Newbie asks "What's in a bin?"

I've been pointed in many helpful directions by many helpful people.
I've gotten to point of "I know I saw that answer "SOMEWHERE" ;]"

A FFT bin has a "width"   (/resolution -- is there a difference?)

Some reference I read commented on "smear" between bins when signal 
frequency was not identically equal to the center frequency of a bin.
[ Comment now if this is in error ]


Assume for sake of argument two cases for a resistive load of 1 Ohm.
1. Pure white noise source of 1 watt/hertz.
2. Pure sine wave of 10W at 1.9856789 Hz.

I can see "1" being predictable for a 10 Hz bandwidth.

What happens for an FFT for a bin width of 6 Hz centered at 4 Hz.

PS. Can someone tell me what the question I'm actually asking is?

Richard Owlett wrote:

> I've been pointed in many helpful directions by many helpful people.
> I've gotten to point of "I know I saw that answer "SOMEWHERE" ;]"
> 
> A FFT bin has a "width"   (/resolution -- is there a difference?)

Nope.  Crucial thing to understand here.  An FFT point is a 
sample of a continuous function of frequency just as the 
time domain point is a sample of a continuous function of time.

> 
> Some reference I read commented on "smear" between bins when signal 
> frequency was not identically equal to the center frequency of a bin.
> [ Comment now if this is in error ]

What happens is that when there are frequencies in the 
sequence being transformed that are not multiples of 1/2T 
then discontinuites occur at the ends which contribute 
across the spectrum and are a kind of smear.

Bob
-- 

"Things should be described as simply as possible, but no 
simpler."

                                              A. Einstein

Bob Cain wrote:
> Richard Owlett wrote:
> 
>> I've been pointed in many helpful directions by many helpful people.
>> I've gotten to point of "I know I saw that answer "SOMEWHERE" ;]"
>>
>> A FFT bin has a "width"   (/resolution -- is there a difference?)
> 
> 
> Nope.  Crucial thing to understand here.  An FFT point is a sample of a 
> continuous function of frequency just as the time domain point is a 
> sample of a continuous function of time.
> 
>>
>> Some reference I read commented on "smear" between bins when signal 
>> frequency was not identically equal to the center frequency of a bin.
>> [ Comment now if this is in error ]
> 
> 
> What happens is that when there are frequencies in the sequence being 
> transformed that are not multiples of 1/2T then discontinuites occur at 
> the ends which contribute across the spectrum and are a kind of smear.
> 
> Bob

Ahhhh, you avoided my question ;{
What happens when signal frequency is *NOT* equal to center of bin?

"Richard Owlett" <rowlett@atlascomm.net> wrote in message
news:102ipt37f2arm65@corp.supernews.com...
> Bob Cain wrote:
> > Richard Owlett wrote:
> >
> >> I've been pointed in many helpful directions by many helpful people.
> >> I've gotten to point of "I know I saw that answer "SOMEWHERE" ;]"
> >>
> >> A FFT bin has a "width"   (/resolution -- is there a difference?)
> >
> >
> > Nope.  Crucial thing to understand here.  An FFT point is a sample of a
> > continuous function of frequency just as the time domain point is a
> > sample of a continuous function of time.
> >
> >>
> >> Some reference I read commented on "smear" between bins when signal
> >> frequency was not identically equal to the center frequency of a bin.
> >> [ Comment now if this is in error ]
> >
> >
> > What happens is that when there are frequencies in the sequence being
> > transformed that are not multiples of 1/2T then discontinuites occur at
> > the ends which contribute across the spectrum and are a kind of smear.
> >
> > Bob
>
> Ahhhh, you avoided my question ;{
> What happens when signal frequency is *NOT* equal to center of bin?

It gets smeared between the two bins that would otherwise be adjacent to it.
ie if you have a bin centred at 4kHz and another at 6kHz and you FFT a 5kHz
signal then the power from the signal will be evenly split between the 4kHz
and 6kHz bins.  If it was closer to 6kHz than 4kHz then the power from the
signal will be primarily shown in the 6kHz bin, the 4kHz will receive
less...
If it's exactly 6kHz then it will entirely fall in the 6kHz bin.

