Inverse Marx generator

No. We're using the Xilinx Spartan6 DSP48 blocks mostly, and they don't saturate. We just allow 4 bits of headroom in the 48 bit multiply/accumulate. The front-end ADC calibration is a small offset add and a fractional multiply, so nothing there can go wrong go wrong go wrong.

The filters were a bear. I rashly offered 1 Hz through 100 KHz 8-pole Bessel and Butterworth (500 Ks/s ADC rate), one filter per channel, and the customer countered with *two* filters per channel, 32 total on the chip. The coefficients get nasty nasty on both ends. We wound up with 18-bit filter coefficients (to use the 18x18 MAC block, 48 bit accumulator) and a sorta-floating-point shifter in *five* bit steps. It's massively pipelined and interleaved and rather clever; Rob is doing all 32 filters in about 280 slices, down from 6500 in a previous design that just wouldn't fit. All constantly dancing around the oddities and bugs of the Xilinx tools.

We also added a post-filter interpolator that lets the user think he's triggering the ADC, when it's actually running at a constant 500 KHz. The user gets, simultaneously, one always-available, filtered "realtime" value of each input, for operator display maybe, and a separate filtered, triggered, FIFOd path for formal data logging.

Lotta work.

John

Cute--sounds very useful. I'm a bit leery of fooling the user like that--you might get one who takes that too seriously and winds up with reasonable-looking wrong answers.

I just did an ultrahigh precision laser wavelength locker in 100% analog, including the acquisition aid. Took me back, it did, but it was the right method for the situation.

Cheers

Phil Hobbs

It really ought to work, and act just like a triggered ADC with maybe

A couple of fast opamps can outrun 100 watts of DSP.

John

The *only* people? How about those who think their model is real, even at the fringes and beyond? How about those who don't believe in real components? How many FTL circuits have we seen here?

Seems thin.

Newbies will take note that Larkin has NOT responded to this request.

Would someone out there like to mathematically prove that charge is NOT conserved in Larkin's folly (and yet energy is ?:-) ...Jim Thompson

How about people who simulate a circuit and don't understand what it's actually doing?

John

OK, John, Where's your mathematical analysis? ...Jim Thompson

I DID say "amateurs". I have also recited here my admonishing a former employee about applying common sense to Spice results.

"FTL"?? We have lots of those ;-)

Thin? What is thin, other than Larkin's BS ?:-) ...Jim Thompson

How about you, Win Hill? What do you think? ...Jim Thompson

But you limited your comments to the subset of amateurs who don't understand what the ".OPTIONS" settings do. My point is that the universe of "amateurs" is *much* larger than that. The real scary ones are the ones who believe the FTL simulations because "the computer said so".

Yes, and it's scary.

It's thin complaining about lack of Spice results to avoid embarrassment. If there were simulations there would be no embarrassment. No?

Charge is conserved even in this situation - by reversing the polarity of one you have also reversed the polarity of charge so of course when you connect them the charges add up to zero.

The energy within the two components however is not conserved - after the connection the electrical energy is zero it having been released as heat, radiation and sound etc.

kevin

I'm wondering why C1 would want to donate more than half it's charge to c2. A vision of two connected propane bottles equalizing comes to mind.

confused, in Seattle...

mike

energy

^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^

_why_.

After closing the switch [beginning at the closure of the switch] to discharge the caps was indeed very clear.

Just to help clarify, you may assume trivial switching losses (or not), then continue with a clear explanation showing reasonable causation. Or with (some) math if you prefer.

The questions still are: Where did the energy go? You mumbled something. How did it get [go to] there? No answer at all yet. ^^^^^^^ clarification added.

--- make one propane bottle full (B1), with gas under pressure, say 100 psi, and the other (B2) full also, but with gas at 1 atmosphere.

connect between them a quick-acting valve and a vaneaxial fan with a heavy rotor.

Open the valve quickly and as gas moves from B1 into B2 it'll start the fan rotating.

Now, when the pressure in the bottles gets to be the same, the rotor's inertia will keep it turning and the fan will turn into a pump, sucking gas out of B1 and forcing it into B2 under pressure.

Eventually, when the rotor gives up all its energy it'll stop, with B2 being under pressure, at 100 psi and B1 being at 1 atmosphere.

At that moment, the pressure differential will cause the fan to start rotating, but in the opposite direction, forcing gas into B1.

When the pressure in B1 reaches 100 psi the fan will stop and the cycle will begin anew.

So, in the real world, when the charge on the caps equalizes, the magnetic field around the choke starts to decay and, when it does, sucks charge out of one cap and forces it into the other, back and forth, forever, if the system was perfect.

Hint: bottle 2 is at -40C.

Liquified gas is a bad example.

Tim

--=20 Deep Friar: a very philosophical monk. Website:

R12 was nice because you could tell you had a full tank if it was 72 psi at 72 F.

No liquefied or pressurized-until-liquidous gas is ever 'bottled' completely full.

You closed a contact between two capacitors. A calculable amount of energy was lost. There was a spark, there was some noise, a nearby radio made a tic noise, maybe the room temp rose a bit. If you want to know exactly how all that happened, ask a physicist. I'm just a circuit designer.

John

You keep evading analyzing YOUR circuit with INDUCTOR. Where did the CHARGE go that you state wasn't conserved.

John Larkin, You're a fake and a coward. ...Jim Thompson