Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: Fourier transforms



Suppose I do shift the origin, as suggested by
JohnM. Then A(k) alone would be sufficient. But
John's qualification that for any even function
f(x) (i.e. symmetric with respect to the y axis)
the phases are "restricted to 0 and pi" might
not be immediately obvious.

To make this obvious I would use the following
argument. Suppose the sum representing the
f(x) in the "k space" contains a pair of terms with
phases equal to +20 and -20 degrees. One might
think that the sum of terms with such phases would
still produce an even function. That is why one may
be tempted to reject the "zero or pi phases" idea.

But the sum would be even only if each of the
two cosine terms had the same k. In reality, each
k appears only once and that is why each term
must be even. The only way to make this possible
is to accept that the phase is either 0 or pi.

Once again, I am thinking aloud (i.e. learning while
composing a message). I know that all this is trivial
to many. But I also know that many have been lost,
as I was, on the way from Fourier series (representing
periodic functions) to Fourier transforms. Otherwise
I would not bother you with this stuff.
Ludwik Kowalski

On Wednesday, October 15, 2003 John Mallinckrodt wrote:

Just a few comments on Ludwik's discussion:

The negative coefficients convey phase information. Ludwik's example
uses a square pulse of height 7 between x = 2 and 5. Because this
function is neither odd nor even, a Fourier transform will
necessarily involve sinusoidal functions of myriad phases. In fact,
the function he describes cannot be expanded in terms of cosines
alone because the result of doing so would necessarily be an even
function. The set of all cosine functions do not span the infinite
dimensional space of all real functions of one variable. He would
need to add sine functions as well.

Having found the expansion in terms of coefficients

A(k)=Integral of f(x)*cos(kx)*dx

and

B(k)=Integral of f(x)*sin(kx)*dx

he could then combine the A(k) and B(k) to find an amplitude C(k) and
a phase F(k) for each value of k and then expand the function as

Integral of C(k)*sin(kx+F(k))*dk

Alternatively, he could simply shift his origin so that the function
is a square pulse between x = -1.5 and +1.5 in which case B(k) = 0
and the cosine functions would suffice with phases restricted to 0
and pi. The contributions with phase = pi would be indicated by the
negative coefficients.

It's easy to see that all values of |k| up to |k| = pi/(halfwidth of
the pulse) will be positive so that the "width" of the central
maximum of the transform A(k) is inversely proportional to the pulse
width.

John Mallinckrodt mailto:ajm@csupomona.edu
Cal Poly Pomona http://www.csupomona.edu/~ajm

--