Monday 27 July 2015

fourier transform - Power Spectrum: Definition


I am new to the study of time series. Recently I have asked a question about the covariance of real and imaginary part of a real(in time domain) stochastic time series and I have received an answer for it. The problem is that for continuous time series the variance of each point in frequency domain is infinite. I've been told there that its the reason that they use power spectrum. Now the confusion comes from the fact that I don't know whether power spectrum is time averaged or not? More precisely as wiki says: $$S_{xx}(\omega)=\lim\limits_{T\to \infty}\mathbf{E} \left[ | \hat{x}_T(\omega) |^2 \right] = \lim\limits_{T\to \infty}\mathbf{E} \left[ \frac{1}{T} \int\limits_0^T x^*(t) e^{i\omega t}\, dt \int\limits_0^T x(t') e^{-i\omega t'}\, dt' \right] = \lim\limits_{T\to \infty}\frac{1}{T} \int\limits_0^T \int\limits_0^T \mathbf{E}\left[x^*(t) x(t')\right] e^{i\omega (t-t')}\, dt\, dt$$ OR $$S_{xx}(\omega)=\mathbf{E} \left[ | \hat{x}_T(\omega) |^2 \right] = \mathbf{E} \left[ \int\limits_0^\infty x^*(t) e^{i\omega t}\, dt \int\limits_0^\infty x(t') e^{-i\omega t'}\, dt' \right] = \int\limits_0^\infty \int\limits_0^\infty \mathbf{E}\left[x^*(t) x(t')\right] e^{i\omega (t-t')}\, dt\, dt$$


For example in Wiener-Khintchine theorem as far as I can see there is no time averaging: $$r_{xx} (\tau) = \int_{-\infty}^\infty S_{xx}(f) e^{2\pi i\tau f} df$$


And is there any difference when the signal is discrete?




Answer



The first definition of the power spectrum is the correct one (at least for causal signals, otherwise you need symmetrical integration limits). The second one assumes that the Fourier transform of $x(t)$ exists, which is generally not the case for random signals. The fact that the power spectrum and the autocorrelation function form a Fourier transform pair is no contradiction with the first definition of the power spectrum. This can be shown as follows:


$$S_x(\omega)=\lim_{T\rightarrow\infty}E\left\{ \frac{1}{T}\left| \int_{-T/2}^{T/2}x(t)e^{-j\omega t}dt \right|^2 \right\}=\\= \lim_{T\rightarrow\infty}E\left\{ \frac{1}{T} \int_{-T/2}^{T/2}x^*(t)e^{j\omega t}dt \int_{-T/2}^{T/2}x(t')e^{-j\omega t'}dt' \right\}$$


The inverse Fourier transform of the power spectrum $S_x(\omega)$ should equal the autocorrelation function $R_x(\tau)$:


$$\mathcal{F}^{-1}\{S_x(\omega)\}=\frac{1}{2\pi}\int_{-\infty}^{\infty}S_x(\omega)e^{j\omega \tau}d\omega=\\= \lim_{T\rightarrow\infty}E\left\{ \frac{1}{2\pi T} \int_{-\infty}^{\infty}\int_{-T/2}^{T/2}x^*(t)e^{j\omega t}dt \int_{-T/2}^{T/2}x(t')e^{-j\omega t'}dt'e^{j\omega \tau}d\omega \right\}=\\= \lim_{T\rightarrow\infty}E\left\{ \frac{1}{T} \int_{-T/2}^{T/2}x^*(t) \int_{-T/2}^{T/2}x(t')\left[\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{j\omega (t+ \tau-t')}d\omega\right]dtdt' \right\} $$ The expression in brackets equals $\delta(t+\tau-t')$, which gives


$$\mathcal{F}^{-1}\{S_x(\omega)\}= \lim_{T\rightarrow\infty}E\left\{ \frac{1}{T} \int_{-T/2}^{T/2}x^*(t) x(t+\tau)dt \right\}=\\ =\lim_{T\rightarrow\infty} \frac{1}{T} \int_{-T/2}^{T/2}E\{x^*(t) x(t+\tau)\}dt=\\ = \lim_{T\rightarrow\infty} \frac{1}{T} \int_{-T/2}^{T/2}R_x(\tau)dt= R_x(\tau)\lim_{T\rightarrow\infty} \frac{1}{T} \int_{-T/2}^{T/2}dt=R_x(\tau) $$


which shows that the chosen definition of the power spectrum is compatible with the fact that the power spectrum and the autocorrelation function are Fourier transform pairs. A similar derivation can be done for discrete-time signals.


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