---
title: "System output using Fourier transforms"
subject: "Signals and Systems"
chapter: 5
key: "semester_4/signals_and_systems:output-via-fourier-transform"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/output-via-fourier-transform
site: BCT Engineering
---

# System output using Fourier transforms

Signals and Systems · Chapter 5: Continuous-Time Systems

- Asked in: 2069 Bhadra, 8 marks
- Video: [Using Fourier Transforms to find the output of an LTI system](https://www.youtube.com/watch?v=neZ1FvH9BCg) — John Buck
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/output-via-fourier-transform
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5

## Question

A LTI system has input x(t) = e^(−t)u(t) and impulse response h(t) = e^(t)u(−t). Find output y(t) of the system using Fourier transform of x(t) and h(t).

## Short answer

Using Fourier transforms, X(w)=1/(1+jw) and H(w)=1/(1-jw), so Y(w)=1/(1+w^2). Inverse transforming gives y(t)=0.5 e^(-t)u(t)+0.5 e^t u(-t), which equals 0.5 e^(-|t|); the output is two-sided because h is anti-causal.

## Answer

For an LTI system, convolution in time becomes multiplication in frequency:

$$
y(t) = x(t) * h(t) \;\Longleftrightarrow\; Y(\omega) = X(\omega)\,H(\omega)
$$

#### Step 1: Fourier transform of the input

$$
\begin{aligned}
X(\omega) &= \int_{0}^{\infty} e^{-t}e^{-j\omega t}dt = \left[\frac{e^{-(1+j\omega)t}}{-(1+j\omega)}\right]_0^{\infty} \\
&= \frac{1}{1 + j\omega}
\end{aligned}
$$

#### Step 2: Fourier transform of the impulse response

$h(t) = e^{t}u(-t)$ is non-zero only for $t < 0$:

$$
\begin{aligned}
H(\omega) &= \int_{-\infty}^{0} e^{t}e^{-j\omega t}dt = \left[\frac{e^{(1-j\omega)t}}{1-j\omega}\right]_{-\infty}^{0} \\
&= \frac{1}{1 - j\omega}
\end{aligned}
$$

#### Step 3: Output spectrum

$$
Y(\omega) = \frac{1}{(1+j\omega)(1-j\omega)} = \frac{1}{1 + \omega^2}
$$

#### Step 4: Inverse transform (partial fractions)

$$
\frac{1}{(1+j\omega)(1-j\omega)} = \frac{A}{1+j\omega} + \frac{B}{1-j\omega}
$$

Multiply out: $1 = A(1-j\omega) + B(1+j\omega)$. Comparing terms: $A + B = 1$ and $-A + B = 0$, so $A = B = \tfrac{1}{2}$.

$$
Y(\omega) = \frac{1}{2}\cdot\frac{1}{1+j\omega} + \frac{1}{2}\cdot\frac{1}{1-j\omega}
$$

Using the pairs from Steps 1 and 2:
- $\frac{1}{1+j\omega} \leftrightarrow e^{-t}u(t)$
- $\frac{1}{1-j\omega} \leftrightarrow e^{t}u(-t)$

$$
y(t) = \frac{1}{2}e^{-t}u(t) + \frac{1}{2}e^{t}u(-t)
$$

**Answer:**

$$
y(t) = \frac{1}{2}e^{-|t|}, \quad -\infty < t < \infty
$$

**Check:** the standard pair $e^{-a|t|} \leftrightarrow \frac{2a}{a^2+\omega^2}$ with $a = 1$ gives $\frac{1}{2}e^{-|t|} \leftrightarrow \frac{1}{1+\omega^2}$. Also $y(0) = \frac{1}{2\pi}\int\frac{d\omega}{1+\omega^2} = \frac{1}{2\pi}\cdot\pi = 0.5$.

```text
          y(t)
          0.5
          /\
       .-'  '-.
 ___.-'        '-.___
 ---------+----------> t
          0
```

The output is two-sided because $h(t)$ is anti-causal (non-zero for $t < 0$).

## Check yourself

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

1. How is the output found using Fourier transforms?
   A. Y(w) = X(w) + H(w)
   B. Y(w) = X(w) H(w), then take the inverse transform
   C. Y(w) = X(w) / H(w)
   D. Y(w) = X(w) * H(w)
   Answer: B. Convolution in time becomes multiplication in frequency. (Key point: Convolution theorem)

2. What is X(w) for x(t) = e^(-t) u(t)?
   A. 1/(1 - jw)
   B. 1/(1 + w^2)
   C. 1/(1 + jw)
   D. jw/(1 + jw)
   Answer: C. The integral of e^(-(1+jw)t) from 0 to inf equals 1/(1+jw). (Key point: Transform of the input)

3. h(t) = e^t u(-t) exists only for t < 0. What is H(w)?
   A. 1/(1 + jw)
   B. 1/(1 - jw)
   C. 1/(1 - w^2)
   D. -1/(1 + jw)
   Answer: B. The integral of e^((1 - jw)t) from -inf to 0 equals 1/(1 - jw). (Key point: Transform of h(t))

4. What is y(t)?
   A. e^(-t) u(t)
   B. e^(-|t|)
   C. (1/2) e^(-|t|)
   D. (1/2) e^(-t) u(t)
   Answer: C. Y(w) = 1/(1 + w^2) = 0.5 [1/(1+jw) + 1/(1-jw)], which inverts to 0.5 e^(-|t|), a two-sided signal. (Key point: Output)

## Close variants set in other exams

- [y(t) for h = e^(−t)u(t−3), x = u(t−2)](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aconvolution-delayed-exponential-step) — Signals and Systems, chapter 5, asked 5×

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