Question 2 of 6: Bode Diagrams of Two Transfer Functions

Question 2: Bode Diagrams of Two Transfer Functions (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two open-loop transfer functions, each already in a factored form with unity leading coefficients:

Find. Magnitude (dB) and phase (degrees) versus $\log\omega$ for each transfer function: the asymptote slopes, the corner frequencies, the exact values at the corners, and the resulting shape of the two curves.

Approach. Put each factor in the time-constant (Bode) form $(1+s/p)$ so that the low-frequency asymptote reads directly off the gain constant, add the individual asymptotic contributions ($\pm20$ dB/decade per real factor, $\pm90^\circ$ per real factor), then correct the asymptotes at each corner with the exact $-3$ dB / $+3$ dB deviation of a first-order factor.

Part (a): $G(s)=1/[s^{2}(s+1)]$

1. Identify the Bode constant and the corner. The denominator factor $(s+1)$ is already in time-constant form, so the Bode gain is $K_{B}=1$ and the single corner frequency is $\omega_{c}=1$ rad/s. The double pole at the origin contributes $1/(j\omega)^{2}$ at every frequency.
2. Build the magnitude asymptotes. Below the corner only the double integrator acts, giving a slope of $-40$ dB/decade passing through $0$ dB at $\omega=1$ rad/s. Above the corner the pole at $-1$ adds a further $-20$ dB/decade, so the asymptote steepens to $-60$ dB/decade. Sample values on the low-frequency asymptote: $$|G(j0.1)| = \frac{1}{0.1^{2}\sqrt{1+0.1^{2}}} = 99.5 \;\Rightarrow\; 39.96\ \text{dB}.$$ At the corner the exact magnitude falls $3$ dB below the asymptote's $0$ dB, so $$\boxed{\;|G(j1)| = \tfrac{1}{\sqrt2}= -3.01\ \text{dB},\qquad \text{slope } -40 \to -60\ \text{dB/decade at }\omega=1\ \text{rad/s}. \;}$$
3. Locate the gain crossover. Setting $|G(j\omega)|=1$ requires $\omega^{2}\sqrt{\omega^{2}+1}=1$, which solves numerically to $\omega_{gc}=0.8688$ rad/s — slightly below the corner, as the $-3$ dB correction implies.
4. Build the phase curve. The double integrator contributes a constant $-180^\circ$ and the real pole a further $-\tan^{-1}\omega$. Hence $\angle G = -180^{\circ}-\tan^{-1}\omega$, which starts at $-180^\circ$, passes through $-225^{\circ}$ at the corner $\omega=1$, and approaches $-270^{\circ}$ at high frequency. The phase is below $-180^\circ$ at every frequency, so the phase margin is negative. Evaluated at the gain crossover found in step 3 — not at the corner — $\angle G(j0.8688) = -180^{\circ}-\tan^{-1}(0.8688) = -220.99^{\circ}$, so $\mathrm{PM} = 180^{\circ}+\angle G = -40.99^{\circ}$. This plant cannot be stabilised by proportional gain alone and needs phase lead.

Part (b): $G(s)=s/(s+1)^{2}$

The printed denominator $(s+1)(s+1)$ is a repeated first-order factor, so $G(s)=s/(s+1)^{2}$: a single differentiating zero at the origin against a double pole at $-1$ rad/s.

1. Set the low-frequency behaviour. For $\omega\ll1$ the denominator is approximately unity and $G\approx j\omega$, so the magnitude asymptote rises at $+20$ dB/decade through $0$ dB at $\omega=1$ rad/s, and the phase sits at $+90^{\circ}$. At $\omega=0.1$ the asymptote reads $-20$ dB.
2. Set the high-frequency behaviour. For $\omega\gg1$, $G\approx j\omega/(j\omega)^{2}=1/(j\omega)$, so the magnitude falls at $-20$ dB/decade and the phase settles at $-90^{\circ}$. The net slope change at the corner is $-40$ dB/decade, because both repeated poles break at the same frequency.
3. Evaluate the peak. The magnitude is $|G(j\omega)| = \omega/(1+\omega^{2})$, whose maximum is found from $d/d\omega\,[\omega/(1+\omega^{2})]=0 \Rightarrow \omega=1$ rad/s. There $$\boxed{\;|G(j1)| = \tfrac{1}{2} = -6.02\ \text{dB},\qquad \angle G(j1)=90^{\circ}-2\tan^{-1}(1)=0^{\circ}. \;}$$ The exact curve therefore lies $6$ dB below the intersection of the two asymptotes, which is the standard correction for a repeated first-order pole (twice the usual 3 dB) and is the single most important number to place on the sketch.
4. Build the phase curve. $\angle G = 90^{\circ}-2\tan^{-1}\omega$, which decreases monotonically from $+90^{\circ}$ through $0^{\circ}$ at the corner to $-90^{\circ}$. Because the phase never reaches $-180^{\circ}$, this plant has an infinite gain margin — the exact opposite of part (a), and a useful contrast to draw in the answer.