320 Homework 4, 12(c)

[Jeffrey Traer Bernstein]

x odd <-> X odd is the actual proof i.e. x is odd iff X is odd (iff = if and
only if i.e x if y and y if x). This is true by virtue of the previous proof
(x even iff X even) here is my reasoning:

If x is even iff X is even, then in order for X to be not even (that is odd
or neither odd or even) its even component must be equal to zero (remember
that all functions can be expressible as a sum of even and odd functions and
the DFT is linear (think superposition...)). Since an even component is
required for a function (and its transform) to be either even or neither
even or odd a function with no even component must in fact be odd.
Therefore, if x is odd its transform, X, must also be odd. The converse, if
X is odd then x is odd, follows from the same principles and the linearity
of the IDFT.

Sounds good to me! Please poke holes if possible though!

Jeffy B

----- Original Message -----
From: "Hiroko Shiraiwa" <shiraiwa@Stanford.EDU>
To: "MA/MST" <mamst@ccrma.Stanford.EDU>
Sent: Saturday, October 26, 2002 12:50 PM
Subject: 320 Homework 4, 12(c)


> Still stuck on homework #4 12(c)?
> (odd -> odd proof)
> Right now I got an idea how to prove it.
>
> Use the phase shift properties. Odd function is
> phase shift of 2/N samples from an even function.
> Then X_o[n] = W^k(2*N)_N X_e[n]. You know X_e[n] is
> even, then you just need to prove W^k(2*N)_N is an
> odd function. Finally you can say X_o[n] is a
> multiplication of odd and even function: odd function.
>
> Do you think it works??
>
> Hiroko
> --
> Hiroko Shiraiwa <shiraiwa@stanford.edu>
>
> _______________________________________________
> mamst mailing list
> mamst@ccrma.stanford.edu
> http://ccrma-mail.stanford.edu/mailman/listinfo/mamst
>