Hi Martin:
Thanks.
If I add a block(Vector to Stream) after FFT block.Then the
sample rate become r4=r3*N.
Am I right?Thank you very much.
Best regards,
xd
At 2014-10-06 21:38:54, "Martin Braun" <martin.br...@ettus.com> wrote:
>Obviously I flipped the fraction. It's N/(N+CP).
>
>M
>
>On 10/06/2014 11:29 AM, xianda wrote:
>>
>> 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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>>
>>
>>
>
>
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