Re: Re: New President Obama
/*
Well, he vowed for some points that make a lot of sense at long last, emphasis on science for the environment, education aiming at technology, business ethics, that stuff.
Looks good, really want to see him suceeding (for the sake of all)
*/
Rodolfo
Bob Cordell said:
/*
Well, he vowed for some points that make a lot of sense at long last, emphasis on science for the environment, education aiming at technology, business ethics, that stuff.
Looks good, really want to see him suceeding (for the sake of all)
*/
Rodolfo
Off topic
/*
Of course we all, including me, really want to see him succeeding.
Let that be perfectly clear.
*/
/*
Of course we all, including me, really want to see him succeeding.
Let that be perfectly clear.
*/
janneman said:
Hi Jan,
The problem I have with your simplified transformation is that it hides certain aspects of HEC. Let me try to make my point clear in a different way:
We know already a couple of things about HEC:
1- It's based on error feedback. So it is a feedback system.
2- It reduces the distortion by a substantial amount.
3- Apart from peculiarities of MOSFETs, emitter/source followers etc, it's prone to HF instabilities.
To me, precisely these three properties can only lead to one conclusion: the mechanism of distortion reduction must rely on a high gain NFB loop somewhere in the system.
In your transformation, this kind of loop is invisible. So it doesn't reflect the reality. That's my problem.
Regards,
Edmond.
Edmond, clear.
Your three points are correct, but does that automagically mean that it MUST be a high gain nfb loop? That's a pretty wide ranging statement; is that documented somewhere? I may have missed it.
I have done some Mathcad work on Hec with first order lp systems for a, b and K. What you see is hf peaking, the magnitude of which depends on the ratio of the -3dB frequencies of a and b versus K. When all freq are identical the peaking is minimal at +3dB. Possibly, with higher order systems, there may actually be oscillations but so far I haven't seen them.
In the next weeks I plan to work more on that as I find the time.
At any rate, when my transformation C, which I think is legal, would lead to a system that no longer corresponds to reality, then we're in REAL trouble. Not just me, but a lot of real engineers as well!
Jan Didden
Your three points are correct, but does that automagically mean that it MUST be a high gain nfb loop? That's a pretty wide ranging statement; is that documented somewhere? I may have missed it.
I have done some Mathcad work on Hec with first order lp systems for a, b and K. What you see is hf peaking, the magnitude of which depends on the ratio of the -3dB frequencies of a and b versus K. When all freq are identical the peaking is minimal at +3dB. Possibly, with higher order systems, there may actually be oscillations but so far I haven't seen them.
In the next weeks I plan to work more on that as I find the time.
At any rate, when my transformation C, which I think is legal, would lead to a system that no longer corresponds to reality, then we're in REAL trouble. Not just me, but a lot of real engineers as well!
Jan Didden
Hi Jan,
You didn't miss anything. That 'pretty wide ranging statement' I just invented it by myself. I believe it holds, because my imagination fails short to devise a different mechanism that fits the bill.
Looking forward to your Mathcad results.
Regards,
Edmond.
You didn't miss anything. That 'pretty wide ranging statement' I just invented it by myself. I believe it holds, because my imagination fails short to devise a different mechanism that fits the bill.
Looking forward to your Mathcad results.
Regards,
Edmond.
janneman said:
As I see it, there is an important difference between A and C. Although the overall transfer function Vout/Vin is the same, the FB network is fundamentally different. In A, there is a PFB loop. In B you break that loop and add an inverse K1 function, 1/K1. So now there is no PFB loop (with potential for gain > 1).
The FB loop gain in A is: -S1.S2.K1 / (1 - S1.S2)
The FB loop gain in C is: -S1.S2.(K1 - 1)
Because A doesn't have an inverse K block it requires excess gain to emulate this function.
Attachments
Edmond Stuart said:
This is true, to try to visualize it let's consider 2 cases:
1. Ideal elements, then there is perfect cancellation equivalent to infinite loop gain stemming from a division by 0 in the loop transfer.
2. Bandwith limited elements, which is the more interesting situation insofar the other is of virtual interest only.
It can be easily seen that the circuit maps to an integrator fed by the error between input and output of the main amplifier to be corrected.
That being the case, whenever there is an error signal present, the integrator output ramps one way or the other to cancel. This is why Brian insists rightly that it is a servo loop (which is not the same to say it is uninteresting).
As long as there is no error, the integrator output remains quiescent at the value which leads once amplified to the correct output.
Note that this does not imply necessarily a classical 6dB/oct slope and infinite DC gain (1/s) but more generally (and practically) a very high DC gain and a more or less monotonical negative slope in frequency response.
Again this shows why large bandwidth is mandatory (which conjugately maps to large DC gain), the integrator output slews towards the value required for correct output only so fast, making for an incorrect output in between.
