Transform coding gain of two channel Haar Wavelet transform (first stage)

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My interpretation

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I'm attempting to do this in MATLAB.

I know that for discrete Random Processes going into an LTI system the autocorrelation function of the output is $R_y[n] = h[n]\star h[-n]\star R_x[n]$ where $\star$ is convolution.
The two tap Haar filters are: h1 = [0.707 0.707] and h2 = [0.707 -0.707] which are orthonormal to each other.

I also know that transform coding gain is defined this way:
$$G = \frac{1}{N}\frac{\displaystyle\sum_{i=0}^{N-1} \sigma_i^2}{\left(\displaystyle\prod_{i=0}^{N-1} \sigma_i^2\right)^{\frac 1N}}$$

where $N$ is the number of samples (3 in this case?) and $\sigma_i$ is the variance of the $i$th transform coefficient.

This is my code:

h1 = 1/sqrt(2).*[1 1];  
h2 = 1/sqrt(2).*[1 -1];  
y1 = conv(x,conv(h1,fliplr(h1)))  
y1 = 4.6500   14.3000   19.3000   14.3000    4.6500  
y2 = conv(x,conv(h2,fliplr(h2)))
y2 = -4.6500    4.3000    0.7000    4.3000   -4.6500

After decimation:

Y1 = 4.6500   19.3000    4.6500
Y2 = -4.6500    0.7000   -4.6500

Calculated coding gains:

GTC1 = 9.533/(417.3134)^(1/3) = 1.2757
GTC2 = -2.8667/(15.1358)^(1/3) = -1.1589

My questions:

Am I calculating $y_1$ and $y_2$ correctly? As I understand it these are supposed to be the autocorrelation functions for a WSS Gaussian random process (because WSS into LTI gives WSS out). After decimation, can i use $Y_1$ and $Y_2$ to calculate coding gain? I'm getting a negative coding gain for the second channel ($Y_2$), does that make sense?

EDIT: Just realized that the Haar filters need not be 2 tap filters. Now I am more lost than before.

Answer 1

I am not sure why you're trying to do this in MATLAB. This can be done quite easily with pen and papers.

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Another way of seeing what you're doing here is the following:

1. Build vectors of size 2 from the sequence $X$ $$[X(0),~X(1)], [X(2),~X(3)], [X(4),~X(5)]\dots$$
2. Apply an orthogonal transform on each vector to obtain the corresponding transform coefficients $Y$.

The order 2 orthogonal transform you're using can mathematically be written as $$\begin{bmatrix}Y_0 \\ Y_1\end{bmatrix} = \frac{1}{\sqrt{2}}\begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix}\begin{bmatrix}X(n) \\ X(n+1)\end{bmatrix}$$ where $Y_0$ and $Y_1$ are the transform coefficients and $X(n)$ and $X(n+1)$ are successive (hence correlated) samples from the process to be compressed. From this, you can easily compute $\sigma^2_0$ (the variance of $Y_0$) and $\sigma^2_1$ (the variance of $Y_1$). Indeed, because $X$ is zero-mean (and so are the transform coefficients), you know that $$\sigma^2_i = R_{Y_i}(0) = E\{Y_iY^*_i\}$$ for $i = 1, 2$. Then, you simply need to plug in the expression of $Y_i$ as a function of $X(n)$ and $X(n+1)$ to find $\sigma^2_i$ as a function of $R_X(0)$ and $R_X(\pm1)$. At the end, you should obtain $$\sigma^2_0 = R_X(0) + R_X(1) = 19.3,$$ $$\sigma^2_1 = R_X(0) - R_X(1) = 0.7.$$ You can clearly see from this the energy compaction in the first transform coefficient, which will contributes to the coding gain. You now have everything you need to compute the coding gain (note that $N$ in your expression for the coding gain is the number of transform coefficients, 2 in this case).