Re: [USRP-users] how to know the number of losted samples when overflow occurs ?

2018-06-13 Thread Martin Braun via USRP-users
Matis,

you also need to keep in mind that via USB, we can only receive a
certain amount of samples at a time, and it's not a factor of 16384.

-- M

On 06/08/2018 04:18 PM, Marcus Müller via USRP-users wrote:
> Hi Matis,
> 
> the time stamp of the first buffer after a disruption should indeed be
> accurate, and describe the time at the first sample in that packet.
> So, I'd say that, yes, you can use that to know how much you've lost.
> Thus, let me ask you why you came to the conclusion that it is false?
> 
> Best regards,
> Marcus
> 
> On Sat, 2018-05-05 at 01:34 +0200, Matis Alun via USRP-users wrote:
>> Hi,
>>
>> I remark that the "time_spec" passed to the "recv" method of an
>> rx_streamer is not in
>> exact relation
>> with the length of the received vectors. For example, when receiving
>> buffers of length
>> 16384 at Fs=362319 Hz,
>> the buffer length should be equal to (16384+1)/362319=0.0452225
>> seconds, while the time
>> spec difference
>> between two successive calls to recv is 0.0483315 and sometimes
>> 0.033402 (with a mean
>> value equals to 0.0452225).
>>
>> My first idea was that the time_spec returned by the recv method was
>> the time stamp of the
>> first sample of the
>> buffer, but I realize that it is false. Is it exact ?
>>
>> As a consequence, how can I know the number of losted samples when an
>> overflow occurs ?
>>
>> thanks.
>>
>> matis
>>
>>
>>
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> 
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Re: [USRP-users] how to know the number of losted samples when overflow occurs ?

2018-06-08 Thread Marcus Müller via USRP-users
Hi Matis,

the time stamp of the first buffer after a disruption should indeed be
accurate, and describe the time at the first sample in that packet.
So, I'd say that, yes, you can use that to know how much you've lost.
Thus, let me ask you why you came to the conclusion that it is false?

Best regards,
Marcus

On Sat, 2018-05-05 at 01:34 +0200, Matis Alun via USRP-users wrote:
> Hi,
> 
> I remark that the "time_spec" passed to the "recv" method of an
> rx_streamer is not in
> exact relation
> with the length of the received vectors. For example, when receiving
> buffers of length
> 16384 at Fs=362319 Hz,
> the buffer length should be equal to (16384+1)/362319=0.0452225
> seconds, while the time
> spec difference
> between two successive calls to recv is 0.0483315 and sometimes
> 0.033402 (with a mean
> value equals to 0.0452225).
> 
> My first idea was that the time_spec returned by the recv method was
> the time stamp of the
> first sample of the
> buffer, but I realize that it is false. Is it exact ?
> 
> As a consequence, how can I know the number of losted samples when an
> overflow occurs ?
> 
> thanks.
> 
> matis
> 
> 
> 
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> USRP-users mailing list
> USRP-users@lists.ettus.com
> http://lists.ettus.com/mailman/listinfo/usrp-users_lists.ettus.com

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[USRP-users] how to know the number of losted samples when overflow occurs ?

2018-05-04 Thread Matis Alun via USRP-users
Hi,

I remark that the "time_spec" passed to the "recv" method of an rx_streamer is 
not in
exact relation
with the length of the received vectors. For example, when receiving buffers of 
length
16384 at Fs=362319 Hz,
the buffer length should be equal to (16384+1)/362319=0.0452225 seconds, while 
the time
spec difference
between two successive calls to recv is 0.0483315 and sometimes 0.033402 (with 
a mean
value equals to 0.0452225).

My first idea was that the time_spec returned by the recv method was the time 
stamp of the
first sample of the
buffer, but I realize that it is false. Is it exact ?

As a consequence, how can I know the number of losted samples when an overflow 
occurs ?

thanks.

matis



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