Bevan Weiss wrote:

> "Richard Owlett" <rowlett@atlascomm.net> wrote in message
> news:102ipt37f2arm65@corp.supernews.com...
> 
>>Bob Cain wrote:
>>
>>>Richard Owlett wrote:
>>>
>>>
>>>>I've been pointed in many helpful directions by many helpful people.
>>>>I've gotten to point of "I know I saw that answer "SOMEWHERE" ;]"
>>>>
>>>>A FFT bin has a "width"   (/resolution -- is there a difference?)
>>>
>>>
>>>Nope.  Crucial thing to understand here.  An FFT point is a sample of a
>>>continuous function of frequency just as the time domain point is a
>>>sample of a continuous function of time.
>>>
>>>
>>>>Some reference I read commented on "smear" between bins when signal
>>>>frequency was not identically equal to the center frequency of a bin.
>>>>[ Comment now if this is in error ]
>>>
>>>
>>>What happens is that when there are frequencies in the sequence being
>>>transformed that are not multiples of 1/2T then discontinuites occur at
>>>the ends which contribute across the spectrum and are a kind of smear.
>>>
>>>Bob
>>
>>Ahhhh, you avoided my question ;{
>>What happens when signal frequency is *NOT* equal to center of bin?
> 
> 
> It gets smeared between the two bins that would otherwise be adjacent to it.
> ie if you have a bin centred at 4kHz and another at 6kHz and you FFT a 5kHz
> signal then the power from the signal will be evenly split between the 4kHz
> and 6kHz bins.  If it was closer to 6kHz than 4kHz then the power from the
> signal will be primarily shown in the 6kHz bin, the 4kHz will receive
> less...
> If it's exactly 6kHz then it will entirely fall in the 6kHz bin.
> 
> 

You are about right in what you will see numerically but the 
contents of the "bins" on either side of the signal 
frequency, and all others for that matter, is still a 
consequence of the discontinuities at the end that result 
from such an in-between signal and the true numerical result 
will be a function of the window applied.  If you take the 
cross product of a signal at one pure frequency with that at 
another the result is zero.  The DFT is the result of an 
inner product of all the discrete frequencies in the DFT 
with the signal itself.


Bob
-- 

"Things should be described as simply as possible, but no 
simpler."

                                              A. Einstein

Richard:

I think I have something useful to add here. I was involved in a project
(optical spectrum analyzer) where my job was to *exactly* determine the
center frequency of the signal, based on the FFT'd "spectrum". When the
exact frequency did not lie exactly on a bin, or even if it did, the result
was smeared due to windowing, and the fact that it did not lie exactly on a
bin.

Then, the question that I answered was, how to *exactly* recover the center
frequency of the very narrowband optical signal using the FFT results. At
first, due to input from a PhD who didn't know very much, I was told to just
take the first moment of the FFT feature. However, please note that the
"FFT" results were those from the data presented to the user, which were
after taking the sqr-root. That is, in the "magnitude" domain.

However, we discovered that there was a "sinusoidal error" that varied with
how far the true laser frequency was from a given bin. This led me to the
true answer--the way to derive exact frequency from an FFT result is to take
the first moment of the FFT bins, but in the MAGNITUDE SQUARED domain (i.e.,
raw FFT results). To do it in the magnitude domain leads to sinusoidal
error, MAX at 1/4 distance between bins, min at 1/2 and at bin centers.

So, the answer to your question is that a "bin", in the magnitude-squared
domain, represents a linearly-interpolateble sample of the energy of the
signal at that exact bin frequency.

Hope this helps,
Jim Gort

"Richard Owlett" <rowlett@atlascomm.net> wrote in message
news:102ipt37f2arm65@corp.supernews.com...
> Bob Cain wrote:
> > Richard Owlett wrote:
> >
> >> I've been pointed in many helpful directions by many helpful people.
> >> I've gotten to point of "I know I saw that answer "SOMEWHERE" ;]"
> >>
> >> A FFT bin has a "width"   (/resolution -- is there a difference?)
> >
> >
> > Nope.  Crucial thing to understand here.  An FFT point is a sample of a
> > continuous function of frequency just as the time domain point is a
> > sample of a continuous function of time.
> >
> >>
> >> Some reference I read commented on "smear" between bins when signal
> >> frequency was not identically equal to the center frequency of a bin.
> >> [ Comment now if this is in error ]
> >
> >
> > What happens is that when there are frequencies in the sequence being
> > transformed that are not multiples of 1/2T then discontinuites occur at
> > the ends which contribute across the spectrum and are a kind of smear.
> >
> > Bob
>
> Ahhhh, you avoided my question ;{
> What happens when signal frequency is *NOT* equal to center of bin?
>
>
>
>
>

Richard:

I think I have something useful to add here. I was involved in a project
(optical spectrum analyzer) where my job was to *exactly* determine the
center frequency of the signal, based on the FFT'd "spectrum". When the
exact frequency did not lie exactly on a bin, or even if it did, the result
was smeared due to windowing, and the fact that it did not lie exactly on a
bin.

