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Signals and Systems β Energy Spectral Density (ESD) and Autocorrelation Function
Energy Spectral Density
The distribution of energy of a signal in the frequency domain is called the energy spectral density (ESD) or energy density (ED) or energy density spectrum. It is denoted by $\psi (\omega )$ and is given by,
$$\mathrm{\psi (\omega )=\left | X(\omega ) \right |^{2}}$$
Autocorrelation
The autocorrelation function gives the measure of similarity between a signal and its time delayed version. The autocorrelation function of an energy signal x(t) is given by,
$$\mathrm{R(\tau )=\int_{-\infty }^{\infty}x(t)\:x^{*}(t-\tau )\:dt}$$
Where, the parameter $\tau$ is called the delayed parameter.
Relation between ESD and Autocorrelation Function
The autocorrelation function $R(\tau$) and the energy spectral density (ESD) function $\psi (\omega )$ form a Fourier transform pair, i.e.,
$$\mathrm{R(\tau )\overset{FT}{\leftrightarrow}\psi (\omega )}$$
Proof
The autocorrelation of a function x(t) is defined as,
$$\mathrm{R(\tau )=\int_{-\infty }^{\infty }x(t)\:x^{*}(t-\tau )dt}$$
Replacing $x^{*}(t-\tau )$ by its inverse transform, we get,
$$\mathrm{R(\tau )=\int_{-\infty }^{\infty }x(t)\left [\frac{1}{2\pi} \int_{-\infty }^{\infty }X(\omega )\:e^{j\omega(t-\tau )}d\omega \right ]^{*}\:dt}$$
$$\mathrm{\Rightarrow R(\tau )=\frac{1}{2\pi} \int_{-\infty }^{\infty }x(t)\left [\int_{-\infty }^{\infty }X^{*}(\omega )\:e^{-j\omega(t-\tau )}d\omega \right ]\:dt}$$
By interchanging the order of the integration, we get,
$$\mathrm{R(\tau )=\frac{1}{2\pi} \int_{-\infty }^{\infty }X^{*}(\omega) \left [\int_{-\infty }^{\infty }x(t)\:e^{-j\omega\tau }dt \right ]e^{j\omega \tau }\:d\omega }$$
$$\mathrm{\Rightarrow R(\tau )=\frac{1}{2\pi} \int_{-\infty }^{\infty }X^{*}(\omega)\:X(\omega )e^{j\omega \tau }d\omega }$$
$$\mathrm{\because X^{*}(\omega) X(\omega )=\left | X(\omega )\right |^{2}=\psi (\omega)}$$
$$\mathrm{\therefore R(\tau ) =\frac{1}{2\pi}\int_{-\infty }^{\infty}\left | X(\omega ) \right |^{2}\:e^{j\omega \tau }d\omega =\frac{1}{2\pi}\int_{-\infty }^{\infty}\psi (\omega) e^{j\omega \tau }d\omega=F^{-1}\left [ \psi (\omega) \right ]}$$
Hence,
$$\mathrm{F\left [R(\tau ) \right ]=\psi (\omega )}$$
Or, it can be written as
$$\mathrm{R(\tau )\overset{FT}{\leftrightarrow}\psi (\omega )}$$
Thus, it proves that the autocorrelation function $R(\tau)$ and the ESD function $\psi (\omega )$ forms the Fourier transform pair.
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