Consider a DT signal $x[n]$. Define the two-point moving average signal : $$ y[n]:=\frac{1}{2}x[n]+\frac{1}{2}x[n-1] $$ Define : $$ \langle x[n]\rangle:=\lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^{N}x[n] $$ One can note that $\langle x^{2}[n]\rangle$ is the total (mean) power of a DT signal. The goal is to interpret $\langle y^{2}[n]\rangle$. I proceeded as follow : \begin{align*} \langle y^{2}[n]\rangle &=\left\langle\left(\frac{1}{2}x[n]+\frac{1}{2}x[n-1]\right)^{2}\right\rangle \\ &=\frac{1}{2}\langle x^{2}[n]\rangle+\frac{1}{4}\langle x^{2}[n-1]\rangle+\frac{1}{2}\langle x[n]x[n-1]\rangle\\ &=\frac{1}{4}\mu_{x}^{2}+\frac{1}{4}\mu_{x}^{2}+\frac{1}{2}\langle x[n]x[n-1]\rangle \\ &=\frac{1}{2}\mu_{x}^{2}+\frac{1}{2}\langle x[n]x[n-1]\rangle \end{align*} where $\mu_{x}^{2}$ is the total DC power of the discrete-time signal $x[n]$. We stated in class that $\langle x[n]x[n-1]\rangle$ represents resemblence (?) I did not quite understand what was meant by this statement. I hope someone can assist me in understanding this interpretation.
1 Answer
$⟨x[n]x[n−1]⟩ $ represents a correlation between the sample $x[n]$ and the previous sample $x[n-1]$. The sum of a sample and it's previous divided by two is a two sample moving average. A moving average is a low pass filter such that samples that are indeed correlated over time (specifically low frequency signals) will pass through but samples that are not correlated (specifically high frequency signals) will be attenuated.
Note the simple example if we pass a signal that is all ones through the process: $(1+1)/2=1$ so every output will be $1$. This represents the lowest possible frequency or DC and we see how it passes completely through. At the other extreme is the highest possible frequency for the sampled system: 1, -1, 1, -1 .... Here we get $1-1)/2)=0$ or complete attenuation.
With further detail the transfer function is given as:
$$H(z)=\frac{1 + z^{-1}}{2}$$
as the z transform of $y[n] = 0.5x[n] + 0.5x[n-1]$. The frequency response is $H(z)$ when z is the unit circle, or $z=e^{j\omega}$ with $\omega$ going from $-\pi$ to $\pi$ associated with frequency in Hz going from $-f_s/2$ to $f_s/2$, where $f_s$ is the sampling rate. Thus the magnitude of the frequency response is:
$$|H(j\omega)|=\frac{|1+e^{-j\omega}|}{2}$$
Noting that $e^{-j\omega}$ is simply a phasor rotating on a complex plane as we sweep frequency, we can intuitively see the frequency response (both magnitude and phase) with help from the graphic below for those less familiar with signal processing.
