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To make the question clear I will ask this with an example.

Let's say we have a LP filter with a -3dB attenuation at cutoff frequency of 1kHz. And lets say we are given the steepness of the filter as 100 dB/decade.

Does that mean the filter will attenuate 100 dB/decade right after -3dB point?

If so, the following possibilities for Bode plot points confuses me when I try to use dB/decade definition:

-103 dB at 10 kHz? or

-100 dB at 10 kHz? or

-100 dB at 11 kHz? or

-103 dB at 11 kHz? or

Which one of the above is exactly correct according to the definition dB/decade? What confuses me is should we take decade as multiplication or addition to the cut off frequency? And should we go down from -3dB or zero for the attenuation?

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    \$\begingroup\$ It would be -100dB at 10kHz. The dB/decade is asymptotic. \$\endgroup\$ – Hearth Mar 19 at 15:39
  • \$\begingroup\$ Asymptotic isn't the word I would use. "Piecewise linear approximation with a corner at the cutoff frequency" is a better description. \$\endgroup\$ – Scott Seidman Mar 19 at 15:42
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    \$\begingroup\$ @ScottSeidman What I mean is that the actual behavior is asymptotic to the piecewise linear approximation. I did not really articulate that very well. \$\endgroup\$ – Hearth Mar 19 at 15:45
  • \$\begingroup\$ Any more questions? \$\endgroup\$ – Sunnyskyguy EE75 Mar 19 at 18:36
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    \$\begingroup\$ @user1999 Unless you know Q, 2f is hard to estimate but 4f is ok to estimate thus (order=) 5x -6dB/oct x 2 oct (=4f) = -60dB is closer. and 10f is exact as I showed If you need more then choose a Chebychev or higher order \$\endgroup\$ – Sunnyskyguy EE75 May 23 at 14:32
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Answer

n order fc=1kHz  10kHz   100kHz
1       -3dB    -20dB   -40dB  
2       -3dB    -40dB   -80dB
3       -3dB    -60dB  -120dB
4       -3dB    -80dB  -160dB
5       -3dB   -100dB  -200dB    

These are the exact values within 0.01 dB with parts with 0% tolerance error.

The standard signal filter passband BW is defined by the -3dB BW. The asymptotes go thru the flat intersection (e.g. 0dB) but actual is shown above. These are not estimates but are using perfect parts.

Details

The stopband attenuation vs frequency slope above cutoff (-3dB) attenuation [dB]\$ = -6n_{dB/octave f} = -20n _{dB/decade}\$ per nth order of filter, where n is the number of independant reactors, ( here just the number of C's)

We can estimate the attenuation at 1 decade up to be pretty accurate and closer in as the shape factor by Q and order of filter > 1

However the maximally flat frequency only produces critical damping at 2nd order and higher orders have more ringing.

enter image description here

For the advanced reader....

The above image shows two modes for both Bessel vs Butterworth:

  1. Sine-wave input log-sweep with envelope response
  2. Square-wave input at 80Hz while corner freq was 1kHz, to see the difference in damping & ringing.
    • Bessel is best in time-domain for zero overshoot but lowest Q or least steep corner frequency and thus lowest component tolerance sensitivity but also worst for image reject ratios for ADC near the corner. So it depends on your criteria.
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    \$\begingroup\$ Unfortunately, Bessel filters have not "this maximally flat response". Instead they are optimized for maximally flat group delay. \$\endgroup\$ – LvW Mar 20 at 18:40
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    \$\begingroup\$ Sunnyskyguy, sorry to say, but many errors in your revised version (critically damping, shape factor, overshoot, overall Q, meaning of fc). \$\endgroup\$ – LvW Mar 20 at 18:47
  • \$\begingroup\$ Both comments above apply to an earlier version of the answer. \$\endgroup\$ – LvW Mar 21 at 10:32
  • \$\begingroup\$ @LvW I disagree "The question is how to interpret the meaning of dB/decade." which is fundamentally clear only in my answer with the table. (-1) for your negative bias contributions \$\endgroup\$ – Sunnyskyguy EE75 Mar 21 at 13:29
  • \$\begingroup\$ what do you disagree with? I did not comment on your "table". And it was not me who downvoted your contribution (which, however, does not answer the question). \$\endgroup\$ – LvW Mar 21 at 14:45
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Look at this filter - it has 20 dB per decade roll-off: -

enter image description here

So, you can choose to use either the real graph of the filter (in red) or use the simpler straight line approximations where the worst error you will get is 3.01 db. This assumes that the damping ratio of the filter is \$\sqrt{0.5}\$.

If you have a filter that is under-damped compared to the above it might produce a peak in the response close to the cut-off frequency and so you have to decide on the merits of simplicity versus accuracy: -

enter image description here

Q is inversely proportional to damping ratio (\$\zeta\$) and, as you should be able to see, apart from critical frequencies around the cut-off point, the straight line approximation holds reasonable for various damping ratios.

To answer your question, the attenuation at 10 kHz is approximately 100 dB.

Does that mean the filter will attenuate 100 dB/decade right after -3dB point?

No, it doesn't - it will attenuate only at this rate as the red line in the graph above starts to merge into the blue line but there will always be a small error. The red line is the real characteristic and the blue line is a useful engineer's approximation to the characteristic.

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  • \$\begingroup\$ Im asking something different and more fundamental. The question is how to interpret the meaning of dB/decade. Thats why I gave those examples in question. Could you also give some hint related to the comments under the question? I dont get what they mean. But have a feeling that the comments are directly related to the reason behind why the answer is 10kHz 100dB. Ignoring the accuracy for now and considering a 1st order LP for ease. \$\endgroup\$ – user1999 Mar 19 at 17:28
  • \$\begingroup\$ 1st order -20dB/decade or -20dB at 10fc , -40dB at 100fc -60B at 1e3 Hz \$\endgroup\$ – Sunnyskyguy EE75 Mar 20 at 16:54
  • \$\begingroup\$ @Andy aka, I have spent some minutes (or more) thinking about your new finding regarding the product of all pole-Qs for Butterworth. I am afraid, it is not too easy to explain the math behind this. I think, it is because (a) the poles are always equally distributed along the unit circkle \$\endgroup\$ – LvW Mar 20 at 18:23
  • \$\begingroup\$ (continued) ...and (b) the pole-Q is directly related to the real part of the pole [1/2cos(phi)]. Hence, the product of two Qs involves the product of cos(x)cos(y), which can be expressed by cos(x-y) + cos(x+y). I have checked this for n=4 - and it works. For larger n values it becomes rather complicated. I must admit, I was really surprised about your finding....I could not find any mentioning of this Butterworth property in textbooks. \$\endgroup\$ – LvW Mar 20 at 18:30
  • \$\begingroup\$ @LvW I used a spreadsheet and altered n in even numbers and let excel do the hard work. It's probably as tricky to prove as the original butterworth idea! \$\endgroup\$ – Andy aka Mar 20 at 18:45

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