Rodolfo
Hi Rodolfo,
Thx for explanation, but does my conjecture hold? Is there indeed only one possibility, i.e. a high gain NFB loop?
(provided, of course, that the 3 points in post 3623 are true and we live in an analogue-only world)
Regards,
Edmond.
edit: I'm not entirely sure and Jan probably even less.
Thx for explanation, but does my conjecture hold? Is there indeed only one possibility, i.e. a high gain NFB loop?
(provided, of course, that the 3 points in post 3623 are true and we live in an analogue-only world)
Regards,
Edmond.
edit: I'm not entirely sure and Jan probably even less.
Edmond Stuart said:
Well I think there is a fundamental truth at work here. Jan's circuit C is like perfect pre-distortion when S1.S2=1. He could have redrawn the diagram with the 1/K block in front of the K block to achieve the same overall transfer function. So there is a reality here, that a FB loop can only perfectly pre-distort if there is infinite excess gain available (and it is stable).What are your thoughts?
Edmond Stuart said:
Yes, and appart from the formalisms that show it, a possible heuristics to grasp why is:
You have something that distorts what is intended to be reproduced, but in an analoge only world you can only know how after the fact.
You try to compensate "immediately" by a reasonable amount derived from comparison, the faster the better.
This "immediately" translates to bandwidth in processing the error, which in turn translates in the real world to gain, the faster the reaction, the higher the equivalent gain.
Rodolfo
PS the reference to the analoge world resides in that while in principle it should be possible to construct an inverse transfer for a given forward transfer, because of causality in the forward path this inverse transfer must violate causality.
The trick in the digital world is to delay the main path as much as required and "cheat" causality.
traderbam said:
What are your thoughts?
The question seems indeed fundamental.
BTW, you can leave the pos fb pick-off point ahead of K and instead move the nfb pick-off from Vo to before K and scale it with K. The math is a bit more involved, but you still get a system with the same cl gain but yet again different loop gain and fb factor.
I am not comfortable with your and Rodolfo's reasoning that by definition to try to reach for perfect correction, the only way is (infinite) high nfb. I understand the need for fast reaction and wide bandwidth (in contrast to feedforward), but why does high speed reaction necessitate high gain in the loop?
Jan Didden
Hi Rodolfo,
That's a clear cut view. IOW, QED.
Thx again!
@Brian.
You know already we share the same opinion and at the moment, I have nothing special to add to this topic.
Regards,
Edmond.
That's a clear cut view. IOW, QED.
Thx again!
@Brian.
You know already we share the same opinion and at the moment, I have nothing special to add to this topic.
Regards,
Edmond.
ingrast said:
Yes, and it has to react faster than immediately if there is any time delay in the plant 😱. Time delays and phase shifts in the plant and the feedback network itself force the FB loop to asymptotically reduce the output error rather than cancel it. Pre-distortion and post-distortion can, in theory, cancel output error but they require knowledge of the plant and/or the possibility of making an inverse plant block. FB has no such knowledge and is therefore compelled to use ignorant gain to reduce the error.
janneman said:
Conceded, it is not intuitively obvious.
Probably the best approach is to note that to null an error you need something that keeps moving as long as there is a difference, and only stops when the difference is zeroed.
Now something like this belongs to the integrator class - not surprisingly popping up in HEC - and this in turn leads to the gain - bandwidth conjugation in the real world.
Rodolfo
I think we had already agreed that Hec cannot perfectly null the error precisely due to the need for infinite bandwith. This was also shown by van der Kooy and Lipshitz. As such, perfect Hec only occurs at DC when all the summers are exactly =1.
If we could build summers with infinite bandwidth, we could null the error in a Hec system exactly at all frequencies.
If we could build infinite gain amps, we could null the error in a gnfb system exactly at all frequencies.
Isn't that the underlying duality of Hec versus gnfb? Imperfect, because you neither can have infinite bandwidth in Hec nor infinite gain in gnfb?
Jan Didden
If we could build summers with infinite bandwidth, we could null the error in a Hec system exactly at all frequencies.
If we could build infinite gain amps, we could null the error in a gnfb system exactly at all frequencies.
Isn't that the underlying duality of Hec versus gnfb? Imperfect, because you neither can have infinite bandwidth in Hec nor infinite gain in gnfb?
Jan Didden
janneman said:
Yes, they converge but asimptotically.
The trick is looking the HEC way, one can at least imagine an alternate design strategy which **may** end up better.
Your mileage may vary.
Rodolfo
janneman said:
I don't think it does. The speed of a reaction is affected by time delay and by inertia. Your diagrams don't include these factors at all. Therefore, the need for infinite excess gain in A is independent of bandwidth considerations.
traderbam said:
I was thinking of the actual implementation. In gnfb one would push the forward gain (while keeping it stable). In Hec, one would push the summer bandwidth, (while keeping it stable).
Jan Didden