[Top][All Lists]

 From: Martin Braun Subject: Re: [Discuss-gnuradio] sample rate Date: Mon, 06 Oct 2014 15:38:54 +0200 User-agent: Mozilla/5.0 (X11; Linux x86_64; rv:24.0) Gecko/20100101 Thunderbird/24.6.0

```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" <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
>>
>>
>>_______________________________________________