\pm j1.94$ in the right half plane.

Answer: $\omega_{pc}$ – none (phase always below −180°); $\omega_{gc}\approx2.34$ rad/s (2.5 rad/s asymptotic); PM ≈ −77.6°; GM negative; closed loop unstable.