# Frequency Derivative Property of Fourier Transform

Signals and SystemsElectronics & ElectricalDigital Electronics

## Fourier Transform

The Fourier transform of a continuous-time function can be defined as,

$$\mathrm{X(\omega)=\int_{−\infty }^{\infty}\:X(t)e^{-j\omega t}\:dt}$$

## Differentiation in Frequency Domain Property of Fourier Transform

Statement − The frequency derivative property of Fourier transform states that the multiplication of a function X(t) by in time domain is equivalent to the differentiation of its Fourier transform in frequency domain. Therefore, if

$$\mathrm{X(t)\overset{FT}{\leftrightarrow}X(\omega)}$$

Then, according to frequency derivative property,

$$\mathrm{t\cdot x(t)\overset{FT}{\leftrightarrow}j\frac{d}{d\omega}X(\omega)}$$

Proof

From the definition of Fourier transform, we have,

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

Differentiating the above equation on both sides with respect to ω, we get,

$$\mathrm{\frac{d}{d\omega}X(\omega)=\frac{d}{d\omega}\left [ \int_{−\infty }^{\infty}x(t)e^{-j\omega t}\:dt \right ]}$$

$$\mathrm{\Rightarrow\:\frac{d}{d\omega}X(\omega)=\int_{−\infty }^{\infty} x(t)\frac{d}{d\omega}\left [e^{-j\omega t} \right ]dt}$$

$$\mathrm{\Rightarrow\:\frac{d}{d\omega}X(\omega)=\int_{−\infty }^{\infty} x(t)(-jt)e^{-j\omega t}dt}$$

$$\mathrm{\Rightarrow\:\frac{d}{d\omega}X(\omega)=-j\int_{−\infty }^{\infty}t\cdot x(t)e^{-j\omega t}dt=-jF[tx(t)]}$$

Therefore,

$$\mathrm{F[tx(t)]=j\frac{d}{d\omega}X(\omega)}$$

Or, it can be represented as

$$\mathrm{t\cdot x(t)\overset{FT}{\leftrightarrow}j\frac{d}{d\omega}X(\omega)}$$

## Numerical Example

Using frequency derivative property of Fourier transform, find the Fourier transform of function $[te^{-2t}\:u(t)]$.

Solution

Given

$$\mathrm{x(t)=te^{-2t}u(t)}$$

Let,

$$\mathrm{x_{1}(t)=e^{-2t}u(t)}$$

By the definition of Fourier transform of a single sided exponential function, we have,

$$\mathrm{F[e^{-at}u(t)]=\frac{1}{a+j\omega}}$$

Therefore, for the function $X_{1}(t)$ , we have,

$$\mathrm{X_{1}(\omega)=F[e^{-2t}u(t)]=\frac{1}{2+j\omega}}$$

Now, by using the frequency derivative property $[i.e., t\cdot x(t)\overset{FT}{\leftrightarrow}j\frac{d}{d\omega}X(\omega)]$ of Fourier transform, we get,

$$\mathrm{F[te^{-2t}u(t)]=j\frac{d}{d\omega}F[e^{-2t}u(t)]}$$

$$\mathrm{\Rightarrow\:F[te^{-2t}u(t)]=j\frac{d}{d\omega}\left (\frac{1}{2+j\omega} \right )=j\frac{-1(j)}{(2+j\omega)^2}}$$

Therefore, the Fourier transform of the given function is,

$$\mathrm{F[te^{-2t}u(t)]=\frac{1}{(2+j\omega)^2}}$$

Or, it can also be written as,

$$\mathrm{te^{-2t}u(t)\overset{FT}{\leftrightarrow}\frac{1}{(2+j\omega)^2}}$$

Updated on 02-Dec-2021 12:01:25