---
title: "Frequency derivative property; t·e^(−at)u(t)"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:ft-frequency-derivative"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-frequency-derivative
site: BCT Engineering
---

# Frequency derivative property; t·e^(−at)u(t)

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2076 Baisakh, 3+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-frequency-derivative
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

State and prove frequency derivative property of fourier transform. Use same property to find fourier transform of x(t) = t·exp(−a·t)u(t), a > 0

## Short answer

Frequency derivative property: t·x(t) ↔ j dX(jω)/dω. Proof: differentiate X(jω)=∫ x(t)e^{-jωt}dt to get dX/dω = ∫[-jt x(t)]e^{-jωt}dt. For x(t)=t e^{-at}u(t), X(jω)=j d/dω[1/(a+jω)] = 1/(a+jω)^2.

## Answer

#### Frequency derivative (differentiation in frequency) property

**Statement:** If $x(t) \leftrightarrow X(j\omega)$, then

$$
-jt\,x(t) \;\longleftrightarrow\; \frac{dX(j\omega)}{d\omega}
\qquad\text{or}\qquad
t\,x(t) \;\longleftrightarrow\; j\frac{dX(j\omega)}{d\omega}
$$

**Proof:** Start from the definition

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

Differentiate both sides with respect to $\omega$ (the derivative can be taken inside the integral):

$$
\begin{aligned}
\frac{dX(j\omega)}{d\omega} &= \int_{-\infty}^{\infty} x(t)\,\frac{d}{d\omega}\left(e^{-j\omega t}\right)dt \\
&= \int_{-\infty}^{\infty} \left[-jt\,x(t)\right]e^{-j\omega t}\,dt
\end{aligned}
$$

The right side is the FT of $-jt\,x(t)$. Multiplying both sides by $j$ gives $t\,x(t) \leftrightarrow j\,\frac{dX(j\omega)}{d\omega}$. Hence proved.

#### FT of $x(t) = t\,e^{-at}u(t)$, $a>0$

Let $x_1(t) = e^{-at}u(t)$. Its FT is

$$
X_1(j\omega) = \int_0^\infty e^{-(a+j\omega)t}dt = \frac{1}{a+j\omega}
$$

Now $x(t) = t\,x_1(t)$, so by the property:

$$
\begin{aligned}
X(j\omega) &= j\frac{d}{d\omega}\left[\frac{1}{a+j\omega}\right] \\
&= j\cdot\frac{-j}{(a+j\omega)^2} \\
&= \frac{1}{(a+j\omega)^2}
\end{aligned}
$$

**Answer:** $t\,e^{-at}u(t) \;\longleftrightarrow\; \dfrac{1}{(a+j\omega)^2}$, with $|X(j\omega)| = \dfrac{1}{a^2+\omega^2}$ and $\angle X(j\omega) = -2\tan^{-1}(\omega/a)$.

**Check:** $X(0) = 1/a^2$, and the area $\int_0^\infty t e^{-at}dt = 1/a^2$. They match.

## Check yourself

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

1. Which pair is the frequency derivative property of the Fourier transform?
   A. t x(t) <-> j dX(jω)/dω
   B. t x(t) <-> -j dX(jω)/dω
   C. t x(t) <-> jω X(jω)
   D. t x(t) <-> dX(jω)/dt
   Answer: A. Differentiating X(jω) w.r.t. ω brings down -jt, so -jt x(t) <-> dX/dω; multiplying by j gives t x(t) <-> j dX/dω. (Key point: Property statement)

2. What is the key step in the proof?
   A. Differentiate x(t) w.r.t. t and integrate by parts
   B. Differentiate the definition integral w.r.t. ω inside the integral; d/dω of e^(-jωt) is -jt e^(-jωt)
   C. Substitute τ = t - t0 in the integral
   D. Replace e^(-jωt) by its Taylor series and drop higher terms
   Answer: B. Only the exponential depends on ω, so differentiating it gives the factor -jt inside the integral. (Key point: Proof step)

3. What is the FT of e^(-at)u(t), a > 0, used as the starting point?
   A. 1/(a - jω)
   B. a/(a^2 + ω^2)
   C. 1/(a + jω)
   D. 1/(a^2 + ω^2)
   Answer: C. The integral of e^(-(a+jω)t) from 0 to ∞ equals 1/(a+jω). (Key point: Base pair)

4. What is the FT of t e^(-at)u(t)?
   A. 1/(a + jω)
   B. j/(a + jω)^2
   C. 2a/(a^2 + ω^2)
   D. 1/(a + jω)^2
   Answer: D. j d/dω[1/(a+jω)] = j · (-j)/(a+jω)^2 = 1/(a+jω)^2, and X(0) = 1/a^2 matches the area. (Key point: Final result)

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