---
title: "Derive the CT Fourier transform pair"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:ctft-derivation"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ctft-derivation
site: BCT Engineering
---

# Derive the CT Fourier transform pair

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2073 Bhadra, 8 marks; 2071 Magh, 7 marks; 2071 Bhadra, 8 marks (asked 3 times)
- Video: [Lecture 7: Fourier Transform Pair: Non-periodic Function Part 1 of 2](https://www.youtube.com/watch?v=KqJ86Yt3DT8) — civilengineeringprof
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/ctft-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Derive the expression for Fourier transform equation and inverse Fourier transform equation for continuous time aperiodic signals.

## Short answer

Treat the aperiodic signal as a periodic signal with period tending to infinity. The Fourier series sum becomes an integral, yielding the analysis equation X(w) = integral x(t) e^{-jwt} dt and the synthesis equation x(t) = (1/(2pi)) integral X(w) e^{jwt} dw.

## Answer

The Fourier transform of an aperiodic signal is obtained by treating the signal as a periodic signal whose period tends to infinity, and then taking the limit of its Fourier series.

#### Step 1: Periodic extension
Let $x(t)$ be aperiodic and of finite duration, with $x(t) = 0$ for $|t| > T_1$. Build $\tilde{x}(t)$ by repeating $x(t)$ every $T$ seconds, with $T > 2T_1$ so the copies do not overlap. Over $-T/2 < t < T/2$, $\tilde{x}(t) = x(t)$, and as $T \to \infty$, $\tilde{x}(t) \to x(t)$ for every $t$.
```text
 x(t):          ___/\___
 x~(t):  _/\_______/\_______/\_
           |<--T-->|
```

#### Step 2: Fourier series of the periodic extension
With $\omega_0 = 2\pi/T$:
$$
\tilde{x}(t) = \sum_{k=-\infty}^{\infty}a_k e^{jk\omega_0 t}, \qquad a_k = \frac{1}{T}\int_{-T/2}^{T/2}\tilde{x}(t)e^{-jk\omega_0 t}dt
$$
Inside $|t| < T/2$, $\tilde{x}(t) = x(t)$, and $x(t) = 0$ outside this range. So the limits can be extended to infinity:
$$
a_k = \frac{1}{T}\int_{-\infty}^{\infty}x(t)e^{-jk\omega_0 t}dt
$$

#### Step 3: Define the envelope X(ω)
$$
X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt \quad\Rightarrow\quad a_k = \frac{1}{T}X(k\omega_0)
$$
The Fourier series coefficients are equally spaced samples of $X(\omega)$, scaled by $1/T$.

#### Step 4: Substitute in the synthesis equation
Using $\frac{1}{T} = \frac{\omega_0}{2\pi}$:
$$
\tilde{x}(t) = \sum_{k=-\infty}^{\infty}\frac{1}{T}X(k\omega_0)e^{jk\omega_0 t} = \frac{1}{2\pi}\sum_{k=-\infty}^{\infty}X(k\omega_0)e^{jk\omega_0 t}\,\omega_0
$$

#### Step 5: Limit T → ∞
- $\omega_0 = 2\pi/T \to d\omega$ (the spectral lines merge).
- $k\omega_0 \to \omega$, a continuous variable.
- The sum $\sum(\cdot)\omega_0$ becomes the integral $\int(\cdot)d\omega$. This is the area under $X(\omega)e^{j\omega t}$, approximated by rectangles of width $\omega_0$.
- $\tilde{x}(t) \to x(t)$.

```text
 X(w) e^{jwt}
      _|_|_
    _|     |_          area of each strip
  _|  w0    |_   =   X(k w0) e^{jk w0 t} w0
 ---+-+-+-+-+-+--- w
```

Therefore
$$
\boxed{x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega \quad \text{(inverse FT / synthesis)}}
$$
$$
\boxed{X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt \quad \text{(FT / analysis)}}
$$
In terms of $f$ (with $\omega = 2\pi f$): $X(f) = \int x(t)e^{-j2\pi ft}dt$ and $x(t) = \int X(f)e^{j2\pi ft}df$.

