On Mon, 24 Mar 2008 19:51:50 -0700 (PDT), bulegoge@columbus.rr.com
wrote:
(snipped by Lyons)
>
>Actually, I think that it still would miss that sine wave. You have
>shifted the pahse and converted the sinusoidal wave from a pure sine
>wave to a combined sine/cosine - sinusoidal wave. It will still miss
>the sin componant of this new sinusoidal wave, but catch the cosine
>componant of the sinusoidal wave at fs/2.
Hi,
Oh shoot. I totally misinterpretted what
you meant by the word "miss". I thought that
word pertained to time-domain samples.
Now I'm beginning to believe that your phrase
"misses energy that is sinusoidal" actually
means something having to do with a DFT's
frequency-domain samples.
[-Rick-]
Reply by steve●March 25, 20082008-03-25
On Mar 24, 11:35�pm, Jerry Avins <j...@ieee.org> wrote:
> steve wrote:
> > On Mar 23, 10:52 pm, buleg...@columbus.rr.com wrote:
> >> I am trying to build up some flash programs to demonstrate the DFT and
> >> I am pondering something.
>
> >> I am looking at a 16 point DFT and am focusing on the eigth (or ninth,
> >> depending on how you look at it) basis function: �e^-j2pi*8*n/16.
> >> This is also the basis function that is at the folding frequency of
> >> the DFT (half the sample rate).
>
> >> what I am realizing is that the DFT will catch cosine energy at one
> >> half the sampling frequency but that it entirely misses sine energy at
> >> one half the sampling frequency.
>
> >> Any energy that is higher than one half the sampling frequency is
> >> folded back down, and everything lower is correctly caught by the
> >> DFT. �However, this sine energy that is exactly one half the sampling
> >> frequency seems to be more like a blind spot. �It is not seen or
> >> folded at all.
> >> But cosine energy at exactly half the sampling frequency would get
> >> properly caught by the DFT.
>
> >> Any comment?
>
> >> Thanks
>
> >> Brent
>
> > It certainly is a blind spot, as well as the multiples of fs/2 as
> > others have indicated. There are other quirks near the boundaries of
> > the DFT/FFT (DC, fs/2,fs, etc) too that will result in the wrong
> > answer, bin 0 of an FFT will have a different gain then all the other
> > "normal" bins, and the 1/f sampling period required to detect an f Hz
> > signal (time bandwidth product) or even to detect a f Hz difference in
> > two high frequency components (you need to sample for 1 second to
> > distinguish between a 1000000hz signal and a 1000001hz signal). It all
> > comes out in the math, but it's not really highlighted very well in
> > any texts.
>
> Just as you need to sample for about a second to catch a 1Hz component,
> you need to sample the same length of time to resolve 1n Fs/2 - 1 Hz
> component. The high end can be worked around by sampling faster, but the
> low end isn't so amenable.
>
> Jerry
> --
> Engineering is the art of making what you want from things you can get.
> �����������������������������������������������������������������������- Hide quoted text -
>
> - Show quoted text -
I don't follow that, but I'll give it some thought
Reply by Jerry Avins●March 25, 20082008-03-25
steve wrote:
> On Mar 23, 10:52 pm, buleg...@columbus.rr.com wrote:
>> I am trying to build up some flash programs to demonstrate the DFT and
>> I am pondering something.
>>
>> I am looking at a 16 point DFT and am focusing on the eigth (or ninth,
>> depending on how you look at it) basis function: e^-j2pi*8*n/16.
>> This is also the basis function that is at the folding frequency of
>> the DFT (half the sample rate).
>>
>> what I am realizing is that the DFT will catch cosine energy at one
>> half the sampling frequency but that it entirely misses sine energy at
>> one half the sampling frequency.
>>
>> Any energy that is higher than one half the sampling frequency is
>> folded back down, and everything lower is correctly caught by the
>> DFT. However, this sine energy that is exactly one half the sampling
>> frequency seems to be more like a blind spot. It is not seen or
>> folded at all.
>> But cosine energy at exactly half the sampling frequency would get
>> properly caught by the DFT.
>>
>> Any comment?
