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From: Rune Allnor on 3 May 2010 13:32
On 3 Mai, 18:56, dvsarwate <dvsarw...(a)gmail.com> wrote:
> On May 3, 10:05 am, Rune Allnor <all...(a)tele.ntnu.no> claimed:
>
>
>
> > Because the DFT works on finite sequences of discrete data.
>
> I am sure there are those on this forum who will
> dispute that claim and say that DFTs implicitly
> assume periodic sequences, but let me ask with
> respect to the specific point that I raised about
> the OP's question.  The OP asked
>
> >> given a discrete signal x[n] of N samples, the DFT X[k] will also have N
> >> samples. The limits of the summation of the DFT can be for k that goes from
> >> -N to (N/2)-1 or from 0 to N-1, correct? DOes it give the same result?

The OP obviously haven't got all the details right.
There aren't N samples between -N and (N/2)-1.
Even if N is odd.

But then, if we were to require that kind of
exact phrasings in *anything* this particular
OP posted, (s)he wouldn't get many responses.

>  I
> ask this because I have finally after all these years
> figured out summation notation but am still a bit
> hazy about matrices.

I have come to understand you are hazy about a lot
of things. Matrices is a fundamental tool of maths
that usually is taught some where around the first
semester of the freshman year of college.

I am sorry on your behalf that you missed that.

Rune
From: robert bristow-johnson on 3 May 2010 16:09
On May 3, 8:57 am, "fisico32" <marcoscipioni1(a)n_o_s_p_a_m.gmail.com>
wrote:
>
> given a discrete signal x[n] of N samples, the DFT X[k] will also have N
> samples. The limits of the summation of the DFT can be for k that goes from
> -N to (N/2)-1 or from 0 to N-1, correct?

i think you made a typo error here. didn't you mean to say that k
goes from
-N/2 to (N/2)-1 or from 0 to N-1? if that is what you meant to say, i
would say you are "correct".

> Does it give the same result?
>
> More generally, I guess we can pick any arbitrary integer n0, and make the
> limits go from n0+N to ( ((-n0+N)/2)-1 ). IF n0=-N, then the limits become
> 0 and N-1....

again, fixing your typo, you can make the limits go from n0 to n0+N-1
for any integer n0.


DFT:

n0+N-1
X[k] = SUM{ x[n] * exp(-j*2*pi*n*k/N }
n0

for k0 <= k < k0+N



iDFT:

k0+N-1
x[n] = (1/N) SUM{ X[k] * exp(+j*2*pi*n*k/N }
k0

for n0 <= n < n0+N


for either the DFT or iDFT definitions, n0 and k0 can be whatever
integers your heart desires. this is because there is no mathematical
difference between the DFT and the DFS ("Discrete Fourier Series").
they are just two different names or labels for the exact same
operation. whether it's DFT/iDFT or DFS/iDFS, either map a discrete
and periodic sequence of period N to another discrete and periodic
sequence of the same period. that is the DFT and iDFT (or DFS/iDFS)
assume a periodic extension of the data input to it:


x[n+N] = x[n] for all n

X[k+N] = X[k] for all k


in practice, because your computer memory does not go on forever, you
might have to use modulo arithmetic to impose this periodicity
explicitly in the indices of your x[n] or X[k] sequences. you would
replace every n with

n <-- mod(n,N) = n - N*floor(n/N)

and every k with

k <-- mod(k,N) = k - N*floor(k/N)

where floor(u) is the largest integer no greater than u. if you do
that, you need not concede the inherent periodic extension that the
DFT and iDFT imposes. but, in my opinion, the above modulo operations
explicitly impose such a periodic extension anyway. for me it's just
easier to say x[n+N]=x[n] and X[k+N]=X[k].

and the mapping is invertible. really the DFT and iDFT are sorta the
same thing because they can be redefined to be "unitary" by adjusting
the constant in front of the summation:

DFT:

n0+N-1
X[k] = sqrt(1/N) SUM{ x[n] * exp(-j*2*pi*n*k/N }
n0

for k0 <= k < k0+N



iDFT:

k0+N-1
x[n] = sqrt(1/N) SUM{ X[k] * exp(+j*2*pi*n*k/N }
k0

for n0 <= n < n0+N


and +j and -j are qualitatively the same thing. both have equal claim
to be the sqrt(-1). again n0 and k0 can be whatever integer you like.


