Active filters and natural frequency vs cut-off frequency
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Posted by on October 16, 2007, 5:09 pm
Please log in for more thread options I have a simple question (I think). How do I find the cut-off frequency of an active filter that has already been designed? In other words, one that I only have the schematic of and no design information. Specifically it's a two-pole low pass sallen-key filter. I can derive the transfer function from the circuit, and I can put the transfer function into the standard second order form. From this I can find the natural frequency and the damping ratio. However, I don't know how to find the cut off frequency. For the longest time I though the cut off frequency was the natural frequency, but I now realize they're two different things, with potentially very different values. Is there a mathematical relationship between the cut off frequency, natural frequency and damping ratio? Any help or direction would be greatly appreciated. Thanks | |||||||||||||||||||
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Posted by Martin Griffith on October 16, 2007, 5:47 pm
Please log in for more thread options PatrickNee2@gmail.com wrote: or the otherway around is to get filter pro (free and quick)from TI.com and enter in your doofers Martin | |||||||||||||||||||
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Posted by Vladimir Vassilevsky on October 16, 2007, 6:53 pm
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PatrickNee2@gmail.com wrote: > Hello:
> > I have a simple question (I think). How do I find the cut-off > frequency of an active filter that has already been designed? In other > words, one that I only have the schematic of and no design > information. Specifically it's a two-pole low pass sallen-key filter. You need to solve the equation |H(s)| = 1/sqrt(2) > I can derive the transfer function from the circuit, and I can put the
> transfer function into the standard second order form. If you know how to do that, solving for S should not be a problem for you. > Is there a mathematical relationship between the cut off frequency,
> natural frequency and damping ratio? Certainly. It should not be difficult to derive it. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com | |||||||||||||||||||
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Posted by Tom Bruhns on October 16, 2007, 8:01 pm
Please log in for more thread options wrote:
> PatrickN...@gmail.com wrote:
> > Hello:
>
> > I have a simple question (I think). How do I find the cut-off
> > frequency of an active filter that has already been designed? In other > > words, one that I only have the schematic of and no design > > information. Specifically it's a two-pole low pass sallen-key filter. >
> You need to solve the equation |H(s)| = 1/sqrt(2) > > > I can derive the transfer function from the circuit, and I can put the
> > transfer function into the standard second order form. >
> If you know how to do that, solving for S should not be a problem for you. > > > Is there a mathematical relationship between the cut off frequency,
> > natural frequency and damping ratio? >
> Certainly. It should not be difficult to derive it. > > Vladimir Vassilevsky > DSP and Mixed Signal Design Consultanthttp://www.abvolt.com I trust that you did not literally mean "|H(s)| = 1/sqrt(2)", but rather solve for the w which satisfies |H(jw)| = |H(0)|/sqrt(2). That is we need to solve for an s that represents a sinusoidal signal, and we want to find the point at which the amplitude response is 3dB down from the DC response. That, of course, assumes a low-pass circuit. For a high-pass filter, you'd use H(infinity) as the comparison point, and for a band-pass, you'd find the maximum response and look for upper and lower 3dB points from that. Note to OP: realize that a second order response can have very significant frequency response peaking. If you put the second-order response in the right form, the damping can be read by inspection. Spend a little time considering the where the roots of the second- order equation lie as you vary k between, say, +1 and -1 in s^2 + 2*k*s*w + w^2. For ease of visualization, try w=1 to start with. Consider the equation to get one side of a right triangle, when the hypotenuse is 1 and the other side is k: does this look like a form you can put your quadratic into? Cheers, Tom | |||||||||||||||||||
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Posted by on October 17, 2007, 12:20 pm
Please log in for more thread options Thanks for all the responses, they were very helpful.
I guess I should reword my question a little. It's not that I don't understand what the cut off frequency is. My confusion was that I though the natural frequency was the cut-off frequency. In other words, I though the transfer function would be 0.707 at the natural frequency. What I've seen recently, is that this statement is close to being true if the system is under damped, and further from true if the system is over damped. >From this observation, I figured there would be some elegant
relationship between the cut-off frequency, natural frequency and
damping factor, but from this conversation it appears there isn't any. Concerning the post by Vladimir, you're overestimating my intelligence. I can solve this equation |H(s)| = 1/sqrt(2), by plotting |H(s)| and looking for 0.707, but I don't know how to do it analytically, (mainly because of the absolute value part). Is there an easy way to do this? Are there any tech notes/books you can point me towards? Thanks | |||||||||||||||||||
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>
>I have a simple question (I think). How do I find the cut-off
>frequency of an active filter that has already been designed? In other
>words, one that I only have the schematic of and no design
>information. Specifically it's a two-pole low pass sallen-key filter.
>
>I can derive the transfer function from the circuit, and I can put the
>transfer function into the standard second order form. From this I can
>find the natural frequency and the damping ratio. However, I don't
>know how to find the cut off frequency. For the longest time I though
>the cut off frequency was the natural frequency, but I now realize
>they're two different things, with potentially very different
>values.
>
>Is there a mathematical relationship between the cut off frequency,
>natural frequency and damping ratio?
>
> Any help or direction would be greatly appreciated.
>
>Thanks