Tracking, disturbance rejection and pole sensitivity

Controller $D(s) = \dfrac{160(s+4)}{s+30}$, plant $P(s) = \dfrac{1}{s(s+2)}$.

Stability first

Characteristic equation $1 + DP = 0$:

$$s(s+2)(s+30) + 160(s+4) = s^3 + 32s^2 + 220s + 640 = 0$$

Routh array:

No sign change, so the closed loop is stable and the final value theorem can be used.

i) Tracking a step reference

$$G(s) = DP = \frac{160(s+4)}{s(s+2)(s+30)}$$

There is one pole at the origin, so the system is Type 1: $K_p = \infty$.

$$e_{ss} = \frac{1}{1 + K_p} = 0$$

Yes, the system tracks a step reference with zero steady state error.

ii) Rejecting a step disturbance $w$

With $r = 0$:

$$\frac{Y(s)}{W(s)} = \frac{P}{1 + DP} = \frac{s+30}{s^3 + 32s^2 + 220s + 640}$$

For a unit step $W(s) = 1/s$:

$$y_{ss} = \lim_{s\to 0} s\cdot\frac{1}{s}\cdot\frac{s+30}{s^3+32s^2+220s+640} = \frac{30}{640} = 0.0469$$

No. The disturbance enters after the controller, and the integrator is in the plant, not in the controller. So a steady error of $0.0469w$ (about 4.7% of the disturbance) remains.

iii) Sensitivity to the plant pole at $-2$

Let the pole be at $-a$, with $a = 2$: $P = \dfrac{1}{s(s+a)}$, $T = \dfrac{DP}{1+DP}$. By the chain rule:

$$S_a^T = S_P^T \cdot S_a^P$$ $$\begin{aligned} S_P^T &= \frac{1}{1 + DP} = \frac{s(s+2)(s+30)}{s^3 + 32s^2 + 220s + 640} \\ S_a^P &= \frac{\partial P}{\partial a}\cdot\frac{a}{P} = \frac{-1}{s(s+a)^2}\cdot a\,s(s+a) = \frac{-a}{s+a} = \frac{-2}{s+2} \end{aligned}$$

Answer:

$$S_a^T = \frac{-2s(s+30)}{s^3 + 32s^2 + 220s + 640}$$

At DC ($s = 0$) the sensitivity is zero, so a change in the pole location has no effect on the steady state output. Its effect grows at higher frequencies.