Bode Magnitude and Phase Asymptotes for Two Open-Loop Transfer Functions: 22-Mec-A3 System Analysis and Control, May 2017, Q6

Question 6: Bode Magnitude and Phase Asymptotes for Two Open-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 type 1 open-loop transfer functions sharing the same pole set $\{0,-1,-5,-10\}$, the second carrying an additional real zero at $s=-2$. Semi-logarithmic graph paper is the only permitted aid, so the sketches are to be built from straight-line asymptotes rather than point-by-point evaluation.

Find. The straight-line magnitude and phase asymptotes of each $L(j\omega)$, with the correct low-frequency asymptote, corner frequencies, slopes between corners and limiting phase.

Approach. Convert each transfer function to Bode (time-constant) form so the low-frequency gain can be read directly, tabulate the corner frequencies with the slope change each contributes, then draw the magnitude as a broken line of $\pm20\ \text{dB/dec}$ steps and the phase as a sum of two-decade ramps centred on each corner.

Part (a)

Put the transfer function into Bode form. Each real factor is normalised so that its constant term is unity, and the leftover constants collect into the Bode gain:$$L(s)=\frac{1}{s(s+1)(s+5)(s+10)}=\frac{1}{(1)(5)(10)}\cdot\frac{1}{s\,(1+s)(1+s/5)(1+s/10)}$$so the Bode gain is $K_{B}=1/50=0.02$ and the transfer function is type 1 (one free integrator).
Draw the low-frequency asymptote. Below the first corner every normalised factor is approximately unity, so$$|L(j\omega)|\approx\frac{0.02}{\omega}\quad\Longrightarrow\quad 20\log_{10}|L|\approx-34.0-20\log_{10}\omega\ \text{dB at }\omega=1$$This is a straight line of slope $-20\ \text{dB/dec}$; it crosses $0\ \text{dB}$ where $0.02/\omega=1$, i.e. at$$\boxed{\;\omega_{c}=0.02\ \text{rad/s}\;}$$Because this crossover lies two decades below the first corner, the phase there is essentially $-90^{\circ}$ and the asymptotic phase margin is close to $90^{\circ}$ — the loop is very sluggish but comfortably stable.
Tabulate the corners and slopes. Each simple pole subtracts a further $20\ \text{dB/dec}$ from the slope at its corner frequency:

Frequency band (rad/s)	Active factors	Magnitude slope (dB/dec)	Asymptotic phase
$\omega<1$	integrator only	$-20$	$-90^{\circ}$
$1<\omega<5$	$+$ pole at 1	$-40$	$\to-180^{\circ}$
$5<\omega<10$	$+$ pole at 5	$-60$	$\to-270^{\circ}$
$\omega>10$	$+$ pole at 10	$-80$	$\to-360^{\circ}$

Anchor two points to fix the broken line. Evaluating the asymptotic magnitude at the corners gives $-34.0\ \text{dB}$ at $\omega=1\ \text{rad/s}$ and $-80.0\ \text{dB}$ at $\omega=10\ \text{rad/s}$, falling to $-160\ \text{dB}$ by $\omega=100\ \text{rad/s}$ once the final $-80\ \text{dB/dec}$ slope is established.
Construct the phase asymptote. The integrator holds a constant $-90^{\circ}$ at all frequencies. Each pole adds a further $-90^{\circ}$, drawn as a straight ramp from one decade below its corner to one decade above (i.e. a slope of $-45^{\circ}$/decade over two decades). Summing the three ramps takes the phase from $-90^{\circ}$ at low frequency to$$\lim_{\omega\to\infty}\angle L(j\omega)=-90^{\circ}-3(90^{\circ})=-360^{\circ}$$

Asymptotic magnitude of L(s) = 1/[s(s+1)(s+5)(s+10)]. The low-frequency asymptote is 0.02/ω (−20 dB/dec); the slope steepens to −40, −60 and −80 dB/dec at the corners ω = 1, 5 and 10 rad/s.

Asymptotic phase of the same L(s): −90° at low frequency, each pole contributing a further −90° over a two-decade ramp centred on its corner, ending at −360°.

Part (b)

Normalise the extra zero into Bode form. The numerator factor becomes $s+2=2(1+s/2)$, which multiplies the Bode gain by 2:$$L(s)=\frac{2}{50}\cdot\frac{(1+s/2)}{s\,(1+s)(1+s/5)(1+s/10)}\quad\Longrightarrow\quad K_{B}=0.04$$The system remains type 1, so the low-frequency asymptote is again a single $-20\ \text{dB/dec}$ line, but lifted by $20\log_{10}2=6.02\ \text{dB}$ relative to part (a).
Fix the low-frequency crossover. Setting $0.04/\omega=1$ gives$$\boxed{\;\omega_{c}=0.04\ \text{rad/s}\;}$$twice that of part (a), as the doubled Bode gain requires.
Tabulate the corners, now including the zero. The zero at $\omega=2\ \text{rad/s}$ adds $20\ \text{dB/dec}$ to the slope, so the magnitude asymptote flattens back before the remaining poles steepen it again:

