Filter design, relationship between energy and stop-band ripple I read some example of design LPF which I didn't understand something. 
The stop-band in that example is $\frac { 22 }{ 25 } $  from the over-all frequency, and I want to filter some white noise.
After we found out in that example that the energy of the white noise in the stop band area is $0.22\%$ from the overall noise before filtering, they used the equation:
$$\frac { 22 }{ 25 } { \delta  }_{ s }^{ 2 }=0.0022$$ 


Why ${ \delta  }_{ s }^{ 2 }$? 
How does is related to power/energy relationship?


The full example in the book of Boaz Porat page 247, can I upload a picture from a book?
$\delta_s$ is the stop-band (hence the ${}_s$ subscript) attenuation.

A signal in the stop-band with amplitude $1$ would have amplitude $\delta_s$ after filtering.

Since power goes quadratic with amplitude, and if you feed in white noise, $\frac{22}{25}$ of the original energy will be in stop band, and will be reduced by a factor of $\delta_s^2$.