>>
>> Thanks
>>
>> Brent
>
> It certainly is a blind spot, as well as the multiples of fs/2 as
> others have indicated. There are other quirks near the boundaries of
> the DFT/FFT (DC, fs/2,fs, etc) too that will result in the wrong
> answer, bin 0 of an FFT will have a different gain then all the other
> "normal" bins, and the 1/f sampling period required to detect an f Hz
> signal (time bandwidth product) or even to detect a f Hz difference in
> two high frequency components (you need to sample for 1 second to
> distinguish between a 1000000hz signal and a 1000001hz signal). It all
> comes out in the math, but it's not really highlighted very well in
> any texts.
Just as you need to sample for about a second to catch a 1Hz component,
you need to sample the same length of time to resolve 1n Fs/2 - 1 Hz
component. The high end can be worked around by sampling faster, but the
low end isn't so amenable.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by ●March 24, 20082008-03-24
On Mar 24, 7:09�am, Rick Lyons <R.Lyons@_BOGUS_ieee.org> wrote:
> On Sun, 23 Mar 2008 20:55:36 -0700 (PDT), buleg...@columbus.rr.com
> wrote:
>
> � �(Snipped by Lyons)
>
> >I understood that all the books claimed that it needs to be greater
> >than twice the sampling period. �I am looking at how the DFT uses the
> >basis functions to find cosine enrgy and sine energy at each discrete
> >frequency. I see that it can find the cosine energy up to and
> >including the frequency that is half the sampling frequency, but it
> >totally misses energy that is sinusoidal at half the sampling
> >frequency. �Of course, higher frequencies get folded back down.
>
> >Of course, a frequency component is rarely phased to be a perfect sine
> >or cosine, so in practice you will be messed up if you don't sample at
> >>2 times the highest frequency component.
>
> Hi,
> � The DFT "misses" the sinewave at Fs/2 because the
> time samples of that sinewave are all zero-valued
> samples. �You're performing a DFT on a string of
> zeros! �The DFT "catches" a cosine at Fs/2 because the
> time samples of that cosine are alternating +1 & -1's.
> (Non-zero-valued samples.)
>
> Add a little phase shift to your time-domain Fs/2
> sinewave by changing your sinewave definition from:
>
> � �Sig = sin(2*pi*(Fs/2)*Time) �
>
> � � � �= sin(2*pi*(Fs/2)/Fs) = sin(pi) = 0
>
> to:
>
> � �Sig = sin(2*pi*(Fs/2)/Fs + pi/8).
>
> Now your DFT will not "miss" that sinewave.
>
> But as you probably know, in normal spectrum analysis
> using the DFT we would be VERY* worried if our
> time signal has appreciable signal energy in the
> vicinity of Fs/2 Hz. �Such high-freq energy would
> indicate that our Fs sample rate was too low relative
> to the bandwidth of some sampled analog signal.
>
> Good Luck,
> [-Rick-]
Actually, I think that it still would miss that sine wave. You have
shifted the pahse and converted the sinusoidal wave from a pure sine
wave to a combined sine/cosine - sinusoidal wave. It will still miss
the sin componant of this new sinusoidal wave, but catch the cosine
componant of the sinusoidal wave at fs/2.
Reply by Rick Lyons●March 24, 20082008-03-24
On Mon, 24 Mar 2008 08:24:45 -0700 (PDT), dbell <bellda2005@cox.net>
wrote:
(Snipped by Lyons)
>
>Hi Rick,
>
>If there is a phase shift as you suggest, there will usually be energy
>measured at Fs/2 (ie. if the shift doesn't make the thing a +-sine
>function) , but you still can't tell anything specific about the
>values of amplitude and phase from the samples alone.
>
>IIRC the phone company spec for anti-aliasing filtering is only a
>minimum of 14 dB attenuation (relative to passband) at 4000 Hz = Fs/2
>for Fs=8000 Hz, which isn't all that far down.
>
>Dirk
Hi Dirk,
14 dB of attenuation! Sheece, you're not kiddin'
about that not being too far down considering that
a healthy human ear has a dynamic range of
(I'm guessing) 100 dB.
See Ya',
[-Rick-]
Reply by steve●March 24, 20082008-03-24
On Mar 23, 10:52�pm, buleg...@columbus.rr.com wrote:
> I am trying to build up some flash programs to demonstrate the DFT and
> I am pondering something.
