Periodogram returning extra values

I'm reading in a small wav-file and just extract the first 512
coefficients from that file. I then proceed to find the periodogram
PSD estimate of that 'frame'. In the following code 'x' is the vector
containing the 512 coefficients:

h = spectrum.periodogram('hann');
hopts = psdopts(h,x);  % Default options based on the signal x
set(hopts,'NFFT',512,'Fs',Fs,'SpectrumType','onesided');
psd(h,x,hopts);

However when I view the properties of the object by:

objproperties = psd(h,x,hopts)

the length of the Data vector and (obviously) frequency vector is 257,
rather than the expected 256. I'm using MatLab 7.5.0.
Hope someone can clarify this for me. Thanks :)

>I'm reading in a small wav-file and just extract the first 512
>coefficients from that file. I then proceed to find the periodogram
>PSD estimate of that 'frame'. In the following code 'x' is the vector
>containing the 512 coefficients:
>
>h = spectrum.periodogram('hann');
>hopts = psdopts(h,x);  % Default options based on the signal x
>set(hopts,'NFFT',512,'Fs',Fs,'SpectrumType','onesided');
>psd(h,x,hopts);
>
>However when I view the properties of the object by:
>
>objproperties = psd(h,x,hopts)
>
>the length of the Data vector and (obviously) frequency vector is 257,
>rather than the expected 256. I'm using MatLab 7.5.0.
>Hope someone can clarify this for me. Thanks :)
>

The length of 257 is correct-- An N point DFT of a real signal returns N/2
+ 1 values.  Here is a link on the topic.
Regards,
Steve

http://www.dspguide.com/ch12/1.htm 

SteveSmith wrote:
(snip)

> The length of 257 is correct-- An N point DFT of a real
> signal returns N/2 + 1 values.  Here is a link on the topic.

> http://www.dspguide.com/ch12/1.htm 

This is going to get darn close to the topic of another
thread related to Nyquist.  Isn't that N/2+1st point the one
at the Nyquist frequency?

Now, to look at it another way without including the DFT,
consider sampling a band limited and time limited signal.

If the duration of the signal is N Fs, and the signal is
not periodic, how many sample points do we need to be able
to recreate the signal?  IT seems to me that N+1 points
are needed to properly include the end points.  A signal
that isn't time limited doesn't have end points.

-- glen

On Oct 30, 11:03 am, "SteveSmith" <Steve.Smi...@SpectrumSDI.com>
wrote:
> >I'm reading in a small wav-file and just extract the first 512
> >coefficients from that file. I then proceed to find the periodogram
> >PSD estimate of that 'frame'. In the following code 'x' is the vector
> >containing the 512 coefficients:
>
> >h = spectrum.periodogram('hann');
> >hopts = psdopts(h,x);  % Default options based on the signal x
> >set(hopts,'NFFT',512,'Fs',Fs,'SpectrumType','onesided');
> >psd(h,x,hopts);
>
> >However when I view the properties of the object by:
>
> >objproperties = psd(h,x,hopts)
>
> >the length of the Data vector and (obviously) frequency vector is 257,
> >rather than the expected 256. I'm using MatLab 7.5.0.
> >Hope someone can clarify this for me. Thanks :)
>
> The length of 257 is correct-- An N point DFT of a real signal returns N/2
> + 1 values.  Here is a link on the topic.
> Regards,
> Steve
>
> http://www.dspguide.com/ch12/1.htm

The N point DFT of a real signal returns a real value at DC, N/2 - 1
complex values, and an imaginary value at fs/2; a total of N values
from N real samples.

