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Re: [Discuss-gnuradio] cycle/period detection of a cyclic/periodic trans

From: kaleem ahmad
Subject: Re: [Discuss-gnuradio] cycle/period detection of a cyclic/periodic transmitter
Date: Wed, 22 Apr 2009 05:19:20 -0700 (PDT)

Thanks Firas,

It is clear and I will implement it and will come back soon hopefully with
successfull results.

Best Regards

Firas A. wrote:
> Hi,
>> On Wed, 4/22/09, kaleem ahmad <address@hidden> wrote:
>> Thanks Firas,
>> But I can simply tell that it always transmit a fix packet (with 
>> bitrate=500kbps) of 1220 micro sec including preamble. It means that 
>> after every T ms (T is repetition cycle time and can be a fixed
>> value from range: 1ms...200ms, I selected 10ms for above given data) it
>> transmits a 1220 microsec long signal.
> If your transmitter works for 1220 microsec (1.22 msec) and it repeat the
> transmission (for example) every 10 msec, then your calculations is wrong.
> If you want to sense the time of a signal, you have to run your FFT frames
> with a rate at least equal to required minimum signal time. For example in
> your case, If the signal to be detected has a 1.22 msec then you have to
> collect the data at rate of 500K (USRP decimation =128). The 512 FFT
> length will be 1.024 msec. This means (after using appropriate spectrum
> threshold value) you need to count number of FFT frames that the desired
> signal is exist in it.
> So, back to our example (signal with duty 1.22msec and repetition of
> 10msec), if data rate is 500k, and you used 512 FFT points, then you will
> see this signal once in every 10 FFT frames. The resolution will be 1.024
> msec. 
> However, if you received the signal with data rate of 8MHz (USRP
> decimation =8), and you computed 512 FFT points then your resolution will
> be 64 usec, which means that the signal will be ON for 19 FFT frames and
> OFF for 137 FFT frames.
> Is that clear ?
> Best Regards,
> Firas
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