#### Remarks
- An aperiodic signal has a **continuous spectrum**, while a periodic signal has a line spectrum.
- **Existence (Dirichlet conditions):** $x(t)$ is absolutely integrable ($\int|x(t)|dt < \infty$), and it has a finite number of maxima, minima and finite discontinuities in any finite interval. Signals with finite energy also have a transform, converging in the mean-square sense.
- **Example:** for $x(t) = 1$ when $|t| < T_1$ (0 otherwise), $X(\omega) = \frac{2\sin\omega T_1}{\omega}$. Its samples $\frac{1}{T}X(k\omega_0) = \frac{\sin k\omega_0 T_1}{k\pi}$ are exactly the coefficients of the periodic square wave.

## Check yourself

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

1. How is the CT Fourier transform derived?
   A. By taking the limit T → ∞ of the Fourier series of a periodic extension of x(t)
   B. By differentiating x(t)
   C. By sampling x(t)
   D. By applying Laplace transform at s = 0
   Answer: A. The aperiodic signal is regarded as a periodic one whose period tends to infinity. (Key point: Approach)

2. What is the relation between the series coefficients a_k and X(ω)?
   A. a_k = X(kω0)
   B. a_k = (1/T) X(kω0)
   C. a_k = T X(kω0)
   D. a_k = X(ω0)/k
   Answer: B. a_k = (1/T)∫x(t)e^{-jkω0 t}dt = (1/T) X(kω0). (Key point: Relation of a_k to X(ω))

3. As T → ∞, which is true?
   A. ω0 → dω, kω0 → ω and the sum becomes an integral
   B. ω0 grows without bound
   C. The spectral lines move apart
   D. X(ω) becomes zero
   Answer: A. The spectral lines merge into a continuous spectrum. (Key point: Limit effects)

4. The inverse Fourier transform and spectrum of an aperiodic signal are:
   A. x(t) = (1/2π)∫X(ω)e^{jωt}dω with a continuous spectrum
   B. x(t) = ∫X(ω)e^{jωt}dω with a line spectrum
   C. x(t) = Σ X(kω0)e^{jkω0 t} with a continuous spectrum
   D. x(t) = (1/2π)Σ X(ω) with a line spectrum
   Answer: A. Aperiodic signals have a continuous spectrum; periodic ones have a line spectrum. (Key point: Result and spectrum type)

## Related in Signals and Systems

- [Fourier transform of CT periodic signals](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aft-periodic-signal-derivation) — Signals and Systems, chapter 3, asked 3×
- [Parseval's theorem for CT aperiodic signals](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Act-parseval-aperiodic) — Signals and Systems, chapter 3, asked 5×
- [Inverse Fourier transform of δ(ω)](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Ainverse-ft-delta-omega) — Signals and Systems, chapter 3
- [Derive the DTFT pair](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtft-pair-derivation) — Signals and Systems, chapter 3, asked 3×
- [Derive exponential Fourier series (CT)](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actfs-exponential-derivation) — Signals and Systems, chapter 2, asked 15×
- [Derive trigonometric Fourier series (CT)](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actfs-trigonometric-derivation) — Signals and Systems, chapter 2, asked 3×

## Related in other subjects

- [Autocorrelation and ESD as Fourier transform pair](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aautocorrelation-esd-fourier-pair) — Communication Systems, chapter 1
- [Properties of continuous-time Fourier series](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Afourier-series-properties) — Data Communication, chapter 3
- [Fourier transform property: Time shifting](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aft-property-time-shifting) — Communication Systems, chapter 1
- [Derive time shifting property of z-transform](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Az-time-shifting-derive) — Digital Signal Analysis and Processing, chapter 2