Power Spectral Density (PSD) and Autocorrelation Function

Signals and SystemsElectronics & ElectricalDigital Electronics

Power Spectral Density

The distribution of average power of a signal in the frequency domain is called the power spectral density (PSD) or power density (PD) or power density spectrum. The power spectral density is denoted by $\mathit{S\left (\omega \right )}$ and is given by,

$$\mathrm{\mathit{S\left (\omega \right )\mathrm{=}\lim_{\tau \rightarrow \infty }\frac{\left | X\left (\omega \right ) \right |^{\mathrm{2}}}{\tau }}}$$

Autocorrelation

The autocorrelation function gives the measure of similarity between a signal and its time-delayed version. The autocorrelation function of power (or periodic) signal $\mathit{x\left ( t \right ) }$ with any time period T is given by,

$$\mathrm{\mathit{R\left(\tau \right)=\lim_{T\rightarrow \infty }\mathrm{\frac{1}{\mathit{T}}}\int_{-\left(T/\mathrm{2}\right)}^{T/\mathrm{2}}x\left(t\right)\:x^{*}\left(t-\tau \right)\:dt}}$$

Where, $\tau$ is called the delayed parameter.

Relation between PSD and Autocorrelation Function

The power spectral density function $\mathit{S\left(\omega\right )}$ and the autocorrelation function $\mathit{R\left(\tau \right)}$of a power signal form a Fourier transform pair, i.e.,

$$\mathrm{\mathit{R\left(\tau \right)\overset{FT}{\leftrightarrow}S\left(\omega\right)}}$$

Proof - The autocorrelation function of a power signal $\mathit{x\left ( t \right ) }$ in terms of exponential Fourier series coefficients is given by,

$$\mathrm{\mathit{R\left(\tau \right)=\sum_{n=-\infty }^{\infty } C_{n}\:C_{-n}\:e^{jn\omega _{\mathrm{0}}\tau }}\:\:\:\:\:\:...(1)}$$

Where,$\mathit{C_{n}}$ and $\mathit{C_{-n}}$ are the exponential Fourier series coefficients.

$$\mathrm{\mathit{\because C_{n}\:C_{-n}=\left | C_{n} \right |^{\mathrm{2}}}}$$

Therefore, Eqn.(1) can be written as,

$$\mathrm{\mathit{R\left(\tau\right)=\sum_{n=-\infty }^{\infty }\left | C_{n} \right |^{\mathrm{2}}\:e^{jn\omega _{\mathrm{0}}\tau }}\:\:\:\:\:\:...(2)}$$

By taking the Fourier transform on both sides of eq. (2), we get,

$$\mathrm{\mathit{F\left [ R\left ( \tau \right ) \right ]=F\left [\sum_{n=-\infty }^{\infty }\left | C_{n} \right |^{\mathrm{2}}e^{jn\omega _{\mathrm{0}}\tau } \right ]=\int_{-\infty }^{\infty }\left [ \sum_{n=-\infty }^{\infty }\left | C_{n} \right |^{\mathrm{2}}e^{jn\omega _{\mathrm{0}}\tau } \right ]e^{-j\omega \tau }\:d\tau}}$$

By interchanging the order of integration and summation on RHS of the above expression, we have,

$$\mathrm{\mathit{F[R(\tau )] =\sum_{n=-\infty }^{\infty }\left|C_{n}\right| ^{\mathrm{2}}\int_{-\infty}^{\infty} e^{jn\omega _{0}\tau} e^{-j\omega \tau } \:d\tau = \sum_{n=-\infty }^{\infty }\left|C_{n}\right| ^{\mathrm{2}}\int_{-\infty}^{\infty} e^{-j\tau (\omega -n\omega _{0}) } \:d\tau }}$$

$$\mathrm{\mathit{\because \int_{-\infty }^{\infty}e^{-j \tau(\omega -n\omega _{0}) }\:d\tau=\mathrm{2}\pi \delta (\omega -n\omega _{0})}}$$

$$\mathrm{\mathit{\therefore F\left [ R\left ( \tau \right ) \right ]=\mathrm{2}\pi \sum_{n=-\infty }^{\infty }\left | C_{n} \right |^{\mathrm{2}}\delta (\omega -n\omega _{\mathrm{0}})}\:\:\:\:\:\:...(3)}$$

The RHS of Eqn. (3) is the power spectral density (PSD) of the power function $\mathit{x\left(t\right)}$. Therefore,

$$\mathrm{\mathit{F\left [ R\left(\tau \right)\right ]=S\left(\omega\right)}}$$

Or, it can also be represented as,

$$\mathrm{\mathit{R\left(\tau\right)\overset{FT}{\leftrightarrow}S\left(\omega\right)}}$$

Hence, it proves that the autocorrelation function $\mathit{R\left(\tau\right )}$ and PSD function $\mathit{S\left (\omega \right )}$ of a power signal form the Fourier transform pair.

raja
Updated on 17-Dec-2021 07:06:40

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