Exercises — Bode plot — magnitude and phase vs frequency

WHAT we do: a flat line with zero phase means no poles, no zeros, no integrators — only a constant gain $K$. WHY: poles/zeros/integrators all bend the curve; only a plain number $K$ leaves it flat with $0^{\circ }$ phase. Convert the dB back to a ratio. From $|K{|}_{\text{dB}}=20{\log }_{10}K$: $26=20{\log }_{10}K\Rightarrow {\log }_{10}K=1.3\Rightarrow K=10^{1.3}\approx \mathrm{19.95.}$ Answer: $G(s)=K\approx 20$ (a pure gain of about 20).

WHAT/WHY: slope $-20$ dB/dec everywhere (no corner, no flattening) plus constant $-90^{\circ }$ phase is the fingerprint of a single integrator $1/s$. Check the level: $|1/(j\omega ){|}_{\text{dB}}=-20{\log }_{10}\omega$, which is $0$ dB at $\omega =1$. ✓ Answer: $G(s)=\frac{1}{s}$.

WHAT/WHY: "flat, then bends down $-20$ dB/dec, phase $0^{\circ }\to -90^{\circ }$" is a single first-order pole (a basic Low-pass filter). The bend location is the corner ${\omega }_c$. Answer: $G(s)=\frac{1}{1+s/10}$, corner ${\omega }_c=10$ rad/s. At ${\omega }_c$ the magnitude is $-3$ dB and the phase is $-45^{\circ }$.

(a) DC gain. At $\omega \to 0$ the $(1+s/20)$ term is $1$, so $|G|=50$. $20{\log }_{10}50=20(1.699)=33.98\text{ dB}\approx 34\text{ dB}.$ (b) At the corner $\omega =20$. The pole contributes exactly $-3$ dB there, so $33.98-3.01=30.97\text{ dB}\approx 31\text{ dB}.$ (c) At $\omega =200$ (one decade above the corner). On the asymptote each decade above the corner drops $20$ dB. From $\omega =20$ to $\omega =200$ is one decade $\Rightarrow -20$ dB off the flat level: $33.98-20=13.98\text{ dB}\approx 14\text{ dB}.$

WHAT: add the phase of each factor (angle of a product = sum of angles).

• Constant $100$: $0^{\circ }$.
• Integrator $1/s$: $-90^{\circ }$ (always).
• Zero $(1+j\omega /2)$ at $\omega =2$: $\arctan (2/2)=\arctan (1)=+45^{\circ }$. $\angle G(j2)=0-90^{\circ }+45^{\circ }=-45^{\circ }.$

(a) $20{\log }_{10}(0.5)=20(-0.301)=-6.02$ dB $\approx -6$ dB. (Halving amplitude $=-6$ dB — worth memorising.) (b) $-40=20{\log }_{10}r\Rightarrow {\log }_{10}r=-2\Rightarrow r=10^{-2}=0.01$.

Corners: the integrator has no corner (its $-20$ dB/dec starts at $\omega \to 0$); the two poles corner at $\omega =10$ and $\omega =100$. Start slope: the lone integrator gives $-20$ dB/dec at the lowest frequencies. Each pole adds $-20$ dB/dec once you pass its corner.

Read the figure below (alt-text: log-frequency Bode magnitude, blue exact curve with dashed orange straight-line asymptotes; two red dots mark the corners at 10 and 100 rad/s). The blue curve is the true magnitude; the dashed orange lines are our hand-sketch asymptotes. Follow left to right: the line falls gently ($-20$ dB/dec) until the first red dot at $\omega =10$, where it kinks steeper to $-40$ dB/dec; at the second red dot at $\omega =100$ it kinks again to $-60$ dB/dec. Each red dot is a pole "switching on." Notice the blue curve sits a hair below each kink — that is the $-3$ dB rounding the straight lines ignore.

WHAT: walk the asymptote levels band by band until it reaches $0$ dB. Below $\omega =10$ the asymptote is the integrator line $|G{|}_{\text{dB}}=20{\log }_{10}(1000/\omega )$.

