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

["Donald Polvani" <dgpolvani@verizon.net>]

On Wednesday, November 23, 2016 4:53 PM, John Denker wrote:

> > FWHM = f_U - f_L = f_0/Q
>
> We all agree that is correct in the high-Q limit.
>
> However, for smallish Q, it's messier.  The quartic has at most two positive
> roots, and it's easy enough to solve in terms of the square roots of square
> roots, but it's ugly.  The cleanest form I've been able to obtain is
>    	Δϕ =   √(ϕ^2_peak + 1/|G|_peak)
> 	     - √(ϕ^2_peak - 1/|G|_peak)   for Q ≥ Qmin
>
> where the normalized FWHM is Δϕ = FWHM / f0.  I checked this graphically
> and numerically for rather small Q.

I don't understand why  FWHM = f_U - f_L = f_0/Q isn't correct for all Q big or small.  I solved the quartic gain function equation exactly.  There was no approximation made that Q had to be "big".  Specifically, I solved:

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

Therefore

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

Therefore:

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

Equation (1) can be simply solved with the quadratic formula to yield:

f_1 = f_0*(1 + sqrt(1 + 4*Q^2))/(2*Q)
f_2 = f_0*(1  - sqrt(1 + 4*Q^2))/(2*Q)

Equation (2) can be similarly solved to yield:

f_3 = -f_0*(1 + sqrt(1 + 4*Q^2))/(2*Q)
f_4 = -f_0*(1  - sqrt(1 + 4*Q^2))/(2*Q)

As we all agree now, the upper and lower edges (f_U and f_L) of the FWHM are given by f_1 and f_4, respectively.  Roots f_2 and f_3 pertain to negative frequency solutions.  Note, no approximations have been made in the solutions with respect to Q.  Both f_U , f_L, and G(F) vary smoothly (and to me, reasonably) with Q as Q is varied from 0 to infinity.  I've computed and plotted the normalized functions, f_U/f_0, f_L/f_0, (f_U/f_0 - f_L/f_0), and G(f)/f_0  for Q = 0.001 to Q = 1000.  They all behave reasonably.

What am I missing here?

Don Polvani

Adjunct faculty, Physics, Retired
Anne Arundel Community College
Arnold, MD 21012