For 2 same waves traveling parallel to each other:
If one wave travel an extra distant d , as shown in the figure below. the falling-behind waves going to have a delay \tau of arrival. And this delay \tau infers phase shift \phi , vice versa: \boxed{ \text{distant traveled } d \leftrightarrow \text{ delay } \tau \leftrightarrow \text{phase shift } \phi } Give a wave with frequency \omega , wavelength \lambda , distance traveled d , and time to travel d is \tau , then the phase shift \phi is: \phi =& \omega \tau\\ =& 2\pi \nu \tau\\ &(\text{given: } c = \lambda \nu = \frac{d}{\tau} \to \nu = \frac{d}{\tau \lambda} )\\ =& 2\pi \frac{d}{\lambda}
Figure
0.1: Integral of Exponential
0.2: Correlation of Power Signals
signal x(t) is a (Finite) Power Signal if x(t) has finite average power P_{avg} (reference) :
P_{avg} = \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} |x(t)|^2 dt , \; \; \text{where } 0 < P_{avg} < \inftyIf the power signal is periodic, we can get rid of the \infty term:
P_{avg} = \frac{1}{2T} \int_{-T}^{T} |x(t)|^2 dt , \; \; \text{where } 0 < P_{avg} < \inftyAnd the Correlation of two power signals: x(t),y(t) , is (reference):
r_{xy}(\tau) = \frac{1}{2T} \int_{-T}^{T} x(t) y^*(t+\tau) dtor in discrete time:
r_{xy}[m] = \frac{1}{N} \sum_{n=\langle N \rangle } x[n] y^*[n+m]0.3: Power Spectrum Density
Starting from the general formula for power:
\text{Power} = \frac{\text{energy}}{\text{time}}Suppose we have a continuous signal that span from across all time, x(t) .
To analyze this infinite signal, we can crop out part of x(t) and make the following signal:
x_T(t) = \begin{cases} x(t), & |t| \le T \\ 0, & else \end{cases}Now we can find the power of this x_T(t) using the following:
P_T = \frac{1}{2T} \int_{-T}^{T} |x_T(t)|^2 dtand as T \to \infty :
P = P_{\infty} = \lim\limits_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} |x_T(t)|^2 dtNext using Parseval's Theorem:
\int_{-\infty}^{\infty} |x(t)|^2 dt = \int_{-\infty}^{\infty} |X(\nu)|^2 d\nuWe can rewrite the equation for Power P as:
{\color{blue} P} = \lim\limits_{T\to\infty} \int_{-T}^{T} \frac{|X(\nu)|^2}{2T}d\nuIf we want to make the Power Spectrum Density(PSD, S_{x_T}(\nu) ) to behave like a probability density function (PDF) as shown below:
\int_{-\infty}^{\infty} p(y) dy = 1We can make the following comparison:
so finally we have the expression for PSD:
S_{x_T}(\nu) = \lim\limits_{T\to\infty} \frac{|X(\nu)|^2}{2T}and if x(t) is of Random Process, then
S_{x_T}(\nu) = \lim\limits_{T\to\infty} \frac{E[|X(\nu)|^2]}{2T}0.4: Wiener Khinchin Theorem
Using Wiener-Khinchin Theorem:
&\begin{matrix*}[l] \text{given:} & \\ & x(t): \text{a sampled power signal from a Random Process } \\ & X(\omega): \text{Fourier Transform of } x(t) \\ & S_x(\omega): \text{(Time-average) Power Spectral Density } =\lim\limits_{T\to\infty} \frac{1}{2T} E[|X_T(\omega)|^2] , \text{ where } x_T(t ) = \begin{cases} x(t) , & |t| \le T \\ 0 , & |t| > T \end{cases} \ \\ & r_{xx}(\tau): \text{Correlation of itself(auto-correlation)} = \lim\limits_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} x_T(t) x_T^*(t+\tau) dt \\ \end{matrix*} \\ &\begin{matrix*}[l] \text{then:} & \\ & F\{ r_{xx} (\tau) \} = S_x(\omega) = R_{xx}(\omega) = \int_{-\infty}^{\infty} r_{xx}(\tau)e^{-j\omega \tau} d\tau \\ & F^{-1}\{S_x(\omega) \} = r_{xx} (\tau) = \frac{1}{2\pi} \int_{-\infty}^{\infty} S_x(\omega)e^{j\omega \tau} d\omega \\ \end{matrix*} \\