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| Author |
Message |
CW
Guest
|
 Posted: Fri Dec 15, 2006 7:18 pm Post subject: sampling theorem with dirac |
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I'm confused
Say i have a signal
model that looks like
r(t) = s(t) * h(t) where * is convolution and h(t)
is a filter. I then
sample the signal by multiplication with a dirac p(t) =
sum_n
delta(t-nT), where delta is the dirac delta function. So according to
Oppenhiem's book i have
rp(t) = r(t)p(t)
rp(t) = r(t) sum_n
delta(t-nT)
so substitute in for r(t) (because i want to express rp(t)
in terms of
the signal s(t)) we have
rp(t) = int s(t-tau) h(tau)
dtau sum_n delta(t-nT)
but what can i do with my delta function now? i
know that really r(t) =
r(nT) but that means that my multiplication with the
dirac function
(p(t)) has suddenly transformed from a time domain
multiplication to a
time domain convolution...does this make any sense?
CW |
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Rune Allnor
Guest
|
| Posted: Fri Dec 15, 2006 8:58 pm Post subject: Re: sampling theorem with dirac |
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|
CW skrev:
| Quote: |
I'm confused
Say i have a signal model that looks like
r(t) = s(t) * h(t) where * is convolution and h(t) is a filter. I then
sample the signal by multiplication with a dirac p(t) = sum_n
delta(t-nT), where delta is the dirac delta function.
|
No, you don't. Your
sampled signal is a sum of convolutions:
x[n] = sum_{n} integral
d(t-nT)x(t) dt
| Quote: |
So according to
Oppenhiem's book i have
rp(t) =
r(t)p(t)
rp(t) = r(t) sum_n delta(t-nT)
|
Then Oppenheim is wrong
[*].
| Quote: |
so substitute in for r(t) (because i want to express rp(t) in
terms of
the signal s(t)) we have
rp(t) = int s(t-tau) h(tau) dtau
sum_n delta(t-nT)
but what can i do with my delta function now? i know
that really r(t) =
r(nT) but that means that my multiplication with the
dirac function
(p(t)) has suddenly transformed from a time domain
multiplication to a
time domain convolution...does this make any sense?
|
I think the whole
discussion is plain wrong.
I don't know why Oppenheim might have
introduced the
multiplication of Diracs in the first place; possibly to
postpone certain philosophical discussions of continuous
functions vs
discrete series.
Continuous functions and discrete series are two
completely
different types of mathematical objects. The discrete series
x[n] has no value for non-integer n. The value is not 0. It does
not
exist. The continuous function x(t) is defined for all t.
The mathematical
function
x_m = integral x(t) d(t-mT) dt (1)
extracts the
instantaneous value of x(t) at time t=mT. That's
all there is to it. All you
need to form a discrete time sequence
is the well-defined Kronecker Delta
function
d[n] = 1, n=0; 0 otherwise. (2)
x[n] = sum_m x_m
d[n-m]. (3)
Dirac's delta function is never used to form the discrete
sequence x[n], it is only used to extract instantaneous
values from
x(t). However, if you insert the integral (1)
for x_m in (2), there are no
established notational tools
to designate the d(t+mT) inside the integral as
*Dirac's*
delta function, and the d[n] outside the integral as
*Kronecker's* delta. This is what Oppenheim either
did not want to get
into, or just didn't understand.
By doing dodging this very relevant
dicussion, however,
Oppenheim introduced other problems regarding the Dirac
functions, that are both avoidable and far more damaging.
Lots of
people have tried to make intuitive sense (as opposed
to formal sense) of
the Dirac function, and lots of people have
failed. Getting these things
wrong can seriously mess up
the understanding of both DSP and maths.
Google for "Airy R. Bean" and get an impression of what
I mean.
Rune
[*] The Dirac function belongs to a class of functions
that are known as "distributions", which is a concept
one does not
encounter very often. The only definition
I have seen of the concept of a
"distribution" states that
"A distribution is a function that ONLY
appears as part of
an integrand." (Papoulis' book on the Fourier integral,
1961).
Any occurence of the Dirac function OUTSIDE and integral
sign
is plain wrong, according to this definition. |
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CW
Guest
|
| Posted: Fri Dec 15, 2006 11:50 pm Post subject: Re: sampling theorem with dirac |
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Rune Allnor wrote:
| Quote: |
CW skrev:
I'm confused
Say i have a signal model
that looks like
r(t) = s(t) * h(t) where * is convolution and h(t) is a
filter. I then
sample the signal by multiplication with a dirac p(t) = sum_n
delta(t-nT), where delta is the dirac delta function.
