information in frequency domain

Sirs,

Conceptually, I am not able to fully understand how time domain information
gets converted into frequency domain.

If the following question is repetitive, please forgive me please.

Anyway, question follows below ...

Whenever the term frequeny components is mentioned, we are normally
referring to something that happens over a period of time (less frequently
- low freqyency; more frequently - high frequency)

With restect to a signal represnted in time domain, let us take a single
point. It appears that one cannot deduce what frequency component a single
point contains unless other points are observed.

For example, if I look at a single point in a time domain, it is not
possble to guess the dc value unless additional points of interest are
observed. Imagine the case where a discrete sine is added with constant DC
value.

Given this, I am having hard time understanding how each frequency
component term in discrete fourier transform can independently extract the
associated frequency values from each of samples. It is not obvious to me.
Can someone please understand this ...

Thanks a lot, Manish

>Sirs,
>
>Conceptually, I am not able to fully understand how time domain
information
>gets converted into frequency domain.
>
>If the following question is repetitive, please forgive me please.
>
>Anyway, question follows below ...
>
>Whenever the term frequeny components is mentioned, we are normally
>referring to something that happens over a period of time (less
frequently
>- low freqyency; more frequently - high frequency)
>
>With restect to a signal represnted in time domain, let us take a single
>point. It appears that one cannot deduce what frequency component a
single
>point contains unless other points are observed.
>
>For example, if I look at a single point in a time domain, it is not
>possble to guess the dc value unless additional points of interest are
>observed. Imagine the case where a discrete sine is added with constant
DC
>value.
>
>Given this, I am having hard time understanding how each frequency
>component term in discrete fourier transform can independently extract
the
>associated frequency values from each of samples. It is not obvious to
me.
>Can someone please understand this ...
>
>Thanks a lot, Manish
>

A simlictic view is this:
in both cases of fft or ifft the relation is that all samples contribute to
produce one sample in the other and for every point.

for example one fft bin set to 1, others set to zeros produces sine wave
all along time domain with frequency decided by bin location.
similarly, one sine wave in time domain produces one point at 1 and also
all other points at zero.

For time signals it is the rate of change that matters. for frequency
domain it is the bin location that decides.

Kadhiem

"manishp" <58525@dsprelated> writes:

> Sirs,
>
> Conceptually, I am not able to fully understand how time domain information
> gets converted into frequency domain.
> [...]

Manish,

Let's use an analogy: 

A point in 2D space can be represented using the standard basis (E =
{(1,0), (0,1)}) or any other basis. Representation of the point
does not change the position of the point - it's the same information no
matter how it's represented.

It's the same thing with a signal. That signal can be represented in the
time domain or the frequency domain. They are two different basis for
representing the same information. 

--Randy

NOTE: The frequency domain (via Fourier transform) does not really span
exactly the same space, but the functions it doesn't cover are fairly
pathological and not encountered in real life as far as I know. See the
"Dirichlet conditions."
-- 
Randy Yates
Digital Signal Labs
http://www.digitalsignallabs.com

On 12/15/12 7:31 AM, manishp wrote:
>
> For example, if I look at a single point in a time domain, it is not
> possible to guess the dc value unless additional points of interest are
> observed.

it's not, if only one point is *observed*.  but if this one point is 
observed to be non-zero, and the other points are observed to be zero, 
that's a different situation.

> Imagine the case where a discrete sine is added with constant DC value.

i'm imagining.

> Given this, I am having hard time understanding how each frequency
> component term in discrete fourier transform can independently extract the
> associated frequency values from each of samples. It is not obvious to me.

it's in the math.  and since it's DFT, the math is pretty simple.  no 
integrals needed.

> Can someone please understand this ...

i dunno if i can understand how you frame this, but i understand the 
issue.  i think so, anyway.


-- 

r b-j                  rbj@audioimagination.com

"Imagination is more important than knowledge."

Sirs,

Thank you. This is very useful.

At the same time, I would like to give more detailed eample to make sure I
understand better.

