2.7 Digital correlation  (Page 2/2)

$$\begin{array}{l}{R}_{xv}^{-}(m)={\displaystyle \sum _{n=-\infty }^{\infty }\left[a^{n}u(n)\right]\left[b^{n-m}u(n-m)\right]}\\ \\ ={\displaystyle \sum _{n=-\infty }^{\infty }{a}^n{b}^{n-m}u(n)u(n-m)}\end{array}$$

Where the formula of infinite geometric serics ( Equation ) has been used. Since m<0, we can write

$$R_{xv}^{-}(m)=\frac{1}{1-ab}{b}^{-m}u(m-1)$$

• For $m\ge 0$ , v(n) is shifted to the right, the summation lower limit is n = m :

$$R_{\text{xv}}^{+}(m)=\sum _{n=m}^{\infty }{a}^n{b}^{n-m}$$

Let’s make a change of variable k = n – m to get

$$\begin{array}{l}{R}_{xv}^{+}(m)={\displaystyle \sum _{k=0}^{\infty }{a}^{k+m}{b}^k}=a^m{\displaystyle \sum _{k=0}^{\infty }{(ab)}^k}\\ \\ =\frac{1}{1-ab},\left|ab\right|<0\end{array}$$

Where the formula finite geometric serics ( Equation ) has been used. Since m $$ 0, we can write

$$R_{\text{xv}}^{+}(m)=\frac{1}{1-\text{ab}}{a}^{m}u(m)$$

On combining the two parts, the overall cross-correlation results

$$R_{\text{xv}}(m)=R_{\text{xv}}^{-}(m)+R_{\text{xv}}^{+}(m)=\frac{1}{1-\text{ab}}[b^{-m}u(m-1)+a^{m}u(m)]$$

Auto-correlation

Auto-correlation of a signal x(n) is the cross-correlation with itself :

or equivalently

At m = 0 (no shifting yet) the auto-correlation is maximum because the signal superimposes completely with itself. The correlation decreases as m increases in both directions.

The auto-correlation is an even symmetric function of m :

Solution

We have

$$R_{\text{xx}}(m)=\sum _{n=-\infty }^{\infty }x(n)x(n-m)=\sum _{n=-\infty }^{\infty }{a}^n{a}^{n-m}u(n)u(n-m)$$

Since $R_{\text{xx}}(m)$ iseven symmetric we need to compute only the $R_{\text{xx}}^{+}(m)$ for m $$ 0 then generalize the result for the correlation.

For $m\ge 0$

$$R_{\text{xx}}^{+}(m)=\sum _{n=m}^{\infty }{a}^n{a}^{n-m}$$

Make a change of varible k = n – m as in previous example :

$$R_{\text{xx}}^{+}(m)=\sum _{k=0}^{\infty }{a}^{k+m}{a}^k=a^m\sum _{k=0}^{\infty }{a}^{\mathrm{2k}}=\frac{a^m}{1-a^2}$$ ,

Above result is for $m\ge 0$ . Now for all m we just write $\mid m\mid$ for m because of the even symmetry of the auto-correlation. So

$$R_{\text{xx}}(m)=\frac{a^{\mid m\mid }}{1-a^2}$$

Correlation and data communication

Consider a digital signal x(n) transmitted to the far end of the communication channel. It reaches the receiver $n_0$ samples later, becoming x(n - n ${}_0$ ), and it is also added with random noise z(n). Thus the total signal at the receiver is

$$y(n)=x(n-1)+z(n)$$

Now let’s look at the cross-correlation betwwen y(n) and x(n) :

The result shows that the cross-correlation consists of two compoments : The auto-correlation $R_{\text{xx}}(m-m_0)$ of the transmitted signal but shifted in time, and the cross-correlation $R{}_{\text{xz}}(m)$ between the transmitted signal x(n) and corrupting noise z(n). The meaning is that $R_{\text{xx}}(m-m_0)$ is usually larger than $R{}_{\text{xz}}(m)$ and has peak at m = n ${}_0$ , whereas $R{}_{\text{xz}}(m)$ is usually smaller due to the random nature of noise and the independence of the signal and noise. Thus by examining $R_{\text{yx}}(m)$ we know the delay $n_0$ of receiving signal.

Solution

Without going details of evalution, only the results are mentioned :

• Auto-correlation of x(m) : $R_{xx}(m)=\left[12,17,13,39,23,13,17,12\right]$
• Cross-correlation beween x(n) and z(n) : $R_{zx}(m)=\left[-16,1.2,-1.8,-2.6,-2.8,1.7,-0.8,-0.1,2.1\right]$
• Cross-correlation beween y(n) and x(n) : $R_{yy}(m)=\left[-1.6,1.2,10.2,14.4,10.2,21.3,38.2,22.7,12.9\right]$

The highest value 38.2 of $R_{\text{yy}}$ oceurs at index m = 1 as expected.

Correlation of periodic signals

For two period signals x(n) and v(n) having the same period of N indices (samples), the cross-correlation and auto-correlation are defined as

The two correlations also have a period of N samples.

Now let’s look at an application. The signal y(n) arrving at the receiver consists of the transmitted signal x(n) and adding noise z(n) :

$$y(n)=x(n)+z(n)$$

The auto-correlation of the received signal for a duration of M samples, M is much greater than N, is

$$R_{yy}(m)=\frac{1}{M}{\displaystyle \sum _{n=0}^{M-1}y(n)y(n-m)}$$

On replacing the expression of y(n) into above auto-correlation, we obtain

Because the signal x(n) is periodic with period N, the auto-correlation $R_{\text{xx}}$ is also periodic with peaks at m = 0, N, 2N ... The cross-correlation $R_{\text{xz}}$ (m) are Rzx(m) of the signal and noise are rather small because the signal and noise are uncorrelated. The last term Rzz(m) is the auto-correlation of noise, it has peak at m = 0 and decays fast to zero due to its random nature. Thus it remains $R_{\text{xx}}$ the largest. This feature allows us to detect the periodic signal x(n) even if the adding noise has amplitude comparable to that of the signal or even much higher. This method of correlation has been used to determine the pitch (fundamental frequency) of voice and music buried in noise.