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Re: Orthogonal Signals Question

by "dvsarwate@[EMAIL PROTECTED] " <dvsarwate@[EMAIL PROTECTED] > Oct 9, 2008 at 11:44 AM

On Oct 9, 9:03=A0am, "westocl" <cwest...@[EMAIL PROTECTED]
> wrote:
> Is the sole criterion for two signals to be orthogonal a cross
correlatio=
n
> type relation****p? Could two signals that have some frequency
cancellatio=
n
> when added together still be viewed as orthogonal?
>
> For example. if x(t) has fourier transform X(f), which has a magnitude
of
> 1 over some BW, B. And y(t) has fourier transform Y(f), has a magnitude
o=
f
> 1 over the same band.
>
> If the sole requirement for signals to be orthogonal is int(x(t)*y(t))dt
=
=3D
> 0 over a time interval and x(t)^2 + y(t)^2 =3D (x(t)+y(t))^2., this is
an
> enery/power like criterion' and the equation doesnt necessarily say
> X(f)+Y(f) have to equal 1.41 over that the bandwith B. The equation just
> alludes to the energy in the 2 singals needing to be preseved, not
> necessarily the energy in any one particular frequency.

You seem to have an unusual definition of orthogonality.
If [x(t)]^2 + [y(t)]^2 =3D [x(t) + y(t)]^2 for all t, then it must
be that for each value of t, x(t)y(t) =3D 0, that is, at least
one of the two signals is 0.  If this is so, then obviously
int(x(t)*y(t))dt =3D 0 over *any* time interval because the
integrand is always 0.  A less restrictive (and very much
more widely accepted) definition of orthogonality (for all
time or over some specified time interval) is that the
inner product int(x(t)*y(t))dt have value 0 where the integral
is over all time, or over the specified time interval, whichever
is appropriate.
 




 4 Posts in Topic:
Orthogonal Signals Question
"westocl" <c  2008-10-09 09:03:15 
Re: Orthogonal Signals Question
Ikaro <ikarosilva@[EMA  2008-10-09 08:09:27 
Re: Orthogonal Signals Question
"dvsarwate@[EMAIL PR  2008-10-09 11:44:51 
Re: Orthogonal Signals Question
glen herrmannsfeldt <g  2008-10-11 20:03:21 

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