DFT X[ ] independent variable
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DFT X[ ] independent variable

 
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Vec
Guest
Posted: Tue Jan 04, 2005 7:58 am    Post subject: DFT X[ ] independent variable Reply with quote
Hello

I need help, in DFT, the frequency domain’s independent variable can be
refereed to in many ways, one way is being a fraction of the sampling rate.
could some one please explain why X[ ] independent variable runs between
0 and 0.5, if you say because discrete data can only contain frequencies
between DC and ½ the sampling rate, then if the sampling rate (x[ ] has)
128 points then 128 * ½ is 64 then why X [ ] has 65 samples?
I must be missing something.

Thanks
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Tim Wescott
Guest
Posted: Tue Jan 04, 2005 7:58 am    Post subject: Re: DFT X[ ] independent variable Reply with quote
Vec wrote:

Quote:
Hello

I need help, in DFT, the frequency domain’s independent variable can be
refereed to in many ways, one way is being a fraction of the sampling rate.
could some one please explain why X[ ] independent variable runs between
0 and 0.5, if you say because discrete data can only contain frequencies
between DC and ½ the sampling rate, then if the sampling rate (x[ ] has)
128 points then 128 * ½ is 64 then why X [ ] has 65 samples?
I must be missing something.

Thanks

The discrete data can only contain frequencies between DC and 1/2 the

sampling rate _inclusive_, and points 0 through 64 is a set with 65 members.

Note: Your rule only applies when you are sampling real data. If you
are sampling complex data (or I & Q data) then you need all 128 points.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
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Fred Marshall
Guest
Posted: Tue Jan 04, 2005 7:58 am    Post subject: Re: DFT X[ ] independent variable Reply with quote
"Tim Wescott" <tim@wescottnospamdesign.com> wrote in message
news:10tjikljh67jh9d@corp.supernews.com...
Quote:
Vec wrote:

Hello

I need help, in DFT, the frequency domain’s independent variable can be
refereed to in many ways, one way is being a fraction of the sampling
rate.
could some one please explain why X[ ] independent variable runs between
0 and 0.5, if you say because discrete data can only contain frequencies
between DC and ½ the sampling rate, then if the sampling rate (x[ ] has)
128 points then 128 * ½ is 64 then why X [ ] has 65 samples?
I must be missing something.

Thanks

The discrete data can only contain frequencies between DC and 1/2 the
sampling rate _inclusive_, and points 0 through 64 is a set with 65
members.

Note: Your rule only applies when you are sampling real data. If you are
sampling complex data (or I & Q data) then you need all 128 points.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

With no disagreement, said a slightly different way:

A DFT means the temporal data is sampled and the spectrum repeats at periods
of fs.
-fs/2 to fs/2 covers one period of the spectrum.
If the temporal data is real, then the spectral data is even.
Accordingly, the spectral data from zero to fs/2 is a mirror image of the
data from zero to -fs/2, and the data from fs/2 to fs is the same as the
data from -fs/2 to zero.
So, all of the information in the spectral data is present from zero to
fs/2.

If we normalize the sequences so that the time sampling interval is assumed
to be 1 second, then fs=1Hz and fs/2 is 0.5Hz. This is handy because often
we are dealing with regularly spaced "temporal" samples or maybe just
regularly spaced unlabeled samples - in which case using a sample interval
of "1.0" is convenient. Then the spectrum from 0 to 0.5 is all we need.

Fred
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Matt Timmermans
Guest
Posted: Tue Jan 04, 2005 7:58 am    Post subject: Re: DFT X[ ] independent variable Reply with quote
"Vec" <vec2u@valoudari.edu.mx> wrote in message
news:41d9b8ed$1_1@news.iprimus.com.au...
Quote:
[...] (x[ ] has) 128 points then 128 * ½ is 64 then why X [ ] has 65
samples?
I must be missing something.

I assume you're using real (i.e., not complex) input data. In the DFT
output, the frequency samples are complex numbers, i.e., each contains two
real numbers, except for the DC and fs/2 which can have only one non-zero
real component each. The total count of real numbers in X[] is then 1 +
63*2 + 1 = 128, as you would expect.

--
Matt
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Fred Marshall
Guest
Posted: Tue Jan 04, 2005 7:58 am    Post subject: Re: DFT X[ ] independent variable Reply with quote
"Vec" <vec2u@valoudari.edu.mx> wrote in message
news:41d9b8ed$1_1@news.iprimus.com.au...
Quote:
Hello

I need help, in DFT, the frequency domain’s independent variable can be
refereed to in many ways, one way is being a fraction of the sampling
rate.
could some one please explain why X[ ] independent variable runs between 0
and 0.5, if you say because discrete data can only contain frequencies
between DC and ½ the sampling rate, then if the sampling rate (x[ ] has)
128 points then 128 * ½ is 64 then why X [ ] has 65 samples?
I must be missing something.

Thanks


And I might have added:
While fs/2 is the mirroring point in frequency, it need not have a sample
located there.
If the number of time samples is even, e.g. 64, then so is the number of
frequency samples out of a DFT.
If the number of frequency samples is 64, the frequency sampling interval is
fs/64.
64 samples covers the spectral range zero to fs-fs/64 and F(0)=F(fs) so
F(fs) need not be included because it would be redundant.
Note that the range 0 to fs is split into 64 intervals that are spanned by
65 samples if fs is included.
If it is in 64 intervals then fs/2 is in exactly 32 intervals which is
spanned by 33 samples if fs/2 is included. The remaining 31 samples run
from fs/2+fs/64 to fs-fs/64.
SO: If the number of samples is even, as is often the case for a DFT, then
there is a sample located at fs/2.

If the number of time samples is odd, e.g. 65, then so is the number of
frequency samples out of a DFT.
If the number of frequency samples is 65, the frequency sampling interfval
is fs/65.
65 samples covers the spectral range zero to fs-fs/65 and F(0)-F(fs) as
above.
Note that the range 0 to fs is split into 65 intervals that are spanned by
66 samples if fs is included.
If it is in 65 intervals, then fs/2 lies in the middle of the 33rd interval
and is not spanned by an integral number of samples.
SO: If the number of samples is odd, then there is not a sample located at
fs/2 and the nearest sample to fs/2 is at fs/2-fs/2N. The spectrum of a
real temporal sequence still mirrors at fs/2.

Fred
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Steve Drake
Guest
Posted: Thu Jan 06, 2005 4:05 am    Post subject: Re: DFT X[ ] independent variable Reply with quote
On Tue, 04 Jan 2005 08:20:11 -0800, Vec <vec2u@valoudari.edu.mx>
wrote:

Quote:
Hello

I need help, in DFT, the frequency domain’s independent variable can be
refereed to in many ways, one way is being a fraction of the sampling rate.
could some one please explain why X[ ] independent variable runs between
0 and 0.5, if you say because discrete data can only contain frequencies
between DC and ½ the sampling rate, then if the sampling rate (x[ ] has)
128 points then 128 * ½ is 64 then why X [ ] has 65 samples?
I must be missing something.

If you look at the DFT it runs around in a circle so the values repeat
after going from 0 to pi. The sample point are points on the circle.

Regards,

Steve Drake
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