>
> I am looking at a 16 point DFT and am focusing on the eigth (or ninth,
> depending on how you look at it) basis function: �e^-j2pi*8*n/16.
> This is also the basis function that is at the folding frequency of
> the DFT (half the sample rate).
>
> what I am realizing is that the DFT will catch cosine energy at one
> half the sampling frequency but that it entirely misses sine energy at
> one half the sampling frequency.
>
> Any energy that is higher than one half the sampling frequency is
> folded back down, and everything lower is correctly caught by the
> DFT. �However, this sine energy that is exactly one half the sampling
> frequency seems to be more like a blind spot. �It is not seen or
> folded at all.
> But cosine energy at exactly half the sampling frequency would get
> properly caught by the DFT.
>
> Any comment?
>
> Thanks
>
> Brent
It certainly is a blind spot, as well as the multiples of fs/2 as
others have indicated. There are other quirks near the boundaries of
the DFT/FFT (DC, fs/2,fs, etc) too that will result in the wrong
answer, bin 0 of an FFT will have a different gain then all the other
"normal" bins, and the 1/f sampling period required to detect an f Hz
signal (time bandwidth product) or even to detect a f Hz difference in
two high frequency components (you need to sample for 1 second to
distinguish between a 1000000hz signal and a 1000001hz signal). It all
comes out in the math, but it's not really highlighted very well in
any texts.
Reply by ●March 24, 20082008-03-24
On Mar 24, 4:58�pm, "Ron N." <rhnlo...@yahoo.com> wrote:
> On Mar 24, 3:54 am, buleg...@columbus.rr.com wrote:
>
> > The half frequency basis function (sine componant) is all zeros as you
> > stated above.
>
> Not only Fs/2, but at an infinite number of frequencies:
> any integer multiple of Fs/2, including Fs, (3/2)*Fs, etc.
> including 0*Fs/2.
>
> > Whereas, I read a lot about frequencies folding above
> > fs/2 and that the DFT is good (theoretically) up to fs/2, then what is
> > happening at fs/2?
>
> Below Fs/2, the representation is correct, above Fs/2
> the default representation will be wrong (aliased), between
> slightly above and slightly below Fs/2, the answer will be
> in transition between right and wrong, or statistically half
> right (depending on the phase distribution).
>
> IMHO. YMMV.
> --
> rhn A.T nicholson d.0.t C-o-M
Thanks for your insight. It is very helpful.
Reply by Ron N.●March 24, 20082008-03-24
On Mar 24, 3:54 am, buleg...@columbus.rr.com wrote:
> The half frequency basis function (sine componant) is all zeros as you
> stated above.
Not only Fs/2, but at an infinite number of frequencies:
any integer multiple of Fs/2, including Fs, (3/2)*Fs, etc.
including 0*Fs/2.
> Whereas, I read a lot about frequencies folding above
> fs/2 and that the DFT is good (theoretically) up to fs/2, then what is
> happening at fs/2?
Below Fs/2, the representation is correct, above Fs/2
the default representation will be wrong (aliased), between
slightly above and slightly below Fs/2, the answer will be
in transition between right and wrong, or statistically half
right (depending on the phase distribution).
IMHO. YMMV.
--
rhn A.T nicholson d.0.t C-o-M
Reply by dbell●March 24, 20082008-03-24
On Mar 24, 7:09�am, Rick Lyons <R.Lyons@_BOGUS_ieee.org> wrote:
> On Sun, 23 Mar 2008 20:55:36 -0700 (PDT), buleg...@columbus.rr.com
> wrote:
>
> � �(Snipped by Lyons)
>
> >I understood that all the books claimed that it needs to be greater
> >than twice the sampling period. �I am looking at how the DFT uses the
> >basis functions to find cosine enrgy and sine energy at each discrete
> >frequency. I see that it can find the cosine energy up to and
> >including the frequency that is half the sampling frequency, but it
> >totally misses energy that is sinusoidal at half the sampling
> >frequency. �Of course, higher frequencies get folded back down.
>
> >Of course, a frequency component is rarely phased to be a perfect sine
> >or cosine, so in practice you will be messed up if you don't sample at
> >>2 times the highest frequency component.
