# 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?

• You can upload a scan of that page and link to it in the question. Someone else (with higher reputation) can add it to your question as a figure. – Matt L. Aug 27 '16 at 20:38

$\delta_s$ 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 $\delta_s^2$.