Ok, can you explain it to me ? My amazement is that this all rather predates solid state transistors. — unenlightened
Are you looking at the 9th canon where he constructs an ever deepening series of nested a's and b's? Page 55 in my version?
If so, you just take the whole right hand expression of a & b as = r. and use J2 in reverse. — unenlightened
Wow, if someone implemented something like that we could have computers and an internet!
Sorry, couldn't resist. — wonderer1
I'm sure I don't understand how a circuit has a memory, still. — Moliere
I didn't mean feedback necessarily, just the view that process might be seen as fundemental, not substance. — Count Timothy von Icarus
The story of a hole in a state of flow with an innumerable number of other holes towards ~Q: We start at 5 V and move through R1 to TR1 because the voltage at Q is lower than the voltage at ~Q (assuming we're already in a steady state), then we go through the unmarked resistor on the other side of the transistor, up through R3 and out ~Q. If you touch "Set" to the zero volts line than you ground the flow causing the voltage to switch over to R4-T2-R2-Q. — Moliere
[tl;dr]More accurately than looking at it as a switch, we might look at the path from collector to emitter as a resistor with a resistance of about 42 Ohms when the transistor is in the 'on state', and as an open circuit when the transistor is in the 'off state'. In that case, if we suppose the resistance of R1 and R4 to be 1000 Ohms, then we have an explanation for why the flip-flop schematic shows a voltage at Q of 0.2 volts for the blue state. In the blue state the 5 volts of the flip-flop power supply gets divided between the 1000 Ohm resistance of R1 and the 42 Ohm resistance of the 'on stated' TR1. In the red state the voltage at Q simply is the +5 Volts of the power supply. (Ignoring for the sake of simplicity, the the relatively low base currents flowing through R3 and R4. The ambitious reader who is into that sort of thing can assume that R3 and R4 have a resistance of 100000 Ohms, and look up linear circuit analysis, and calculate voltages out to more decimal places. However, for pragmatic purposes we can ignore current through R3 and R4, and just consider whether the voltage at the transistor bases are above or below 0.7 Volts to know whether a transistor is off or on.)
And for this discussion we can ignore the resistance of the unlabeled resistors altogether and treat them as open circuits. They are for practical details engineers need to worry about but, not of any help in looking at things in the simple voltage focused model of the flip-flop schematic.[/tl;dr]
That helps me understand the feedback part very well -- so thank you again for taking the time. When Set is grounded the voltage from R3 no longer gives the voltage necessary for the transistor to be in the "on" state, but the parallel circuit through R2 does so the circuit flips over to Tr2. Since Tr1 is now off that means 5V goes to Q as the path of least resistance. The same holds for reset and the blue state. — Moliere
Based on the website I linked it looks like Q and ~Q are out of phase with one another. So the memory comes from being able to output an electrical current at inverse phases of one another? How do we get from these circuits to a logic? And the phase shift is perhaps caused by subtle manipulations of the transistor? — Moliere
At this point it is pragmatic to jump up a level in abstraction and think in terms of logic gates instead of transistor circuits. — wonderer1
I think what I'm wanting to settle, for myself, is whether or not the circuits are in turn being interpreted by us, or if they are performing logical operations. — Moliere
What makes Q and ~Q different other than one is on the left side, and the other on the right side? Do we just arbitrarily choose one side to be zero and the other side to be 1? Or do the logical circuits which have a threshhold for counting do it differently?
To my mind the circuit still doesn't really have a logical structure anymore than a stop light has the logical structure of Stop/Go without an interpretation to say "red means stop, green means go". So are we saying "Q means 1, and ~Q means 0"? — Moliere
The SR f!ip-flop circuit is symmetrical, so it is somewhat arbitrary which output is chosen to be Q and ~Q. However, the Set pin is defined as the input that can cause Q to produce a 1 (5V) output. So one could swap Q and ~Q, but to be consistent with the conventions for SR flip-flops one would also need to swap which input is labeled S and which R. So like the stoplight it is a matter of convention. — wonderer1
Also, flip-flops themselves don't perform logical operations. They just serve as memories that can be used to provide inputs to logic gates (or combinations thereof), and store outputs from logic gates.
