---
title: "Output of LTI with h(t) = e^(-at)u(t) and e^(at)u(-t)"
subject: "Data Communication"
chapter: 2
key: "semester_5/data_communication:convolution-exp-u-and-u-neg"
source: https://www.bctengineer.com/q/semester_5/data_communication/convolution-exp-u-and-u-neg
site: BCT Engineering
---

# Output of LTI with h(t) = e^(-at)u(t) and e^(at)u(-t)

Data Communication · Chapter 2: Signals and Systems

- Asked in: 2076 Asoj, 10 marks; 2073 Shrawan, 3+6+1 marks (asked 2 times)
- Video: [Time domain - tutorial 8: LTI systems, impulse response & convolution](https://www.youtube.com/watch?v=Y-h7qoTVQtw) — SP hacks with Iman
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/data_communication/convolution-exp-u-and-u-neg
- Whole chapter: https://www.bctengineer.com/md/semester_5/data_communication/chapter-2

## Question

Sketch the output of LTI system having impulse response h(t) = e^(-at)u(t) (a>0) and h(t) = e^(at)u(-t) (a>0).

## Short answer

For a unit impulse input, the output is the impulse response itself. h(t)=e^{-at}u(t) is causal, decaying from 1 at t=0; stable. h(t)=e^{at}u(-t) is anticausal, rising to 1 at t=0; stable.

## Answer

For an LTI system the output is y(t) = x(t) * h(t). No input is given, so take the input as a unit impulse, x(t) = δ(t): the output is then h(t) itself. Below, each h(t) is sketched, and the output for a unit step input is also found as an extra illustration.

#### Case 1: h(t) = e^(-at)u(t), a > 0 (causal)
Impulse response (output for δ(t)):
```text
h(t)
1 |*
  | *
  |  **
  |    ****_____
  +-----------------> t
  0
```
It decays from 1 at t = 0 towards 0. It is causal and stable, since ∫₀^∞ e^(-at) dt = 1/a < ∞.

Step response, s(t) = ∫₀^t e^(-aτ) dτ = (1 - e^(-at))/a for t ≥ 0:
```text
s(t)
1/a |        _______
    |     _--
    |   /
    | /
    +-----------------> t
```

#### Case 2: h(t) = e^(at)u(-t), a > 0 (anticausal)
```text
h(t)
          * 1
        * |
     **   |
_****     |
----------+--------> t
          0
```
The impulse response exists only for t ≤ 0 and grows towards 1 at t = 0. It is anticausal (non-causal) and stable, since ∫_{-∞}^0 e^(at) dt = 1/a.

Step response: s(t) = ∫_{-∞}^{t} e^(aτ)u(-τ) dτ.
- For t ≤ 0: s(t) = e^(at)/a (rising to 1/a at t = 0).
- For t > 0: s(t) = 1/a (constant).
```text
s(t)
1/a |           ______________
    |        /
    |    __/
    +--------+--------------> t
             0
```
The output begins before the input is applied (at t < 0), which shows that the system is non-causal and cannot be realised in real time.

## Check yourself

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

1. The output for impulse input is
   A. integral of h
   B. h(t) itself
   C. zero
   D. derivative of h
   Answer: B. y = delta * h = h. (Key point: Output)

2. e^(-at)u(t) is
   A. anticausal
   B. unstable
   C. causal and stable
   D. causal, unstable
   Answer: C. Integral is 1/a. (Key point: Case 1)

3. e^(at)u(-t) with a>0 is
   A. causal
   B. unstable
   C. causal stable
   D. anticausal and stable
   Answer: D. Lives on t<=0, integral 1/a. (Key point: Case 2)

4. Step response of case 1 is
   A. (1 - e^(-at))/a
   B. e^(-at)
   C. 1/a for t<0
   D. e^(at)/a
   Answer: A. Integral from 0 to t. (Key point: Step response)

## Close variants set in other exams

- [LTI output for h(t)=e^(-2t), rectangular input](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Alti-output-rect-input) — Data Communication, chapter 2, asked 3×

## Related in Data Communication

- [Define LTI system and impulse response](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Alti-impulse-response-define) — Data Communication, chapter 2, asked 2×
- [LTI output for h(t)=e^(-at)u(t), x(t)=u(t)-u(t-2)](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Alti-output-u-minus-u2) — Data Communication, chapter 2

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