--- title: "Bode plot of 4(s+4)/[(s+2)(s²+2s+4)]" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:bode-4-s4-s2-s2-2s" source: https://www.bctengineer.com/q/semester_3/control_systems/bode-4-s4-s2-s2-2s site: BCT Engineering --- # Bode plot of 4(s+4)/[(s+2)(s²+2s+4)] Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2068 Baisakh (old course), 8 marks - Video: [1. Bode Plot Solved Example: Steps, Procedure, Parameters, and Stability](https://www.youtube.com/watch?v=lTnqZVNReF8) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/bode-4-s4-s2-s2-2s - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question Draw bode plot of a system having open loop transfer function G(s) = 4(s+4)/[(s+2)(s²+2s+4)]. Also analyze the stability. ## Short answer From the Bode plot of G(s)=2(1+0.25s)/[(1+0.5s)(1+0.5s+0.25s²)], gain crossover occurs at ωgc≈2.59 rad/s with PM≈43.2°. Phase approaches −180° only at infinity, so GM=∞ dB. The closed-loop system is stable with good relative stability. ## Answer #### 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 | Range | Slope | Value at end | |---|---|---| | $\omega<2$ | 0 dB/dec | 6.02 dB | | $2<\omega<4$ | $0-20-40=-60$ dB/dec | $6.02-60\log2 = -12.04$ dB | | $\omega>4$ | $-60+20=-40$ dB/dec | at $\omega=10$: $-12.04-40\log2.5=-27.96$ dB | 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 $$ \phi(\omega)=\tan^{-1}\frac{\omega}{4}-\tan^{-1}\frac{\omega}{2}-\tan^{-1}\frac{2\zeta(\omega/\omega_n)}{1-(\omega/\omega_n)^2} $$ | $\omega$ (rad/s) | 0.1 | 0.5 | 1 | 2 | 2.59 | 4 | 10 | |---|---|---|---|---|---|---|---| | $|G|$ (dB, exact) | 6.02 | 6.09 | 6.22 | 3.98 | 0 | $-9.1$ | $-27.3$ | | $\phi$ | $-4.3^\circ$ | $-21.8^\circ$ | $-46.2^\circ$ | $-108.4^\circ$ | $-137^\circ$ | $-164.7^\circ$ | $-178.7^\circ$ | ```text 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For G(s) = 4(s+4)/[(s+2)(s^2+2s+4)], what is the DC gain K and the initial slope? A. K = 2 (6.02 dB), slope 0 dB/dec B. K = 4, slope -20 dB/dec C. K = 1, slope 0 dB/dec D. K = 2, slope -20 dB/dec Answer: A. At s = 0, G = 16/8 = 2, so the system is type 0 with 20 log 2 = 6.02 dB. (Key point: Bode form constants) 2. For s^2 + 2s + 4, what are wn and zeta? A. wn = 4, zeta = 0.25 B. wn = 2, zeta = 1 C. wn = 2, zeta = 0.5 D. wn = 4, zeta = 0.5 Answer: C. wn^2 = 4 gives wn = 2, and 2 zeta wn = 2 gives zeta = 0.5. (Key point: Damping of quadratic) 3. Between w = 2 and w = 4 rad/s, what is the asymptotic slope? A. -20 dB/dec B. -40 dB/dec C. -60 dB/dec D. 0 dB/dec Answer: C. The simple pole (-20) and the quadratic pole pair (-40) both start at 2 rad/s, giving -60 dB/dec; the zero at 4 then adds +20. (Key point: Slope between 2 and 4) 4. The system has GM = infinity and PM = 43.2 degrees. What is the conclusion? A. Unstable B. Marginally stable C. Stable D. Cannot be determined Answer: C. The phase reaches -180 degrees only at infinite frequency, so GM is infinite, and PM is positive, so the closed loop is stable. (Key point: Stability conclusion) ## Related in Control Systems - [Bode plot of 40(s+2)/[(2+s+25s²)(1+2s)s]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abode-40-s2-2-s-25s2) — Control Systems, chapter 6, asked 3× - [Transfer function from asymptotic Bode plot](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-from-bode-asymptote) — Control Systems, chapter 6, asked 2× - [Determining transfer function from Bode plot](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abode-to-tf) — Control Systems, chapter 6, asked 2× - [Transfer function from Bode plot with example](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abode-to-tf-example) — Control Systems, chapter 6, asked 2× - [Bode plot of 48(1+s)/[s²(1+3s)(1+0.5s)(2+0.2s)]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abode-48-1s-s2-1-3s) — Control Systems, chapter 6 - [Transfer function from field asymptotic Bode plot](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-from-bode-engineer) — Control Systems, chapter 6 ## Related in other subjects - [DT transfer function; drawing a Bode plot](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adt-transfer-function-bode) — Signals and Systems, chapter 6 - [Bode plot: 64(S+2)/(S(S²+0.5S)(S²+3.2S+64))](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Abode-64s2-s-s2-05s-s2) — Electrical Circuit Theory, chapter 5, asked 4× - [Bode plot: 200(S+1)/(S(S+5)(S²+2S+100))](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Abode-200s1-s-s5-s2-2s) — Electrical Circuit Theory, chapter 5, asked 4× - [Bode plot: 10 S(S+3)/((S+1)(S²+2S+16))](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Abode-10s-s3-s1-s2-2s) — Electrical Circuit Theory, chapter 5, asked 4×