---
title: "Fourier transform of a CT signal"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:ft-unspecified-ct-signal"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-unspecified-ct-signal
site: BCT Engineering
---

# Fourier transform of a CT signal

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2078 Chaitra, 6 marks
- Video: [Signals & Systems - Fourier Transforms - working examples - 1 - UNIT II](https://www.youtube.com/watch?v=wpK9qzqPjbk) — Dr.P.Prasanna Murali krishna
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-unspecified-ct-signal
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Obtain the Fourier transform of the continuous-time signal. [The signal is not printed on the paper.]

## Short answer

Assuming x(t)=e^{-a|t|}, a>0, split the integral into t<0 and t>0, integrate e^{at} and e^{-at} with e^{-jωt}. The Fourier transform is X(jω)=2a/(a^2+ω^2).

## Answer

The signal was not printed on the paper, so a common exam signal is assumed here: the two-sided exponential $x(t) = e^{-a|t|}$, $a > 0$. The same method applies to any other signal.

**Definition used:**

$$
X(j\omega) = \int_{-\infty}^{\infty} x(t)\,e^{-j\omega t}\,dt
$$

**Step 1: split the signal.** $e^{-a|t|} = e^{at}$ for $t<0$ and $e^{-at}$ for $t>0$:

$$
X(j\omega) = \int_{-\infty}^{0} e^{at}e^{-j\omega t}dt + \int_{0}^{\infty} e^{-at}e^{-j\omega t}dt
$$

**Step 2: integrate each part.**

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

Both limits at $\pm\infty$ go to zero because $a>0$.

**Step 3: add.**

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

**Result:**

- $X(j\omega)$ is real, positive and even, because $x(t)$ is real and even. So the phase is $0$ for all $\omega$.
- Peak value $X(0) = \frac{2}{a}$ (equal to the area under $x(t)$); it falls to half at $\omega = \pm a$.
- A larger $a$ (faster decay in time) gives a wider spectrum: time and frequency widths are inversely related.

```text
  X(jw)
 2/a ^
     |    ..
     |  .    .
 1/a |.--------.   (half value at w = +/- a)
    .|          .
 ..  |            ..
-----+-------------------> w
    -a   0    a
```

If the question paper shows a different signal, use the same steps: write $x(t)$ piece by piece, apply the definition (or known pairs and properties such as shifting and scaling), and simplify.

## Check yourself

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

1. To find the FT of e^(-a|t|), what is the first step?
   A. Split the integral into t<0 (e^(at)) and t>0 (e^(-at)) parts
   B. Differentiate the signal
   C. Use Parseval's relation
   D. Take the Laplace transform
   Answer: A. The absolute value makes the signal piecewise, so the integral is split at t = 0. (Key point: Splitting the signal)

2. The FT of e^(-a|t|), a>0, is:
   A. 1/(a + jω)
   B. 2a/(a^2 + ω^2)
   C. 2/(a^2 + ω^2)
   D. a/(a^2 + ω^2)
   Answer: B. 1/(a - jω) + 1/(a + jω) = 2a/(a^2 + ω^2). (Key point: Result)

3. Why is X(jω) real and even?
   A. x(t) is real and even
   B. x(t) is causal
   C. x(t) is odd
   D. a is positive
   Answer: A. A real even signal has a real even transform, so the phase is 0. (Key point: Why real and even)

4. What happens to the spectrum if a is increased?
   A. It gets narrower
   B. It gets wider, since faster decay in time means wider bandwidth
   C. It is unchanged
   D. Its peak rises to 2/a
   Answer: B. Time and frequency widths are inversely related; the peak 2/a falls and the half-value points ±a move out. (Key point: Width relation)

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