Latching relays require only a single pulse of control power to operate the switch persistently. Another pulse applied to a second set of control terminals, or a pulse with opposite polarity, resets the switch, while repeated pulses of the same kind have no effects. Magnetic latching relays are useful in applications when interrupted power should not affect the circuits that the relay is controlling. — wiki
Wait, I think i see what you are doing - treating each line as an equation, and then substituting the right back in for f. — unenlightened
You don't want to do that! Each line is a result for a combination of an and b. There is no working shown, and almost none to do. so for (2):– — unenlightened
and the re-entered f can be ignored.
We can now find, by the rule of dominance, the values which f may take in each possible case of a,b.
You're making the very simple, complicated. these are not equations, but evaluations of f for all the possible values of a & b. the right hand side in each case is the result of simplifying the left — unenlightened
We may take the evident degree of this indeterminacy to
classify the equation in which such expressions are equated.
Equations of expressions with no re-entry, and thus with no
unresolvable indeterminacy, will be called equations of the
first degree, those of expressions with one re-entry will be called
of the second degree, and so on.
It is evident that Jl and J2 hold for all equations, whatever
their degree. It is thus possible to use the ordinary procedure
of demonstration (outlined in Chapter 6) to verify an equation
of degree > 1. But we are denied the procedure (outlined
in Chapter 8) of referring to the arithmetic to confirm a demonstration
of any such equation, since the excursion to infinity
undertaken to produce it has denied us our former access to a
complete knowledge of where we are in the form. Hence it
was necessary to extract, before departing, the rule of demonstration,
for this now becomes, with the rule of dominance,
a guiding principle by which we can still find our way.
A consequence is acceptable because we decided the rules.
All we need to show is that it follows through them.
But demonstrations of any but the simplest consequences
in the content of the primary arithmetic are repetitive and
tedious, and we can contract the procedure by using theorems,
which are about, or in the image of, the primary arithmetic.
For example, instead of demonstrating the consequence above,
we can use T2.
T2 is a statement that all expressions of a certain kind, which
it describes without enumeration, and of which the expression
above can be recognized as an example, indicate the marked
state. Its proof may be regarded as a simultaneous demonstration
of all the simplifications of expressions of the kind it
describes.
But the theorem itself is not a consequence. Its proof does
not proceed according to the rules of the arithmetic, but follows,
instead, through ideas and rules of reasoning and counting
which, at this stage, we have done nothing to justify.
Thus if any person will not accept a proof, we can do no
better than try another. A theorem is acceptable because what
it states is evident, but we do not as a rule consider it worth
recording if its evidence does not need, in some way, to be
made evident. This rule is excepted in the case of an axiom,
which may appear evident without further guidance. Both
axioms and theorems are more or less simple statements about
the ground on which we have chosen to reside.
Since the initial steps in the algebra were taken to represent
theorems about the arithmetic, it depends on our point of view
whether we regard an equation with variables as expressing a
consequence in the algebra or a theorem about the arithmetic.
Any demonstrable consequence is alternatively provable as
a theorem, and this fact may be of use where the sequence of
steps is difficult to find.
But we are denied the procedure (outlined
in Chapter 8) of referring to the arithmetic to confirm a demonstration
of any such equation, since the excursion to infinity
undertaken to produce it has denied us our former access to a
complete knowledge of where we are in the form.
OK I think all that GSB is saying in that paragraph, in simplified terms, is that re-entry doesn't allow us to use the arithmetic to solve for the value of an expression because re-entry creates an infinite sequence which doesn't allow us to substitute the marked or unmarked state for all cases within an expression (given that the expression is infinite). — Moliere
We see, in such a case, that the theorems of representation no longer hold, since the arithmetical value of e' is not, in every possible case of a, b, uniquely determined. — P 57
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