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Time Differentiation Property of Fourier Transform
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 $j\omega$ 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 $n^{th}$ 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 $\left [ X(\omega)=\frac{j\omega}{(1+j\omega)^{2}} \right]$
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 $[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,
$$\mathrm{F^{-1}[j\omega \cdot X_{1}(\omega)]=\frac{d}{dt}x_{1}(t)=\frac{d}{dt}[t\:e^{-t}u(t)]}$$
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