Bode Diagram and Unconditional Stability of a Second-Order Servomechanism: 22-Mec-A3 System Analysis and Control, December 2019, Q5

Question 5: Bode Diagram and Unconditional Stability of a Second-Order Servomechanism (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. Unity-feedback servomechanism with $G(s)=\dfrac{16}{s\left(2+s\right)}$.

Find. (a) the magnitude and phase characteristics plotted against frequency; (b) whether any finite increase of forward gain can destabilise the loop, with justification.

Approach. Put $G(s)$ into time-constant (Bode) form to expose the gain factor and corner frequency, sketch the straight-line asymptotes and the phase curve, then argue stability from the phase asymptote and confirm it with the closed-loop characteristic equation.

Convert to time-constant form. Factor the 2 out of the linear term: $$G(s)=\frac{16}{s\left(2+s\right)}=\frac{16}{2s\left(1+0.5s\right)}=\frac{8}{s\left(1+0.5s\right)}.$$ The Bode gain is $K_B=8\ \text{s}^{-1}$, there is one pole at the origin, and the corner frequency is $\omega_c = 1/0.5 = 2\ \text{rad}\,\text{s}^{-1}$.
Lay out the magnitude asymptotes. Below the corner the integrator dominates: $\left|G\right|_{\text{dB}} \approx 20\log_{10}\!\left(8/\omega\right)$, a $-20$ dB/decade line passing through $20\log_{10}8 = 18.1$ dB at $\omega=1$ and crossing 0 dB at $\omega=8\ \text{rad}\,\text{s}^{-1}$. Beyond the corner the pole at $-2$ adds a further $-20$ dB/decade, giving $\boxed{\;-20\ \text{dB/dec below }2\ \text{rad/s},\ -40\ \text{dB/dec above}\;}$ The two asymptotes intersect at the corner at $20\log_{10}(8/2)=12.0$ dB; the true curve there lies 3 dB below that, at $\left|G(j2)\right|=8/(2\sqrt{2})=2.83$, i.e. $9.0$ dB.
Lay out the phase. $\angle G(j\omega) = -90^\circ - \tan^{-1}\!\left(0.5\omega\right)$, so the phase starts at $-90^\circ$ at low frequency, passes $-135^\circ$ at the corner $\omega=2\ \text{rad}\,\text{s}^{-1}$, and approaches $-180^\circ$ asymptotically as $\omega\to\infty$ — without ever attaining it.
Locate the gain crossover and the phase margin. Setting $\left|G(j\omega)\right|=1$ gives $16 = \omega\sqrt{\omega^{2}+4}$, i.e. $u^{2}+4u-256=0$ with $u=\omega^{2}$, whence $u=14.12$ and $$\omega_{gc}=3.76\ \text{rad}\,\text{s}^{-1}, \qquad \angle G(j\omega_{gc}) = -90^\circ-\tan^{-1}(1.879) = -152.0^\circ,$$ so the phase margin is $\boxed{\;\text{PM}=180^\circ-152.0^\circ=28.0^\circ\;}$ and, since the phase never reaches $-180^\circ$, the gain margin is infinite.
(b) Argue from the phase asymptote. Raising the forward gain by a factor $K_f$ shifts the magnitude curve bodily upward by $20\log_{10}K_f$ dB and moves the gain crossover to a higher frequency, but it leaves the phase curve untouched. Instability by the Nyquist/Bode criterion requires $\left|G\right|=1$ at a frequency where the phase is $-180^\circ$ or beyond. Here the phase exceeds $-180^\circ$ at every finite frequency, so no finite gain can create that condition: $\boxed{\;\text{the system can never be made unstable by increasing the forward gain}\;}$
Confirm algebraically. With forward gain $K_f$ the characteristic equation is $$s\left(s+2\right)+16K_f = s^{2}+2s+16K_f = 0,$$ a second-order polynomial whose coefficients are all positive for every $K_f>0$; the roots are $s=-1\pm\sqrt{1-16K_f}$ and their real part is $-1$ once $K_f>1/16$. The closed-loop poles simply run up the vertical line $\sigma=-1$ and never cross into the right half plane. At the printed gain ($K_f=1$) the loop is $\omega_n=\sqrt{16}=4\ \text{rad}\,\text{s}^{-1}$, $\zeta=2/(2\times4)=0.25$.

Bode diagram of G(s) = 16/[s(2 + s)]: magnitude falls at −20 dB/decade to the corner at 2 rad/s and at −40 dB/decade beyond it; the phase runs from −90° to an asymptote of −180° that it never reaches.

The plotted curves show exactly what the algebra predicts. The magnitude passes through 0 dB at 3.76 rad/s while the phase there is still 28° short of the critical −180° line, and the phase trace flattens toward −180° without touching it. The practical consequence is worth stating: although this loop cannot be made unstable, large gain still drives $\zeta$ toward zero, so the response becomes violently oscillatory and long-settling — unusable in service even though it is formally stable. Any real machine also has neglected high-frequency lags (amplifier, tachometer filtering, structural modes) which add phase beyond −180° and destroy this guarantee.

Question 5 — frequency-response summary
Quantity	Value
Bode (time-constant) form	$G(s)=8/\left[s\left(1+0.5s\right)\right]$
Bode gain $K_B$	$8\ \text{s}^{-1}$ ($18.1$ dB at $\omega=1$)
Corner frequency	$2\ \text{rad}\,\text{s}^{-1}$
Asymptote slopes	$-20$ dB/dec, then $-40$ dB/dec
Gain crossover $\omega_{gc}$	$3.76\ \text{rad}\,\text{s}^{-1}$
Phase margin	$28.0^\circ$
Gain margin	infinite
Closed loop at $K_f=1$	$\omega_n=4\ \text{rad}\,\text{s}^{-1}$, $\zeta=0.25$
(b) Can it be unstable?	No — stable for every finite forward gain

Topic: Bode construction from time-constant form, and the unconditional stability of loops with fewer than three poles.

Key relations: $\left|G\right|_{\text{dB}}=20\log_{10}\left|G\right|$; slope $-20n$ dB/decade per $n$ poles; $\angle G=-90^\circ-\tan^{-1}(\omega T)$ for $K/[s(1+Ts)]$; $\text{PM}=180^\circ+\angle G(j\omega_{gc})$; instability requires $\left|G\right|\ge1$ at $\angle G\le-180^\circ$.

Why this works: Each simple pole contributes at most $-90^\circ$ of phase, so a loop with one integrator and one lag can only approach $-180^\circ$ in the limit. Since changing the forward gain translates the magnitude curve vertically but leaves the phase curve fixed, the Nyquist plot of such a loop can be scaled arbitrarily without ever encircling the $-1$ point. Three or more lags — as in Question 6 — break this immediately, which is exactly why that loop does have a critical gain.

Common pitfall: Sketching the asymptotes from the raw form $16/[s(2+s)]$ and reading the low-frequency line as $20\log_{10}16$; the gain factor must first be normalised so the lag reads $(1+0.5s)$, giving $K_B=8$. A second is confusing “stable at any gain” with “acceptable at any gain” — the damping ratio still collapses as $1/\sqrt{K_f}$.

Source: Ogata, Modern Control Engineering, 5th ed., Ch. 7.2 (Bode diagrams) and 7.6 (stability from frequency response); Nise, Control Systems Engineering, 8th ed., Ch. 10.

Question 5 of 6: Bode Diagram and Unconditional Stability of a Second-Order Servomechanism

Question 5: Bode Diagram and Unconditional Stability of a Second-Order Servomechanism (25 marks)