--- title: "Bode plot of 48(1+s)/[s²(1+3s)(1+0.5s)(2+0.2s)]" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:bode-48-1s-s2-1-3s" source: https://www.bctengineer.com/q/semester_3/control_systems/bode-48-1s-s2-1-3s site: BCT Engineering --- # Bode plot of 48(1+s)/[s²(1+3s)(1+0.5s)(2+0.2s)] Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2069 Chaitra, 10 marks - Video: [Bode plot in control system](https://www.youtube.com/watch?v=qnsznGf70eU) — 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/bode-48-1s-s2-1-3s - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question Draw the bode plot for transfer function G(s) = 48(1+s)/[s²(1+3s)(1+0.5s)(2+0.2s)]; from the graph determine (i) Phase crossover frequency (ii) Gain crossover frequency (iii) P.M (iv) G.M (v) Stability of the system. ## Short answer From the Bode plot of G(s)=24(1+s)/[s²(1+3s)(1+0.5s)(1+0.1s)], phase remains below −180° at all finite frequencies, so no phase crossover exists. Gain crossover ωgc≈2.34 rad/s gives PM≈−77.6°. Gain margin is negative (effectively −∞ dB); the closed-loop system is unstable. ## Answer #### 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 | Range | Factor added | Net slope | |---|---|---| | $\omega<0.333$ | $24/s^2$ | $-40$ dB/dec | | $0.333$–$1$ | pole $1/(1+3s)$ | $-60$ dB/dec | | $1$–$2$ | zero $(1+s)$ | $-40$ dB/dec | | $2$–$10$ | pole $1/(1+0.5s)$ | $-60$ dB/dec | | $>10$ | pole $1/(1+0.1s)$ | $-80$ dB/dec | $$ \begin{aligned} \omega=0.1:&\ 20\log(24/0.01) = 67.6\ \text{dB}\\ \omega=0.333:&\ 67.6-40\log(3.33)=46.7\ \text{dB}\\ \omega=1:&\ 46.7-60\log 3 = 18.1\ \text{dB}\\ \omega=2:&\ 18.1-40\log 2 = 6.0\ \text{dB}\\ \omega=10:&\ 6.0-60\log 5 = -35.9\ \text{dB} \end{aligned} $$ #### Step 3: Phase plot $$ \phi(\omega) = -180^\circ + \tan^{-1}\omega - \tan^{-1}3\omega - \tan^{-1}0.5\omega - \tan^{-1}0.1\omega $$ | $\omega$ | 0.1 | 0.2 | 0.5 | 1 | 2 | 4 | 10 | |---|---|---|---|---|---|---|---| | $\phi$ | $-194^\circ$ | $-207^\circ$ | $-227^\circ$ | $-239^\circ$ | $-253^\circ$ | $-275^\circ$ | $-307^\circ$ | ```text 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. After writing G(s) = 48(1+s)/[s^2(1+3s)(1+0.5s)(2+0.2s)] in time-constant form, what are K and the initial slope? A. K = 48, slope -20 dB/dec B. K = 24, slope -40 dB/dec C. K = 24, slope -20 dB/dec D. K = 48, slope -40 dB/dec Answer: B. Factoring 2 out of (2+0.2s) gives K = 24, and the double pole at the origin gives -40 dB/dec. (Key point: Time-constant form and type) 2. What is the net slope between w = 1 and w = 2 rad/s? A. -60 dB/dec B. -40 dB/dec C. -20 dB/dec D. -80 dB/dec Answer: B. After the pole at 0.333 the slope is -60; the zero at 1 adds +20, giving -40 dB/dec until the pole at 2. (Key point: Slope changes) 3. Why does this system have no phase crossover frequency? A. The magnitude is always above 0 dB B. The phase is always below -180 degrees because tan^-1(3w) > tan^-1(w) C. The phase is always above -180 degrees D. The system has no poles Answer: B. The lag from the pole at 0.333 always exceeds the lead from the zero at 1, so the phase never returns to -180 degrees. (Key point: Phase crossover frequency) 4. With PM about -77.6 degrees, what is the conclusion? A. Stable with large margin B. Marginally stable C. Unstable, two closed-loop poles in the RHP D. Stable but oscillatory Answer: C. A negative PM (and negative GM) means unstable; the characteristic equation has two RHP roots. 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