Skip to main content

Speed of response increases with gain

Short answer

For a first-order system, the closed-loop time constant τ_cl = τ/(1+K) decreases as gain K increases, reducing settling time and speeding response. For a second-order system, ω_n = √K grows with gain, so rise and peak times reduce, though overshoot may increase.

Show that the speed of response increases with the increase of the gain of the system.

Speed of response is measured by the time constant (first order) or natural frequency (second order); both improve as the loop gain KK rises.

First-order system

With unity feedback around G(s)=Kτs+1G(s) = \dfrac{K}{\tau s+1}:

C(s)R(s)=Kτs+1+K=K/(1+K)τ1+Ks+1\frac{C(s)}{R(s)} = \frac{K}{\tau s+1+K} = \frac{K/(1+K)}{\dfrac{\tau}{1+K}s+1} τcl=τ1+K,ts≈4τcl=4τ1+K\tau_{cl} = \frac{\tau}{1+K}, \qquad t_s \approx 4\tau_{cl} = \frac{4\tau}{1+K}

As KK increases, τcl\tau_{cl} decreases, so the response reaches its final value faster. Example (τ=1\tau = 1 s): K=1K = 1 gives τcl=0.5\tau_{cl} = 0.5 s; K=9K = 9 gives 0.10.1 s.

Second-order system

For G(s)=Ks(s+a)G(s) = \dfrac{K}{s(s+a)}: ωn=K\omega_n = \sqrt{K}. Rise time and peak time tp=π/(ωn1−ζ2)t_p = \pi/(\omega_n\sqrt{1-\zeta^2}) decrease as KK rises (though overshoot increases).

Hence speed of response increases with gain.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Control Systems chapter 1 questions →