Active filters and natural frequency vs cut-off frequency

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

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 Consultant

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

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

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

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

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

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:

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

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