From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Trig Problem
Date: 27 Apr 1995 19:22:49 GMT
In article <3nj3rp$7ld$1@mhadg.production.compuserve.com>,
Mark Whittington <100566.3172@CompuServe.COM> wrote:
>I wonder if any can help with a seemingly intractable pron^?^?^?^?^?^?^[[Dblem^R^R^R.
You really need to fix that back-space key. This is how your post looked
(I edited it to appear just as odd on your screen).
>We need to know the maximum possible amplitude of two superimposed sine waves
>of variable phase(p), amplitude(A) and angular frequency(f).
>
>In other words, the maximum possible amplitude (between 0 and A1+A2) of
>
>A1.SIN(f1.t + P1) + A2.SIN(f2.t + P2)
It depends on whether the ratio f2/f1 is rational or not, but essentially
the answer is A1+A2.
If f2/f1 = n/m in lowest terms, then f1/m = f2/n is some number, f, say.
In that case, your function looks like
A1.SIN(m.u + P1) + A2.SIN(n.u + P2)
where u = f.t. In this case, the function is periodic with period 2 pi
(for u); you can compute the maximum for all t just by finding the maximum
for u in [0, 2 pi]. There are only a finite number of local maxima here
(something like n*m, I guess), so you can look for them graphically and
compute them numerically. Alternatively, you can go algebraic: use
the angle-addition formulas to write the above expression as a (very large)
polynomial in sin(u) and cos(u) (involving A1, A2, and sin and cos
of P1 and P2). Maple will expand these for you, for example. Then from
cos^2(x) = 1 - sin^2(x), you can write the whole thing as a function
Q1(s) + sqrt(1-s^2) Q2(s)
where Q1 and Q2 are polynomials in s (=sin(u)). This should be
even easier to maximize numerically, since there are no trig
function-calls.
If f2/f1 is irrational, then there are values of t
you can choose which will make _both_ of f1.t + P1 and f2.t + P2 be
as close as you like to pi/2 + 2.k.pi for some integer k (i.e. k1 or k2);
thus, you can make each of the SIN terms as close to 1 as you like,
giving a value for the sum close to A1+A2.
Indeed, in this case the set of integer multiples (f2/f1).k is dense in
the unit circle, so you can choose an integer k making this product
be as close as desired to the following magic number:
-(P2/2pi - 1/4) + (f2/f1) (P1/2pi - 1/4)
(mod 1), say
(f1/f2).k = n - (P2/2pi - 1/4) + (f2/f1) (P1/2pi - 1/4) + epsilon
for some integer n and very small epsilon. Then we let
t = (1/f1).( -P1 + pi/2 + 2.k.pi ),
so that SIN(f1.t + P1) = SIN(pi/2) = 1. On the other hand,
f2.t+P2 = (f2/f1).k.2pi + (f2/f1)(-P1+pi/2) + P2
= n.2pi + pi/2 + 2pi.epsilon
so SIN(f2.t+P2) = SIN(pi/2 + 2pi.epsilon) is as close to 1 as you want.
The case in which f2/f1 = n/m is rational but with m and n large
resembles the case with f2/f1 irrational: the multiples (f2/f1).k will
occupy m evenly-spaced points on the unit circle R/Z, so that you
can get rather close to the magic number, though you can't make epsilon
as small as you want even by taking large k.
You can discover a relation between epsilon and k by working with
rational approximations to f2/f1, for example using the continued-fraction
expansion of f2/f1.
dave