Frequency band (rad/s)	Corner reached	Magnitude slope (dB/dec)	Asymptotic phase
$\omega<1$	integrator only	$-20$	$-90^{\circ}$
$1<\omega<2$	pole at 1	$-40$	$\to-180^{\circ}$
$2<\omega<5$	zero at 2	$-20$	recovering
$5<\omega<10$	pole at 5	$-40$	$\to-180^{\circ}$
$\omega>10$	pole at 10	$-60$	$\to-270^{\circ}$

Anchor the broken line. The asymptotic magnitude is $-28.0\ \text{dB}$ at $\omega=1\ \text{rad/s}$ and $-40.0\ \text{dB}$ at the zero corner $\omega=2\ \text{rad/s}$, reaching $-120\ \text{dB}$ at $\omega=100\ \text{rad/s}$ on the final $-60\ \text{dB/dec}$ slope.
Construct the phase asymptote. The zero contributes $+90^{\circ}$ over its own two-decade ramp (from $0.2$ to $20\ \text{rad/s}$), partially cancelling the lag of the pole at $1\ \text{rad/s}$. The high-frequency limit is therefore$$\lim_{\omega\to\infty}\angle L(j\omega)=-90^{\circ}-3(90^{\circ})+90^{\circ}=-270^{\circ}$$which is $90^{\circ}$ less lag than part (a) at every high frequency.

Asymptotic magnitude of L(s) = (s+2)/[s(s+1)(s+5)(s+10)]. The zero at ω = 2 rad/s lifts the slope back from −40 to −20 dB/dec before the poles at 5 and 10 rad/s drive it to −60 dB/dec.

Asymptotic phase of the same L(s). The zero returns +90° over its own two-decade ramp, so the phase recovers towards −90° near 2 rad/s before falling to −270°.

Comparing the two sketches shows exactly what a real zero buys in a feedback design. It raises the gain at every frequency above its corner and, more importantly, it returns $90^{\circ}$ of phase, so the high-frequency asymptote stops at $-270^{\circ}$ instead of running to $-360^{\circ}$. That is the mechanism behind lead compensation: place a zero somewhere near the intended crossover and the phase margin improves without any change to the plant itself.

Quantity	(a) $1/[s(s+1)(s+5)(s+10)]$	(b) $(s+2)/[s(s+1)(s+5)(s+10)]$
Bode gain $K_{B}$	$0.02$	$0.04$
System type	1	1
Low-frequency asymptote	$0.02/\omega$, $-20$ dB/dec	$0.04/\omega$, $-20$ dB/dec
Gain crossover $\omega_{c}$	$0.02\ \text{rad/s}$	$0.04\ \text{rad/s}$
Corner frequencies (rad/s)	poles 1, 5, 10	pole 1, zero 2, poles 5, 10
Slope sequence (dB/dec)	$-20,-40,-60,-80$	$-20,-40,-20,-40,-60$
Magnitude at $\omega=1$ rad/s	$-34.0$ dB	$-28.0$ dB
Phase, $\omega\to0$ / $\omega\to\infty$	$-90^{\circ}$ / $-360^{\circ}$	$-90^{\circ}$ / $-270^{\circ}$

Topic. Straight-line (asymptotic) Bode construction for factored open-loop transfer functions.

Key relations. Bode form $L(s)=K_{B}\prod(1+s/z_{i})\big/\left[s^{N}\prod(1+s/p_{j})\right]$; low-frequency asymptote $|L|\approx K_{B}/\omega^{N}$ with slope $-20N$ dB/dec; each simple pole (zero) changes the slope by $-20$ ($+20$) dB/dec at its corner and the phase by $-90^{\circ}$ ($+90^{\circ}$) over a two-decade ramp; $\omega_{c}$ where $|L|=1$.

Why this works. Because the logarithm turns products into sums, the log-magnitude of a factored transfer function is the sum of the log-magnitudes of its factors, and each first-order factor has only two asymptotes — flat below its corner and $\pm20$ dB/dec above it. Sketching therefore reduces to adding slope increments in the right order, which is exactly why semi-logarithmic paper is the only aid a candidate needs. The phase behaves the same way, except that the transition is gradual: each factor spreads its $90^{\circ}$ over the two decades bracketing its corner, and at the corner itself the true phase is exactly $\pm45^{\circ}$.

Common pitfall. Sketching from the un-normalised factors, so that the low-frequency asymptote is drawn as $1/\omega$ instead of $K_{B}/\omega$ — that is a $34\ \text{dB}$ error in part (a) and misplaces the crossover by more than a decade. The second common slip is treating the phase corner as a step of $90^{\circ}$ at the break frequency rather than a ramp beginning a decade earlier.

Source. Ogata, Modern Control Engineering, 5th ed., §7.2 (Bode diagrams); Franklin, Powell & Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed., §6.1; Nise, Control Systems Engineering, 8th ed., §10.2–10.3.

Question 6 of 6: Bode Magnitude and Phase Asymptotes for Two Open-Loop Transfer Functions

Question 6: Bode Magnitude and Phase Asymptotes for Two Open-Loop Transfer Functions (25 marks)

Part (a)

Part (b)