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Re: [perseus_SDR] Re: Fierce competition

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  • Leif Asbrink
    Hello Nico, ... It is one pixel for each data point of the 2MHz sampling frequency. Yellow is I and magenta is Q. The white is the sum of the squares (power)
    Message 1 of 48 , Jul 9, 2013
      Hello Nico,

      > Now seriously, what is the horizontal scale of the
      > discharge on Fig. 1?
      It is one pixel for each data point of the 2MHz sampling
      frequency. Yellow is I and magenta is Q. The white is the
      sum of the squares (power)

      If Linrad had been properly calibrated for the converter
      that was in front of the Perseus the pulses would have
      been a little narrower.

      > I mean which is the minimum interval between two consecutive
      > pulses in a discharge? which is the total time interval,
      > on average, of the sequence created by a rain drop?
      I have asked for recordings, but there is no interest
      in the HAM community. I only have one single recording.
      I would guess the frequency depends on the size of the
      droplet (capacitance)

      > and which is the time interval between to consecutive
      > drops?
      This I know from the old days. In heavy static rain the
      pulses are randomly distributed and the average rate
      could be 10000 per second and more.

      > A very wide IF bandwidth is needed only if one wants to
      > resolve each pulse in a sequence but this is not strictly
      > required if one wants (or accepts) to suppress its
      > effects just in a narrow bandwidth, i.e. in that of the
      > signal of interest and if the noise bursts are
      > sufficiently apart in time (that's to say the noise is
      > really impulsive).
      But this condition is not met in static rain.
      Look here:
      With details here:
      I used this analog blanker for a decade and it worked beautifully,
      but a requirement was that the bandwidth was several MHz
      to make pulses short enough to allow short duration of
      the gate pulse. (about 0.5 microseconds)

      I did never observe the pulse trains during static rain,
      presumably because I could not see individual traces
      on my old analog oscilloscope. There was just a very large
      number of pulses.

      There is a minimum bandwidth below which the static rain
      behaves like white noise. Then it can not be removed.
      Look at figure 2. Pulses are about 100 times longer
      because the pulse trains are not resolved. In low intensity
      static rain like that figure one can apply a blanker, but
      just 10 times more droplets would make the horizontal signal
      like white noise.
      I think

      > In this case there's no need for high rate processing,
      > the blanker can be operated at a sample rate which is
      > just slightly higher than that strictly required to
      > demodulate the signal as i.e. I showed some months ago.
      > One just needs a narrowband energy detector and a
      > narrowband spectral estimator which reconstructs the
      > effects of the offending noise pulse sequence *only*
      > in the band of interest.
      But this can not be done on white noise.

      > But ok, I agree that if one wants to suppress it
      > everywhere, maybe because he would like to show a
      > clean FFT and discover where the narrowband (and weak)
      > signals really are, there's no other choice then
      > increasing the IF bandwidth so that each individual
      > pulse can be resolved. And it is also easier to
      > implement then the narrowband method.
      I do not agree. If the pulse repetition rate is so low that
      one can separate the pulses in the signal bandwidth,
      a conventional blanker will work perfectly well
      provided that strong signals are notched out from the
      blanker passband.


    • Leif Asbrink
      Hello Nico, ... I am afraid you apply a conventional model which is not applicable in the QRN-fighting context. Consider a sampling rate of 4 MHz. Apply a
      Message 48 of 48 , Jul 18, 2013
        Hello Nico,

        > If one computes the number of taps of a FIR decimation
        > filter with a decent performance (say 0.1 dB in-band ripple
        > and 100 dB alias image rejection) he discover a simple
        > rule of thumb:
        > N =(about) 4*D/(1-B/Fco)
        > where:
        > N is the required decimation filter number of taps
        > D is the decimation factor
        > B/Fco is ratio between the desired output alias free bandwidth and the output sampling frequency.
        > Since after filtering the decimator takes one output every D
        > input samples, the output impulse response is no more
        > than N/D samples long, that's to say:
        > N/D =(about) 4/(1-B/Fco)
        > Note that the length of the output impulse response
        > *does not* depend on the output sampling frequency, but just on the B/Fco ratio.
        > If such a ratio is high the output pulse can be quite long.

