Re: [Phys-L] weighting in the wings ... damped harmonic oscillator ... bandwidth ... algebra ... bug hunting

[David Bowman <David_Bowman@georgetowncollege.edu>]

Regarding Don Polvani's work checking:

> Regarding David Bowman's 11/21/16 post for the series R,L, C
> circuit, I can't get his result for the FWHM using his gain
> function (G(f)).  Using G(f) = 1/(1 + Q^2*(f/f_0 - f_0/f)^2) =
> 1/2, I get a quartic equation with the following four roots for
> f:
>
> f_1 = f_0*(1 + sqrt(1 + 4*Q^2))/(2*Q)
> f_2 = f_0*(1  - sqrt(1 + 4*Q^2))/(2*Q)
> f_3 = -f_0*(1 + sqrt(1 + 4*Q^2))/(2*Q)
> f_4 = -f_0*(1  - sqrt(1 + 4*Q^2))/(2*Q)
>
> For the upper edge of the FWHM, we require f_U > f_0.  The only
> root which does this is f_1, so:
>
> f_U =f_1 =  f_0*(1 + sqrt(1 + 4*Q^2))/(2*Q)
>
> For the lower edge of the FWHM, we require 0 < f_L  < f_0. The
> only root which does this is f_4, so:
>
> f_L = f_4 = -f_0*(1 - sqrt(1 + 4*Q^2))/(2*Q)
>
> Therefore:
>
> FWHM = f_U - f_L = f_0/Q (and not f_0*sqrt(1 + 4*Q^2)/Q which
> was David Bowman's result) .
>
> If my work is correct, and accepting that the Nyquist bandwidth
> is B = pi*f_0/(2*Q), the ratio of B to the FWHM is given by:
>
> B/FWHM = pi/2
>
> This is the same result that John Denker gave in starting this
> thread on 11/19/16.
>
> What am I doing wrong?
>
> Don Polvani

Nothing.  Nothing at all.  Mea culpa.  I lost half of the 4 roots, mistakenly subtracted f_1 - f_2, and (to my great embarrassment) never noticed that f2 was negative.  The reason I didn't find half of the roots is that I stupidly solved

1 = Q*(f/f_0 - f_0/f)

instead of solving

1 = (Q*(f/f_0 - f_0/f))^2

as I should have.

Therefore, in the words of Gilda Radner's Emily Litella, "Never mind."

I should have realized that when my result contradicted JD's I should have double & triple checked mine.

Thanks Don!

Dave Bowman