No, you don't.
Your sampled signal is a sum of convolutions:
x[n] = sum_{n} integral
d(t-nT)x(t) dt
|
I'm
sorry, but this doesn't make sense at the moment. I'm not trying
to be
difficult, i'd just like to learn. I've read the rest of your
email and i
think my sticking points (which Oppenheim/Wilsky/Young's
book shows and also
http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem
the section just prior to 'Concise summary of the mathematical proof'
shows as well) is the fact that a sampled continuous time signal is
given as
using your notation now
xs(t) = xc(t) . sum_n
d(t-nT)
xs(t) = sampled continuous time signal (which sounds like an
oxymoron
to me)
xc(t) = continuous time signal
first of all, do
you agree with this?
if you're saying that i'm starting on the wrong
thread here, then how
does it fit so neatly when i want to perfectly
reconstruct xc(t) by
doing the following
x_rec(t) = int xs(t-tau)
h(tau) dtau
x_rec(t) = int xc(t-tau) sum_n d(t-nT-tau) h(tau) dtau
int d(t-nT-tau) h(tau) = h(t-nT)
x_rec(t) = sum_n x(nT) h(t-nT)
where h(t) = sinc(t/T). This is probably where i'm going wrong.
CW
| Quote: |
So according to
Oppenhiem's book i have
rp(t) =
r(t)p(t)
rp(t) = r(t) sum_n delta(t-nT)
Then Oppenheim is wrong
[*].
so substitute in for r(t) (because i want to express rp(t) in terms
of
the signal s(t)) we have
rp(t) = int s(t-tau) h(tau) dtau sum_n
delta(t-nT)
but what can i do with my delta function now? i know that
really r(t) =
r(nT) but that means that my multiplication with the dirac
function
(p(t)) has suddenly transformed from a time domain multiplication
to a
time domain convolution...does this make any sense?
I think the
whole discussion is plain wrong.
I don't know why Oppenheim might have
introduced the
multiplication of Diracs in the first place; possibly to
postpone certain philosophical discussions of continuous
functions vs
discrete series.
Continuous functions and discrete series are two
completely
different types of mathematical objects. The discrete series
x[n] has no value for non-integer n. The value is not 0. It does
not
exist. The continuous function x(t) is defined for all t.
The mathematical
function
x_m = integral x(t) d(t-mT) dt (1)
extracts the
instantaneous value of x(t) at time t=mT. That's
all there is to it. All you
need to form a discrete time sequence
is the well-defined Kronecker Delta
function
d[n] = 1, n=0; 0 otherwise. (2)
x[n] = sum_m x_m
d[n-m]. (3)
Dirac's delta function is never used to form the discrete
sequence x[n], it is only used to extract instantaneous
values from
x(t). However, if you insert the integral (1)
for x_m in (2), there are no
established notational tools
to designate the d(t+mT) inside the integral as
*Dirac's*
delta function, and the d[n] outside the integral as
*Kronecker's* delta. This is what Oppenheim either
did not want to get
into, or just didn't understand.
By doing dodging this very relevant
dicussion, however,
Oppenheim introduced other problems regarding the Dirac
functions, that are both avoidable and far more damaging.
Lots of
people have tried to make intuitive sense (as opposed
to formal sense) of
the Dirac function, and lots of people have
failed. Getting these things
wrong can seriously mess up
the understanding of both DSP and maths.
Google for "Airy R. Bean" and get an impression of what
I mean.
Rune
[*] The Dirac function belongs to a class of functions
that are known as "distributions", which is a concept
one does not
encounter very often. The only definition
I have seen of the concept of a
"distribution" states that
"A distribution is a function that ONLY
appears as part of
an integrand." (Papoulis' book on the Fourier integral,
1961).
Any occurence of the Dirac function OUTSIDE and integral
sign
is plain wrong, according to this definition. |
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CW
Guest
|
| Posted: Sat Dec 16, 2006 12:08 am Post subject: Re: sampling theorem with dirac |
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|
Rune Allnor wrote:
| Quote: |
CW skrev:
I'm confused
Say i have a signal model
that looks like
r(t) = s(t) * h(t) where * is convolution and h(t) is a
filter. I then
sample the signal by multiplication with a dirac p(t) = sum_n
delta(t-nT), where delta is the dirac delta function.
No, you don't.
