Low pass filter design

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cane21

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Hi

I just need some clarification regarding a LPF design.

I have a signal f(t)= 4 sin(500t)+ sin(15000t) and want to filter out the high frequencies by at least 20dB.

I’m confused about the 20dB, does that just refer to the roll-off rate?

Would a RC or RLC LPF be better in this case?

Thank you.

- - - Updated - - -

I’m also unclear as to how to determine what the cut-off frequency should be.
 

Hi,

No this 20dB is not the roll off rate.

****
The aim is
* to leave "4 sin(500t) " amplitude as is (as good as possible)
* to attenuate "sin(15000t)" amplitude by 20dB

20dB is a factor of 10
20dB attenuation is -20dB which means a factor of 1/10

The quotient of both frequencies is 15000/500 = 30.
Because 30 >> 10 this means a first order filter should be sufficient.
RC is good.

Cutoff frequency needs to be less than 1/10 of the higher frequency.

I recommend to use a simulation tool to play around. Like LTspice.

Klaus
 
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    cane21

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I’m also unclear as to how to determine what the cut-off frequency should be.
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1. Roll off are usual slopes (the linear part) of the dB vs log f : they are easy to remember 6 dB /octave or 20 dB /decade. Double them for 2nd order and quad them for a third order filer.

2. The two frequencies are well separated (how do I know?) and hence a first order filter is ok. (you want a 20 dB loss and that means if I use 1000Hz at the corner frequency, 10kHz will be 20dB less)

3. If the upper freq is less than 10kHz, you better use a second order filter. If the two fellows are close by, use a digital filter.

I see KlausST has explained the same solution very succinctly. I also suggest you play with the dB and log f plots and try to work out simple examples by hand (with a calculator).
 
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    cane21

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If a Simple Butterworth Filter will attenuate 20dB@15k Ohmega so the Cut-off frequency will be around 1.5k Ohmega.
Because simple one pole filter 20dB/decade slope.
 

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