Bode diagrams of two loop transfer functions: 22-Mec-A3 System Analysis and Control, May 2018, Q6

Question 6: Bode diagrams of two loop 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 (loop) transfer functions, to be sketched as magnitude in decibels and phase in degrees against $\log\omega$.

Find. Asymptotic Bode magnitude and phase plots for each, with the corner frequencies, low- and high-frequency asymptotes, and any stability margins that the plots reveal.

Approach. Convert each transfer function to time-constant (Bode) form so the low-frequency asymptote is read directly from the constant, identify the corner frequency and slope change contributed by each factor, then accumulate the slopes and the phase contributions.

Part (a) — a non-minimum-phase second-order loop

Normalise to Bode form. Divide numerator and denominator by their constant terms:$$GH(s)=\frac{s-10}{s^{2}+6s+10}=\frac{-10\left(1-\dfrac{s}{10}\right)}{10\left(1+\dfrac{6s}{10}+\dfrac{s^{2}}{10}\right)}=\frac{-\left(1-s/10\right)}{1+0.6s+0.1s^{2}} .$$The Bode gain is $-1$, so the low-frequency magnitude asymptote is $20\log_{10}1=0$ dB and the low-frequency phase is $180^{\circ}$ because of the minus sign.
Identify the factors. The quadratic $s^{2}+6s+10$ has $\omega_{n}=\sqrt{10}=3.162$ rad/s and $\zeta=6/\left(2\sqrt{10}\right)=0.949$. Since $\zeta>1/\sqrt2$ there is no resonant peak — the magnitude simply rolls off at $-40$ dB/dec past 3.16 rad/s. The numerator contributes a right-half-plane zero at $s=+10$: it lifts the slope by $+20$ dB/dec at $\omega=10$ rad/s exactly like an ordinary zero, but its phase contribution is $-\arctan(\omega/10)$ instead of $+\arctan(\omega/10)$.
Assemble the asymptotes. The magnitude is flat at 0 dB out to 3.16 rad/s, breaks to $-40$ dB/dec there, and returns to $-20$ dB/dec beyond 10 rad/s. Exact values on the curve are $-0.64$ dB at 1 rad/s, $-5.15$ dB at 3.16 rad/s, $-17.67$ dB at 10 rad/s and $-39.96$ dB at 100 rad/s.
Assemble the phase. Starting at $+180^{\circ}$, the RHP zero and the pole pair both drive the phase down, giving a total swing of $180^{\circ}\rightarrow-90^{\circ}$:$$\angle GH(j\omega)=180^{\circ}-\arctan\frac{\omega}{10}-\arctan\frac{6\omega}{10-\omega^{2}} .$$The measured values are $140.6^{\circ}$ at 1 rad/s, $72.5^{\circ}$ at 3.16 rad/s, $-11.3^{\circ}$ at 10 rad/s and $-80.9^{\circ}$ at 100 rad/s, tending to $-90^{\circ}$.
Read the stability implication. The magnitude is exactly 0 dB at DC, where the phase is $180^{\circ}$ — the critical point. Closing this loop confirms it: $s^{2}+6s+10+(s-10)=s^{2}+7s=s(s+7)$, so$$\boxed{\,\text{the closed loop has a pole exactly at }s=0\,}$$The RHP zero has cancelled the DC term of the characteristic polynomial, leaving a free integrator: the loop is marginally stable, and a step reference produces a ramping output rather than a settled one.

Question 6(a) — Bode magnitude and phase for $GH=(s-10)/\left(s^{2}+6s+10\right)$. The phase begins at $+180^{\circ}$ because the Bode gain is negative.

Part (b) — a type-1 loop with three corners

Normalise to Bode form. Factor each binomial so that its constant term is unity:$$GH(s)=\frac{30(s+8)}{s(s+2)(s+4)}=\frac{30\times 8\left(1+s/8\right)}{s\times 2\times 4\left(1+s/2\right)\left(1+s/4\right)}=\frac{30\left(1+s/8\right)}{s\left(1+s/2\right)\left(1+s/4\right)} .$$The Bode constant is 30, not the printed 30 by coincidence but because $30\times8/(2\times4)=30$. The single pole at the origin makes this a type-1 loop, so the velocity error constant is $K_{v}=30$ s$^{-1}$ and the ramp error is $1/K_{v}=0.0333$.
Draw the magnitude asymptotes. At low frequency $\left|GH\right|\approx 30/\omega$, a $-20$ dB/dec line passing through $20\log_{10}30=29.5$ dB at $\omega=1$ rad/s. The slope then steepens at each corner: $-40$ dB/dec past $\omega=2$, $-60$ dB/dec past $\omega=4$, and back to $-40$ dB/dec past the zero at $\omega=8$. Exact magnitudes are 28.4 dB at 1 rad/s, 19.8 dB at 2 rad/s, 8.5 dB at 4 rad/s and $-4.8$ dB at 8 rad/s.
Draw the phase. The integrator contributes a constant $-90^{\circ}$, and each remaining factor swings $\mp90^{\circ}$ about its corner:$$\angle GH(j\omega)=-90^{\circ}-\arctan\frac{\omega}{2}-\arctan\frac{\omega}{4}+\arctan\frac{\omega}{8} .$$The phase starts at $-90^{\circ}$, falls through $-180^{\circ}$ once at $\omega_{pc}=5.657$ rad/s, reaches a minimum of $-185.79^{\circ}$ at $\omega=12.03$ rad/s, and is then pulled back up towards $-180^{\circ}$ from below by the zero, approaching it asymptotically and never recrossing. Because the phase is unwrapped by construction here, never read it from a wrapped arctangent — the wrap would hide the crossing entirely.
Extract the margins. The phase crossover is available in closed form. Writing $GH(j\omega)=N/D$ with $N=240+j30\omega$ and $D=-6\omega^{2}+j\omega\left(8-\omega^{2}\right)$, the loop gain is real (phase $-180^{\circ}$) exactly when $\operatorname{Im}\left\{N\overline{D}\right\}=0$:$$-240\omega\left(8-\omega^{2}\right)-180\omega^{3}=-\omega\left(1920-60\omega^{2}\right)=0\ \Longrightarrow\ \omega_{pc}^{2}=32,$$so$$\omega_{pc}=\sqrt{32}=4\sqrt2=5.657\ \text{rad/s},$$in agreement with a numerical solution of $\angle GH=-180^{\circ}$. There $\left|GH\right|=1.25$, so the gain margin is$$\text{GM}=\frac{1}{1.25}=0.80\quad(-1.94\ \text{dB}).$$The gain crossover is at $\omega_{gc}=6.252$ rad/s, where the phase is $-181.6^{\circ}$, so$$\boxed{\,\text{GM}=-1.94\ \text{dB},\qquad \text{PM}=-1.6^{\circ}\,}$$Both margins are negative.
Confirm with Routh. Scaling the loop gain by $g$, the characteristic polynomial is $s^{3}+6s^{2}+(8+30g)s+240g$, and the Routh condition $6(8+30g)>240g$ requires $g<48/60=0.80$ — precisely the gain margin read off the Bode plot. At the nominal $g=1$ the closed-loop poles are $-6.158$ and $0.079\pm j6.242$, so$$\boxed{\,\text{the closed loop is unstable as it stands}\,}$$and the loop gain must be reduced by at least 20 % (or the plot reshaped with a lead network) before the design is usable.

