---
title: "Multiplication by t ↔ differentiation"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:ft-multiplication-by-t"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-multiplication-by-t
site: BCT Engineering
---

# Multiplication by t ↔ differentiation

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2081 Asoj, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-multiplication-by-t
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Show that the multiplication of a signal x(t) by t is equivalent to differentiation of its Fourier Transform.

## Short answer

Multiplication by t corresponds to j times the derivative of the Fourier transform: t x(t) ↔ j dX(ω)/dω. This is proven by differentiating the Fourier integral with respect to ω and then multiplying by j.

## Answer

#### Statement (differentiation in frequency)
If $x(t) \xleftrightarrow{F} X(\omega)$, then
$$
t\,x(t) \xleftrightarrow{F} j\frac{dX(\omega)}{d\omega}
$$

#### Proof
Start from the definition:
$$
X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt
$$
Differentiate both sides with respect to $\omega$. Differentiation can be taken inside the integral, since the limits do not depend on $\omega$:
$$
\frac{dX(\omega)}{d\omega} = \int_{-\infty}^{\infty}x(t)\frac{\partial}{\partial\omega}e^{-j\omega t}dt = \int_{-\infty}^{\infty}(-jt)\,x(t)e^{-j\omega t}dt
$$
Multiply both sides by $j$, using $j(-j) = 1$:
$$
j\frac{dX(\omega)}{d\omega} = \int_{-\infty}^{\infty}t\,x(t)e^{-j\omega t}dt = F\{t\,x(t)\}
$$
So multiplying by $t$ in time is the same as differentiating in frequency (and multiplying by $j$). Repeating the step gives $t^n x(t) \leftrightarrow j^n\frac{d^nX(\omega)}{d\omega^n}$.

#### Example
$e^{-at}u(t) \leftrightarrow \frac{1}{a + j\omega}$, so
$$
t\,e^{-at}u(t) \leftrightarrow j\frac{d}{d\omega}\left(\frac{1}{a + j\omega}\right) = j\cdot\frac{-j}{(a + j\omega)^2} = \frac{1}{(a + j\omega)^2}
$$

## Check yourself

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

1. If x(t) <-> X(ω), then t x(t) <-> ?
   A. dX/dω
   B. j dX/dω
   C. -j dX/dω times 2π
   D. X(ω)/(jω)
   Answer: B. Differentiating X(ω) gives -j times the transform of t x(t), so t x(t) <-> j dX/dω. (Key point: Statement)

2. What step is central to the proof?
   A. Differentiating the transform integral with respect to ω, bringing out (-jt)
   B. Integrating the transform by parts in time
   C. Using the shifting property
   D. Taking the conjugate of X
   Answer: A. d/dω of e^(-jωt) is -jt e^(-jωt), so dX/dω = ∫(-jt)x(t)e^(-jωt)dt. (Key point: Key step of proof)

3. What is the transform of t^n x(t)?
   A. j^n d^nX/dω^n
   B. (-j)^n X(ω)
   C. n j dX/dω
   D. X(ω)^n
   Answer: A. Repeating the step n times gives j^n times the nth derivative. (Key point: Repeated application)

4. Using the property, the transform of t e^(-at) u(t) is:
   A. 1/(a + jω)
   B. 1/(a + jω)^2
   C. -1/(a + jω)^2
   D. a/(a + jω)^2
   Answer: B. j d/dω [1/(a + jω)] = j(-j)/(a + jω)^2 = 1/(a + jω)^2. (Key point: Example)

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