Question 2 of 6: Two-Phase AC Servomotor Positioning System

Question 2: Two-Phase AC Servomotor Positioning System (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. A unity-feedback position loop (Fig 1.0). The error detector is a difference amplifier of gain $K_a=10$; the two-phase induction motor plus load has the standard servomotor transfer function $\dfrac{\Theta(s)}{V(s)} = \dfrac{K_m}{s\left(T_m s + 1\right)}$, where $K_m$ is the motor gain constant (rad/s per volt) and $T_m$ the motor time constant. The feedback path is unity.

Find. (a) the undamped natural frequency $\omega_n$ and damping ratio $\zeta$ of the closed loop; (b) the percent overshoot and the time to peak for a unit step demand.

Approach. Close the loop, force the characteristic polynomial into the standard second-order form $s^{2}+2\zeta\omega_n s+\omega_n^{2}$ to read off $\omega_n$ and $\zeta$, then apply the standard underdamped step-response formulae for overshoot and peak time.

1. Form the open-loop transfer function. The amplifier and motor are in cascade, so $$G(s) = K_a\cdot\frac{K_m}{s\left(T_m s+1\right)} = \frac{K_aK_m}{s\left(T_m s+1\right)}, \qquad H(s)=1.$$ Writing $K = K_aK_m$ for the loop gain keeps the algebra compact.
2. Close the loop and clear the denominator. For unity negative feedback $\dfrac{C(s)}{R(s)} = \dfrac{G}{1+G}$, hence $$\frac{C(s)}{R(s)} = \frac{K}{s\left(T_ms+1\right)+K} = \frac{K/T_m}{s^{2}+\dfrac{1}{T_m}s+\dfrac{K}{T_m}}.$$ Dividing through by $T_m$ is the step that exposes the standard form.
3. Match the standard second-order denominator. Comparing with $s^{2}+2\zeta\omega_n s+\omega_n^{2}$ term by term gives $\omega_n^{2}=K/T_m$ and $2\zeta\omega_n = 1/T_m$, so $$\boxed{\;\omega_n = \sqrt{\frac{K_aK_m}{T_m}}, \qquad \zeta = \frac{1}{2\sqrt{K_aK_mT_m}}\;}$$ Note the structural result: raising the amplifier gain raises $\omega_n$ as $\sqrt{K_a}$ but lowers $\zeta$ as $1/\sqrt{K_a}$ — speed is bought with damping, and the product $\zeta\omega_n = 1/(2T_m)$ is fixed by the motor alone.
4. Evaluate for the representative motor. With $K_a=10$, $K_m=0.2\ \text{rad}\,\text{s}^{-1}\text{V}^{-1}$ and $T_m=0.5\ \text{s}$, the loop gain is $K = 10\times0.2 = 2.0\ \text{s}^{-1}$, so $$\omega_n = \sqrt{\frac{2.0}{0.5}} = 2.00\ \text{rad}\,\text{s}^{-1}, \qquad \zeta = \frac{1}{2\sqrt{2.0\times0.5}} = 0.500.$$ The loop is underdamped, as a position servo of this type normally is.
5. Percent overshoot. For an underdamped second-order step response the first peak exceeds the final value by $$M_p = \exp\!\left(\frac{-\pi\zeta}{\sqrt{1-\zeta^{2}}}\right) = \exp\!\left(\frac{-\pi(0.500)}{\sqrt{1-0.250}}\right) = \exp(-1.8138) = 0.1630,$$ that is $\boxed{\;\text{PO} = 16.3\%\;}$ Overshoot depends on $\zeta$ alone, so this figure holds for every motor whose constants satisfy $K_aK_mT_m = 1$.
6. Time to peak. The first maximum occurs half a damped period after the step, $$t_p = \frac{\pi}{\omega_d} = \frac{\pi}{\omega_n\sqrt{1-\zeta^{2}}} = \frac{\pi}{2.00\sqrt{0.750}} = \frac{\pi}{1.7321} = \boxed{\;1.81\ \text{s}\;}$$

The simulated response confirms both figures: the output first crosses its final value at 1.21 s, peaks at 1.163 (16.3 % above unity) at t = 1.81 s, and settles inside a 2 % band after roughly 4 s. A servo that overshoots by one sixth of its commanded travel is acceptable for many positioning duties but would be trimmed — by reducing amplifier gain or adding rate feedback — wherever mechanical stops or backlash make overshoot expensive.