Folks:

I received an e-mail yesterday as follows:

> I got one question about the computer homework 3 in DSP.
> When you ask us to show that Q can be given approximately by
> sqrt(L/C)/2R, I really can not make use of your assumption
> ".....assume Q is large", where you define Q is the ratio of 3 dB
> bandwidth to center frequency. Can you give more hints about the
> relation of your assumption to the approximation of Q (without solving
> the quadratic)?

And my answer:

Well, you may have noticed, when solving for the 3 dB down points, that you had an equation of the form aw + b + c/w = 0. Let w = w0 + dw, where w0 is the center frequency, and dw/w0 is small. Then c/w is approximately equal to c/w0. This greatly simplifies things.

You are correct about the definition of Q - I made an error in the assignment. Q is of course the ratio of center frequency to 3 dB bandwidth. I have updated the PDF file on the Web to reflect this. Thank you for pointing that out.

Tom

Saturday, 2 May 1998

The solution is up on the Web site.

The bugs continue to plague me. As several of you pointed out, the correct expression for Q is sqrt(L/C)/R. My high-Q approximations were rampant, and wrong. However, I do employ this approximation (correctly) in the solution. It's still easier than solving the quadratic! I didn't take points off for deriving my faulty Q.

Mostly this problem was very well done. Many people obtained the correct system response, albeit for a Q that was effectively twice what I had asked for (which I ignored). The most common error was assuming that the desired boost/cut in dB was the required gain of the filter. A few forgot that the FFT returns a mirror image; this always looks odd on a logarithmic frequency scale, and should have been discarded.

The equalizer is indeed approximately graphic, with minor deviations occurring because of filter overlap. The use of second-order sections simplifies the arithmetic needed to compute the filter outputs, as only four multiply/adds are needed for each filter.

Tom