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# Discussion Groups | Comp.DSP | a question about the DFT

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# a question about the DFT - lmuri - 2012-06-28 23:15:00

```I am a newbie in the field.

I am working on a problem on a DSP book.

Under what condition, DFT of a time series generates all real value. A
simple matlab code shows that if I have a time series x(n)
n=0,1,2,3,...,N-1 then I can build another time series,
y(n) = x(n) when n=0,...,N-1 and y(n) = x(2N-2-n) when n=N,N
+1,...,2N-3. However I do not know how to prove it.

Thank you for help.

ML
```
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# Re: a question about the DFT - mnentwig - 2012-06-29 07:16:00

```you need a hermitian symmetric spectrum (positive and negative frequencies
are conjugated).

prove via 2 cos(x) = exp(ix) + exp(-x)

Intuitively, each bin in the IFFT controls a rotating exp phasor. Positive
frequencies spin forwards, negative frequencies spin backwards. The
imaginary parts cancel, only the real part remains.

Proof or not, I'd recommend to make an effort to understand it intuitively.
It will help to apply FFTs to real-world problems.
Think in terms of positive -and- negative frequencies (=> complex-valued
exp-phasors instead of real-valued (co)sines), otherwise some rather basic
problems like aliasing or nonlinear distortion will turn into a nightmare.
```
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# Re: a question about the DFT - robert bristow-johnson - 2012-06-29 12:13:00

```On 6/29/12 7:16 AM, mnentwig wrote:
> you need a hermitian symmetric spectrum (positive and negative frequencies
> are conjugated).
>
> prove via 2 cos(x) = exp(ix) + exp(-x)
>

small typo.  can you spot the missing symbol?

--

r b-j                  r...@audioimagination.com

"Imagination is more important than knowledge."

```
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# Re: a question about the DFT - 2012-06-29 12:49:00

```On Friday, June 29, 2012 11:13:34 AM UTC-5, robert bristow-johnson wrote:
> On 6/29/12 7:16 AM, mnentwig wrote:
> > you need a hermitian symmetric spectrum (positive and negative frequencies
> > are conjugated).
> >
> > prove via 2 cos(x) = exp(ix) + exp(-x)
> >
>
> small typo.  can you spot the missing symbol?
>

yes. I learn matlab now. you miss * for multiply sign

matlab code is exp(i*x) not exp(ix). you miss too the first one there need 2*cos(x) not cos(x).

you learn matlab too, it is good.

Steve
```
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# Re: a question about the DFT - robert bristow-johnson - 2012-06-29 13:26:00

```On 6/29/12 12:49 PM, s...@gmail.com wrote:
> On Friday, June 29, 2012 11:13:34 AM UTC-5, robert bristow-johnson wrote:
>> On 6/29/12 7:16 AM, mnentwig wrote:
>>> you need a hermitian symmetric spectrum (positive and negative frequencies
>>> are conjugated).
>>>
>>> prove via 2 cos(x) = exp(ix) + exp(-x)
>>>
>>
>> small typo.  can you spot the missing symbol?
>>
>
>
> yes. I learn matlab now. you miss * for multiply sign
>

nope, that's not it.

--

r b-j                  r...@audioimagination.com

"Imagination is more important than knowledge."

```
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# Re: a question about the DFT - 2012-06-29 16:46:00

```Thank you for the help. I will try to make it out.

ML

On Friday, June 29, 2012 7:16:33 AM UTC-4, mnentwig wrote:
> you need a hermitian symmetric spectrum (positive and negative frequencies
> are conjugated).
>
> prove via 2 cos(x) = exp(ix) + exp(-x)
>
> Intuitively, each bin in the IFFT controls a rotating exp phasor. Positive
> frequencies spin forwards, negative frequencies spin backwards. The
> imaginary parts cancel, only the real part remains.
>
> Proof or not, I'd recommend to make an effort to understand it intuitively.
> It will help to apply FFTs to real-world problems.
> Think in terms of positive -and- negative frequencies (=> complex-valued
> exp-phasors instead of real-valued (co)sines), otherwise some rather basic
> problems like aliasing or nonlinear distortion will turn into a nightmare.

```
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# Re: a question about the DFT - Rick Lyons - 2012-06-30 04:25:00

```On Fri, 29 Jun 2012 12:13:34 -0400, robert bristow-johnson
<r...@audioimagination.com> wrote:

>On 6/29/12 7:16 AM, mnentwig wrote:
>> you need a hermitian symmetric spectrum (positive and negative frequencies
>> are conjugated).
>>
>> prove via 2 cos(x) = exp(ix) + exp(-x)
>>
>
>small typo.  can you spot the missing symbol?

Hi Robert,
Yep.  There's a missing "i" in front
of the "-x."

(What is this "i" business?  Real men use "j.")

I produced a simple algebra derivation, using
only four input samples (N=4), to prove the
necessary conjugate-symmetric DFT input samples
(mentioned by Markus Nentwig) that yield a
real-only DFT result.

It appears, to me, that another restriction,
along with conjugate-symmetric inputs, is that
input samples x(0) and x(N/2) must both be real-only.

I don't remember ever thinking about that
'real-only x(0) and x(N/2)' restriction before.

See Ya',
[-Rick-]
```
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# Re: a question about the DFT - Nasser M. Abbasi - 2012-06-30 04:55:00

```On 6/30/2012 3:25 AM, Rick Lyons wrote:

>
> (What is this "i" business?  Real men use "j.")
>

And true warriors use SQRT(-1). Not these wimpy symbols i and j.

--Nasser

```
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# Re: a question about the DFT - glen herrmannsfeldt - 2012-06-30 06:03:00

```Nasser M. Abbasi <n...@12000.org> wrote:

(snip)
> And true warriors use SQRT(-1). Not these wimpy symbols i and j.

Or CSQRT(-1.).

or, I suppose SQRT((1.,0.))

-- glen
```
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# Re: a question about the DFT - mnentwig - 2012-06-30 11:32:00

```>> small typo.  can you spot the missing symbol?
yes, now that you mention it.
Sorry for the confusion, good that it was spotted early.
The second "exp" needs an "i". There is one forwards and one backwards
rotating phasor.

The midpoint sample in an even-sized FFT is a special case. It's true, it
needs to be real for a real-valued time domain signal.

And yes, electrical engineers prefer "j" as "i" gets mixed up with current.

>> Real men use "j."
That may well be. On the other hand, I haven't yet met an imaginary man
:-)

-markus
```
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