---
title: "y(t) for h = e^(−t)u(t−3), x = u(t−2)"
subject: "Signals and Systems"
chapter: 5
key: "semester_4/signals_and_systems:convolution-delayed-exponential-step"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-delayed-exponential-step
site: BCT Engineering
---

# y(t) for h = e^(−t)u(t−3), x = u(t−2)

Signals and Systems · Chapter 5: Continuous-Time Systems

- Asked in: 2082 Kartik, 5 marks; 2071 Magh, 8 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-delayed-exponential-step
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5

## Question

Find the output y(t) of LTI system if the impulse response h(t) = e^(−t)u(t−3) and input x(t) = u(t−2).

## Short answer

For x(t)=u(t-2) and h(t)=e^{-t}u(t-3), output y(t) = 0 for t<5, and y(t)=e^{-3}-e^{-(t-2)} for t≥5. Overlap occurs when τ>2 and τ<t-3, requiring t>5. Integration on [2, t-3] gives y(t)=e^{-3}(1-e^{-(t-5)})u(t-5).

## Answer

For an LTI system, $y(t) = x(t) * h(t) = \displaystyle\int_{-\infty}^{\infty} x(\tau)\,h(t - \tau)\,d\tau$.

Given $x(t) = u(t-2)$ and $h(t) = e^{-t}u(t-3)$.

#### Step 1: Write both functions in $\tau$
$$
x(\tau) = u(\tau - 2), \qquad h(t-\tau) = e^{-(t-\tau)}\,u(t - \tau - 3)
$$
- $x(\tau) \neq 0$ when $\tau > 2$.
- $h(t-\tau) \neq 0$ when $t - \tau > 3$, i.e. $\tau < t - 3$.

```text
 x(tau)=1 for tau>2      h(t-tau) for tau<t-3
          ___________   ___________
         |              decaying   |
 --------+-------->   <------------+------ tau
         2                        t-3
 overlap: 2 < tau < t-3  (exists only if t-3 > 2)
```

#### Step 2: Range of $t$
The overlap $2 < \tau < t - 3$ exists only when $t - 3 > 2$, i.e. **$t > 5$**.
- For $t < 5$: no overlap, $y(t) = 0$.

#### Step 3: Evaluate for $t > 5$
$$
\begin{aligned}
y(t) &= \int_{2}^{t-3} 1\cdot e^{-(t-\tau)}\,d\tau = e^{-t}\int_{2}^{t-3} e^{\tau}\,d\tau \\
&= e^{-t}\left[e^{\tau}\right]_{2}^{t-3} = e^{-t}\left(e^{t-3} - e^{2}\right) \\
&= e^{-3} - e^{-(t-2)}
\end{aligned}
$$

#### Result
$$
y(t) = \left(e^{-3} - e^{-(t-2)}\right)u(t-5)
$$
**Answer: $y(t) = 0$ for $t < 5$, and $y(t) = e^{-3} - e^{-(t-2)}$ for $t \ge 5$.**

Checks:
- At $t = 5$: $e^{-3} - e^{-3} = 0$ (continuous start).
- As $t \to \infty$: $y \to e^{-3} \approx 0.0498$.
- Shift view: $u(t)*e^{-t}u(t) = (1 - e^{-t})u(t)$; here $h(t) = e^{-3}\,e^{-(t-3)}u(t-3)$, so total shift is $2 + 3 = 5$: $y(t) = e^{-3}(1 - e^{-(t-5)})u(t-5)$, which is the same expression.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For x(τ) = u(τ-2) and h(t-τ) = e^{-(t-τ)}u(t-τ-3), the nonzero overlap is:
   A. 2 < τ < t - 3
   B. 2 < τ < t + 3
   C. τ > t - 3
   D. -∞ < τ < 2
   Answer: A. x is nonzero for τ > 2 and h(t-τ) for τ < t - 3. (Key point: Limits of overlap)

2. From which t is y(t) non-zero?
   A. t > 2
   B. t > 5
   C. t > 3
   D. t > 1
   Answer: B. The overlap exists only when t - 3 > 2. (Key point: Starting time)

3. What is y(t) for t > 5?
   A. 1 - e^{-t}
   B. e^{-3} + e^{-(t-2)}
   C. e^{-3} - e^{-(t-2)}
   D. e^{-(t-5)}
   Answer: C. ∫_2^{t-3} e^{-(t-τ)} dτ = e^{-t}(e^{t-3} - e^2). (Key point: Result)

4. As t -> ∞, y(t) approaches:
   A. 0
   B. 1
   C. e^{-2}
   D. e^{-3} ≈ 0.0498
   Answer: D. The e^{-(t-2)} term vanishes leaving e^{-3}. (Key point: Final value)

## Close variants set in other exams

- [Output for x = e^(−at)u(t), h = u(t)](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aconvolution-exponential-step) — Signals and Systems, chapter 5, asked 4×
- [Output of e^(−2t)u(t) to a rectangular input](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aconvolution-exponential-rect-input) — Signals and Systems, chapter 5, asked 4×
- [System output using Fourier transforms](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aoutput-via-fourier-transform) — Signals and Systems, chapter 5, asked 3×

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