Re: windowing question
- From: "Rune Allnor" <allnor@xxxxxxxxxxxx>
- Date: 25 Apr 2006 11:13:15 -0700
Fred Marshall skrev:
<robert.w.adams@xxxxxxxxxxx> wrote in message
news:1145980027.476608.310590@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Suppose that I have a continous-time signal that consists of a finite
number of Dirac impulses at integer time points. For example, lets say
the signal starts at t=0 and lasts for 100 seconds, with 100 dirac
impulses, one per second.
It's easy to take the Fourier Transform of this signal, since it is
time-limited.
Now lets apply a rectangular window to this signal, such that only 50
seconds of the signal are passed. There are 2 ways to compute the
resulting spectrum;
1) The rule says that multiplication in the time domain is convolution
in the freq domain, so I should be able to convolve the sinc function
of the windowing function with the original spectrum to get the new
spectrum. If the window function is not symettric about 0 then the sinc
function will be complex, but I think the theory still holds.
2) Or I can just simply take the Fourier Transform of the
time-truncated signal (much easier).
These two methids should give the same result.
Now here is the "twist". The window function can "end" at any time
between 50 seconds and 51 seconds. In other words, it doesn't matter if
the window ends at 50.01 seconds or 50.99 seconds; the input signal is
0 in this time-range anyway.
It's obvious that by using method "2" above, you get the correct
spectrum. But using method "1" above, the Fourier Transform of the sinc
function will change slightly depending on the exact length of the
window. But somehow, you have to get the same answer.
Can anyone explain this using calculation method "1" ??
Bob,
Good question!
Consider that the DFT or FFT version treats the time function as periodic -
otherwise the frequency transform wouldn't be discrete.
Wrong version. The way I read the first post, we are talking about
a continuous-time signal that by coincidence happens to contain
a periodic train of pulses. Substitude the Diracs wuth a square
pulse train, and the example is still good, as long as the two
different window limits are kept inside one "0 section" of the train.
But then, the question is a tough one. I'm with RBJ in that
the FT/IFT pair is 1-to-1, so we are faced with one out of
two objectives.
1) Show that the two integrals come up with different FTs
2) Show that the different limits are insignificant as long
as they vary inside a "0 section" of the signal to be
transformed.
I believe the 2nd approach is the one to pursue. If we select
some real numbers such that
0 < a <= b < c <= d
where
0 is the start of the signal
a is the end of the non-zero part of the signal
b is the first FT integration limit
c is the second integration limit
d is the duration of the signal
and x(t) == 0 for t >a. We don't care what the signal looks
like between t == 0 and t == a, only that it is continuous.
My argument is that no matter how we play with the
intergration limits, we end up with an integral that
looks like
integral_0^d x(t) dt = integral_0^a x(t) dt + integral_a^d 0 dt
== integral_0^a x(t) dt
I think this is the key: No matter how you play with the
limits inside the section where x(t) == 0, the non-vanishing
contribution from the integral stays the same.
For What It's Worth,
Rune
.
- References:
- windowing question
- From: robert . w . adams
- Re: windowing question
- From: Fred Marshall
- windowing question
- Prev by Date: Re: Coding Gain
- Next by Date: Re: windowing question
- Previous by thread: Re: windowing question
- Next by thread: Re: windowing question
- Index(es):
Relevant Pages
|
Loading
