That did it! I was using my K backwards. dumb. So I got the following

Choosing R1= 4k C=20.8nf
Plugging this back into the equation yields \( R1*C= \frac{2}{2.4*10^4}\)

.000083 = .000083

And Then, If R1=4k R2=1k

\(\frac{1}{(4000)(1000)((20.8*10^-9)^2)}\)

yields:

5.78*10^8 = 5.76*10^8

pretty close taking in round off errors and such.

 

Very good.

That was not so bad was it.

You can probably imagine how challenging it would be to solve a third-order filter. These filter design problems get tougher exponentially as the filter order increases.

You can see why most designers resort to filter design software just to keep from going insane.

hgmjr

 

Yea, I don't want to think about it right now ha ha.

Thank you a thousand times.

 

I'm glad you had a positive experience. We at AAC are here to help when we can.

hgmjr