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)?
⚑ Report issue
Found a wrong answer, translation, image or source problem? Send it to editorial review.
Answer & Explanation
Correct AnswerC. 0
Quick 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 / Key Relation
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
Detailed Explanation
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+) = lims→∞ sY(s)
= lims→∞ [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.
Quick Trick
Delay = 2 s; at t=0 the signal has not started → 0.
Why Other Options Are Wrong
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.
Continue learningRelated PYQs & Exam ContextExam path, knowledge path and deeper practice links.
