Bode Diagrams of Two Open-Loop Transfer Functions: 23-Mechatronics-A1 Systems Dynamics and Controls, Undated paper, Q5

Question 5: Bode Diagrams of Two Open-Loop Transfer Functions

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Given. Two open-loop transfer functions: (a) one integrator plus an underdamped quadratic factor; (b) a double integrator plus one real pole at $s=-10$.

Find. The asymptotic and corrected (actual) Bode magnitude and phase plots for each.

Approach. Rewrite each transfer function in Bode (time-constant) form to read off the DC gain and break frequencies, draw the straight-line magnitude asymptote, then correct near each break using the exact quadratic/first-order corrections and plot the running phase.

Part (a) — Bode form. $$G(s)H(s)=\frac{150}{s(s^2+2s+9)}=\frac{K}{s\!\left(\dfrac{s^2}{\omega_n^2} +\dfrac{2\zeta s}{\omega_n}+1\right)},\qquad K=\frac{150}{9}=16.7,\ \ \omega_n=3\ \text{rad/s},\ \ \zeta=\tfrac13=0.333$$
Part (a) — asymptote. $-20$ dB/decade line through $20\log_{10} 16.7=24.4$ dB at $\omega=1$ rad/s (the integrator alone), breaking to $-60$ dB/decade at $\omega_n=3$ rad/s (integrator's $-20$ plus the quadratic's additional $-40$).
Part (a) — correction near the break. Since $\zeta=0.333< 0.707$ the quadratic factor peaks: the correction at $\omega=\omega_n$ is $-20\log_{10}(2\zeta)=+3.5$ dB above the asymptote, and the true resonant peak $M_r=1/(2\zeta\sqrt{1-\zeta^2})=1.59$ ($+4.0$ dB) occurs slightly below $\omega_n$ at $\omega_r=\omega_n\sqrt{1-2\zeta^2}=2.65$ rad/s. Phase runs from $-90^\circ$ ($\omega\ll \omega_n$, integrator alone) through exactly $-180^\circ$ at $\omega=\omega_n$, toward $-270^\circ$ as $\omega\to\infty$. $$\boxed{K=16.7,\ \omega_n=3\ \text{rad/s},\ \zeta=0.333,\ M_r\approx+4.0\ \text{dB at}\ \omega_r=2.65\ \text{rad/s}}$$
Part (b) — Bode form. $$G(s)H(s)=\frac{50}{s^2(s+10)}=\frac{K}{s^2(s/10+1)},\qquad K=\frac{50}{10}=5,\quad \text{break at}\ \omega=10\ \text{rad/s}$$
Part (b) — asymptote and correction. $-40$ dB/decade line (double integrator) through $20\log_{10}5=14.0$ dB at $\omega=1$ rad/s, breaking to $-60$ dB/decade at $\omega=10$ rad/s. A simple real pole has no resonance: the standard correction is $-3$ dB exactly at the break and $\mp1$ dB at half/double the break frequency. Phase starts at $-180^\circ$ (double integrator, constant for $\omega\ll10$), runs through $-225^\circ$ at the break, and approaches $-270^\circ$ as $\omega\to\infty$. $$\boxed{K=5,\ \text{break at}\ \omega=10\ \text{rad/s (monotonic, no resonance)}}$$

Figure Q5(a) — asymptotic (dashed) vs. actual (solid) Bode magnitude and phase for $150/[s(s^2+2s+9)]$.

Figure Q5(b) — asymptotic (dashed) vs. actual (solid) Bode magnitude and phase for $50/[s^2(s+10)]$.

	$K$	Break(s)	Asymptotic slopes	Phase range
(a)	16.7	$\omega_n=3$	$-20\to-60$ dB/dec	$-90^\circ\to-270^\circ$
(b)	5	$\omega=10$	$-40\to-60$ dB/dec	$-180^\circ\to-270^\circ$

Concept & Theory

Topic: Constructing Bode magnitude/phase plots from pole-zero (time-constant) form.

Key relations: each factor contributes $\mp20$ dB/decade per order at its break frequency and $\mp90^\circ$ per order ($\mp180^\circ$ for a quadratic factor) of total phase swing, summed algebraically; underdamped quadratic correction $-20\log_{10}(2\zeta)$ at $\omega=\omega_n$ and true peak $M_r=1/(2\zeta\sqrt{1-\zeta^2})$ at $\omega_r=\omega_n\sqrt{1-2\zeta^2}$ when $\zeta<1/\sqrt2$.

Why this works: a rational $G(j\omega)$ is a product of first/second order factors, and log-magnitude turns the product into a sum of straight-line asymptotes — Bode's log-log convenience — while phase is likewise additive term by term.

Common pitfall: forgetting the double slope/phase penalty of a repeated pole at the origin ($s^2$ is a double integrator: $-40$ dB/dec and a flat $-180^\circ$, not $-20$ dB/dec$/-90^\circ$); applying the quadratic-peak formula when $\zeta\ge1/\sqrt2$ (no resonant peak exists in that regime, the curve rolls off monotonically).

Source: Nise, Control Systems Engineering, Ch. 10 (frequency response); Ogata, Modern Control Engineering, §7-3/7-4 (Bode diagrams).

Question 5 of 6: Bode Diagrams of Two Open-Loop Transfer Functions

Question 5: Bode Diagrams of Two Open-Loop Transfer Functions