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Re: [Discuss-gnuradio] sample rate

 From: xianda Subject: Re: [Discuss-gnuradio] sample rate Date: Mon, 6 Oct 2014 17:29:29 +0800 (CST)

Hi Martin:
Thanks.But I'm confused about it.
First, data goes through the HPD, which takes out the CP. The outgoing rate of that block is thus
r2 = (N+CP)/N * r

Now r2>r.But according to the downsampling,when the sample become less than before,the sample rate become smaller than before.Now the data goes through the HPD,the sample become less ,but the r2>r?Maybe I'm wrong.Can you help me?Thanks.
I think maybe this:r2=r*N/(N+CP)
Best regards,
xd

At 2014-10-06 17:11:18, "Martin Braun" <address@hidden> wrote: >The question is ill-formed, the concept of a 'sampling rate' does not >really apply here. > >However, to an item rate, you can do the math yourself: > >Assume N is the FFT length, and CP the length of cyclic prefix in >samples. r is the incoming sampling rate. > >First, data goes through the HPD, which takes out the CP. The outgoing >rate of that block is thus > >r2 = (N+CP)/N * r > >Actually, it's output is already OFDM symbols ("FFT-ready"). So, the >output item rate is > >r3 = r2/N > >Then, the packet header is removed. This further reduces the rate, but >let's ignore that. It's also dependent on the packet length. > >Since the FFT doesn't change the rate, what you want is most likely r3. > >M > > >On 10/06/2014 10:36 AM, xianda wrote: >> Hi all: >> Thanks in advance. >> Example:gnuradio/gr-digital/examples/ofdm/rx_ofdm.grc. >> If I set the sample rate of usrp equal to 1Msps,then what is >> the sample rate of the output of the FFT block?Thank you very much. >> Best regards, >> xd > > >_______________________________________________ >Discuss-gnuradio mailing list >address@hidden >https://lists.gnu.org/mailman/listinfo/discuss-gnuradio

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