Active filters and natural frequency vs cut-off frequency

[[email protected]]

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.

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

 

[Martin Griffith]

On Tue, 16 Oct 2007 14:09:35 -0700, in sci.electronics.design

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.

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

My way is to load the circuit into LTspice (free) and click simulate,
or the otherway around is to get filter pro (free and quick)from
TI.com and enter in your doofers


Martin

 

[Vladimir Vassilevsky]

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

 

[Tom Bruhns]

You need to solve the equation |H(s)| = 1/sqrt(2)


If you know how to do that, solving for S should not be a problem for you.


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

 

[Al]

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.

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

It's what you specify it be. For example, the cutoff frequency could be
- 3dB from the center frequency. If the skirts are very steep, the
cutoff frequency could be very close to the center frequency.

In an audio filter that has its cutoffs specified as say, 20Hz and
20KHz, those could be -3 dB below the reference frequency of 1KHz. It
then becomes a marketing game.

Al

 

[[email protected]]

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

 

[[email protected]]

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

 

[[email protected]]

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

 

[Tom Bruhns]

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.


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

The -3dB point of a high pass or low pass Butterworth filter is indeed
the radius of the circle on which the poles lie...for your second
order case, it's the w of s^2 + sqrt(2)*w*s + w^2. But if the filter
isn't Butterworth, the 3dB point won't be at that frequency. (By
"frequency" here, I mean the radian frequency, 2*pi*f; and I really
should be using an omega-sub-zero symbol instead of just w...).

H(s) for a sinusoidal input is H(j*w), and in general the answer is a
complex number, with real and imaginary parts. |H(s)| is just the
magnitude of that complex number, or sqrt(real^2 + imaginary^2). Are
you comfortable enough with complex arithmetic to do that? I can show
you just some answers, but if you can do the math, or bootstrap
yourself into being able to do it, I think you'll get a lot more
insight.

For a low-pass, normalized to omega-sub-zero = 1, with damping d, the
numerator is a constant (the DC gain) and the denominator is s^2 +
2*d*s + 1. The denominator, evaluated for s=0, is just 1. When the
denominator magnitude is sqrt(2), the response will be 3dB down from
the DC response. Evaluating for s = jw -- that is, for sinusoid
excitation -- the denominator is (1-w^2) + j*2*d*w. The magnitude is
sqrt(real^2+imag^2), but we don't even have to take the square root:
we know that we want the square of the magnitude to be sqrt(2)
squared, or just 2. So find the w that satisfies 2 = (1-w^2)^2 +
(2*d*w)^2 -- that turns out to be just a quadratic in w^2 that you
can solve for and have a fairly simple expression for the -3dB
frequency in terms of the damping factor, d. Elegant, I don't know,
but it's not too bad. Let me know what you get and I'll compare it
with what I got.

Cheers,
Tom

 

[[email protected]]

The -3dB point of a high pass or low pass Butterworth filter is indeed
the radius of the circle on which the poles lie...for your second
order case, it's the w of s^2 + sqrt(2)*w*s + w^2. But if the filter
isn't Butterworth, the 3dB point won't be at that frequency. (By
"frequency" here, I mean the radian frequency, 2*pi*f; and I really
should be using an omega-sub-zero symbol instead of just w...).

H(s) for a sinusoidal input is H(j*w), and in general the answer is a
complex number, with real and imaginary parts. |H(s)| is just the
magnitude of that complex number, or sqrt(real^2 + imaginary^2). Are
you comfortable enough with complex arithmetic to do that? I can show
you just some answers, but if you can do the math, or bootstrap
yourself into being able to do it, I think you'll get a lot more
insight.

For a low-pass, normalized to omega-sub-zero = 1, with damping d, the
numerator is a constant (the DC gain) and the denominator is s^2 +
2*d*s + 1. The denominator, evaluated for s=0, is just 1. When the
denominator magnitude is sqrt(2), the response will be 3dB down from
the DC response. Evaluating for s = jw -- that is, for sinusoid
excitation -- the denominator is (1-w^2) + j*2*d*w. The magnitude is
sqrt(real^2+imag^2), but we don't even have to take the square root:
we know that we want the square of the magnitude to be sqrt(2)
squared, or just 2. So find the w that satisfies 2 = (1-w^2)^2 +
(2*d*w)^2 -- that turns out to be just a quadratic in w^2 that you
can solve for and have a fairly simple expression for the -3dB
frequency in terms of the damping factor, d. Elegant, I don't know,
but it's not too bad. Let me know what you get and I'll compare it
with what I got.

Cheers,
Tom

Tom:

Thanks a lot for that post. That really cleared it up for me. I don't
know why I couldn't do this, because now that you've explained it, it
seems very obvious. I put the equation into MathCad and solved it.
The result I got wasn't the elegant relationship I was hoping for but
at least I understand it know.

wc=(wn^2-2d*wn^2+(4d*wn^4+2-4wn^2*d*wn^2)^0.5)^0.5

normalized to wn:
wc=(1-2d^2+(2-4d^2+4d^4)^0.5)^0.5

Where:
wc = cut off frequency (rad)
wn = natural frequency (rad)
d = damping ratio

 

[[email protected]]

Tom:

Thanks a lot for that post. That really cleared it up for me. I don't
know why I couldn't do this, because now that you've explained it, it
seems very obvious. I put the equation into MathCad and solved it.
The result I got wasn't the elegant relationship I was hoping for but
at least I understand it know.

wc=(wn^2-2d*wn^2+(4d*wn^4+2-4wn^2*d*wn^2)^0.5)^0.5

normalized to wn:
wc=(1-2d^2+(2-4d^2+4d^4)^0.5)^0.5

Where:
wc = cut off frequency (rad)
wn = natural frequency (rad)
d = damping ratio

This might be too late and not useful, but I'm in the middle of
comparing filter designs, and I found this:
http://en.wikipedia.org/wiki/Sallen_Key_filter

Cutoff_freq = 1 / ( 2 * PI * sqrt(r1*r2*c1*c2) )

Your topology may vary, and I just might be full of it.
Cheers,