Assertion–Reason (Control Systems): For a stable linear time-invariant system driven by a sinusoidal input, the steady-state response depends on initial conditions; moreover, the steady-state frequency response is obtained by substituting s = jω into the transfer function.
-
ABoth A and R are correct and R is correct explanation of A
-
BBoth A and R are correct but R is not correct explanation of A
-
CA is correct but R is wrong
-
DR is correct but A is wrong
-
EBoth A and R are wrong
Answer
Correct Answer: R is correct but A is wrong
Explanation
Introduction / Context:In classical control systems, we frequently compare time-domain intuition (initial conditions, transients, steady state) with frequency-domain tools (Bode, Nyquist). This assertion–reason item tests whether steady-state sinusoidal response depends on initial conditions and recalls how frequency response is constructed from a transfer function.
Given Data / Assumptions:
- System is linear, time-invariant (LTI) and stable.
- Input is a pure sinusoid of fixed frequency ω.
- Transfer function G(s) exists and poles lie in the open left-half plane.
Concept / Approach:For a stable LTI system, the total response = transient response + forced (steady-state) response. Transients depend on initial conditions but decay to zero as t → ∞. The steady-state sinusoidal output has the same frequency as the input, scaled by |G(jω)| and phase-shifted by ∠G(jω). Therefore steady state is independent of initial conditions. Frequency response is indeed obtained by evaluating G(s) on the jω axis (s = jω).
Step-by-Step Solution:
Write output y(t) = y_tr(t) + y_ss(t).For stability, y_tr(t) → 0 as t → ∞; only y_ss(t) remains.Compute phasor: Y(jω) = G(jω) X(jω).Back to time domain: y_ss(t) = |G(jω)| A sin(ωt + ∠G(jω)).Verification / Alternative check:
Simulate any stable first/second-order system with different initial conditions; long-time output waveforms coincide in amplitude/phase, confirming independence from initial conditions.Why Other Options Are Wrong:
Options claiming A is correct contradict the decay of transients in stable LTI systems.Saying both are wrong ignores the standard definition of frequency response (s = jω) and steady-state behavior.Common Pitfalls:
Confusing transient overshoot (initial conditions matter) with the ultimate periodic output (they do not).Final Answer:
R is correct but A is wrong