passive filters

[jaymee]

I have an assignment to make a high pass filter which has a cut off point of 10kHz at -3dB, I have managed to do this, but the question asks for the roll off to be -12db/octave.

I tried putting 4 of my filters together, making a fourth order filter? the slope seems right but the -3db point is no longer at the desired frequency.

I tried again with after re calculating, the fc using a similar formula to the one you use for a second order filter, ie using:


1 / 2.pi.fc.C1.C2.C3.C4 = R1.R2.R3.4

then fourth rooted the answer to get R, where C1=C2=C3=C4=10nF R=1997ohms

need theaching please!

 

[Laplace]

Normal roll off for a simple first-order filter is 6 dB/octave or 20 dB/decade. It would seem the problem is to design a second-order filter. You should derive the transfer function of your second-order network, and from that calculate the attenuation in dB at any frequency. Use that result to select component values to give 3 dB attenuation at the corner frequency. As an added bonus, verify that the roll off is 12 dB/octave.

 

[jaymee]

Thanks for your reply, I had it in my head that a second order filter has 6db / octave, so I'd need 2 hence the 'fourth' order I created.

Im not sure what you mean by derive the transfer function? or how to do it?

Im fairly sure though that the 'attenuation' calculation is derived from this formula:

1 / 2pi. √(fo.R1.R2.C1.C2)

although Ive never used the term attenuation before?

Thanks again

 

[Laplace]

Well then I would ask, can you get 12 dB/octave attenuation from the formula that you have provided?

To derive a transfer function in the steady-state frequency domain, do a circuit analysis where the passive elements are replaced by their reactive impedance, R=R, L=jwL, C=1/(jwC). Then use complex algebra to get the expression for (Vo/Vi). The attenuation in dB is then 20log(abs(Vo/Vi)).