obtaining continuous time signal bin energy

Randy Yates <yates@ieee.org> writes:
> [...]
> Not at all. The sinusoids I gave were only two out of an infinite
> number of phases, +pi, and -pi, 

Correction: +pi/2 and -pi/2.
-- 
%  Randy Yates                  % "Remember the good old 1980's, when 
%% Fuquay-Varina, NC            %  things were so uncomplicated?"
%%% 919-577-9882                % 'Ticket To The Moon' 
%%%% <yates@ieee.org>           % *Time*, Electric Light Orchestra
http://home.earthlink.net/~yatescr

>> .. Counterexample: x[n] = A*cos(n*pi). This is a "Nyquist frequency"
that is reconstructable for both A = 1 and A = -1.
>> ... Didn't you just contradict yourself?

Well, I did. Let me correct my explanation:

On the Nyquist limit (plus/minus), the FFT resolves the cos() signal, but
it does not have a "match" for a sin() term at the highest frequency. All
of the FFT's rotating phasors are orthogonal to the sin(), meaning that
the  FFT of the sin() is 0.

Sampled at the Nyquist limit, the cos() term for the most negative and the
most positive frequency cannot be separated - they happen to be the same.

Before I send a signal (which represents samples of a continuous-time
waveform) into an FFT, I would need to make a decision:

- Does the [-1, 1, -1, 1, ...] term represent a _backwards_ rotating
phasor? In this case the lowpass filter / reconstruction filter _includes_
the most negative frequency and _excludes_ the most positive one.
- Does it represent a _forward_  rotating phasor instead? In this case,
it's the other way round.
- Does it represent a _real_-valued signal (which is symmetric on the
spectrum)?
In this case it's the sum of both, reconstructed as
1/2(exp(-...t)+exp(...t)) This is the usual assumption.

My own solution: "don't load the FFT to the last bin"....

Cheers

Markus 

On 15 Sep., 00:18, Jerry Avins <j...@ieee.org> wrote:
> Brian Zhang wrote:
>
>    ...
>
> > Besides, I also feel confused about Nyquist Limit.  For a sine wave
> > sampled twice per period, if sampling is started at 1 or -1, we get 1,
> > -1, 1, -1 ..., while if sampling is started at zero crossing, we
> > always get zeros.  I really don't get it.
>
> The Nyquist limit applied to reconstruction is a "not a as great as"
> one. You seem to be familiar with aliasing above Nyquist. *At* the
> Nyquist limit, Fs/2 aliases with DC.

Not quite: Fs gets aliased to DC, Fs/2 stays right there.

For a n-point DFT where n is even, both the DC and the Nyquist bins
have zero imaginary content for real "time-domain" inputs. For
sampling of bandlimited periodic signals, this means that a Nyquist
term is allowed, as long as its phase is zero, as Randy already
pointed out. In this case, perfect and unambiguous reconstruction is
possible.

And you know what that means for the phase shifter: for DC and
Nyquist, shifting the phase by p radians is not distinguishable from
scaling by cos(p) :-))).

Regards,
Andor

Andor <andor.bariska@gmail.com> writes:

> Fs/2 stays right there.

Really? Then why does sin(n*pi) disappear?

Randy Yates <yates@ieee.org> writes:

> p.kootsookos@remove.ieee.org (Peter K.) writes:
> 
> > Randy Yates <yates@ieee.org> writes:
> >
> >> Not true. Counterexample: x[n] = A*cos(n*pi). This is a "Nyquist frequency" that
> >> is reconstructable for both A = 1 and A = -1.
> >> 
> >> Unless I misunderstand you, which is very possible since you use
> >> terminology and language that is very unconventional.
> >> 
> >> What is true, however, is that a sinusoid at the Nyquist frequency
> >> with arbitrary phase cannot be reconstructed.
> >
> > Didn't you just contradict yourself?
> 
> Not at all. The sinusoids I gave were only two out of an infinite
> number of phases, +pi, and -pi, or equivalently, +/- cos(n*pi). 
> 
> I.e., in my example the phases were specific, not arbitrary.

OK, that means that (given your correction) you're saying that

x = -1*cos([0:10]*pi) 

is distinguishable from 

y = -365*cos([0:10]*pi + acos(1/365))

Doesn't it?  