Dale B. Dalrymple
http//:dbdimages.com

On Oct 30, 1:31 pm, Glen Herrmannsfeldt <g...@ugcs.caltech.edu> wrote:

> (snip)
>
>
> This is going to get darn close to the topic of another
> thread related to Nyquist.  Isn't that N/2+1st point the one
> at the Nyquist frequency?
>
> Now, to look at it another way without including the DFT,
> consider sampling a band limited and time limited signal.
>
> If the duration of the signal is N Fs, and the signal is
> not periodic, how many sample points do we need to be able
> to recreate the signal?  IT seems to me that N+1 points
> are needed to properly include the end points.  A signal
> that isn't time limited doesn't have end points.
>
> -- glen

The signal may be time limited. The DFT doesn't care. The DFT is a
circular convolution. It has no end points.

The N + 1 'points' don't represent the same number of values. Some are
single values and some are complex with two values.

Dale B. Dalrymple

dbd wrote:
(snip, I wrote)

>>Now, to look at it another way without including the DFT,
>>consider sampling a band limited and time limited signal.

>>If the duration of the signal is N Fs, and the signal is
>>not periodic, how many sample points do we need to be able
>>to recreate the signal?  IT seems to me that N+1 points
>>are needed to properly include the end points.  A signal
>>that isn't time limited doesn't have end points.

> The signal may be time limited. The DFT doesn't care. The DFT is a
> circular convolution. It has no end points.

> The N + 1 'points' don't represent the same number of values. Some are
> single values and some are complex with two values.

As I said, though, that paragraph was meant not to be related
to the DFT, but to sampling and Nyquist.  For a given length
non-periodic signal how many sample points do you need to
reconstruct a band limited signal?  It seems to me that
for a periodic signal it is N, and for a non-periodic signal
it is N+1 because the ends could have different values.

I was trying to connect the N+1 to the discussion on DFT,
but not to the DFT itself.

-- glen

dbd wrote:
(snip)

> The N point DFT of a real signal returns a real value at DC, N/2 - 1
> complex values, and an imaginary value at fs/2; a total of N values
> from N real samples.

And by symmetry, the DFT of an imaginary signal should return an
imaginary value at DC, N/2-1 complex values, and a real value at fs/2.

I don't believe that there is a mathematical symbol for 'less than
and partially equal to'.

-- glen

On Oct 30, 4:12 pm, Glen Herrmannsfeldt <g...@ugcs.caltech.edu> wrote:
> ...
> For a given length
> non-periodic signal how many sample points do you need to
> reconstruct a band limited signal?  It seems to me that
> for a periodic signal it is N, and for a non-periodic signal
> it is N+1 because the ends could have different values.
> ...
> -- glen

What non-periodic band limited signal can you reconstruct from N+1
sample points?

Doesn't Nyquist deal with bandwidths and rates, not numbers of
samples?

Dale B. Dalrymple

dbd wrote:
(snip)

> What non-periodic band limited signal can you
> reconstruct from N+1 sample points?

> Doesn't Nyquist deal with bandwidths and 
> rates, not numbers of samples?

One that is N divided by the appropriate sampling rate long.

-- glen

On 31 Okt, 00:12, Glen Herrmannsfeldt <g...@ugcs.caltech.edu> wrote:

>&#4294967295;For a given length
> non-periodic signal how many sample points do you need to
> reconstruct a band limited signal?

The question is irrelevant. You don't reconstruct a signal
form its PSD. The PSD lacks phase information.

>&#4294967295;It seems to me that
> for a periodic signal it is N, and for a non-periodic signal
> it is N+1 because the ends could have different values.

When you include the n=0 and n=N/2 coeffcients (N even) to form
an N/2+1 PSD estimate from a real-valued signal, the reason is
that there are N/2+1 distinct coeffcients in the PSD estimate.
The n=0 and n = N/2 coefficients appear one time each, all the
others appear twice.

Returning the n = N/2 coefficient makes perfect sense since
it is needed for a complete decription of the PSD (its relative
magnitude might e.g. tell something about the quality of an anti-
alias filter).

And anyway, it is the analyst's responsibilty, not the implementer
of the PSD estimator routine, to make the decision on whether to
include the n = N/2  coeffcient in this or that analysis,
depending on the context.

Rune