Disturbance sensitivity and effect of gain

Which system is more sensitive to disturbance

The open-loop system is more sensitive to disturbance.

Let the disturbance $D(s)$ enter between two forward blocks $G_1$ (controller) and $G_2$ (plant).

• Open loop: $C_D(s) = G_2(s)D(s)$.
• Closed loop (feedback $H$), with $R = 0$ (superposition):

$$C_D(s) = \frac{G_2(s)}{1 + G_1(s)G_2(s)H(s)}\,D(s)$$

The disturbance output is divided by $(1+G_1G_2H)$. If $|G_1G_2H| \gg 1$, $C_D \approx \dfrac{D}{G_1H}$, which is very small for large $G_1$.

Ratio of disturbance outputs:

$$\frac{C_{D,\,cl}}{C_{D,\,ol}} = \frac{1}{1+G_1G_2H}$$

Since $|1+G_1G_2H| > 1$ (usually $\gg 1$), the closed-loop output is much less affected. Example: $G_1G_2H = 49$ means the disturbance effect is only 2% of that in open loop. The open loop has no way to sense the change in output, so the disturbance passes fully to the output.

Effect of gain on the response

Take unity feedback with $G(s) = \dfrac{K}{s(s+2)}$:

$$\frac{C(s)}{R(s)} = \frac{K}{s^2+2s+K}, \quad \omega_n = \sqrt{K}, \quad \zeta = \frac{1}{\sqrt{K}}$$

As the gain $K$ increases:

1. Speed increases: $\omega_n$ rises, and rise time and peak time fall.
2. Steady-state error falls: $e_{ss}$ for a step ($1/(1+K_p)$) or a ramp ($1/K_v$) decreases.
3. Damping falls: $\zeta$ decreases, so overshoot and oscillation increase.
4. Disturbance and sensitivity effects fall as $1/(1+GH)$.
5. Stability may be lost: for third- and higher-order systems, e.g. $\dfrac{K}{s(s+1)(s+2)}$, Routh's test shows instability for $K > 6$.

So the gain is chosen as a compromise between accuracy/speed and relative stability.