Bode design of a plant zero location for a target gain-crossover: 22-Mec-A3 System Analysis and Control, May 2013, Q6

Question 6: Bode design of a plant zero location for a target gain-crossover (equal value)

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. Open-loop transfer function in a unity-feedback loop

$$G(s)=\frac{3\,(s+1)}{s\,(s+a)}.$$

Target: gain-crossover frequency $\omega_{gc}=100$ rad/s (where $|G(j\omega_{gc})|=1$, i.e. $0$ dB).

Find. the value of $a$ that places the gain crossover at $\omega=100$ rad/s.

Bode magnitude of $3(s+1)/[s(s+a)]$, drawn for the feasible worked case $a=2.69$, whose $0$ dB crossing sits exactly at $\omega_{gc}=2$ rad/s. The $-20$ dB/dec integrator slope is flattened by the zero at $\omega=1$ and steepened again by the pole at $\omega=a$; because the high-frequency asymptote falls as $3/\omega$, no $a\ge0$ can push the crossing beyond the ceiling $\omega_{gc,\max}=3.15$ rad/s (reached as $a\to0$), and raising $a$ only moves it lower. At the requested $\omega=100$ rad/s the magnitude is at most $\approx-30$ dB — the target crossover is not attainable with this plant (see the callout).

Approach. Write the exact magnitude, set it to unity at the target frequency, and solve for $a$; check the result is physically realisable before reporting it.

Magnitude condition. At $s=j\omega$, $$|G(j\omega)|=\frac{3\sqrt{\omega^{2}+1}}{\omega\sqrt{\omega^{2}+a^{2}}}=1\;\Longrightarrow\;a^{2}=\frac{9(\omega^{2}+1)}{\omega^{2}}-\omega^{2}.$$ This is the design equation: it returns the pole location $a$ that puts the $0$ dB crossing at any chosen $\omega$.
Evaluate at $\omega=100$ rad/s. Substituting, $$a^{2}=\frac{9(10001)}{10000}-10000=9.0009-10000=\boxed{-9991}.$$ The required $a^{2}$ is negative, so no real value of $a$ exists that yields a $100$ rad/s crossover.
Why it fails — the magnitude ceiling. At high frequency the zero at $1$ cancels the integrator and $|G|\to 3/\omega$, so the largest magnitude the plant can present at $100$ rad/s (achieved as $a\to0$) is $$|G(j100)|_{\max}=\frac{3\sqrt{10001}}{100\cdot100}=0.0300\;\Rightarrow\;-30.5\ \text{dB}.$$ The curve is already $30$ dB below $0$ dB at $100$ rad/s, and increasing $a$ only pushes it lower. The maximum crossover this plant can ever reach (set by $\omega^{4}-9\omega^{2}-9=0$) is $$\omega_{gc,\max}=\sqrt{\tfrac12\!\left(9+\sqrt{117}\right)}=3.15\ \text{rad/s}.$$
Method demonstrated at a feasible target. The design equation is used exactly as intended once the target lies below the ceiling. For example, to place the crossover at $\omega_{gc}=2$ rad/s, $$a^{2}=\frac{9(5)}{4}-4=7.25\;\Rightarrow\;a=2.69.$$ The same one-line formula delivers $a$ whenever $\omega_{gc}\le3.15$ rad/s.
What the printed data would require. If instead the numerator gain were a free constant $K_n$ (rather than $3$), the design equation is $a^{2}=K_n^{2}(1+\omega^{2})/\omega^{2}-\omega^{2}$; a real $a$ at $\omega=100$ rad/s needs $K_n>\omega^{2}/\sqrt{1+\omega^{2}}\approx100$. A crossover at $100$ rad/s is thus consistent with a loop gain about $33\times$ larger than the printed $3$.

Check: As printed (numerator gain $3$, target $\omega_{gc}=100$ rad/s), the problem has no real solution — the plant magnitude at $100$ rad/s is only $\approx-30$ dB for every $a\ge0$, so it can never reach $0$ dB there, and the design equation returns $a^{2}=-9991$. The exam data are internally inconsistent (most likely the loop gain or the target frequency is misprinted — a gain of $\approx100$, or a target near $3$ rad/s, makes the question well-posed). The method above is exact and is shown solved at a feasible target; the boxed conclusion is that the printed target is unattainable with a gain of $3$.

Quantity	Value
Design equation	$a^{2}=9(\omega^{2}+1)/\omega^{2}-\omega^{2}$
Required $a^{2}$ at $\omega=100$	$-9991$ (no real $a$)
Plant magnitude at $100$ rad/s	$\le-30.5$ dB
Max achievable crossover	$\omega_{gc,\max}=3.15$ rad/s
$a$ for a feasible $\omega_{gc}=2$ rad/s	$a=2.69$

Topic: Bode-plot design — placing the gain-crossover frequency by choosing a pole/zero location, and recognising an infeasible specification.

Key relations: gain crossover $|G(j\omega_{gc})|=1$; for $3(s+1)/[s(s+a)]$, $a^{2}=9(\omega^{2}+1)/\omega^{2}-\omega^{2}$; high-frequency asymptote $|G|\to K_n/\omega$ once the zero cancels the integrator.

Why this works: The magnitude condition is a single algebraic equation in $a$; solving it is the analytic equivalent of sliding the pole corner until the $0$ dB crossing lands on the target. The asymptotic slope diagram immediately reveals the feasibility ceiling: because the numerator and the integrator leave a net $-20$ dB/dec tail scaled by the DC gain, the crossover cannot exceed the frequency where that tail alone hits $0$ dB.

Common pitfall: plugging numbers into the design formula and reporting an imaginary $a$ (or its absolute value) without noticing the magnitude at the target is far below $0$ dB. Always sketch the asymptotic magnitude first: the crossover region is set by the DC gain and the corner frequencies, so an out-of-reach target shows up immediately.

Question 6 of 6: Bode design of a plant zero location for a target gain-crossover

Question 6: Bode design of a plant zero location for a target gain-crossover (equal value)