Your sampled signal is a sum of convolutions:
x[n] = sum_{n} integral
d(t-nT)x(t) dt
|
I'm
sorry, but this doesn't make sense at the moment. I'm not trying
to be
difficult, i'd just like to learn. I've read the rest of your
email and i
think my sticking points (which Oppenheim/Wilsky/Young's
book shows and also
http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem
the section just prior to 'Concise summary of the mathematical proof'
shows as well) is the fact that a sampled continuous time signal is
given as
using your notation now
xs(t) = xc(t) . sum_n
d(t-nT)
xs(t) = sampled continuous time signal (which sounds like an
oxymoron
to me)
xc(t) = continuous time signal
first of all, do
you agree with this?
if you're saying that i'm starting on the wrong
thread here, then how
does it fit so neatly when i want to perfectly
reconstruct xc(t) by
doing the following
x_rec(t) = int xs(t-tau)
h(tau) dtau
x_rec(t) = int xc(t-tau) sum_n d(t-nT-tau) h(tau) dtau
int d(t-nT-tau) h(tau) = h(t-nT)
x_rec(t) = sum_n x(nT) h(t-nT)
where h(t) = sinc(t/T). This is probably where i'm going wrong.
CW
| Quote: |
So according to
Oppenhiem's book i have
rp(t) =
r(t)p(t)
rp(t) = r(t) sum_n delta(t-nT)
Then Oppenheim is wrong
[*].
so substitute in for r(t) (because i want to express rp(t) in terms
of
the signal s(t)) we have
rp(t) = int s(t-tau) h(tau) dtau sum_n
delta(t-nT)
but what can i do with my delta function now? i know that
really r(t) =
r(nT) but that means that my multiplication with the dirac
function
(p(t)) has suddenly transformed from a time domain multiplication
to a
time domain convolution...does this make any sense?
I think the
whole discussion is plain wrong.
I don't know why Oppenheim might have
introduced the
multiplication of Diracs in the first place; possibly to
postpone certain philosophical discussions of continuous
functions vs
discrete series.
Continuous functions and discrete series are two
completely
different types of mathematical objects. The discrete series
x[n] has no value for non-integer n. The value is not 0. It does
not
exist. The continuous function x(t) is defined for all t.
The mathematical
function
x_m = integral x(t) d(t-mT) dt (1)
extracts the
instantaneous value of x(t) at time t=mT. That's
all there is to it. All you
need to form a discrete time sequence
is the well-defined Kronecker Delta
function
d[n] = 1, n=0; 0 otherwise. (2)
x[n] = sum_m x_m
d[n-m]. (3)
Dirac's delta function is never used to form the discrete
sequence x[n], it is only used to extract instantaneous
values from
x(t). However, if you insert the integral (1)
for x_m in (2), there are no
established notational tools
to designate the d(t+mT) inside the integral as
*Dirac's*
delta function, and the d[n] outside the integral as
*Kronecker's* delta. This is what Oppenheim either
did not want to get
into, or just didn't understand.
By doing dodging this very relevant
dicussion, however,
Oppenheim introduced other problems regarding the Dirac
functions, that are both avoidable and far more damaging.
Lots of
people have tried to make intuitive sense (as opposed
to formal sense) of
the Dirac function, and lots of people have
failed. Getting these things
wrong can seriously mess up
the understanding of both DSP and maths.
Google for "Airy R. Bean" and get an impression of what
I mean.
Rune
[*] The Dirac function belongs to a class of functions
that are known as "distributions", which is a concept
one does not
encounter very often. The only definition
I have seen of the concept of a
"distribution" states that
"A distribution is a function that ONLY
appears as part of
an integrand." (Papoulis' book on the Fourier integral,
1961).
Any occurence of the Dirac function OUTSIDE and integral
sign
is plain wrong, according to this definition. |
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robert bristow-johnson
Guest
|
| Posted: Sat Dec 16, 2006 12:40 am Post subject: Re: sampling theorem with dirac |
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|
CW wrote:
| Quote: |
Rune Allnor wrote:
CW skrev:
I'm confused
Say i
have a signal model that looks like
r(t) = s(t) * h(t) where * is
convolution and h(t) is a filter. I then
sample the signal by multiplication
with a dirac p(t) = sum_n
delta(t-nT), where delta is the dirac delta
function.