> Does it matter if the signal x[n] is real or complex?

no. but if x[n] is purely real you will get conjugate symmetry a.k.a
"Hermitian symmetry" in X[k], which may or may not be a useful
property.

r b-j
From: dbd on 3 May 2010 17:09
On May 3, 1:09 pm, robert bristow-johnson <r...(a)audioimagination.com>
wrote:
....
>
> and the mapping is invertible.  really the DFT and iDFT are sorta the
> same thing because they can be redefined to be "unitary" by adjusting
> the constant in front of the summation:
....
>
> and +j and -j are qualitatively the same thing.  both have equal claim
> to be the sqrt(-1). again n0 and k0 can be whatever integer you like.
>
....
> r b-j

It isn't clear to me from what you have posted: Do you intend the
concepts of "sorta the same thing" and "qualitatively the same thing"
to apply to items that are functions of terms that are the square
roots of any real number or only the square roots of the same negative
real numbers? (I couldn't find either term in wikipedia or in the new
or old testaments of O&S.)

Dale B. Dalrymple
From: glen herrmannsfeldt on 3 May 2010 18:54
robert bristow-johnson <rbj(a)audioimagination.com> wrote:
(snip)

> and +j and -j are qualitatively the same thing. both have equal claim
> to be the sqrt(-1). again n0 and k0 can be whatever integer you like.

I have seen claims that the EE j is equivalent to -i in physics.

This comes out from the use of exp(jwt) in places where physics
uses exp(-iwt). The distinction comes from looking at a time
and space varying signal at different times and one place, or
at different places and one time. Or consider an oscilloscope
display and a chart recorder.

You can change all +j to -j, but some is not enough.

-- glen
From: robert bristow-johnson on 4 May 2010 19:09
On May 3, 5:09 pm, dbd <d...(a)ieee.org> wrote:
> On May 3, 1:09 pm, robert bristow-johnson <r...(a)audioimagination.com>
> wrote:
> ...
>
>
>
> > and the mapping is invertible.  really the DFT and iDFT are sorta the
> > same thing because they can be redefined to be "unitary" by adjusting
> > the constant in front of the summation:
> ...
>
> > and +j and -j are qualitatively the same thing.  both have equal claim
> > to be the sqrt(-1). again n0 and k0 can be whatever integer you like.
>
> ...
> > r b-j
>
> It isn't clear to me from what you have posted: Do you intend the
> concepts of "sorta the same thing" and "qualitatively the same thing"
> to apply to items that are functions of terms that are the square
> roots of any real number or only the square roots of the same negative
> real numbers? (I couldn't find either term in wikipedia or in the new
> or old testaments of O&S.)

this is what i mean in Wikipedia:

http://en.wikipedia.org/wiki/Imaginary_unit#i_and_.E2.88.92i

+j and -j are qualitatively the same thing. everything you say about
+j is something you can say about -j. but they are quantitatively
*not* the same thing. they are non-zero and are negatives of each
other. they are not the same number. but their roles can be
interchanged. this qualitative equivalence is not true for the real
unit +1 and its negative -1. this is why imaginary numbers are more
different than real numbers than just a matter of convention (the real
axis is qualitatively different than the imaginary axis). but the
difference between +j and -j is merely one of convention. and that
means that (with scaling made so that the transforms are "unitary")
the only difference between the forward and inverse Fourier Transform
(and the DFT vs. iDFT) is merely a matter of convention.


On May 3, 6:54 pm, glen herrmannsfeldt <g...(a)ugcs.caltech.edu> wrote:
> robert bristow-johnson <r...(a)audioimagination.com> wrote:
>
> (snip)
>
> > and +j and -j are qualitatively the same thing. both have equal claim
> > to be the sqrt(-1).
>
....
>
> You can change all +j to -j, but some is not enough.

yup. if all textbooks and literature in math, physics, EE or whatever
science or discipline, where changed so that +j and -j (and +i and -
i) were switched, every fact and theorem would continue to be just as
valid.

r b-j


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