Learning objective: Set the correct integration limits for causal signals, evaluate the convolution integral and cross-check the result using partial fractions.
Understand the idea
For a continuous-time linear time-invariant system, the output is the convolution of the input and impulse response. A frequent error is to treat convolution as ordinary multiplication in time. Multiplication of Laplace transforms corresponds to convolution in time, which gives us a useful independent check.
Worked example: given data
- Input signal x(t) = e⁻²ᵗu(t).
- Impulse response h(t) = e⁻³ᵗu(t).
- Find y(t) = x(t) * h(t) and evaluate y(1 s).
Detailed solution
Set limits from causality
The signals are zero before t = 0. For t ≥ 0, the only contributing values have 0 ≤ τ ≤ t. Therefore y(t) = ∫₀ᵗ e⁻²τ e⁻³(t−τ) dτ.
Separate time dependence
Pull e⁻³ᵗ outside the integral: y(t) = e⁻³ᵗ ∫₀ᵗ eτdτ = e⁻³ᵗ(eᵗ − 1).
Write the simplified output
For t ≥ 0, y(t) = e⁻²ᵗ − e⁻³ᵗ. Including the causal step, y(t) = (e⁻²ᵗ − e⁻³ᵗ)u(t). At t = 0, the integral is zero.
Evaluate at one second
y(1) = e⁻² − e⁻³ ≈ 0.13534 − 0.04979 = 0.08555. The output is positive for t > 0.
Verify in the Laplace domain
X(s) = 1/(s + 2) and H(s) = 1/(s + 3). Thus Y(s) = 1/[(s + 2)(s + 3)] = 1/(s + 2) − 1/(s + 3). Inverse transformation reproduces the time-domain answer.
What the answer means
Because both exponentials decay, the output eventually returns to zero even though it first rises from zero. The final value here is not a DC steady-state step response; the input itself is a decaying exponential. Understanding the type of input prevents incorrect use of the final-value theorem.
Three mistakes to avoid
- Using integration limits from −∞ to +∞ without checking where the signals are nonzero.
- Treating e⁻²ᵗ × e⁻³ᵗ as the time-domain convolution.
- Dropping the unit-step factor, which hides that the response is zero for t < 0.
Try a related question
Replace h(t) with e⁻²ᵗu(t), identical to x(t). What is y(t)?
Show the worked-answer check
The integral becomes ∫₀ᵗ e⁻²τ e⁻²(t−τ)dτ = te⁻²ᵗ for t ≥ 0. Therefore y(t) = te⁻²ᵗu(t).
How to revise this concept
Compare impulse response, step response, and frequency response for a first-order system. First attempt the example on paper without opening the answer, then compare the intermediate steps. Explain each assumption before memorising a formula.
Content note: This is a newly prepared practice example, not an official GATE previous-year question. It was drafted with AI assistance for this educational website. Students should verify calculations and exam requirements against standard textbooks and official sources; educator review is recommended before publication.