Exercises — TVC dynamics — gimbal servo bandwidth, time delay

WHAT we do: match term-by-term to the standard denominator. WHY: the standard form is a template — the numbers slot into named slots, no derivation needed.

• Constant term ${\omega }_n^2=2500\Rightarrow {\omega }_n=\sqrt{2500}===50\text{rad/s}==$.
• Middle term $2\zeta {\omega }_n=50\Rightarrow \zeta =\frac{50}{2\times 50}===0.5==$.

The numerator equals ${\omega }_n^2=2500$, confirming unit DC gain (at $s=0$, $G=2500/2500=1$).

WHAT/WHY: on the $j\omega$ axis (i.e. evaluating the frequency response at $s=j\omega$) a pure delay has magnitude exactly 1 — it only rotates phase. This qualifier matters: off the imaginary axis, in the general complex-$s$ plane, $|e^{-s\tau }|$ is not 1. Frequency-response magnitude and general-$s$ magnitude are different things; the "always 1" claim is a statement about the $j\omega$ axis only.

• (a) $|e^{-j\omega \tau }|===1==$ for any real $\omega$ (the exponent is purely imaginary, so it is a unit-modulus rotation).
• (b) Phase $=-\omega \tau =-30\times 0.01=-0.30\text{rad}$. In degrees $=-0.30\times 57.2958===-17.2^{\circ }==$.

Step 1 — ${\omega }_n$ (WHY first: it sets the scale for everything else). ${\omega }_n=\sqrt{\frac{k+K}{J}}=\sqrt{\frac{4500}{0.5}}=\sqrt{9000}=94.87\text{rad/s}.$

Step 2 — $\zeta$. $\zeta =\frac{c}{2\sqrt{(k+K)J}}=\frac{30}{2\sqrt{4500\times 0.5}}=\frac{30}{2\sqrt{2250}}=\frac{30}{2\times 47.43}=\mathrm{0.316.}$

Step 3 — ${\omega }_B$. Compute $1-2{\zeta }^2=1-2(0.316{)}^2=1-0.2=0.8$. ${\omega }_B=94.87\sqrt{0.8+\sqrt{0.8^2+1}}=94.87\sqrt{0.8+\sqrt{1.64}}=94.87\sqrt{0.8+1.2806}.$ ${\omega }_B=94.87\sqrt{2.0806}=94.87\times 1.4425===136.8\text{rad/s}==.$

Because $\zeta <1/\sqrt{2}$ the system is lightly damped, so ${\omega }_B>{\omega }_n$ (it resonates before rolling off).

WHAT/WHY: delay phase is $\Delta \varphi =-{\omega }_c\tau$ evaluated at crossover ${\omega }_c$ (defined above), because crossover is where the $-180^{\circ }$ stability limit is judged. $\Delta \varphi =-{\omega }_c\tau =-20\times 0.025=-0.50\text{rad}=-0.50\times 57.2958=-28.6^{\circ }.$ The size of the loss (what we "spend") is $==28.6^{\circ }==$ — same magnitude, positive because we are quoting how much margin is consumed. Remaining margin $=50^{\circ }-28.6^{\circ }===21.4^{\circ }==$ — thin (below the $30^{\circ }$ comfort line).

Step 1 — exploit the $\zeta =0.707$ identity. At $\zeta =1/\sqrt{2}$, $1-2{\zeta }^2=0$, so ${\omega }_B={\omega }_n\sqrt{0+\sqrt{0+1}}={\omega }_n.$ Therefore ${\omega }_n=120\text{rad/s}$ directly. This is exactly why we design to 0.707 — the algebra collapses.

Step 2 — back out stiffness. From ${\omega }_n^2=(k+K)/J$: $k+K={\omega }_n^{2}J=120^2\times 0.6=14400\times 0.6===8640\text{N⋅m/rad}==.$

WHAT we do: plug each $\zeta$ into the bandwidth radical. WHY: we want to see how ${\omega }_B$ moves relative to ${\omega }_n$ as damping changes — one variable ($\zeta$) swept, everything else fixed, so the result isolates damping's effect alone.

Use ${\omega }_B={\omega }_n\sqrt{(1-2{\zeta }^2)+\sqrt{(1-2{\zeta }^2{)}^2+1}}$.

• $\zeta =0.5$: WHAT — $1-2{\zeta }^2=0.5$ (positive, so the light damping pushes ${\omega }_B$ above ${\omega }_n$). ${\omega }_B=100\sqrt{0.5+\sqrt{1.25}}=100\sqrt{0.5+1.118}=100\sqrt{1.618}===127.2\text{rad/s}==$.
• $\zeta =0.707$: WHAT — $1-2{\zeta }^2=0$, the term vanishes and the radical collapses to 1. ${\omega }_B=100\sqrt{0+1}===100\text{rad/s}==$.
• $\zeta =1.0$: WHAT — $1-2{\zeta }^2=-1$ (negative, so heavy damping drags ${\omega }_B$ below ${\omega }_n$). ${\omega }_B=100\sqrt{-1+\sqrt{2}}=100\sqrt{-1+1.4142}=100\sqrt{0.4142}===64.4\text{rad/s}==$.

Trend: as damping rises, ${\omega }_B$ falls — heavy damping trades away following-speed for calm settling.

