sampling question.

Hi all,

In a manuals solution, I see when compute the discrete version with
period T of z(t) where

z(t)=x(t) * h(t), and they write:

z(Tn)=x(Tn) * h(Tn). Where * is convolution,

Based on this, can I refer that the following (1) and (2) are equal?

1.  z(t)=x(t)*h(t)

z(n)=z(nT).


2. x(n)=x(nT), h(n)=h(nT), z(n)=x(n)*h(n).

I mean, is z(n) in (1) is equal to z(n) in (2)?

Thanks

>Hi all,
>
>In a manuals solution, I see when compute the discrete version with
>period T of z(t) where
>
>z(t)=x(t) * h(t), and they write:
>
>z(Tn)=x(Tn) * h(Tn). Where * is convolution,
>
>Based on this, can I refer that the following (1) and (2) are equal?
>
>1.  z(t)=x(t)*h(t)
>
>z(n)=z(nT).
>
>
>2. x(n)=x(nT), h(n)=h(nT), z(n)=x(n)*h(n).
>
>I mean, is z(n) in (1) is equal to z(n) in (2)?
>
>Thanks
>
>

The following is from wikipedia
(http://en.wikipedia.org/wiki/Sample_(signal)):

Let z(t) be a continuous signal which is to be sampled, and that sampling
is performed by measuring the value of the continuous signal every T
seconds. Thus, the sampled signal z[n] is given by

z[n] = z(nT) 

with n = 0,1,2,3,....

The sampling frequency or sampling rate fs is defined as the number of
samples obtained in one second, or fs = 1 / T. The sampling rate is
measured in hertz or in samples per second.

Hope that helps!
John 

On Aug 8, 4:20 pm, "A.E lover" <aelove...@gmail.com> wrote:
> Hi all,
>
> In a manuals solution, I see when compute the discrete version with
> period T of z(t) where
>
> z(t)=x(t) * h(t), and they write:
>
> z(Tn)=x(Tn) * h(Tn). Where * is convolution,
>
> Based on this, can I refer that the following (1) and (2) are equal?
>
> 1.  z(t)=x(t)*h(t)
>
> z(n)=z(nT).

z[n] = z(nT)

the convention in many DSP texts (and a convention that i support
without reservation, i think it should be the convention in all DSP
papers and lit) is that arguments in brackets [n] are integers.  so
z[n] is a discrete sequence, not a continuous-time function.  in the
olden days they would write

    z
     n

("z sub n") but when we had a family of continuous-time signals in a
multi-input or multi-output device (like you would get in state-
variable control systems), then those signals already had subscripts
and we needed a way of preserving notation so that

    z [n]
     1

is the discrete-time version of

    z (t)
     1

where t = n*T.


r b-j

Hi all,

It seems that you guys  still don't understand my question. Sorry for
making your confusion.
Let's notate (*) a convolution operator, * normal multiplying
operator. z(t),x(t), and h(t) signals in continuous time domain,
z[n],x[n],z[n] are signals in discrete time domain.

I have z(t)=x(t) (*) h(t)  all are in continuous time domain.

Now my question is that if I sample z(t) by sampling rate Fs=1/T  in
order to obtain a discrete signal z[n] , can I write

z(n*T)=x(n*T)  (*)  h (n*T)

and therefore  z[n]=x[n] (*) h[n]

In other words, is sampling after convolution equal to sampling first
and then taking convolution?

Please help.

Thanks



On Aug 8, 4:42 pm, robert bristow-johnson <r...@audioimagination.com>
wrote:
> On Aug 8, 4:20 pm, "A.E lover" <aelove...@gmail.com> wrote:
>
> > Hi all,
>
> > In a manuals solution, I see when compute the discrete version with
> > period T of z(t) where
>
> > z(t)=x(t) * h(t), and they write:
>
> > z(Tn)=x(Tn) * h(Tn). Where * is convolution,
>
> > Based on this, can I refer that the following (1) and (2) are equal?
>
> > 1.  z(t)=x(t)*h(t)
>
> > z(n)=z(nT).
>
> z[n] = z(nT)
>
> the convention in many DSP texts (and a convention that i support
> without reservation, i think it should be the convention in all DSP
> papers and lit) is that arguments in brackets [n] are integers.  so
> z[n] is a discrete sequence, not a continuous-time function.  in the
> olden days they would write
>
>     z
>      n
>
> ("z sub n") but when we had a family of continuous-time signals in a
> multi-input or multi-output device (like you would get in state-
> variable control systems), then those signals already had subscripts
> and we needed a way of preserving notation so that
>
>     z [n]
>      1
>
> is the discrete-time version of
>
>     z (t)
>      1
>
> where t = n*T.
>
> r b-j

"A.E lover" <aelover11@gmail.com> writes:

> Hi all,
>
> It seems that you guys  still don't understand my question. Sorry for
> making your confusion.
> Let's notate (*) a convolution operator, * normal multiplying
> operator. z(t),x(t), and h(t) signals in continuous time domain,
> z[n],x[n],z[n] are signals in discrete time domain.
>
> I have z(t)=x(t) (*) h(t)  all are in continuous time domain.
>
> Now my question is that if I sample z(t) by sampling rate Fs=1/T  in
> order to obtain a discrete signal z[n] , can I write
>
> z(n*T)=x(n*T)  (*)  h (n*T)
>
> and therefore  z[n]=x[n] (*) h[n]
>
> In other words, is sampling after convolution equal to sampling first
> and then taking convolution?

