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This procedure (the calibration) works as long as the mixer's conversion gain is constant; i.e., its RF port isn't driven into saturation during the real measurement.
Yes, it added to the noise figure.
...asinine assumptions......nonsense...I am trying to be civil, but your insults can not stand.
I would quote here, in it's original form, from the book: "Microwave antenna theory and design Di Samuel Silver"
That'll be the Acid Test then.I don't recommend long cables like that. The shortest interval between edges for 44k1 is 1/(44100*128) or 177nS. If you're talking Belden 1694A then 10nS equates to a round trip of 2.46m so a round trip time of 177nS gives a cable length of 17.7 * 1.23 = 21.77 metres would cause exact coincidence of the incident edges and the reflected edges. So with this cable, at 44k1 you're OK. If you're running 48k, however, the distance between edges is ~163nS. 16.3 * 1.23 = 20 metres almost exactly. So if there are any reflections in there, 20 metres is the absolutely worst length at 48k and 10 metres (and 20 metres) are the worst for 96k and 192k. And of course 5 metres at 192k. OTOH few cables have a velocity factor less than 0.66 (a speed of 200,000,000 m/S). That is to say a 1 metre cable has a round trip time of 10nS. The distance between edges for 192k is ~40nS so all sample rates will be accommodated by most cables at a length of 3 metres or less, i.e the reflected edge will return 30nS or less behind the incident.If there are any reflections that is.w
Wakibaki,Let's try to clear up your point here.First - the power spectral density, vn for 75ohm is 1.1147nV/sqrtHz.(Using your formula)Vrms for 175MHz bandwith is vn*sqrtB, 14.75 uV.Now, what is the jitter introduced by this noise floor in a square signal typical for our transmitter? The Hiface is doing 2.5Vpp in less than 2nsec. That is > 1.25V/nsec slew rate, but let's stay with 1.25. That is 1250mV/nsec.In figure 7, cited by You the introduced jitter in function of the slew rate is given.It shows 500psec jitter at 100mV/nsec slew rate. It is inversely proportional with slew rate, so for 1250mV/nsec it is 500psec*100/1250 = 40psec.This the introduced jitter when 50 mVrms noise is present. From Figure 6 cited by You we know that jitter is directly proportional with the noise Vrms value. We have 40psec jitter introduced by 50 mVrms noise, then at 14.75 uVrms (calculated for 75ohm) noise level we have 40psec * 14.75/50000 = 0.0118 psec, 12femtosec jitter.If we attenuate by 6db, then we half the slew rate, so the jitter introduced this way will raise to 24femtosec?Can we agree in this? when attenuating 6db, we introduce +12 femtosec jitter?Also, with 10dB, we are introducing +24 femtosec extra jitter?And all this is Gaussian distributed noise.
We went from a 5 m cable, one that is so long, almost no one wants to use it, to one that is 1 m, which is what everyone wants to use.And.............This is with a very fast source. It has a rise time, around 0.8 pSec. The problem is...........A lot of you guys have transports/sources that are closer to 3-4 pSec. And what is the significance of that?That first reflection...............guess what.............guess where it ends up.............yeah, somewhere in the transition portion of the waveform.
Last, really.. also I don't know if it is permissible - I would quote here, in it's original form, from the book: "Microwave antenna theory and design Di Samuel Silver"Please, moderate it freely if does not fit.Ciao, George
In art's pics the reflection from the DAC lasts about 50nS. I'll take that as a worst case.
Oh, here's another one...He means nanoseconds....
Which scope photo are you discussing? As I recall, when he opened the case, he found ferrite beads in front of a transformer. There are significant differences between an active termination such as a DAC summing point, and a lossy inductive one... It is important to retain the distinction in discussion.It is useful for all to remember (or learn)...An active termination like a dac summing node may not behave very well if it cannot keep up with the node driven slew rate. The internal circuitry responsible for zeroing the node can easily saturate if it is called upon to react faster than design. This was a common circuit problem back in the 70's. Many a test equipment input had to be protected with a diode bridge and pull resistors (or even fast current sources) to prevent overvoltages at inputs which caused long recovery delays. I still have bad dreams over those 1N4148 monolithic bridge chips with silicon nitride tub isolation..
I am confused with this comment. This was cleared up Feb 1, I asked and Pat explained that it was typo.
It's important to actually focus on the evidence being presented.
What he found when he opened the case is irrelevant.
I think you'll agree that the duration is ~50nS (50nS/div, a pulse of ~350nS), you might stretch it to 75 nS but this still does not invalidate my points.
Again, focus on the point. I acknowledge that he has made a typo, but... The cable is 1 metre. The roundtrip time for a 1 metre cable of 0.88 velocity factor is ~7.7nS. The rise time in question is stated to be 3-4pS, but this is absurd, he means nS.
