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True/false: feedback makes response faster

Short answer

True. Negative feedback reduces the time constant. For a first-order plant K/(τs+1) with unity feedback, closed-loop time constant becomes τ/(1+K), making response faster. Example: τ=1 s, K=9 reduces settling time from 4 s to 0.4 s.

State whether the following statement is true or false and justify: Introduction of feedback on the system makes the system response faster.

True (for negative feedback in a stable system). Feedback reduces the time constant and increases the bandwidth.

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

Open loop: for a unit step input

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

Closed loop (unity feedback):

C(s)R(s)=G1+G=Kτs+1+K=K1+Kτ1+Ks+1\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)=K1+K(1−e−t(1+K)/τ),τcl=τ1+Kc(t) = \frac{K}{1+K}\left(1 - e^{-t(1+K)/\tau}\right), \qquad \tau_{cl} = \frac{\tau}{1+K}

The time constant falls from τ\tau to τ/(1+K)\tau/(1+K), so the response is (1+K)(1+K) times faster, at the cost of reduced gain K/(1+K)K/(1+K).

Example: τ=1\tau = 1 s, K=9K = 9. Open loop: τ=1\tau = 1 s, settling time 4τ=44\tau = 4 s. Closed loop: τcl=1/10=0.1\tau_{cl} = 1/10 = 0.1 s, settling time =0.4= 0.4 s, i.e. ten times faster.

✓ Answer checked by Claude Opus 5.5 (high effort)

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