Question 4 of 5: Asymptotic Bode construction and the stability margins of $K/[s(s+1)(s+5)]$

Question 4: Asymptotic Bode construction and the stability margins of $K/[s(s+1)(s+5)]$ (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. The unity-feedback loop $L(s)=K/[s(s+1)(s+5)]$ with $K=5$: a free integrator plus two real first-order poles at 1 and 5 rad/s, and no zeros.

Find. The normalised component-by-component Bode construction and the composite diagram; the gain margin, the phase margin and the two frequencies at which they are measured; and the largest gain for which the closed loop remains stable.

Approach. Normalise the transfer function so every factor is unity at low frequency, which turns the sketch into the sum of four standard templates. Then solve the two crossover conditions exactly: the phase crossover from an arctangent identity and the gain crossover from a cubic in $\omega^{2}$. The maximum gain is the present gain multiplied by the gain margin, and Routh confirms it.

1. Part (a) — normalise before sketching. Factor each pole so that it reads $1+s/\omega_{c}$: $$L(s)=\frac{5}{s(s+1)(s+5)} =\frac{5}{1\times5}\cdot\frac{1}{s\,(1+s)\,(1+s/5)} =\frac{1}{s\,(1+s)\,(1+s/5)} .$$ The normalised gain is therefore $\boxed{1\equiv0\ \text{dB}}$, which is what makes the sketch easy: the whole low-frequency behaviour is carried by the integrator alone.
2. Part (a) — the four component templates. The constant gain is a horizontal 0 dB line with $0^\circ$ phase. The integrator $1/s$ is a straight $-20$ dB/decade line passing through 0 dB at $\omega=1$ rad/s, with a constant $-90^\circ$ phase. Each first-order pole is flat at 0 dB below its break frequency and falls at $-20$ dB/decade above it, with phase sweeping from $0^\circ$ to $-90^\circ$ and passing $-45^\circ$ at the break — at $\omega_{c}=1$ rad/s for one pole and $\omega_{c}=5$ rad/s for the other.
3. Part (a) — add the templates to get the composite. Summing slopes, the asymptotic magnitude falls at $-20$ dB/decade below 1 rad/s, at $-40$ dB/decade between 1 and 5 rad/s, and at $-60$ dB/decade above 5 rad/s; the phase runs from $-90^\circ$ at low frequency to $-270^\circ$ at high frequency. Three exact points calibrate the sketch against the true curve: $$|L(j0.1)|=19.96\ \text{dB},\qquad |L(j1)|=-3.18\ \text{dB},\qquad |L(j5)|=-31.14\ \text{dB} .$$ The asymptotic construction reads 20 dB, 0 dB and $-27.96$ dB at those three frequencies, so the familiar $-3$ dB error appears at each corner and the asymptotes are otherwise accurate to a fraction of a dB.
4. Part (b) — phase crossover in closed form. The phase is $\angle L=-90^\circ-\tan^{-1}\omega-\tan^{-1}(\omega/5)$, so $\angle L=-180^\circ$ requires $\tan^{-1}\omega+\tan^{-1}(\omega/5)=90^\circ$. Two arctangents sum to a right angle exactly when the product of their arguments is one, so $\omega^{2}/5=1$ and $$\boxed{\omega_{pc}=\sqrt5=2.236\ \text{rad/s}}$$ independently of the gain — a change in $K$ moves the magnitude curve, never the phase curve.
5. Part (b) — gain margin. Evaluating the magnitude there, $$|L(j\omega_{pc})|=\frac{5}{2.236\times\sqrt{1+5}\times\sqrt{25+5}} =\frac{5}{2.236\times2.449\times5.477}=\frac{1}{6} ,$$ so the gain margin is $$\boxed{GM=6=15.56\ \text{dB at }\omega_{pc}=2.236\ \text{rad/s}} .$$
6. Part (b) — gain crossover. Setting $|L|=1$ gives $\omega^{2}(1+\omega^{2})(25+\omega^{2})=25$, i.e. with $x=\omega^{2}$, $$x^{3}+26x^{2}+25x-25=0\;\Longrightarrow\;x=0.6074 \;\Longrightarrow\;\boxed{\omega_{gc}=0.7794\ \text{rad/s}} .$$ The other two roots are negative, so the crossover is unique.
7. Part (b) — phase margin. At that frequency $$\angle L=-90^\circ-\tan^{-1}(0.7793)-\tan^{-1}(0.1559)=-136.79^\circ ,$$ hence $$\boxed{PM=180^\circ-136.79^\circ=43.21^\circ\ \text{at } \omega_{gc}=0.779\ \text{rad/s}} .$$ Both margins are comfortable, and a $43^\circ$ phase margin corresponds to a closed-loop damping of roughly $\zeta\approx0.43$, i.e. about 22% overshoot.
8. Part (c) — maximum gain, two ways. The gain margin is by definition the factor by which the gain may be raised before the magnitude reaches unity at $\omega_{pc}$, so $$K_{\max}=K\times GM=5\times6=\boxed{30} .$$ Routh agrees: the closed-loop polynomial is $s^{3}+6s^{2}+5s+K$, which requires $6\times5>K$, i.e. $K<30$. At exactly $K=30$ the loop sustains an undamped oscillation at $\sqrt5=2.236$ rad/s, matching the phase crossover.

It is worth noticing that the frequency-domain and root-locus answers to part (c) are literally the same calculation. The Routh limit $K<30$ and the auxiliary equation $6s^{2}+30=0$, giving $s=\pm j\sqrt5$, reproduce both the maximum gain and the phase crossover frequency; the Bode route reaches them through the arctangent identity instead. Getting two different numbers from the two routes is a reliable signal that one of them contains an arithmetic slip.