No, you don't. Your sampled signal is a sum of convolutions:
x[n] = sum_{n} integral d(t-nT)x(t) dt
I'm sorry, but this
doesn't make sense at the moment. I'm not trying
to be difficult, i'd just
like to learn. I've read the rest of your
email and i think my sticking
points (which Oppenheim/Wilsky/Young's
book shows and also
http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem
the section just prior to 'Concise summary of the mathematical proof'
shows as well) |
i have
to confess that the current version of the section "Mathematical
basis for
the theorem" is mostly my doing (but there was some editing
and compromise
from other editors in doing that). can you, using the
notation of that
Wikipedia article (and section) restate your question?
judging from
Rune's response, this might be another case of "the
dirac-delta function is
not a function but a distribution" argument,
which i don't want to get very
far into (i just treat it like a
function).
i'll re-read your
original post and see if i can decode its meaning.
r b-j |
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Rune Allnor
Guest
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| Posted: Sat Dec 16, 2006 12:57 am Post subject: Re: sampling theorem with dirac |
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|
CW skrev:
| Quote: |
Rune Allnor wrote:
CW skrev:
I'm confused
Say i
have a signal model that looks like
r(t) = s(t) * h(t) where * is
convolution and h(t) is a filter. I then
sample the signal by multiplication
with a dirac p(t) = sum_n
delta(t-nT), where delta is the dirac delta
function.
No, you don't. Your sampled signal is a sum of convolutions:
x[n] = sum_{n} integral d(t-nT)x(t) dt
I'm sorry, but this
doesn't make sense at the moment. I'm not trying
to be difficult, i'd just
like to learn. I've read the rest of your
email and i think my sticking
points (which Oppenheim/Wilsky/Young's
book shows and also
http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem
the section just prior to 'Concise summary of the mathematical proof'
shows as well) is the fact that a sampled continuous time signal is
given as
using your notation now
xs(t) = xc(t) . sum_n
d(t-nT)
xs(t) = sampled continuous time signal (which sounds like an
oxymoron
to me)
xc(t) = continuous time signal
first of all, do
you agree with this? |
No.
You are drifting towards my position when you think that
a "sampled
continuous time signal" is an oxymoron. I think
it is. In my world, a
sampled signal is discrete, not continuous.
| Quote: |
if you're saying that i'm starting on the wrong thread here,
then how
does it fit so neatly when i want to perfectly reconstruct xc(t) by
doing the following |
The discrete samples are the coefficients for the sinc
reconstruction. When done correctly, exactly one sinc
contributes to the
reconstructed signal, xr(t), at t = nT.
Note that just as x[n] is undefined
between samples,
the exact nature of xr(qT) is undefined for non-integer q.
That's the essence of Nyquist's sampling theorem,
one can make xr(t)
well-behaved by contricitng the
bandwidth of x(t).
Rune |
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Rune Allnor
Guest
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| Posted: Sat Dec 16, 2006 1:00 am Post subject: Re: sampling theorem with dirac |
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robert bristow-johnson skrev:
| Quote: |
judging from Rune's response, this might be another case of "the
dirac-delta function is not a function but a distribution" argument,
which i don't want to get very far into (i just treat it like a
function). |
Don't get
into distributions. I was merely trying to make the
point that there are two
different types of deltas involved,
Dirac's and Kronecker's. As far as I am
concerned,
keeping track of which is where is the key to understand
sampling.
Rune |
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CW
Guest
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| Posted: Sat Dec 16, 2006 1:06 am Post subject: Re: sampling theorem with dirac |
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|
<snipped out original>
| Quote: |
i have to confess that the current version of the section
"Mathematical
basis for the theorem" is mostly my doing (but there was some
editing
and compromise from other editors in doing that). can you, using the
notation of that Wikipedia article (and section) restate your question?
judging from Rune's response, this might be another case of "the
dirac-delta function is not a function but a distribution" argument,
which i don't want to get very far into (i just treat it like a
function).
i'll re-read your original post and see if i can decode
its meaning.
r b-j |
Robert, I wasn't trying to point out inconsistencies, just
trying to
understand. This is one of these annoying ones where i know that i
know this stuff (how hard is sampling after all) I just can't spot the
errors in my progression through the equations...here goes...