Reading the figure (s01): the horizontal axis is the damping ratio $\zeta$ and the vertical axis is the bandwidth ratio ${\omega }_B/{\omega }_n$. The red curve is that ratio; it starts above 1 (light damping, fast following), crosses the dashed horizontal line ${\omega }_B/{\omega }_n=1$ exactly at the dotted vertical $\zeta =1/\sqrt{2}$, and continues downward. The three black dots are our three answers ($1.27$, $1.00$, $0.64$) — visual proof that the sole reason ${\omega }_B={\omega }_n$ is special is that we chose to land on that crossing point.

The delay's phase contribution is $\Delta \varphi =-{\omega }_c\tau$ (a lag, hence negative); we quote its magnitude as the loss.

• Before: $|\Delta {\varphi }_1|={\omega }_c\tau =15\times 0.02=0.30\text{rad}=0.30\times 57.2958===17.2^{\circ }==$.
• After: $|\Delta {\varphi }_2|={\omega }_c^{\prime }\tau =25\times 0.02=0.50\text{rad}=0.50\times 57.2958===28.6^{\circ }==$.

Synthesis point: raising gain raised ${\omega }_c$, and since delay phase magnitude is proportional to ${\omega }_c$, the loss grew from $17.2^{\circ }$ to $28.6^{\circ }$ — gain made it worse. On the $j\omega$ axis the magnitude of $e^{-j\omega \tau }$ is 1 at every frequency, so gain cannot "outrun" a delay; only cutting $\tau$ (faster computer / higher sample rate, see Digital control — sampling & computational delay) or lowering bandwidth helps.

$\frac{\tau }{2}=0.02\text{s}$.

• Zero: numerator $1-0.02s=0\Rightarrow s=+\frac{2}{\tau }=+\frac{2}{0.04}===+50{\text{s}}^{-1}==$. This sits in the right half-plane (positive real part).
• Pole: denominator $1+0.02s=0\Rightarrow s=-\frac{2}{\tau }===-50{\text{s}}^{-1}==$ (left half-plane, stable).

The right-half-plane zero at $+50$ is the fingerprint: it makes the output initially move the wrong way (undershoot), which is the destabilising essence of transport delay. See Padé approximation of transport delay for higher orders.

Reading the figure (s02): the horizontal axis is $\text{Re}(s)$ and the vertical axis is $\text{Im}(s)$ — the complex $s$-plane. The shaded band on the right marks the right half-plane. The black "$\times$" at $-50$ is the harmless stable pole; the red open circle at $+50$ is the zero — placing it inside the shaded (right-half) region is exactly the picture of "the output first goes the wrong way." The mirror symmetry about the imaginary axis is why first-order Padé keeps magnitude flat while injecting pure phase lag.

Step 1 — bandwidth floor (Req 1). Need ${\omega }_B={\omega }_n\ge 5\times 12=60\text{rad/s}$.

Step 2 — bending ceiling (Req 2). Need ${\omega }_n<180\text{rad/s}$; a common rule keeps it well below, say $\le 90$. A choice of ${\omega }_n=80\text{rad/s}$ sits comfortably in the window $[60,90]$.

Step 3 — delay check (Req 3). Delay phase magnitude at crossover: $|\Delta \varphi |={\omega }_c\tau =12\times 0.015=0.18\text{rad}=0.18\times 57.2958===10.3^{\circ }==\le 12^{\circ }.✓$ (Note this depends only on ${\omega }_c$ and $\tau$, not on ${\omega }_n$ — so it passes regardless of the ${\omega }_n$ pick.)

Step 4 — hardware from ${\omega }_n=80$, $\zeta =0.707$, $J=0.7$. $k+K={\omega }_n^{2}J=80^2\times 0.7=6400\times 0.7===4480\text{N⋅m/rad}==.$ $c=2\zeta \sqrt{(k+K)J}=2(0.707)\sqrt{4480\times 0.7}=1.414\sqrt{3136}=1.414\times 56===79.2\text{N⋅m⋅s}==.$

Verdict: ${\omega }_n=80$ rad/s gives ${\omega }_B=80\ge 60$ ✓, sits below the $180$ rad/s bending mode ✓, and the delay spends only $10.3^{\circ }\le 12^{\circ }$ ✓. All three requirements met.

The bandwidth floor ($\ge 60$) is satisfied, but ${\omega }_n=170$ places the servo's own $-3$ dB edge only $180-170=10\text{rad/s}$ ($\approx 6\%$) below the bending mode. The servo would drive the flexible structure near resonance; small gain/phase errors turn TVC into an oscillator. The margin to $180$ is far too thin — you would need a notch filter (Structural bending modes & notch filters) or, better, back off to ${\omega }_n\approx 80$. Fast is good; too fast collides with structure.

• L1.1: ${\omega }_n=50$, $\zeta =0.5$. L1.2: mag $=1$, phase $=-17.2^{\circ }$.
• L2.1: ${\omega }_n=94.87$, $\zeta =0.316$, ${\omega }_B=136.8$. L2.2: $28.6^{\circ }$ eaten, $21.4^{\circ }$ left.
• L3.1: ${\omega }_n=120$, $k+K=8640$. L3.2: ${\omega }_B=127.2,100,64.4$.
• L4.1: $17.2^{\circ }\to 28.6^{\circ }$. L4.2: zero $+50$, pole $-50$.
• L5.1: ${\omega }_n=80$ valid, $|\Delta \varphi |=10.3^{\circ }$, $k+K=4480$, $c=79.2$.