Taking an example of a discrete sinewave x1[n] = [1,2,3,4,5,6,5,4,3,2,1]
Now assume this contains a DC component of 2 unit amplitude 
x2[n] =[3,4,5,6,7,8,7,6,5,4,3]

Now, if I were to look at only one sample, say x2[3], which is 6.
It is almost impossible to say that this contains a dc component of 2.
Remember that in the case of x1 also x1[5] also contains a value of 6.

Let me know if I am going in the right direction with my question.

THanks, Manish

>Sirs,
>
>Thank you. This is very useful.
>
>At the same time, I would like to give more detailed eample to make sure
I
>understand better.
>
>Taking an example of a discrete sinewave x1[n] = [1,2,3,4,5,6,5,4,3,2,1]
>Now assume this contains a DC component of 2 unit amplitude 
>x2[n] =[3,4,5,6,7,8,7,6,5,4,3]
>
>Now, if I were to look at only one sample, say x2[3], which is 6.
>It is almost impossible to say that this contains a dc component of 2.
>Remember that in the case of x1 also x1[5] also contains a value of 6.
>
>Let me know if I am going in the right direction with my question.
>
>THanks, Manish
>

Welldone, you are now in the right direction. One sample is just a number
and could be used for anything that is countable.

Kadhiem

On Sat, 15 Dec 2012 06:31:00 -0600, manishp wrote:

> Sirs,
> 
> Conceptually, I am not able to fully understand how time domain
> information gets converted into frequency domain.
> 
> If the following question is repetitive, please forgive me please.
> 
> Anyway, question follows below ...
> 
> Whenever the term frequeny components is mentioned, we are normally
> referring to something that happens over a period of time (less
> frequently - low freqyency; more frequently - high frequency)
> 
> With restect to a signal represnted in time domain, let us take a single
> point. It appears that one cannot deduce what frequency component a
> single point contains unless other points are observed.
> 
> For example, if I look at a single point in a time domain, it is not
> possble to guess the dc value unless additional points of interest are
> observed. Imagine the case where a discrete sine is added with constant
> DC value.
> 
> Given this, I am having hard time understanding how each frequency
> component term in discrete fourier transform can independently extract
> the associated frequency values from each of samples. It is not obvious
> to me. Can someone please understand this ...
> 
> Thanks a lot, Manish

You seem to be lumping a bunch of different things into one concept, 
which would explain your struggling.

First, you need to be clear on the fact that the frequency domain isn't a 
separate thing from the time domain: it's just a different way of looking 
at the same thing.

First-and-a-half, depending on how you define things, there may be more 
than one frequency domain: there are four different ways to do a Fourier 
"something": you can take a continuous-time signal defined over all time 
and get a continuous-frequency signal defined over all frequencies, you 
can take a discrete-time signal defined over all time and get a 
continuous-frequency signal that's uniquely defined over one segment of 
the spectrum (2 * pi * radians), you can take a continuous-time signal 
that's uniquely defined over some finite interval and get a discrete-
frequency spectrum over all frequencies, or you can take a discrete-time 
signal that's uniquely defined over some integer number of samples and 
get a discrete-frequency spectrum that's defined over the same integer 
number of samples.

Second, sampling changes the frequency content of the time-domain data 
you're looking at.  Or, depending on how you look at things, it changes 
the time domain you're looking at and renders direct comparisons 
meaningless.

So now let's get to your signal represented in the time domain.  If you 
have a signal in a discrete time domain with N points, then you can only 
have N discrete frequency points (including negative frequencies).  In 
your example, you consider a signal that's sampled exactly once.  The 
frequency-domain representation of this has just one frequency, and f = 0 
(it's a very boring signal that way).  _After_ sampling it is not 
impossible to guess the DC value of the signal.  _After_ sampling, the 
value of the one point _is_ the DC value of the signal.

Therein lies your confusion: it's not the conversion into the frequency 
domain that gives you trouble in the "one sample" case, it's the action 
of the sampling.  When you take a continuous-time signal and sample it 
just once, it aliases _all_ of that signal's energy down to DC.

So now we get to how each bin of a DFT can extract the value of the 
signal's frequency components.  The answer is, for the signal _after 
sampling_, it does so perfectly.  You did not so state, but given the 
wording of your question I gather that your real problem is that you are 
starting with a continuous-time signal of infinite extent, and you want 
to use the DFT to estimate it's spectral content.

