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Hi ,
      I found one problem in sampling  a continuous time signal. The signal
frequency is 4000Hz. According to the Shanon's sampling theorem, the
sampling frequency should more than or equal to 8000Hz.

     But, in one telecom applications i found that , the zero resting
samples are missing if i do sampling at 8000Hz( Double the Fs).


  What may be the reson for it?
   How can i avoid the same ?

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"Sai Kumar" <s...@in.bosch.com> wrote in message
news:cecvbe$jjt$1...@ns1.fe.internet.bosch.com...
> Hi ,
>       I found one problem in sampling  a continuous time signal. The
signal
> frequency is 4000Hz. According to the Shanon's sampling theorem, the
> sampling frequency should more than or equal to 8000Hz.

Nope!  The sampling frequency must be *greater than* 8000Hz by theory.
That is, greater than 2*4000 if 4000Hz is the highest frequency present in
"the signal".  You need to be careful of the frequency content of shifts,
transitions, etc.

In practice, it should be more like:

2.5*(the highest frequency) or greater
as you're learning....

Fred

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Sai Kumar wrote:
> Hi ,
>       I found one problem in sampling  a continuous time signal. The signal
> frequency is 4000Hz. According to the Shanon's sampling theorem, the
> sampling frequency should more than or equal to 8000Hz.

Look again.  It has to be greater than 8000Hz - equal is not high enough.
> 
>      But, in one telecom applications i found that , the zero resting
> samples are missing if i do sampling at 8000Hz( Double the Fs).

See?

> 
> 
>   What may be the reson for it?

8000 Hz isn't high enough.

>    How can i avoid the same ?

sample at a higher rate.

-- 
Jim Thomas            Principal Applications Engineer  Bittware, Inc
j...@bittware.com  http://www.bittware.com          (703) 779-7770
Unix IS user friendly.  It's just more particular about its friends.

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"Sai Kumar" <s...@in.bosch.com> wrote in message
news:<cecvbe$jjt$1...@ns1.fe.internet.bosch.com>...
> Hi ,
>       I found one problem in sampling  a continuous time signal. The signal
> frequency is 4000Hz. According to the Shanon's sampling theorem, the
> sampling frequency should more than or equal to 8000Hz.
> 
>      But, in one telecom applications i found that , the zero resting
> samples are missing if i do sampling at 8000Hz( Double the Fs).

I don't understand what you mean by "zero resting samples". 

You want to sample with a sampling rate strictly greater than 8000 Hz, 
with a comfortable margin. Consider the following signal s(t), 

   s(t)= sin(t/(2*pi)) 

wich is a sinusoidal at 1 Hz and is phase aligned relative to t=0. Sampling
the signal with 2 Hz sampling rate, starting at t=0.25, yields the following 
sequence of samples: 

  s[0.25] =  1
  s[0.75] = -1
  s[1.25] =  1
  s[1.75] = -1 
   :

while sampling with 2 Hz sampling rate and starting at t=0 yields sequence

  s[0]   = 0
  s[0.5] = 0
  s[1]   = 0
  s[1.5] = 0
   :

In the first case it is easy to reconstruct the sinusoidal, provided one 
knows that the sequence is sampled at the period maxima. The problem is that 
one never knows that kind of things. In the second case, all information 
kept in the signal is lost regardless of knowledge of sampling alignment.

The Nyquist limit is a strict inequality. If you sample a sinusoidal 
at 2*f, information about the signal is lost, since phase and amplitude 
information can no longer be retrieved from the samples.

Rune

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Two samples per cycle is not enough to represent the amplitude of a
sampled sinusoid. If you have exactly 2 samples per cycle then for a
sine wave of amplitude 'E' you get outputs of the form

[ a  -a  a  -a  a -a ...]

where 'a' is in the range -E<=a<=E. Note that you could get 'a'=0 so
all zero outputs. The actual 'a' will depend on both 'E' and the
sampling times relative to the phase of the sinusoid. Without
additional knowledge, determining 'E' from 'a', or reconstructing the
original waveform, is not possible from 2 samples per period.

You must sample >2x, not >=2, times per period.

Dirk A. Bell


"Sai Kumar" <s...@in.bosch.com> wrote in message
news:<cecvbe$jjt$1...@ns1.fe.internet.bosch.com>...
> Hi ,
>       I found one problem in sampling  a continuous time signal. The signal
> frequency is 4000Hz. According to the Shanon's sampling theorem, the
> sampling frequency should more than or equal to 8000Hz.
> 
>      But, in one telecom applications i found that , the zero resting
> samples are missing if i do sampling at 8000Hz( Double the Fs).
> 
> 
>   What may be the reson for it?
>    How can i avoid the same ?

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Sai Kumar wrote:


>  ... The signal
> frequency is 4000Hz. According to the Shanon's sampling theorem, the
> sampling frequency should more than or equal to 8000Hz.

"Or equal" is a common misunderstanding. You discovered why it cannot
be. Good for you!

Jerry
-- 
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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In article <y...@centurytel.net>,
Fred Marshall <fmarshallx@remove_the_x.acm.org> wrote:
>In practice, [the sampling frequency] should be more like:
>
>2.5*(the highest frequency) or greater
>as you're learning....

Actually, in practice, I've found that 2.5 isn't even that good.  At
least 4X the highest frequency works best for me.

Related to this, is an experiment I did with a 2-pole lowpass
critically-damped Butterworth filter using a sampling frequency
Fs and a cutoff frequency Fc.  It started ringing more than usual
when Fc/Fs >= 1/4, and blew up to infinity when Fc/Fs = 1/2 -- this
occurs due to the angular frequency calculated as tan(pi*Fc/Fs) is
undefined when the argument is pi/2.  That told me that my sampling
frequency needed to be AT LEAST 4 times the cutoff.  8X worked best.

-A

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axlq wrote:

> In article <y...@centurytel.net>,
> Fred Marshall <fmarshallx@remove_the_x.acm.org> wrote:
> 
>>In practice, [the sampling frequency] should be more like:
>>
>>2.5*(the highest frequency) or greater
>>as you're learning....
> 
> 
> Actually, in practice, I've found that 2.5 isn't even that good.  At
> least 4X the highest frequency works best for me.

That depends on what you are trying to accomplish. 4X or 10X for servo 
work. CD audio does well enough with 2.2:1

> Related to this, is an experiment I did with a 2-pole lowpass
> critically-damped Butterworth filter using a sampling frequency
> Fs and a cutoff frequency Fc.  It started ringing more than usual
> when Fc/Fs >= 1/4, and blew up to infinity when Fc/Fs = 1/2 -- this
> occurs due to the angular frequency calculated as tan(pi*Fc/Fs) is
> undefined when the argument is pi/2.  That told me that my sampling
> frequency needed to be AT LEAST 4 times the cutoff.  8X worked best.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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Thank u all,

     Thank u all for ur valuable suggestions

sai kumar
"Jerry Avins" <j...@ieee.org> wrote in message
news:410a8bc1$0$2853$6...@news.rcn.com...
> Sai Kumar wrote:
>
>
> >  ... The signal
> > frequency is 4000Hz. According to the Shanon's sampling theorem, the
> > sampling frequency should more than or equal to 8000Hz.
>
> "Or equal" is a common misunderstanding. You discovered why it cannot
> be. Good for you!
>
> Jerry
> -- 
> Engineering is the art of making what you want from things you can get.
> ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
>

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