Sampling Rate

[cpshah99]

Hello Everybody

This is basic question. Generally we say that sampling frequency should b
integer multiple of the (1/T) i.e. 2/T,3/T....


but what is 'sampling rate = 2.5/T'?

In my case sampling frequency is Fs=48KHz, Data Rate=4KHz, center fre
Fc=12KHz.

Help in this respect will be greatly appreciated?

Chintan

[Manolis C. Tsakiris]

>Hello Everybody
>
>This is basic question. Generally we say that sampling frequency shoul
be
>integer multiple of the (1/T) i.e. 2/T,3/T....
>

Hello,

the sampling frequency is exactly 1/Ts, namely the reciprocal of th
sampling period Ts with which a continuous time signal is sampled an
converted to a descrite time signal.

What is T in your context?

Manolis

[cpshah99]

>>Hello Everybody
>>
>>This is basic question. Generally we say that sampling frequency should
>be
>>integer multiple of the (1/T) i.e. 2/T,3/T....
>>
>
>Hello,
>
>the sampling frequency is exactly 1/Ts, namely the reciprocal of the
>sampling period Ts with which a continuous time signal is sampled and
>converted to a descrite time signal.
>
>What is T in your context?
>
>Manolis
>

HI

Thank You very much for replying

sorry for my mistake

In my case T is symbol period.

Now, I am designing Linear Interpolator to remove doppler effect.

Now it says that i/p at the Linear Interpolator is complex baseband signa
at sampling rate of 2.5/T. Also I am using DFE to mitigate multipath. Th
system is as below:

received signal -> BP Filter -> Down Converter -> Linear Interpolator -
DFE


Chintan

[Manolis C. Tsakiris]

>
>HI
>
>Thank You very much for replying
>
>sorry for my mistake
>
>In my case T is symbol period.
>
>Now, I am designing Linear Interpolator to remove doppler effect.
>
>Now it says that i/p at the Linear Interpolator is complex baseban
signal
>at sampling rate of 2.5/T. Also I am using DFE to mitigate multipath
The
>system is as below:
>
>received signal -> BP Filter -> Down Converter -> Linear Interpolator ->
>DFE
>
>
>Chintan
>
***************************************
Hi,

well, it is very likely that the sampling rate, in which the processin
takes place channges, as the signal flows from the input to the ouput.
That means that your system is multirate, i.e. downsampling and upsamplin
occurs.

If it is mentioned that a signal's sampling rate is 2.5/T, that means tha
it was sampled 2.5 faster than the signal with sampling rate 1/T.

Manolis

[cpshah99]

>>
>>HI
>>
>>Thank You very much for replying
>>
>>sorry for my mistake
>>
>>In my case T is symbol period.
>>
>>Now, I am designing Linear Interpolator to remove doppler effect.
>>
>>Now it says that i/p at the Linear Interpolator is complex baseband
>signal
>>at sampling rate of 2.5/T. Also I am using DFE to mitigate multipath.
>The
>>system is as below:
>>
>>received signal -> BP Filter -> Down Converter -> Linear Interpolato
->
>>DFE
>>
>>
>>Chintan
>>
>***************************************
>Hi,
>
>well, it is very likely that the sampling rate, in which the processing
>takes place channges, as the signal flows from the input to the ouput.
>That means that your system is multirate, i.e. downsampling an
upsampling
>occurs.
>
>If it is mentioned that a signal's sampling rate is 2.5/T, that mean
that
>it was sampled 2.5 faster than the signal with sampling rate 1/T.
>
>Manolis
>
***********************

Hi Manolis

Thanks again.

I have designed one system where I am not using interpolator.

Now in that, when I am down converting the signal i m generating sine an
cosine wave where t=[0:length(rec_sig)-1]*Ts, where Ts=1/fs=1/48KHz

Now what I am thinking is that sampling rate =2.5/T is that when I dow
convert, my t should be t=[0:length(y)-1]*Ts, where Ts=1/(2.5*48KHz)

Is this correct?

Chintan

[Manolis C. Tsakiris]

>***********************
>
>Hi Manolis
>
>Thanks again.
>
>I have designed one system where I am not using interpolator.
>
>Now in that, when I am down converting the signal i m generating sin
and
>cosine wave where t=[0:length(rec_sig)-1]*Ts, where Ts=1/fs=1/48KHz
>
>Now what I am thinking is that sampling rate =2.5/T is that when I down
>convert, my t should be t=[0:length(y)-1]*Ts, where Ts=1/(2.5*48KHz)
>
>Is this correct?
>
>Chintan
>
****************************************
When you downsample a signal, the new sampling frequency is less than th
sampling frequency of the original signal, because you retain certai
samples and the others you throw them away. When you upsample a signal
the new sampling frequency becomes higher because you use interpolation t
create additional intermediate samples, which are transmitted faster. Chec
out for example the fractional spaced equalizers.

Consequently your statement is not correct.
If you have a signal sampled at 1/T rate and you downsample it by a facto
of 2.5 then the new sampling rate is 1/(2.5*T)<1/T.

Manolis