So here, in order, is what you are doing:

First, you start with a continuous-time signal of infinite extent.  So 
far, so good.

Second, you sample that signal.  You have now done an approximation: in 
sampling your continuous-time signal you have thrown away an infinite 
amount of information on the spectral content of your original signal.  
This is aliasing.  To make the approximation as accurate (or perhaps 
useful) as possible, you use foreknowledge of the signal to declare that 
its spectral content above Nyquist is insignificant, or you declare that 
its spectral content above Nyquist is useless to know and you filter that 
out.

Third, you truncate the sampled signal.  You have now done _another_ 
approximation: in truncating your signal you have thrown away an infinite 
amount of information about the signal content prior to the start of 
collection, and subsequent to it.  To make the approximation as accurate 
(or as useful) as possible, you use foreknowledge of the signal to 
declare that any chunk of it is the same as any other (i.e., you declare 
the signal to be stationary), or you declare that the signal's non-
stationary nature is immaterial to your measurement and you take steps to 
cancel out the effects of this nonstationary nature (by detrending).  You 
must also take steps to prevent the non-cyclical (or non-wrapping, if 
Rune is listening) nature of the original signal from corrupting your 
results; you do this by windowing.

Fourth, you take the DFT of the signal. THIS IS EXACT.  Yes, the DFT _at 
this point_ is exact -- but it is an exact transform of a signal that is 
a twice-approximated version of what you really wanted information on.

I hope this helps.

-- 
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com

Sirs,

thanks a lot. I can't say I have understood everything but they give me a
direction to think further.

Thanks, Manish

On 12/16/12 12:55 AM, manishp wrote:
> Sirs,
>
> thanks a lot. I can't say I have understood everything but they give me a
> direction to think further.
>
> Thanks, Manish


here's a way to look at it, Manish.  it just two different choices of 
orthogonal basis functions.

first consider the simple Kronecker delta function:

                  { 1    for n = 0
      delta[n] =  {
                  { 0    otherwise

it's the discrete-time unit impulse function.  bang a filter with this, 
and what comes out is the impulse response, h[n].

that impulse happens when n=0, but to generalize, the basis function is

      delta[n-k]

which is an impulse that happens when n=k

so the time domain represents your signal (for N samples) as the sum of 
weighted basis functions

              N-1
     x[n]  =  SUM{ x[k] delta[n-k] }
              k=0

the coefficient for the function in n which is delta[n-k] is x[k] which 
is x[n] evaluated at the constant time, n=k.  x[k] is a constant.


in the frequency domain, the basis function (for the length-N DFT) is

     e^(j*2*pi*k*n/N)

and these basis functions are weighted and summed:

              N-1
     x[n]  =  SUM{ X[k] e^(j*2*pi*k*n/N) }
              k=0

the coefficient the function in n which is e^(j*2*pi*k*n/N) is X[k] 
(this might be scaled by 1/N, depending on the DFT definition).  a 
different basis function and a different constant attached to it.


now there *is* a mathematical difference between the two x[n], but there 
is no difference for 0 <= n < N.  then the two x[n] representations 
agree exactly.  but outside of that interval of n, one representation is 
0 and the other is a periodic extension of what is in between 0 and N. 
we occasionally get into fights here at comp.dsp about whether or not 
this difference is either essential and/or necessary.

i am a strong partisan that it is not essential nor necessary.  and i 
say so by simply redefining the time-domain basis set from this:

     delta[n-k]

for 0 <= k < N

to this:
                   +inf
     comb[n-k]  =  SUM{ delta[n - (k + m N)] }
                  m=-inf

and then, in the time domain

              N-1
     x[n]  =  SUM{ x[k] comb[n-k] }
              k=0

this means that x[n-N] = x[n]  for all n.  x[n] is periodic with period 
N.  and that agrees with the frequency-damain representation above.

is this clear?  just two different sets of basis functions.

-- 

r b-j                  rbj@audioimagination.com

"Imagination is more important than knowledge."

>is this clear?  just two different sets of basis functions.

Thank you very much, Sir