Time Differentiation Property of Fourier Transform

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Fourier Transform

The Fourier transform of a continuous-time function x(t) can be defined as,

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

And the inverse Fourier transform is defined as,

$$\mathrm{x(t)\:=\:\frac{1}{2\pi}\int_{-\infty}^{\infty}\:X(\omega)\:e^{j\omega t}d \omega}$$

Time Differentiation Property of Fourier Transform

Statement

The time differentiation property of Fourier transform states that the differentiation of a function in time domain is equivalent to the multiplication of its Fourier transform by a factor jω in frequency domain. Therefore, if

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

Then, according to the time differentiation property,

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

Proof

From the definition of inverse Fourier transform, we have,

$$\mathrm{x(t) \:=\: \frac{1}{2\pi}\int_{-\infty}^{\infty}\:X(\omega)\:e^{j\omega t}\: d\omega}$$

Taking time differentiation on both sides, we get,

$$\mathrm{\frac{d}{dt}x(t) \:=\: \frac{d}{dt}\left[\frac{1}{2\pi}\: \int_{-\infty}^{\infty}X(\omega)\:e^{j\omega t} d\omega\right ]}$$

$$\mathrm{\Rightarrow\:\frac{d}{dt}x(t) \:=\: \frac{1}{2\pi}\:\int_{-\infty}^{\infty}\:X(\omega)\:\frac{d}{dt}[e^{j\omega t}]d\omega \:=\: \frac{1}{2\pi}\:\int_{-\infty}^{\infty}\:X(\omega)j\omega\: e^{j\omega t}d\omega}$$

$$\mathrm{\Rightarrow\:\frac{d}{dt}x(t) \:=\: j\omega \left[\frac{1}{2\pi}\:\int_{-\infty}^{\infty}\:X(\omega)\:e^{j\omega t}d\omega \right ] \:=\: j\omega \:\cdot\: F^{-1}[X(\omega)]}$$

Therefore,

$$\mathrm{F\left[\frac{d}{dt}x(t) \right] \:=\: j\omega \:\cdot\: X(\omega)}$$

Or, it can also be represented as,

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

In general, the time differentiation property for $\mathrm{n^{th}}$ order differentiation is given by,

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

Numerical Example

Using time differentiation property of Fourier transform, find the inverse Fourier transform of $\mathrm{\left[ X(\omega) \:=\: \frac{j\omega}{(1 \:+\: j\omega)^{2}} \right]}$

Solution

Given

$$\mathrm{X(\omega) \:=\: \frac{j\omega}{(1 \:+\: j\omega)^{2}}}$$

The Fourier transform of single-sided exponential function is defined as,

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

Therefore, for the given function (a=1), we have,

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

Let,

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

Then,

$$\mathrm{x_{1}(\omega) \:=\: \frac{1}{(1 \:+\: j\omega)^{2}}}$$

Now, using time differentiation property $\mathrm{[i.e.,\: \frac{d}{dt}x(t)\overset{FT}{\leftrightarrow}j\omega \:\cdot\: X(\omega)]} $ of Fourier transform, we get,

$$\mathrm{F\left[\frac{d}{dt}x_{1}(t)\right ] \:=\: j\omega \:\cdot\: X_{1}(\omega)}$$

Hence, the inverse Fourier transform of the given function is,

$$\mathrm{F^{-1}[j\omega \:\cdot\: X_{1}(\omega)] \:=\: \frac{d}{dt}x_{1}(t) \:=\: \frac{d}{dt}[t\:e^{-t}u(t)]}$$