Attenuation problem

  • Thread starter EvLer
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  • #1
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I am going blank here...

I have a high-pass 2nd order (butterworth) that has fc = 5500 kHz, fp = 7kHz and fs = 1kHz and asked for attenuation at fc and fp.

I end up with more loss on passband than on stop band! I am actually not sure how to approach this. Any help is very very much appreciated... :cry:
 

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  • #2
berkeman
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What are the definitions for fc, fp and fs for a 2nd order Butterworth HP filter?
 
  • #3
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berkeman said:
What are the definitions for fc, fp and fs for a 2nd order Butterworth HP filter?
oh, sorry... H-P i mean is high-pass filter
fp is pass-band frequency
fs is stop-band
fc is the -3db frequency
second order Butterworth has this "normalized loss function":

H(s) = 1/{s^2 + sqrt(2)s + 1} (ps: sorry about the LaTex, where did they move tutorials !?)

basically, there's this thing about converting high-pass to low-pass, but before anything, the problem asks to find attenuation at fp and fs.
I think i get confused why/when i need to frequency scale.
 
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  • #4
berkeman
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EvLer said:
fp is pass-band frequency
fs is stop-band
fc is the -3db frequency
second order Butterworth has this "normalized loss function":

H(s) = 1/{s^2 + sqrt(2)s + 1}

basically, there's this thing about converting high-pass to low-pass, but before anything, the problem asks to find attenuation at fp and fs.
I think i get confused why/when i need to frequency scale.
It seems like fp and fs would be definition-dependent. There's no such thing as a "stop-band" or a "pass-band", it's all up to how you want to define them. Does your textbook offer any insights into what it wants? Are fs and fp defined in any way related to the H(s) polynomial? Butterworth filters are so rounded that you'll need something more explicit from somewhere to figure out what good values for fp and fs are, IMO.
 

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