Conference rusure::math

601.0. "conditions on/proofs for moving across integral signs" by EAGLE1::BEST (R D Best, Systems architecture, I/O) Wed Oct 22 1986 17:49

  In Simon Haykin's 'Communications Systems' (2nd ed.), there is a proof on
p. 36 that demonstrates that the Fourier transform of the product of two
time functions is the convolution of their Fourier transforms.

g1(t).g2(t)
<-->
integral( -inf, inf, G1( L ).G2( f-L ), L )

where	. means multiply,
	integral( p, q, R, L ) means integrate function R from p to q wrt L.
	and inf means infinity.

The following is extracted from the proof:

"
(step 1 )

	G12( f ) =
integral(
  -inf,
  inf,
  integral( -inf, inf, g1( t ).G2( f' ).exp( j.2.pi.[f-f'].t ), f' ),
  t )

Define lambda = f - f'.  Then, interchanging the order of integration, we
obtain:
(step 2)

	G12( f ) =
integral(
  -inf,
  inf,
  G2( f - lambda ).integral( -inf, inf, g1( t ).exp( -j.2.pi.lambda.t ), t ),
  lambda )

The inner integral is recognised simply as G1( lambda ) and so we may write:
(step 3)

	G12( f ) =
integral(
  -inf,
  inf,
  G1( lambda ).G2( f-lambda ),
  lambda ) "

What I don't understand is what justifies moving G2( f - lambda ) across the
integral sign in going from step 1 to step 2.  Does anyone know:

(a) Where there is a proof that this is OK ?

(b) What are the conditions on G(f) and/or g(t) that are necessary to make this
operation legal ?  Are there conditions on the limits ?

or

(c) Am I completely misunderstanding and the author is really doing something
else that will be obviously legal to the practiced mathematical eye ?

  I have looked through several calc. and adv. calc. books for the
justification and can't find one.  I've seen this done before, but never
seen a proof that it's OK or any mention of whether there are contingent
conditions on the integrand or limits that make it OK.

  To avoid wasting time, here's where I've already looked:

	calculus texts:
	Purcell
	Anton

	advanced calculus:
	Kaplan

		/R Best

    The reason you can do this is because of the infinite limits on the
    integrals;  shifting the frequency origin by any finite amount will
    not effect the total integral, because the tails of the integral
    will still get everything.

    Here's a compelling reason for why convolution is so important -

	suppose you had two linear, time-invariant systems, call them
	X and Y, and imagine that the impulse response of either
	was just a discrete sequence of echos, so if you shock excite
	either system, you'd see its unique reverberation pattern.

	If you now feed X into Y and shock excite X, you can imagine
	the convoluted reverberation pattern that will come out	of Y:
        every echo from X will stimulate a new chain of impulses
	from Y, and by linearity, they will all just superimpose to give
	the total output.  If you work out the summations you get a
	discrete convolution, and if you blend the impulses into a
	continuum you then have a convolution integral...

    - Jim