```cv - Convolve the data in stack 1 with a filter function in stack 3
putting the result in stack 1.					RWW

EXAMPLES:
rt 9;st 3;rt 10;cv	Convolve stack 10 with stack 9.

rt 10;c .stak(1)=1; c .numpt=1;c .expch=1;st 3;rt 10;cv
Set up a null filter in stack 3 and apply it (with
no change in stack 10).

rt 10;c .stak(1)=-1;c .stak(2)=1;c .numpt=2;c .expch=1.5;st 3;rt 10;cv
Set up a first derivative filter and apply it.
Note that expch1 will change.  The change in
expch1 and the spectrum would be different by
about 1 if .expch3 were set to 1.501.

rt 10;c .stak(1)=.25;c .stak(2)=.5;c .stak(3)=.25;c .numpt=3;c .expch=2
st 3;rt 10;cv		Set up a Hanning smoothing function and apply it.
There will be no change in .expch1.

DESCRIPTION:
Cv does a convolution of stack 1 with stack 3 and replaces stack 1 with
the result.  Although the convolution is symmetric, this function is
set up to apply a filter in stack 3 to data in stack 1.  Each channel
in stack 1 is replaced by:

sum [.stak1(ch - i) * stak3(int(expch3 + 0.499999) + i)]
i

where i is limited to values with valid channels in both stacks.  If
the center channel of stack 3 (expch3) is not an integer, then expch1
is adjusted to reflect the fractional channel shift in the spectrum
assuming that expch3 is set to the 'center' of the convolving
function.  .numpt3 must be set to the number of points in the
convolving function.  If you wish to preserve the amplitude of the
spectrum in stack 1, you must normalize the function in stack 3
correctly.
```