# Capacity computation in Costa sheme

I can not understand the following sentences, could you please anyone guide me ? "Comparing with the embedding in audio signals, the usable bandwidth is greatly reduced, from 22.05 kHz to 4 kHz. According to Costa’s result, the corresponding capacity will be reduced to l/5 th accordingly."

The costa formula is Cw,Costa=(1/2)*Log2(1+(var_w/var_N))

I also compete it in MATLAB

((1/2)(log(1+(1/4))/log(2))) / ((1/2)(log(1+(1/22.05))/log(2))) = 5.0311 is it right ?

• We need more context. Where does the sentence come from? It's hard (or close to impossible) to read your formulas, especially the last one. Please try Latex. – Matt L. Apr 7 '14 at 7:19
• I feel it is ok now :) – Ali Bodaghi Apr 7 '14 at 7:27

You really didn't give sufficient context and background information to provide a complete answer, but I'll try anyway. You didn't say so, but I assume that $\sigma_W^2$ is the power of a watermark and that $\sigma_N^2$ is the power of additive noise. I furthermore assume that the "usable bandwidth" is the bandwidth for generating the watermark. If we additionally assume that the watermark signal is white with spectral density $W_0$, then its power inside the usable bandwidth $B$ is given by

$$\sigma_W^2=2W_0B$$

If $B_1=22050Hz$ and $B_2=4000Hz$ we get for the capacity ratio

$$\frac{\log_2(1+\frac{2W_0B_2}{\sigma_N^2})}{\log_2(1+\frac{2W_0B_1}{\sigma_N^2})}$$

which is not a fixed ratio, but it depends on the constants $W_0$ and $\sigma_N^2$. If they are not given, then there is no way to claim that this ratio is $1/5$.

However, if you can assume that $\sigma_W^2/\sigma_N^2\ll 1$ then the capacity can be approximated by

$$\frac{1}{2\log (2)}\frac{\sigma_W^2}{\sigma_N^2}=\frac{1}{2\log (2)}\frac{2W_0B}{\sigma_N^2}$$

and with the same assumptions as above the capacity ratio is equal to the ratio of bandwidths $B_2/B_1=0.18$, which is indeed close to $1/5$.

If any of my assumptions are not correct, please clarify your question.