SarkariResultOutBMS · 2024

Previous Year Question

Given the Laplace transform of a signal as Y(s) = 6e^−2s/(s + 4), what will be the initial value in time domain y(0)?

Correct answer: C

Explanation

The factor e^−2s is a 2 s transport delay. Therefore y(t)=6e^−4(t−2)u(t−2), which is zero for t<2 s. Hence y(0)=0.

Formula

ℒ⁻¹{6/(s + 4)} = 6e^−4tu(t)

e^−asF(s) ↔ f(t − a)u(t − a)

y(t) = 6e^−4(t−2)u(t − 2)

y(0) = 0

Step-by-step solution

Initial Value with Time Delay explanatory diagram

Correct Answer

C — 0

The factor e⁻²ˢ delays the causal signal by 2 seconds, so the waveform has not started at t=0 and its initial value is zero.

What This Question Is Really Testing

The Laplace time-shifting property is the key concept. Multiplication by e^−as does not scale the amplitude; it shifts a causal time-domain waveform to the right by a seconds.

Concept Foundation

Use the second shifting theorem:

e^−asF(s) ↔ f(t − a)u(t − a)

Step 1 — Remove the Delay Factor Temporarily

F(s) = 6/(s + 4)

f(t) = 6e^−4tu(t)

Step 2 — Apply the 2-Second Delay

a = 2 s

Y(s) = e^−2sF(s)

y(t) = f(t − 2)u(t − 2)

y(t) = 6e^−4(t−2)u(t − 2)

Step 3 — Evaluate the Initial Value

At t=0, the delayed unit-step factor is u(−2)=0.

y(0) = 6e⁸·0 = 0

Therefore option C is correct.

Why This Method Works

The unit-step factor u(t−2) enforces causality: the delayed waveform is identically zero before t=2 s.

Independent Verification / Cross-Check

y(0⁺) = lim_s→∞ sY(s)

= lim_s→∞ [6s e^−2s/(s + 4)]

= 0

The initial-value theorem independently confirms the same result.

Concept Extension

e^−as represents delay by a seconds.
Before the delayed start time, a causal signal is zero.
The pole at −4 controls the exponential decay after t=2 s.

Exam Strategy

Whenever you see e^−as, first mark the start time t=a. If the question asks for a value before that time, the delayed causal signal is zero.

Other options

A is the undelayed exponential’s value at t=0 and ignores the 2 s delay. B equals 6/4, which is not y(0). D is the magnitude of the pole location, not the signal value.

Common mistake

Ignoring the delay factor e⁻²ˢ and directly taking the undelayed initial value 6.

Exam tip

e⁻ᵃˢ means a-second delay; a causal delayed signal is zero for 0 ≤ t < a.

Quick method

Delay = 2 s; at t=0 the signal has not started → 0.