320 Homework 4, 12(c)

[Hiroko Shiraiwa]

Sounds good to me too! Unfortunately I am not familiar with the
logical reasoning style (I tend to stick on equations), so that 
I cannot evaluate it with confidence. 

But is it really "x is even iff X is even"?
In my proof of x_ven->X_even, I proved that if x_even then X_even, 
but didn't prove that it is vice versa... ah, of course it is. IDFT is
just a variation of DFT. Then it makes sense. 


Hiroko

> 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
> 
> > 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>