Question 1 of 6: Asymptotic Bode Magnitude Plots

Question 1: Asymptotic Bode Magnitude Plots (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: (a) $G_a(s) = \dfrac{4}{s+2}$, a single real pole at $s=-2$; (b) $G_b(s) = \dfrac{4}{(0.4s+1)(s+1)}$, two real poles written already in time-constant form with $\tau_1 = 0.4\ \text{s}$ and $\tau_2 = 1\ \text{s}$.

Find. The straight-line (asymptotic) log-magnitude characteristic of each function in decibels against $\log_{10}\omega$, with the low-frequency level, every corner frequency, and every asymptote slope identified.

Approach. Put each factor into the normalised Bode form $(1+j\omega\tau)$, read the low-frequency asymptote from the resulting constant, then add one $-20\ \text{dB/decade}$ break at each pole corner $\omega_c = 1/\tau$.

1. Normalise (a) to time-constant form. A Bode plot is drawn from factors of the form $(1+j\omega\tau)$, so the pole binomial must be scaled so that its constant term is unity: $$G_a(s)=\frac{4}{s+2}=\frac{4}{2\left(\tfrac{s}{2}+1\right)}=\frac{2}{0.5s+1}$$ The Bode gain is therefore $K_B = 2$, not 4. This normalisation is the single most common place to lose marks: reading the gain as 4 puts the whole plot 6 dB too high.
2. Low-frequency asymptote of (a). As $\omega \to 0$ the denominator tends to unity, so $$20\log_{10}|G_a| \to 20\log_{10}(2) = \boxed{6.02\ \text{dB}}$$ This horizontal line is the entire plot below the corner.
3. Corner frequency and high-frequency slope of (a). The break occurs where the imaginary part of the normalised factor equals its real part, i.e. at $\omega_c = 1/\tau = 1/0.5 = 2\ \text{rad/s}$. Above it the single pole contributes $-20\ \text{dB/decade}$, so the magnitude falls linearly at that rate. Starting from 6.02 dB at the corner, the asymptote reaches $0\ \text{dB}$ after $6.02/20 = 0.301$ decade, i.e. at $\omega = 2 \times 10^{0.301} = 4\ \text{rad/s}$ — one octave above the corner. The exact curve runs $3.01\ \text{dB}$ below the asymptote at the corner, and it crosses unity gain slightly earlier than the asymptote predicts: $|G_a(j\omega)|=1$ requires $|j\omega+2| = 4$, giving $\omega = \sqrt{12} = 3.46\ \text{rad/s}$.

Part (b) needs no algebraic normalisation because both binomials are already printed in time-constant form, so the constant 4 is the Bode gain.

4. Low-frequency asymptote of (b). With both factors tending to unity, $$20\log_{10}|G_b| \to 20\log_{10}(4) = \boxed{12.04\ \text{dB}}$$
5. Locate both corners and accumulate the slopes. The corner of a $(1+j\omega\tau)$ factor is $\omega_c = 1/\tau$: $$\omega_{c1}=\frac{1}{1}=1\ \text{rad/s},\qquad \omega_{c2}=\frac{1}{0.4}=2.5\ \text{rad/s}$$ The slope is therefore $0\ \text{dB/dec}$ below 1 rad/s, $-20\ \text{dB/dec}$ between 1 and 2.5 rad/s, and $-40\ \text{dB/dec}$ above 2.5 rad/s.
6. Evaluate the asymptote at the second corner. Falling at $-20\ \text{dB/dec}$ from 12.04 dB over the interval $1 \to 2.5\ \text{rad/s}$ (a factor 2.5, i.e. $\log_{10}2.5 = 0.398$ decade): $$12.04 - 20(0.398) = 4.06\ \text{dB}$$ so the second break starts from $4.06\ \text{dB}$ and the $-40\ \text{dB/dec}$ tail runs from there. Solving $|G_b(j\omega)| = 1$ numerically gives the exact unity-gain crossing at $\omega = 2.59\ \text{rad/s}$; the asymptotic construction predicts $2.79\ \text{rad/s}$, the usual few-per-cent optimism when two corners lie close together.