        I am afraid you apply a "conventional" model which is
        not applicable in the QRN-fighting context.

        Consider a sampling rate of 4 MHz.
        Apply a FIR filter that has say 0.1 dB in-band ripple
        and a -1 dB point at say 0.8 MHz. The -20 dB point should
        be at 2 MHz and the -100 dB point at 3.2 MHz. The alias-free
        range (-100 dB) would be +/- 0.8 MHz but a clever DSP software
        could compensate for the fall-off between say 0.8 and 1.6 MHz
        to provide a perfectly flat passband of 3.2 MHz or so. The alias
        suppression at the corner frequencies would be poor. Maybe 20 dB,
        but I do not think that would impair the noise-fighting.

        The useful bandwidth for receiving would be 1.6 MHz only and
        not any improvement over the 2 MHz sampling. The purpose of the
        faster sampling would only be to eliminate certain interference
        sources better.

        > In Perseus the decimation filter has been designed so that
        > the alias-free bandwidth is 80% the output sampling frequency
        > (1.6 MHz when the sampling rate is 2 MS/s) which is a good
        > compromise between the decimation filters complexity and
        > the efficiency of the digital signal processing made on the PC.
        > At such a B/Fco ratio you can expect that each output pulse
        > due to an istantaneous glitch at the receiver input is
        > approximately 4/(1-0.8) = 20 samples long whatever the
        > output sampling frequency is.

        > You can't really resolve it into a single pulse even if
        > the output sampling frequency were 40 MS/s. It will
        > always be 20 samples long.
        In Linrad, the PC software will take the fourier transform of the
        input data stream, divide it by the fourier transform of the
        impulse response of the hardware and multiply it by a "desired
        pulse response" This way the pulse length is made shorter than 20
        samples and at the same time the ~0.1 dB ripple is removed.

        The length of the pulse is determined by the "desired pulse response"
        which depends on the skirt steepness that the user has decided.
        The smart blanker knows the exact shape of the pulse and its length
        so it does not matter that the pulse is long in terms of samples.

        I am aware that very few operators use Linrad and that only
        a very small fraction of the users care to calibrate their
        systems properly. I have tried to explain the theory, but I
        do not think I have been sucessful at all. I am interested
        in static rain at high bandwidth because I have a feeling
        recordings would show a dramatic difference between the
        Linrad blanker and other blankers.

        > Of course 20 samples at 40 MS/s are a 0.5us interval,
        > which is a much shorter time interval than that obtained
        > if the sample rate were 2 MS/s but instead of increasing
        > the output sample rate one can obtain the same result
        > simply relaxing the B/Fco requirement.

        > If the B/Fco ratio were 60% instead of 80% the output
        > pulse lenght would be the half the original, if it were
        > 40% one third and if it were 20% one fourth of it, a
        > mere 5 samples interval (2.5us @ 2MS/s), which is even
        > the half of what one could obtain attempting to double
        > the output sampling frequency (and mantaining the
        > original 80% B/Fco ratio).
        > The penalty is that the the alias free bandwidth
        > is much less than the output sample rate...
        Yes:-) This is what I advocate. 4 MHz sampling and
        40% alias-free bandwidth. I also want the -10 dB point
        to be fairly high, maybe 80% of Nyquist.

        > but who cares if we would just be satisfied to (carefully)
        > clean-up a not-so-wide 200 kHz bandwidth out of a 2 MS/s
        > IQ stream?
        > And if it works, wouldn't it be better than obtaining the
        > same result using 4 MS/s maybe overloading a poor man CPU?
        As far as I undersdtand it is impossible to clean up a 200 kHz
        wide segment of a 2MS/s IQ stream if the (random) secondary
        pulses can not be resolved. From old experience as well as from
        the one and only wideband recording at my disposal a bandwidth
        of 1.6 MHz is marginal. It may or it may not work.

        > BTW, making a new 4MS/s DDC would not be impossible but
        > as I haven't implemented it yet I can't say that what
        > was initially conceived for a much smaller output sample
        > rate could sustain it (in 2008 I was even not sure that
        > the 2 MS/s rate could really work).
        Five years later it is very likely that a factor of two is OK:-)


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