Question 6(b) — Bode plot for $GH=30(s+8)/[s(s+2)(s+4)]$, showing the phase dipping below $-180^{\circ}$ between the pole corners.

The two parts are a deliberate contrast. Part (a) shows that a right-half-plane zero costs phase while adding magnitude, which is why non-minimum-phase plants are hard to control; part (b) shows a conventional type-1 loop whose gain is simply set too high, so that the phase falls through $-180^{\circ}$ before the magnitude has fallen through 0 dB.

Quantity	Part (a)	Part (b)
Bode gain	$-1$ (0 dB, phase $180^{\circ}$)	$30$ (29.5 dB)
System type	0	1 ($K_{v}=30$, $e_{ss}^{\text{ramp}}=0.0333$)
Corner frequencies	$\omega_{n}=3.162$ ($\zeta=0.949$), zero at $+10$	$2,\ 4$ (poles), $8$ (zero)
Slope sequence	$0\rightarrow-40\rightarrow-20$ dB/dec	$-20\rightarrow-40\rightarrow-60\rightarrow-40$ dB/dec
Phase range	$+180^{\circ}\rightarrow-90^{\circ}$	$-90^{\circ}\rightarrow-180^{\circ}$ (minimum $-185.79^{\circ}$ at $\omega=12.03$ rad/s)
Crossovers	0 dB only at DC	$\omega_{gc}=6.252$, $\omega_{pc}=5.657$ rad/s
Margins / closed loop	pole at $s=0$ (marginally stable)	GM $=-1.94$ dB, PM $=-1.6^{\circ}$ — unstable

Topic: Asymptotic Bode construction, non-minimum-phase zeros, and reading stability margins off the frequency response.

Key relations: Time-constant (Bode) form $K\prod(1+s/z)/\left[s^{N}\prod(1+s/p)\right]$; low-frequency asymptote $20\log_{10}K-20N\log_{10}\omega$; $\pm20$ dB/dec per real factor and $\pm40$ dB/dec per quadratic; resonant peak only when $\zeta<1/\sqrt2$; GM $=1/\left|GH(j\omega_{pc})\right|$; PM $=180^{\circ}+\angle GH(j\omega_{gc})$.

Why this works: Normalising to time-constant form is what makes the low-frequency asymptote readable by inspection; leaving the transfer function in pole-zero form gives a DC asymptote that is wrong by the product of the pole and zero magnitudes. A right-half-plane zero is invisible in the magnitude plot — $\left|1-j\omega/10\right|=\left|1+j\omega/10\right|$ — but reverses the sign of the phase contribution, which is precisely why such plants resist high-gain feedback. The Routh condition on the gain-scaled characteristic polynomial reproduces the gain margin exactly, and running that check catches reciprocal slips for free.

Common pitfall: Reading the crossover frequency off the asymptotes rather than the exact curve (the asymptotes over-estimate it whenever two corners are within a decade), and quoting a phase taken from a wrapped $\arctan$ — a wrapped phase jumps from $-180^{\circ}$ to $+180^{\circ}$ and destroys any search for the phase crossover. Also, a healthy-looking gain crossover means nothing on its own: here the phase crossover occurs at a lower frequency than the gain crossover, which is the signature of an unstable loop.

Source: Dorf & Bishop, Modern Control Systems, 14th ed. §8.2–8.4 (Bode construction) and §9.3 (gain and phase margins); Nise, Control Systems Engineering, 8th ed. §10.2–10.7 (frequency response, margins).

Question 6 of 6: Bode diagrams of two loop transfer functions

Question 6: Bode diagrams of two loop transfer functions (25 marks)

Part (a) — a non-minimum-phase second-order loop

Part (b) — a type-1 loop with three corners