I'm
following the progression in this fascinating paper http://bigwww.epfl.ch/publications/unser0001.pdf
which
consists of an input signal f(t) going through a filter h(t) and
then
sampled and then going through another 'reconstruction' filter,
say, g(t)
f(t) r(t) rs(t) f_rec(t)
-->| h(t) |-->X-----> | g(t)
|----->
^
| p(t)
my aim is to look at the signal in the
different parts of fig2
so, replacing letting t=x in that paper and
define the output of the
first filter as r(t), we have
r(t) = int
f(t-tau) h(tau) dtau (1)
Now, we multiply by the delta sampling function
d(t). Now, according
to Oppenheim/Willsky (sigs and Systems) et al, if i
perform this
multiplication, i can write
rs(t) = r(t) . p(t) (2)
where p(t) = sum_n d(t-nT), and rs(t) is the continous time sampled
signal. Subbing (1) into (2) and using p(t) we have
rs(t) = int
f(t-tau) h(tau) dtau . sum_n d(t-nT)
Now, what i'd like to do is verify
that this is correct up to here,
before i continue, because this is where
things start to fall to pieces
for me...my ultimate goal is to express the
reconstructed signal
f_rec(t) after it has been post filtered by g(t) as a
function of the
input signal f(t)....so, how do i look so far??
CW |
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Randy Yates
Guest
|
| Posted: Sat Dec 16, 2006 1:08 am Post subject: Re: sampling theorem with dirac |
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|
"CW" <prada_white@yahoo.ca> writes:
| Quote: |
I'm confused
Say i have a signal model that looks like
r(t) = s(t) * h(t) where * is convolution and h(t) is a filter. I then
sample the signal by multiplication with a dirac p(t) = sum_n
delta(t-nT), where delta is the dirac delta function. So according to
Oppenhiem's book i have
rp(t) = r(t)p(t)
rp(t) = r(t) sum_n
delta(t-nT)
so substitute in for r(t) (because i want to express rp(t)
in terms of
the signal s(t)) we have
rp(t) = int s(t-tau) h(tau)
dtau sum_n delta(t-nT)
but what can i do with my delta function now?
|
Anything you want. That's
a reasonable result (given that we allow
Dirac delta functions out in the
open like this, which I'm OK with).
At the risk of confusing even
further, what is your goal? Are you trying
to follow Oppenheim's derivation?
Are you trying to apply this to some
problem?
--
% Randy Yates %
"She tells me that she likes me very much,
%% Fuquay-Varina, NC % but when I
try to touch, she makes it
%%% 919-577-9882 % all too clear."
%%%%
<yates@ieee.org> % 'Yours Truly, 2095', *Time*, ELO http://home.earthlink.net/~yatescr |
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Rune Allnor
Guest
|
| Posted: Sat Dec 16, 2006 1:12 am Post subject: Re: sampling theorem with dirac |
|
|
CW skrev:
| Quote: |
Rune Allnor wrote:
CW skrev:
I'm confused
Say i
have a signal model that looks like
r(t) = s(t) * h(t) where * is
convolution and h(t) is a filter. I then
sample the signal by multiplication
with a dirac p(t) = sum_n
delta(t-nT), where delta is the dirac delta
function.
No, you don't. Your sampled signal is a sum of convolutions:
x[n] = sum_{n} integral d(t-nT)x(t) dt
I'm sorry, but this
doesn't make sense at the moment. |
That's because it doesn't, you should forget about it.
I
started out writing quick'n dirty answers, but ended
up making a linger
argument. The above is inconsistent
with the conclusions of the longer
"philosophical"
discussion. Forget about the expression above.
Kronecker's delta is missing.
Rune |
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CW
Guest
|
| Posted: Sat Dec 16, 2006 1:12 am Post subject: Re: sampling theorem with dirac |
|
|
Rune Allnor wrote:
| Quote: |
CW skrev:
Rune Allnor wrote:
CW skrev:
I'm confused
Say i have a signal model that looks like
r(t) = s(t) * h(t)
where * is convolution and h(t) is a filter. I then
sample the signal by
multiplication with a dirac p(t) = sum_n
delta(t-nT), where delta is the
dirac delta function.
No, you don't. Your sampled signal is a sum of
convolutions:
x[n] = sum_{n} integral d(t-nT)x(t) dt
I'm
sorry, but this doesn't make sense at the moment. I'm not trying
to be
difficult, i'd just like to learn. I've read the rest of your
email and i
think my sticking points (which Oppenheim/Wilsky/Young's
book shows and also
http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem
the section just prior to 'Concise summary of the mathematical proof'
shows as well) is the fact that a sampled continuous time signal is
given as
using your notation now
xs(t) = xc(t) . sum_n
d(t-nT)
xs(t) = sampled continuous time signal (which sounds like an
oxymoron
to me)
xc(t) = continuous time signal
first of all, do
you agree with this?
