Alt-BEAM Archive
Message #02147
To: beam@corp.sgi.com
From: Wouter Brok w.j.m.brok@stud.tue.nl
Date: Fri, 09 Apr 1999 10:15:05 +0200
Subject: [alt-beam] Re: Linked BiCores; How Does The Slave Cycle?
Hello Charlie,
Very interesting question; It took me a while to figure it out (I tried to
calculate it, which worked out partially); only now I hope I can make it
clear:
>This is my understanding (all caps being equal)...
By the way: have you read the article by Susanne Still and Mark Tilden,
Coupled oscillators and walking control .... I believe?
>1. The masters floating resistor governs the period of both bicores.
To a certain extend, yes. I assumed the master bicore doesn't feel the
slave (zero output inpedance of the invertors), so that master oscillated
nicely in its own way with a duty-cycle of 50%
If the slave has approximately has the same time constant it will have the
same period as the master (with a phase-difference in between it), if the
slave-time-constants are much smaller (i.e. shorter) than it is possible
that it oscillates with a higher frequency than the master.
But lets assume they have the same period.
>2. The linking resistors from the master to the slave governs how fast the
>slave 'follows' the master (creates a phase shift).
True
>Q: I know the phase shift in step two is based on the RC of the slave bicore
>and the linking resistors to the master.
Well, the R's of your slave-RC's are the linking resistors. If you assume
caps are equal (for both master and slave) the phase-shift in step two is
based on the linking resistors only (those are the only variables now).
>But I can't figure out how the
>slave RC's charge/discharge in this circuit? It seems to me that both ends
>of the RC's are at the same potential at the same time (ie +V, or Gnd)? Ex.
>One output of master is high and fed into the slave bicore input, it's
>inverted to low (the same as the other master output)
That's not correct: to oscillate, the capacitor of the slave needs to be
charged and discharged. This happens via the linking-resistors. while a
slave-capacitor is (dis)charging there is a current flowing through the
linking-resistors, so there is a potential-difference over these resistors.
This means that the input of the slave-invertor does not feel +V or Gnd,
but something in between. How fast that voltage rises is dependent on (the
capacitor of course, but we said we would have a look at the resistors, so
on) the resistors. This 'how-fast' also determines how long it takes before
the slave-invertor-input passes its threshold-voltage.
>now both ends of the
>RC are low, and both ends of the other slave RC are at high. What am I
>missing? Does the RC charge/discharge of the slave occur at the time the
>masters output changes state?
mmmm, I find it very hard to explain this; I hope you get it a bit;
For the slave-capacitor to charge there needs to flow a current through the
linking-transistors, right? Well, if that current can flow depends on the
voltage on the master-side of the linking-resistors.
Maybe you should do this: draw one nv-neuron (in a microcore for example;
but just have a look at that one neuron). Normally you will draw an input,
an output and the resistor to ground at one side. I guess you understand
how this work. Assume the resistor isn't connected to ground but to
something that's either Gnd or +V. Try to imagine what happens if the whole
thing is oscillating, but when the pulse (or process, or whatever you want
to call it) is 'in' the neuron you are looking at that one side of your
resistor goes from gnd to +V and back to gnd. Imagine what happens then.
Maybe that gives you some understanding.
>Any clarification of the slaves RC charge/discharge behavior would be
>appreciated.
Well, I tried ... If you still don't get it, mail again.
For the other ones on the list (who took care to read this): maybe any of
you can explain it in a better way.
Wouter Brok
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