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" <martin.br...@ettus.com> 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
>
>
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