Then, the question that I answered was, how to *exactly* recover the center
frequency of the very narrowband optical signal using the FFT results. At
first, due to input from a PhD who didn't know very much, I was told to just
take the first moment of the FFT feature. However, please note that the
"FFT" results were those from the data presented to the user, which were
after taking the sqr-root. That is, in the "magnitude" domain.

However, we discovered that there was a "sinusoidal error" that varied with
how far the true laser frequency was from a given bin. This led me to the
true answer--the way to derive exact frequency from an FFT result is to take
the first moment of the FFT bins, but in the MAGNITUDE SQUARED domain (i.e.,
raw FFT results). To do it in the magnitude domain leads to sinusoidal
error, MAX at 1/4 distance between bins, min at 1/2 and at bin centers.

So, the answer to your question is that a "bin", in the magnitude-squared
domain, represents a linearly-interpolateble sample of the energy of the
signal at that exact bin frequency.

Hope this helps,
Jim Gort

"Richard Owlett" <rowlett@atlascomm.net> wrote in message
news:102ipt37f2arm65@corp.supernews.com...
> Bob Cain wrote:
> > Richard Owlett wrote:
> >
> >> I've been pointed in many helpful directions by many helpful people.
> >> I've gotten to point of "I know I saw that answer "SOMEWHERE" ;]"
> >>
> >> A FFT bin has a "width"   (/resolution -- is there a difference?)
> >
> >
> > Nope.  Crucial thing to understand here.  An FFT point is a sample of a
> > continuous function of frequency just as the time domain point is a
> > sample of a continuous function of time.
> >
> >>
> >> Some reference I read commented on "smear" between bins when signal
> >> frequency was not identically equal to the center frequency of a bin.
> >> [ Comment now if this is in error ]
> >
> >
> > What happens is that when there are frequencies in the sequence being
> > transformed that are not multiples of 1/2T then discontinuites occur at
> > the ends which contribute across the spectrum and are a kind of smear.
> >
> > Bob
>
> Ahhhh, you avoided my question ;{
> What happens when signal frequency is *NOT* equal to center of bin?
>
>
>
>
>

Richard Owlett wrote:

> > Bob
> 
> Ahhhh, you avoided my question ;{
> What happens when signal frequency is *NOT* equal to center of bin?

Think of it this way: when you do a DFT you are separating an N length
signal into N channels each of which has a single frequency with a given
magnitude and phase. Each of those frequencies has an integer number of
cycles in the window of length N. If you sum all those channels you
reconstruct the original signal. If your signal has the same frequency
as one of those channels it will show up only in that bin (actually 2 as
there is both negative and positive freq., but for real signals the
negative frequencies can be ignored). If your signal is a single
frequency that is not one of the component frequencies in the DFT, then
it will show up in all bins. Fortunately, most of the energy will show
up in the vicinity of the bin which has the closest frequency. 

-jim


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jim wrote:
> 
> Richard Owlett wrote:
> 
>>What happens when signal frequency is *NOT* equal to center of bin?
> 
> 
> [snip]  Fortunately, most of the energy will show
> up in the vicinity of the bin which has the closest frequency. 
> 
>

That's what I THOUGHT I read.
Solves my problem as I'll looking at how the "average" energy in a 
group of bins varies with time. Average in quotes as I may be able to 
be lazy and do arithmetic rather than RMS.

Richard Owlett <rowlett@atlascomm.net> wrote in message news:<102ikldrj3erpbc@corp.supernews.com>...
> I've been pointed in many helpful directions by many helpful people.
> I've gotten to point of "I know I saw that answer "SOMEWHERE" ;]"
> 
> A FFT bin has a "width"   (/resolution -- is there a difference?)

There is a practical difference between bin width and resolution. 

The bin width is related to the number N of samples in the data sequence 
you Fourier tranform. Each bin width in the DFT is fs/N where fs is the 
sampling frequency. 

It would be very convenient if we could state that the spectrum 
resolution is directly related to the bin width of the DFT, but it 
isn't. The phenomenon is demosntrated by zero padding, i.e. when 
a data sequence of original lengths M is expanded by appending zeros
to make up a sequence of total length N > M. If you compute the DFT
of this zero padded sequence, you will find that the bin width 
is fs/N, but the resolution, when understood as the ability to 
separate different spectrum lines, is related to fs/M. 

> Some reference I read commented on "smear" between bins when signal 
> frequency was not identically equal to the center frequency of a bin.
> [ Comment now if this is in error ]

This is correct.

[...]
> PS. Can someone tell me what the question I'm actually asking is?

You may find it useful to check out "spectral leakage".

Rune