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

Hi Martin:
```            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
>
>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:
>>          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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