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Re: Fierce competition

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  • Leif Asbrink
    Hello Nico, ... Something like that. ... Yes. I never used a small antenna;-) ... Math is so difficult and I often confuse myself:-( Let me think in a
    Message 1 of 48 , Jul 10, 2013
      Hello Nico,

      > > 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.
      > I've tried to guess this number with a simple model.
      > If each rain drop generates 20/30 discharge pulses (as you
      > showed in fig. 1) then the rate of the drops that discharge
      > to the antenna conductors should be about 300/500 per second.
      Something like that.

      > If we suppose that a discharge occurs only when a drop
      > passes near an antenna conductor, say at a distance of 1 cm,
      > then the cross section of the problem should be just a
      > small fraction of a square meter (section = 1cm times
      > the total antenna lenght in the horizontal plane).
      Yes. I never used a small antenna;-)

      > On average, the rate at which rain drops traverse this
      > section, call it Rd, is:
      > Rd = RainfallRate * CrossSection /Average Drop volume
      > Under very heavy rainfall rates, say 50 mm/hour, and
      > with drops 4 mm in diameter (I think it would be very
      > unlikely that they were much smaller when the rainfall
      > rate is so high), one should experience about 400 events
      > per square meter of cross section.
      Math is so difficult and I often confuse myself:-(

      Let me think in a different way.
      With 50 mm/hour = 0.014 mm/s cubic raindrops with a side length
      of 3 mm would have to fall at a rate of 0.014/3 for each quadratic
      area of 3 by 3 mm (9 square mm)

      That is about 0.005 drops per 10 square mm and second.

      A typical antenna that I have used would have 40 elements and
      a diameter on each of 5 mm. 40m times 5mm is 200000 square mm
      so I would expect 0.005*200000/10=100 events per second.

      > For a 2 m band horizontal polarisation 10 element yagi
      > the cross section would be 0.2 square meter (Area =
      > number of elements * lambda/2 * 2D, where D is the
      > maximum distance at which a discharge can occur) and
      > the drops that discharge on the antenna would be 80
      > per second.
      OK. By a far more primitive reasoning I get a similar result:-)

      > If we reduce the drop diameter to 2 mm we get numbers
      > that match the observed drop discharge rate (300/500 per
      > second) but I think that the occurrences of such rainstorms
      > should be quite rare.
      Maybe the number of pulses from each droplet can be much larger
      some times? I only have a single recording so I know nothing
      about the variability. The charge could be much larger and the
      time constant (droplet capacitasnce and plasma inductance)
      might be very different.

      I have observedon a longwire antenna that was floating that
      arcovers happened at a distance of over 10 mm. That would imply that
      a discharge would happen 10 mm away from the 5 mm element
      making the area 25 mm by 40 m or 5 times larger. The reason why
      I did not see more than 10 mm discharges from the long wire antenna
      was of course that my insulation distance was not greater.
      In static rain the charge could be quite a bit higher, megavolts
      perhaps, for discharges over decimeters.

      > At least in southern Europe :-D
      With warmer climate it should probably be worse;-)

      > > But this condition is not met in static rain.
      > > Look here:
      > > ...
      > My Swedish is a little bit rusted but I think I've understood.
      > And I was also pleasured to see a picture of a young,
      > black bearded Leif in the sixties :-)
      > > 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 see. If just the drop rate were an order of magnitude
      > smaller one could attempt to estimate the effects and
      > subtract them in the band of interest, but to my knowledge
      > this is possible only if each pulse sequence were
      > separated from the others by at least several time
      > intervals of its duration. Which is not this case.

      > > 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.
      > No matter the bandwidth, what makes it impossible to
      > blank a pulse sequence on a small portion of the
      > spectrum is how much each pulse sequence is separated
      > in time from others.
      Two different ways to try to explain the same thing.
      When the bandwith is so narrow that the individual
      pulses are not resolved one has a noise burst with a
      duration of the entire event. That is a lot of time.
      When the rate is low we can gate out the noise bursts at
      a high bandwidth, but it would be far more clever to do it
      at a narrow bandwidth. The bandwidth should be reduced
      until the interference pulse is formed essentialy by the
      filter. There would be no loss because the gate time would not
      increase, but the advantage is that signals and noise outside
      the filter would disappear.

      > If they were separated by tens millisecond and each of
      > them were made by 10 femtosecond pulses randomly distributed
      > over 100 microseconds you would not need to sample the
      > signal at giga samples per second.
      Yes. Agree:-)

      > But they aren't, they are separated by an amount of
      > time which is the same order of magnitude of their lenght.
      Hmmm, with 96 kHz sampling rate the pulses are not
      really separated. They do overlap quite often. One has to
      sample at 2 MHz, but I would love to see data from a
      wideband hardware at 4 MHz. (One can calibrate linrad to
      eliminate the influence of transverter filters.)

      > > 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.
      > With static rain this would work if sequences were
      > separated, not just their constituent pulses.
      Yes. We can formulate it differently, but I think we
      agree on the basic physics:-)

      So, back to the start of this thread. If you could
      supply a bad sbs file that would give a higher
      sample rate where the alias suppression is mediocre to bad
      outside the 1 MHz (or 2 MHz) central region, maybe
      there would be a chance that somebody who suffers from QRM
      might record a file with resolved discharge events of
      different kinds. There seems to be some very odd things
      going on in snow storms....

      > Not with a conventional blanker but with a more
      > sophisticated one of course, but since it looks like
      > that they aren't I can't guess other way other
      > than to resolve each single pulse. So far I agree with you.
      Terra incognito. There are new phenomena that we know
      nothing about. Maybe the professionals have written about it,
      Searching for this kind of specialized information is
      non-trivial however.,,,,


    • 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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