I have not carefully studied all the posts in this thread, so I
apologize if I say something redundant.
Ludwik says he has a square-wave pulse that is f(x) = 7 between x=2 and
x=5. He then says he solves for the fourier cosine coefficients by
integrating from x=2 to x=5.
John Mallinckrodt suggested Ludwik cannot use cosines alone because the
function Ludwik is working on is not an even function. John suggested
moving the square wave pulse so it is centered about zero.
It seems to me John's suggestion would yield an entirely different
situation, that is, the two situations are not even close to equivalent.
Fourier theory assumes the function you are trying to express (as a
fourier series) is a periodic function. If your "function" is constant
at f(x) = 7 from x=2 to x=5 (and zero from x=0 to x=2), then when you
try to integrate over only that domain (from 2 to 5) you are getting the
same result as if you had integrated from 0 to 5. That means the
periodic function you are trying to express actually runs from 0 to 5,
then repeats. That is, the periodic function is f(x)=0 from x = 0 to 2,
5 to 7, 10 to 12, etc. and f(x)=7 from x = 2 to 5, 7 to 10, 12 to 15,
etc.
We would indeed need both sine and cosine terms to represent this
(because it is neither even nor odd), except it is also possible to
express it as only sines or only cosines if a phase constant is
specified for each wavenumber. But to find the phase constants you
first find both the sine and cosine coefficients, so even if you want to
express it as a cosine series you first find both the sine and cosine
coefficients.
If we move the f(x) = 7 "pulse" to center it on the origin, such that
f(x) = 7 for x between -1.5 and +1.5, and we integrate this from -1.5 to
1.5, then we don't really have a pulse. This is just a constant
function. a(0) = 7 and all other a(j) and b(j) are zero.
I also don't understand what Ludwik is doing when he talks about k = 6.1
and k = 5.9. Doesn't fourier analysis involve integer wave numbers?
Michael D. Edmiston, Ph.D.
Professor of Chemistry and Physics
Bluffton College
Bluffton, OH 45817
(419)-358-3270
edmiston@bluffton.edu
-----Original Message-----
From: Ludwik Kowalski [mailto:kowalskil@MAIL.MONTCLAIR.EDU]
Sent: Wednesday, October 15, 2003 1:04 PM
To: PHYS-L@lists.nau.edu
Subject: Fourier transforms
Suppose I have a function f(x) which is a
rectangular pulse between x=2 and x=5,
[say f(x)=7] and zero everywhere else. I want
to find A(k), the Fourrier transform of that f(x).
The formula tells me that for any chosen k
A(k)=Integral of f(x)*cos(kx)*dx
In general the limits of integration are minus
and plus infinity but in practice they refer
to a region in which f(x) is not zero. Thus in
my example xmin=2 and xmax=5. During
the integration k remains a constant.
According to illustrations seen in books
A(k) is a functions which wiggles between
positive and negative values, depending
on k. In this message I was going to ask
how to justify negative A(k) on the intuitive
basis. But the answer came to me right now.
So instead of explaining why I was puzzled
let me share the qualitative explanation;
perhaps it can help somebody.
The function I am integrating wiggles between
-7 and +7. The number of wiggles (cycles)
is different for each k. If k=6, for example, then
the integral is zero; the number of + swings and
the number of - swings are even. But for k equal
to 6.1, for example, the integral is positive because
the added positive area (under the function and x
axis) does not have a negative counterpart. For
k=5.9, on the other hand, the integral becomes
negative because the last positive swing is not
completed and the sum of negative areas is still
larger than the sum of positive areas.
It is really an AHA moment for me; the nonzero
values of A(k) come from situations in which the
cos(k*x) function does not fit the region of x in
which f(x) is not zero. A small change in the width
of my f(x) pulse can change the shape of the A(k)
pulse significantly.
A connection with the uncertainty principle is
sometimes made in chapters describing Fourrier
transforms. The A(k) becomes wider when f(x)
become narrower, and vice versa.
Ludwik Kowalski