Transfer functions: reading response from the poles

How fast, how much overshoot?

• Given a transfer function, an engineer must say how its output will move: how fast, how much overshoot, and whether it settles at all.
• The roots of the denominator are the poles; the roots of the numerator are the zeros. Poles in the left half-plane decay, so the system is stable.

The second-order dictionary

• Poles at $-\zeta\omega_n \pm j\omega_n\sqrt{1 - \zeta^2}$: their distance from the origin is $\omega_n$, and their angle from the negative real axis is $\cos^{-1}\zeta$.
• Overshoot $M_p = e^{-\zeta\pi/\sqrt{1 - \zeta^2}}$ at $t_p = \pi/\omega_d$, and 2% settling after $t_s \approx 4/(\zeta\omega_n)$.

The valve position loop

The frequency response

• With $s = j\omega$, the open loop has $|L| = 1$ at the crossover $\omega_c = \mathbf{48.2}$ rad/s, where the phase is −114.8°: a phase margin of $\mathbf{65.2°}$.
• The rule $\zeta \approx PM/100 = 0.65$ agrees with the 0.7 designed for. Gain alone cannot destabilise this loop, but delay can.

More gain cannot settle faster

• The closed-loop poles solve $0.0096s^2 + s + 0.398K_p = 0$. As $K_p$ rises they move along the root locus 根轨迹: from 0 and −104.2 towards each other, then up and down the vertical line $s = -52.1$.
• Once complex, their real part stays at −52.1, so $t_s = 4/52.1 = \mathbf{77}$ ms for every gain beyond critical. More gain buys speed of rise and overshoot, never faster settling; that needs a faster motor or a lead term.

Worked example: double the gain

• $K_p = 268$: $\omega_n = \sqrt{0.398 \times 268/0.0096} = \mathbf{105.4}$ rad/s and $\zeta = 1/(2 \times 0.0096 \times 105.4) = 0.494$.
• Overshoot rises to $\mathbf{16.8\%}$ and fails the 5% spec, while $t_s$ stays 77 ms. The loop rises sooner and rings more: the design value of 134 was the better trade.

Three response slips

Recap