When a unit step input voltage is applied to a simple RC lag network, what is the time-domain behavior of the output response?
Electronics and Communication Engineering
Automatic Control Systems
Difficulty: Easy
Choose an option
-
Aremains constant at the unit step value
-
Bincreases exponentially from zero to the final value
-
Cdecreases exponentially from 1 to 0
-
Deither (b) or (c) depending on R and C values
-
Eoscillates around the final value
Answer
Correct Answer: increases exponentially from zero to the final value
Explanation
Introduction / Context:Lag networks (first-order RC circuits) are widely used in control and communication systems for shaping frequency response. Understanding their time-domain step response helps correlate time and frequency domain behavior.
Given Data / Assumptions:
- Unit step input u(t) applied at t = 0.
- Lag network: RC with transfer function H(s) = 1 / (1 + sRC).
- Output across capacitor.
Concept / Approach:The step response of a first-order low-pass network is exponential. It starts at 0 V and asymptotically approaches the final steady-state value equal to the input amplitude (here, 1 V).
Step-by-Step Solution:
Transfer function H(s) = 1 / (1 + sRC).Laplace of input: 1/s.Output: Y(s) = (1/s)(1 / (1 + sRC)).Inverse Laplace: y(t) = 1 − e^(−t/RC).Hence, y(t) rises exponentially from 0 to 1 as t → ∞.Verification / Alternative check:
Compare with capacitor charging curve; matches physical intuition.Why Other Options Are Wrong:
Constant from start: wrong, no system can instantaneously jump to final.Decreasing from 1 to 0: describes discharge, not step input.Oscillation: absent in first-order lag networks.Common Pitfalls:
Confusing lag network with lead or resonant circuits; forgetting exponential charging law.Final Answer:
increases exponentially from zero to the final value