Hello:
In the lecture 12 example for the cascaded filter design, you solved for ff1 by using the following equation:
(fs/ff1)^2 = attenuation.
How is this derived? I do not see this in the book.
Can we use this equation in our homework problem 1 and 2 or is it specific to the cascaded Rf-Lb parallel form?
Thanks,
A Dub
I believe the derivation goes as follows: (someone correct me if I am wrong, please :) )
ReplyDeleteAttenuation = Output/Input... when in Bode plot land this will be in log form of H(jw) = 20*log|H(jw)|... since the filter is of the form 1/(1+sw) w being 1/sqrt(LC)... we get that |H(jw)| = 1/(1 + jw/wff1)
20 log (|1/(1 + jw/wff1)|) = -20 log (|1 + jw/wff1|)
since w/wff1 >> 1,
this can be written as -20 log(|jw/wff1|)
= -20 log(w/wff1)
-20log(w/wff1) = -10log((w/wff1)^2)
Since we are already in dB scale, we can say that (w/wff1)^2 = attenuation (in dB)
we want to know where there is 100 dB of attenuation given w = ws, so...
((2*pi*fs)/(2*pi*ff1))^2 = 100 dB.. 2*pi cancels out to give
fs^2/ff1^2 = 100 dB (or whatever attenuation)
You should be able to use this on the HW :)
Hope that helps
Hi Olga:
DeleteThank you for posting that derivation.
Cheers,
A.W.
There are error in Olga's derivation. Please see my comment below - this has all been well explained in textbook section 8.1.6, which was covered in ECEN5797.
DeletePlease look through Section 8.1.6 of the textbook.
ReplyDelete