sampling theorem with dirac

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

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

> 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.

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?

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

> > 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.

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?

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

> > 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.

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?

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

> > 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.

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?

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

> > 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.

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?

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

> > 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.

CW wrote:
> 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

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

robert bristow-johnson skrev:
> 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