Fw: Hello everyone. I am a new here. I have some questions

----- Forwarded Message ----
From: Alex M. [email protected]
To: Tom R. [email protected]
Sent: Friday, March 21, 2008 10:57:21 PM
Subject: Re: [Discuss-gnuradio] Hello everyone. I am a new here. I have
some questions.

Hello Tom !

Thank you for your reply. I have the 2nd edition ( Low Price Edition ).

Here is my question rephrased.

The sifting property for the cts dirac delta fn has an INTEGRAL.
The sifting property for the discrete delta fn has an infinite sum in
it.

So we cannot 1st say as is said in the book that delta is the Dirac
delta function and the coolly proceed and apply the sifting property of
the discrete
delta function as has been done below.

Do you see my question ?

Thank you for your effort,
Ashim.
----- Original Message ----
From: Tom R. [email protected]
To: Alex M. [email protected]
Cc: [email protected]
Sent: Friday, March 21, 2008 10:51:28 PM
Subject: Re: [Discuss-gnuradio] Hello everyone. I am a new here. I have
some questions.

Alex M. wrote:

So I have started reading Discrete Time Signal Processing by Oppenheim
/ Schafer / Buck.

Chapter 4/ Page 168 says : -
Which edition are you using? I have the second edition and this is on
page 142 :slight_smile: I’m assuming your copy is a new edition.
function NOT the continuous Dirac delta function, but just before the
discussion starts it says that delta(t) is the unit impulse function
or the Dirac delta function. Is this a typo ? Shouled this have been
the discrete delta function? If it is’nt, how do the above steps hold?

Thank you,
Alex.

They are still representing all of this in the time domain. The sampling
function is a continuous time signal as is x_c(t) as it x_s(t) where the
discrete time representation is x_s[n] = x_s(nT).

I don’t see this as a typo, and I don’t see how this changes anything.

Tom

Looking for last minute shopping deals? Find them fast with Yahoo!
Search.

  ____________________________________________________________________________________

Be a better friend, newshound, and
know-it-all with Yahoo! Mobile. Try it now.
http://mobile.yahoo.com/;_ylt=Ahu06i62sR8HDtDypao8Wcj9tAcJ

Alex M. wrote:

delta function as has been done below.
Do you see my question ?

Thank you for your effort,
Ashim.

Nope, sorry, I still don’t see the problem. I’m cool with how they
present this in the book. It sounds like your problem is based on the
definition of the Dirac function, and that might be true, but I’ve never
much concerned myself with specifics like that. The outcome of their
math in the sampling process still works.

Tom

This forum is not affiliated to the Ruby language, Ruby on Rails framework, nor any Ruby applications discussed here.

| Privacy Policy | Terms of Service | Remote Ruby Jobs