Yes, provided that your sample rate is high enough that neither x[n] nor
h[n] are aliased.

However, you are obfuscating the issue somewhat by using two different
notations for the same discrete-time signal: x(n*T) and x[n]. They are
equivalent. The "x[n]" (rather than "x(n*T)") notation is preferable,
in my opinion. 
-- 
%  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

Thanks Randy, however I doubt this statement.

Let's assume X(0)=H(0)=1, where X(f), H(f) are the continuous fourier
transform of x(t) and h(t). (x(t), h(t) are in continuous domain).

I take z(t)=x(t)(*) h(t)  first,
so in frequency domain clearly  Z(0)=X(0)*H(0)= 1*1=1.
then I sample z(t) at Fs to obtain z[n]=z(n/Fs).
In frequency domain of Z(w), a DTFT of z[n], we have Z(w=0)=Fs*1=Fs,
according to the equation of relationship between (DTFT of z[n]) and
(CTFT of z(t).


Now consider the case I sample two continuous function x(t) and h(t)
firstly to obtain x[n] and h[n], then convolute them to obtain
z[n]=x[n](*)h[n].

Similarly, X(w=0)=Fs, H(w=0)=Fs,==> Z(w=0)=X(w=0)*H(w=0)= Fs^2.

I see two ways are not equivalent in frequency domain.

What do you think? or am I missing some points?

Thanks

> Yes, provided that your sample rate is high enough that neither x[n] nor
> h[n] are aliased.
>
> However, you are obfuscating the issue somewhat by using two different
> notations for the same discrete-time signal: x(n*T) and x[n]. They are
> equivalent. The "x[n]" (rather than "x(n*T)") notation is preferable,
> in my opinion.
> --
> %  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."
> %%%% <ya...@ieee.org>           %        'Yours Truly, 2095', *Time*, ELO  http://home.earthlink.net/~yatescr

"A.E lover" <aelover11@gmail.com> writes:

> Thanks Randy, however I doubt this statement.
>
> Let's assume X(0)=H(0)=1, where X(f), H(f) are the continuous fourier
> transform of x(t) and h(t). (x(t), h(t) are in continuous domain).
>
> I take z(t)=x(t)(*) h(t)  first,
> so in frequency domain clearly  Z(0)=X(0)*H(0)= 1*1=1.
> then I sample z(t) at Fs to obtain z[n]=z(n/Fs).
> In frequency domain of Z(w), a DTFT of z[n], we have Z(w=0)=Fs*1=Fs,
> according to the equation of relationship between (DTFT of z[n]) and
> (CTFT of z(t).
>
>
> Now consider the case I sample two continuous function x(t) and h(t)
> firstly to obtain x[n] and h[n], then convolute them to obtain
> z[n]=x[n](*)h[n].
>
> Similarly, X(w=0)=Fs, H(w=0)=Fs,==> Z(w=0)=X(w=0)*H(w=0)= Fs^2.
>
> I see two ways are not equivalent in frequency domain.
>
> What do you think? or am I missing some points?

Your thinking is correct and I am the one who made the error. 

Let xd[n] = x(n*Ts), where Ts = 1 / Fs and Fs is the sample rate. Let
Xd(w) be the DTFT of xd[n] and let X(w) be the continuous-time Fourier
transform of x(t), where w = 2 * pi * f. Then

  Xd(w) = Fs * X(w).

This Fs is the factor I had omitted but which you correctly accounted
for. 

This result follows from modeling sampling as a modulation of the input
signal x(t) with the infinite impulse train. See, e.g., [signalsandsystems].

Therefore the two methods are equivalent EXCEPT for a scale factor of Fs,
as you've already discovered.

--Randy

@BOOK{signalsandsystems,
  title = "{Signals and Systems}",
  author = "{Alan~V.~Oppenheim, Alan~S.~Willsky, with Ian~T.~Young}",
  publisher = "Prentice Hall",
  year = "1983"}

-- 
%  Randy Yates                  % "I met someone who looks alot like you,
%% Fuquay-Varina, NC            %             she does the things you do, 
%%% 919-577-9882                %                     but she is an IBM."
%%%% <yates@ieee.org>           %        'Yours Truly, 2095', *Time*, ELO   
http://home.earthlink.net/~yatescr