>
> Hi,
> � The DFT "misses" the sinewave at Fs/2 because the
> time samples of that sinewave are all zero-valued
> samples. �You're performing a DFT on a string of
> zeros! �The DFT "catches" a cosine at Fs/2 because the
> time samples of that cosine are alternating +1 & -1's.
> (Non-zero-valued samples.)
>
> Add a little phase shift to your time-domain Fs/2
> sinewave by changing your sinewave definition from:
>
> � �Sig = sin(2*pi*(Fs/2)*Time) �
>
> � � � �= sin(2*pi*(Fs/2)/Fs) = sin(pi) = 0
>
> to:
>
> � �Sig = sin(2*pi*(Fs/2)/Fs + pi/8).
>
> Now your DFT will not "miss" that sinewave.
>
> But as you probably know, in normal spectrum analysis
> using the DFT we would be VERY* worried if our
> time signal has appreciable signal energy in the
> vicinity of Fs/2 Hz. �Such high-freq energy would
> indicate that our Fs sample rate was too low relative
> to the bandwidth of some sampled analog signal.
>
> Good Luck,
> [-Rick-]
Hi Rick,
If there is a phase shift as you suggest, there will usually be energy
measured at Fs/2 (ie. if the shift doesn't make the thing a +-sine
function) , but you still can't tell anything specific about the
values of amplitude and phase from the samples alone.
IIRC the phone company spec for anti-aliasing filtering is only a
minimum of 14 dB attenuation (relative to passband) at 4000 Hz = Fs/2
for Fs=8000 Hz, which isn't all that far down.
Dirk
Reply by ●March 24, 20082008-03-24
On Mar 24, 7:09�am, Rick Lyons <R.Lyons@_BOGUS_ieee.org> wrote:
> On Sun, 23 Mar 2008 20:55:36 -0700 (PDT), buleg...@columbus.rr.com
> wrote:
>
> � �(Snipped by Lyons)
>
> >I understood that all the books claimed that it needs to be greater
> >than twice the sampling period. �I am looking at how the DFT uses the
> >basis functions to find cosine enrgy and sine energy at each discrete
> >frequency. I see that it can find the cosine energy up to and
> >including the frequency that is half the sampling frequency, but it
> >totally misses energy that is sinusoidal at half the sampling
> >frequency. �Of course, higher frequencies get folded back down.
>
> >Of course, a frequency component is rarely phased to be a perfect sine
> >or cosine, so in practice you will be messed up if you don't sample at
> >>2 times the highest frequency component.
>
> Hi,
> � The DFT "misses" the sinewave at Fs/2 because the
> time samples of that sinewave are all zero-valued
> samples. �You're performing a DFT on a string of
> zeros! �The DFT "catches" a cosine at Fs/2 because the
> time samples of that cosine are alternating +1 & -1's.
> (Non-zero-valued samples.)
>
> Add a little phase shift to your time-domain Fs/2
> sinewave by changing your sinewave definition from:
>
> � �Sig = sin(2*pi*(Fs/2)*Time) �
>
> � � � �= sin(2*pi*(Fs/2)/Fs) = sin(pi) = 0
>
> to:
>
> � �Sig = sin(2*pi*(Fs/2)/Fs + pi/8).
>
> Now your DFT will not "miss" that sinewave.
>
> But as you probably know, in normal spectrum analysis
> using the DFT we would be VERY* worried if our
> time signal has appreciable signal energy in the
> vicinity of Fs/2 Hz. �Such high-freq energy would
> indicate that our Fs sample rate was too low relative
> to the bandwidth of some sampled analog signal.
>
> Good Luck,
> [-Rick-]
Thanks for the reply.
The half frequency basis function (sine componant) is all zeros as you
stated above. Whereas, I read a lot about frequencies folding above
fs/2 and that the DFT is good (theoretically) up to fs/2, then what is
happening at fs/2?
I see that it is a blind spot. It misses the sine componants.
As an example, if you had a time function that had discreet
frequencies up to 7 Hz and samples the wave form at 16 Hz and did a 16
point DFT.
If you added a perfect sine comonant at 8hz, regardless of the
amplitude of the signal you would get the exact same answer for the
DFT.
This blind spot is subtle.