• At $\omega =10$: $20{\log }_{10}(1000/10)=20{\log }_{10}100=40$ dB. Still above $0$.
• Between $10$ and $100$ the slope is $-40$ dB/dec. Solve for where it hits $0$: $0=40-40({\log }_{10}\omega -{\log }_{10}10)\Rightarrow {\log }_{10}\omega -1=1\Rightarrow {\log }_{10}\omega =2\Rightarrow \omega =100.$ That lands exactly at the next corner. Answer: ${\omega }_{gc}\approx 100$ rad/s (asymptote estimate). On the figure above, this is where the dashed orange line meets the grey $0$ dB horizontal.

WHAT: phase margin $=180^{\circ }+\angle L(j{\omega }_{gc})$. Sum the phases at $\omega =100$:

• Integrator: $-90^{\circ }$.
• Pole @10: $-\arctan (100/10)=-\arctan (10)=-84.3^{\circ }$.
• Pole @100: $-\arctan (100/100)=-\arctan (1)=-45^{\circ }$. $\angle L(j100)=-90-84.3-45=-219.3^{\circ }.$ $\text{PM}=180^{\circ }+(-219.3^{\circ })=-39.3^{\circ }.$ A negative phase margin ⇒ the closed loop is UNSTABLE. This plant needs help (see L5). Full details of margins live in Stability margins.

WHAT/WHY: a first-order low-pass rolls off at $-20$ dB/dec above ${\omega }_c$. We need $20$ dB of attenuation at $600$ rad/s. $20$ dB at $-20$ dB/dec = one decade above the corner. So the corner should sit one decade below the noise: ${\omega }_c=\frac{600}{10}=60\text{ rad/s}.$ Check the signal band: at $\omega =6$ rad/s, ratio to corner is $6/60=0.1$, so $|F(j6)|=\frac{1}{\sqrt{1+0.1^2}}\Rightarrow |F{|}_{\text{dB}}=20{\log }_{10}\frac{1}{\sqrt{1+0.1^2}}=-20{\log }_{10}\sqrt{1+0.1^2}.$ (WHY the minus sign appears: ${\log }_{10}(1/x)=-{\log }_{10}x$ — the log of a reciprocal is the negative of the log. Since $F$ is $1$ over a square root, its dB is minus the dB of that square root. Same rule that makes a pole's high-frequency asymptote slope down.) Numerically $\sqrt{1+0.01}=1.004988$, so $|F{|}_{\text{dB}}=-20{\log }_{10}(1.004988)=-0.0432$ dB — essentially zero loss. ✓ Requirement ($\le 3$ dB) easily met. Answer: $F(s)=\frac{1}{1+s/60}$, corner ${\omega }_c=60$ rad/s.

WHAT/WHY: a Lead-lag compensator in lead mode injects phase to rescue a low/negative phase margin. Let $\alpha =z/p$ (with $0<\alpha <1$). Then $\sin {\varphi }_{\max }=\frac{p-z}{p+z}=\frac{1-\alpha }{1+\alpha }.$ Set ${\varphi }_{\max }=45^{\circ }$, so $\sin 45^{\circ }=\frac{1}{\sqrt{2}}=0.7071$: $\frac{1-\alpha }{1+\alpha }=0.7071\Rightarrow 1-\alpha =0.7071(1+\alpha )\Rightarrow 0.2929=1.7071\alpha \Rightarrow \alpha =\mathrm{0.1716.}$ With ${\omega }_m=\sqrt{zp}=100$ and $z=\alpha p$: $\sqrt{\alpha p\cdot p}=100\Rightarrow p\sqrt{\alpha }=100$: $p=\frac{100}{\sqrt{0.1716}}=\frac{100}{0.4142}=241.4\text{ rad/s},z=\alpha p=0.1716\times 241.4=41.4\text{ rad/s}.$ Answer: $z\approx 41.4$ rad/s, $p\approx 241.4$ rad/s. Read the figure below (alt-text: log-frequency phase plot, green curve rising from 0 to a peak then falling; a red dot marks the +45° peak at omega_m = 100; dashed blue and orange verticals mark the zero at 41.4 and pole at 241.4 rad/s). The green curve is the compensator's phase. It climbs after the blue zero line at $41.4$ rad/s, peaks at the red dot ($+45^{\circ }$) exactly at ${\omega }_m=100$ — the geometric middle of the two dashed lines — then eases back down after the orange pole line at $241.4$ rad/s. The bump is the "phase we hand back to the loop" at crossover.

