Effect of feedback on gain and speed of response

Negative feedback reduces the overall gain and increases the speed of response, both by the factor $(1+GH)$.

Effect on gain

$$\text{Open loop: } T = G, \qquad \text{Closed loop: } T = \frac{G}{1+GH}$$

The gain falls by $(1+GH)$. Example: $G = 100$, $H = 0.1$ gives closed-loop gain $100/11 = 9.09$. For large $GH$, $T \approx 1/H$, so the gain depends mainly on the (accurate) feedback element. The lost gain is recovered with an amplifier.

Effect on speed of response

Take a first-order plant $G(s) = \dfrac{K}{\tau s + 1}$.

Open loop: for a unit step input

$$c(t) = K\left(1 - e^{-t/\tau}\right), \qquad \text{time constant} = \tau$$

Closed loop (unity feedback):

$$\frac{C(s)}{R(s)} = \frac{G}{1+G} = \frac{K}{\tau s + 1 + K} = \frac{\dfrac{K}{1+K}}{\dfrac{\tau}{1+K}s + 1}$$ $$c(t) = \frac{K}{1+K}\left(1 - e^{-t(1+K)/\tau}\right), \qquad \tau_{cl} = \frac{\tau}{1+K}$$

The time constant is reduced by $(1+K)$ and the bandwidth increases by the same factor, so the response is faster. Gain and speed are traded: gain $\times$ bandwidth stays constant ($K/\tau$).

Example: $\tau = 1$ s, $K = 9$. Open loop: $\tau = 1$ s, settling time $4\tau = 4$ s. Closed loop: $\tau_{cl} = 1/10 = 0.1$ s, settling time $= 0.4$ s, i.e. ten times faster.