No. You are drifting towards my position when you
think that
a "sampled continuous time signal" is an oxymoron. I think
it
is. In my world, a sampled signal is discrete, not continuous.
if you're
saying that i'm starting on the wrong thread here, then how
does it fit so
neatly when i want to perfectly reconstruct xc(t) by
doing the following
The discrete samples are the coefficients for the sinc
reconstruction. When done correctly, exactly one sinc
contributes to the
reconstructed signal, xr(t), at t = nT.
Note that just as x[n] is undefined
between samples,
the exact nature of xr(qT) is undefined for non-integer q.
That's the essence of Nyquist's sampling theorem,
one can make xr(t)
well-behaved by contricitng the
bandwidth of x(t).
|
Rune, I appreciate
you taking the time to respond. I agree with
everything that you say in the
above, however (there's always a but), I
really need to see the equation for
this one. What you say is true -
and i know it's true, but i'm not convinced
when i plug in the
perceived definitions of the various functions. Perhaps
you could
contribute to the other thread that Robert and I have started,
I've
hopefully restated my problem more succinctly there.
Thanks
CW
> Rune |
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Guest
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| Posted: Sat Dec 16, 2006 1:17 am Post subject: Re: sampling theorem with dirac |
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| Quote: |
[*] The Dirac function belongs to a class of functions
that
are known as "distributions", which is a concept
one does not encounter very
often. The only definition
I have seen of the concept of a "distribution"
states that
"A distribution is a function that ONLY appears as part of
an integrand." (Papoulis' book on the Fourier integral, 1961).
Any
occurence of the Dirac function OUTSIDE and integral
sign is plain wrong,
according to this definition. |
Interesting. How would one represent the Fourier transform of
exp(j*w*t) in this case?
Jason |
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CW
Guest
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| Posted: Sat Dec 16, 2006 1:18 am Post subject: Re: sampling theorem with dirac |
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Randy Yates wrote:
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"CW" <prada_white@yahoo.ca> writes:
I'm confused
Say i have a signal model that looks like
r(t) = s(t) * h(t)
where * is convolution and h(t) is a filter. I then
sample the signal by
multiplication with a dirac p(t) = sum_n
delta(t-nT), where delta is the
dirac delta function. So according to
Oppenhiem's book i have
rp(t)
= r(t)p(t)
rp(t) = r(t) sum_n delta(t-nT)
so substitute in for
r(t) (because i want to express rp(t) in terms of
the signal s(t)) we have
rp(t) = int s(t-tau) h(tau) dtau sum_n delta(t-nT)
but what can
i do with my delta function now?
Anything you want. That's a reasonable
result (given that we allow
Dirac delta functions out in the open like this,
which I'm OK with).
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I
hadn't realised that unleashing this beasty was going to be trouble,
but i'm
fine with the definitions also...
| Quote: |
At the risk of confusing even further, what is your goal? Are
you trying
to follow Oppenheim's derivation? Are you trying to apply this to
some
problem? |
well,
i'm following the derivation as described in the thread http://groups.google.com/group/comp.dsp/browse_frm/thread/5001e736cdc9da42/ee878d5eafc1a1da#ee878d5eafc1a1da
and my ultimate goal is to express the reconstructed signal as a
function of the input signal.
Thanks,
CW
| Quote: |
--
% Randy Yates % "She tells me that she likes me very
much,
%% Fuquay-Varina, NC % but when I try to touch, she makes it
%%%
919-577-9882 % all too clear."
%%%% <yates@ieee.org> % 'Yours Truly,
2095', *Time*, ELO
http://home.earthlink.net/~yatescr |
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Rune Allnor
Guest
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| Posted: Sat Dec 16, 2006 1:22 am Post subject: Re: sampling theorem with dirac |
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cincy...@gmail.com skrev:
| Quote: |
[*] The Dirac function belongs to a class of functions
that
are known as "distributions", which is a concept
one does not encounter very
often. The only definition
I have seen of the concept of a "distribution"
states that
"A distribution is a function that ONLY appears as part of
an integrand." (Papoulis' book on the Fourier integral, 1961).
Any
occurence of the Dirac function OUTSIDE and integral
sign is plain wrong,
according to this definition.
Interesting. How would one represent the
Fourier transform of
exp(j*w*t) in this case?
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You'll have to check the
book by Papoulis.
Rune |
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Guest
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| Posted: Sat Dec 16, 2006 1:23 am Post subject: Re: sampling theorem with dirac |
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With what type of reconstruction? If your signal is
bandlimited, you
can theoretically get perfect reconstruction, in which case
your
reconstructed signal = your input signal. |
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