WHAT/WHY: cascading multiplies transfer functions, so phases add. The lead adds its $+45^{\circ }$ right at $\omega =100$. From Solution 3.3, $\angle L_0(j100)=-219.3^{\circ }$. Add the lead's phase at ${\omega }_m=100$, which is exactly $+45^{\circ }$ by design: $\angle L(j100)=-219.3^{\circ }+45^{\circ }=-174.3^{\circ }.$ $\text{PM}=180^{\circ }+(-174.3^{\circ })=+5.7^{\circ }.$ The loop is now (barely) stable. A single $45^{\circ }$ lead flipped a $-39^{\circ }$ margin to $+6^{\circ }$ — a real compensator would add gain trim and more lead, but the mechanism is exactly this: buy back phase where the gain crosses $0$ dB. This is Loop shaping in one move.

WHAT/WHY: for Transfer functions that are minimum-phase, slope and phase are locked together: each $-20$ dB/dec of slope ⇒ about $-90^{\circ }$ of phase. A $-40$ dB/dec slope is two of those: $\text{phase}\approx 2\times (-90^{\circ })=-180^{\circ }.$ One-line reason: steep roll-off is extra phase lag — they're the same physical fact for a minimum-phase system, so an aggressive filter and a large phase lag are inseparable. That lag is precisely what eats your phase margin and destabilises loops (see Nyquist plot and Stability margins).

Step 1 — force crossover at $\omega =1$. With $K=1$, $L(j\omega )=\frac{1+j\omega /{\omega }_z}{(j\omega {)}^2}.$ At $\omega =1$: $|L|=\frac{|1+j/{\omega }_z|}{1}=\sqrt{1+1/{\omega }_z^2}$. For crossover ($|L|=1$) we'd need this $=1$, forcing ${\omega }_z\to \infty$; instead the standard trick is to place the zero at crossover ${\omega }_z={\omega }_{gc}=1$ to maximise phase there, and accept the small magnitude bump. Take ${\omega }_z=1$. Step 2 — phase at $\omega =1$.

• Double integrator $1/(j\omega {)}^2$: $-180^{\circ }$.
• Zero $(1+j\omega /1)$ at $\omega =1$: $+\arctan (1)=+45^{\circ }$. $\angle L(j1)=-180^{\circ }+45^{\circ }=-135^{\circ }.$ $\text{PM}=180^{\circ }+(-135^{\circ })=+45^{\circ }\ge 30^{\circ }.✓$ Answer: ${\omega }_z=1$ rad/s gives PM $=45^{\circ }$. The PD zero pulled the phase up from the deadly $-180^{\circ }$ (marginal) to a comfortable $-135^{\circ }$ — exactly the "add phase back" lesson from Example 2.

How many decades below a noise tone must a first-order low-pass corner sit for $20$ dB of rejection? ::: One decade. A $45^{\circ }$ lead needs $z/p$ equal to what? ::: $\alpha =\frac{1-\sin 45^{\circ }}{1+\sin 45^{\circ }}\approx 0.172$. What phase does a $-40$ dB/dec minimum-phase slope imply? ::: About $-180^{\circ }$. A negative phase margin means the closed loop is…? ::: Unstable. Adding a PD zero at crossover for a $1/s^2$ plant gives what phase margin? ::: $+45^{\circ }$.