Unless you are using the information that the phase is zero
(i.e. you've got synchronous sampling with the peaks / troughs), I
don't think x and y are distinguishable.

Ciao,

Peter K.

-- 
"And he sees the vision splendid 
of the sunlit plains extended
And at night the wondrous glory of the everlasting stars."

 

Randy Yates <ya...@ieee.org> wrote:
> Andor <andor.bari...@gmail.com> writes:
> > Fs/2 stays right there.
>
> Really? Then why does sin(n*pi) disappear?

That's not aliasing. As I (and you too) said, the phase at Nyquist has
to be zero. Any other phase is interpreted as scaling by cos(p):

sin(n*pi)
= cos(n*pi - pi/2)
= cos(n*pi) cos(-pi/2)
= 0.

Regards,
Andor

Andor <andor.bariska@gmail.com> writes:

> Randy Yates <ya...@ieee.org> wrote:
>> Andor <andor.bari...@gmail.com> writes:
>> > Fs/2 stays right there.
>>
>> Really? Then why does sin(n*pi) disappear?
>
> That's not aliasing. 

I agree - it is not aliasing. But neither is it "staying right there."

I trust we both know what we're talking about, but there is some
semantic problem.
-- 
%  Randy Yates                  % "She's sweet on Wagner-I think she'd die for Beethoven.
%% Fuquay-Varina, NC            %  She love the way Puccini lays down a tune, and
%%% 919-577-9882                %  Verdi's always creepin' from her room." 
%%%% <yates@ieee.org>           % "Rockaria", *A New World Record*, ELO   
http://home.earthlink.net/~yatescr

On Sep 15, 6:53 am, Randy Yates <ya...@ieee.org> wrote:
> Andor <andor.bari...@gmail.com> writes:
> > Fs/2 stays right there.
>
> Really? Then why does sin(n*pi) disappear?

Consider sampling a windowed sinusoid approaching either 0 or
Fs/2.  Within a finite window aperature, the in-phase component
starts looking like a barely modulated constant or alternating
constant, and the out-of-phase component starts approaching
zero, both to bounds close to the noise or quantization level,
until you can't tell the phase any more.  The sin(t) term
disappears because it starts at zero and doesn't have enough
time to inform you of its existence in the given window length.

In the complex FT domain, the real portion of the Sinc wraps
around and adds, the complex portion wraps around and
approaches complete cancellation, due to Hermitian symmetry,
which gives you the same result.


IMHO. YMMV.
--
rhn A.T nicholson d.0.t C-o-M

"Ron N." <rhnlogic@yahoo.com> writes:

> On Sep 15, 6:53 am, Randy Yates <ya...@ieee.org> wrote:
>> Andor <andor.bari...@gmail.com> writes:
>> > Fs/2 stays right there.
>>
>> Really? Then why does sin(n*pi) disappear?
>
> Consider sampling a windowed sinusoid approaching either 0 or
> Fs/2.  

Do you mean "approaching" in frequency? 

> Within a finite window aperature, the in-phase component
> starts looking like a barely modulated constant or alternating
> constant, and the out-of-phase component starts approaching
> zero, 

What do you mean by "in-phase" and "out-of-phase"? We are
performing real sampling.
-- 
%  Randy Yates                  % "She has an IQ of 1001, she has a jumpsuit
%% Fuquay-Varina, NC            %            on, and she's also a telephone."
%%% 919-577-9882                % 
%%%% <yates@ieee.org>           %        'Yours Truly, 2095', *Time*, ELO   
http://home.earthlink.net/~yatescr

On 15 Sep., 20:50, Randy Yates <ya...@ieee.org> wrote:
> Andor <andor.bari...@gmail.com> writes:
> > Randy Yates <ya...@ieee.org> wrote:
> >> Andor <andor.bari...@gmail.com> writes:
> >> > Fs/2 stays right there.
>
> >> Really? Then why does sin(n*pi) disappear?
>
> > That's not aliasing.
>
> I agree - it is not aliasing. But neither is it "staying right there."

Well, it certainly isn't going anywhere either.

>
> I trust we both know what we're talking about, but there is some
> semantic problem.

I'm not quite sure what you are talking about now (I thought before
that I was when you were talking about the zero-phase property of a
Nyquist cosine).

?