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.... but what is 'sampling rate = 2.5/T'? In my case sampling frequency is Fs=48KHz, Data Rate=4KHz, center freq Fc=12KHz. Help in this respect will be greatly appreciated? Chintan

# Sampling Rate

Started by ●March 24, 2008

Reply by ●March 24, 20082008-03-24

>Hello Everybody > >This is basic question. Generally we say that sampling frequency shouldbe>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

Reply by ●March 24, 20082008-03-24

>>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 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

Reply by ●March 24, 20082008-03-24

> >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 basebandsignal>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 processing takes place channges, as the signal flows from the input to the ouput. That means that your system is multirate, i.e. downsampling and upsampling occurs. If it is mentioned that a signal's sampling rate is 2.5/T, that means that it was sampled 2.5 faster than the signal with sampling rate 1/T. Manolis

Reply by ●March 24, 20082008-03-24

>> >>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 Interpolator->>>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 andupsampling>occurs. > >If it is mentioned that a signal's sampling rate is 2.5/T, that meansthat>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 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

Reply by ●March 24, 20082008-03-24

>*********************** > >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 sineand>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 the sampling frequency of the original signal, because you retain certain samples and the others you throw them away. When you upsample a signal, the new sampling frequency becomes higher because you use interpolation to create additional intermediate samples, which are transmitted faster. Check 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 factor of 2.5 then the new sampling rate is 1/(2.5*T)<1/T. Manolis