You may be confusing Fourier transform and Fourier series. The Fourier transform of a function is a function on the reals.
This question is strange enough so I'm beginning to suspect a typo/misinterpretation.
Let be defined by
Show that is admissible and compute its normalization constant.
First of all it is very strange to think of the fourier transform to be a function on the reals as opposed to a sequence, right? (I suspect this might be a typo)
Anyway, I interpret admissible to mean: satisfies the Orthonormal Conjugate Quadrature Filter Conditions.
So first I actually find by taking the inverse fourier transform:
and this is zero at all integer values, which makes it a very strange trivial case (which is not orthonormal).
Any thoughts?
Thanks.
You may be confusing Fourier transform and Fourier series. The Fourier transform of a function is a function on the reals.
Good point.
That gives me
Edit: I spotted a miscalculation...I'll get back to this
which takes on at odd , at even
and at 0, we get .
Is that really a good result. It's technically non-zero at infinitely many integers and I don't like what's going on at 0.
Does this all make sense?
I think I might have cleared some things up for myself.
Continuous wavelet transform - Wikipedia, the free encyclopedia