[Top][All Lists]
[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
Re: [Discuss-gnuradio] cycle/period detection of a cyclic/periodic trans
From: |
Firas Abbas |
Subject: |
Re: [Discuss-gnuradio] cycle/period detection of a cyclic/periodic transmitter |
Date: |
Wed, 22 Apr 2009 05:05:45 -0700 (PDT) |
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