Bob Cordell Interview: Error Correction
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 28th November 2006, 02:59 PM #921 traderbam   diyAudio Member     Join Date: Jan 2002 Location: Earth Mike, Sorry about your head. Where have I gone wrong? We have to make a break in the feedback from the output. Let's call the S1 side of the break Vo' to distinguish it from Vo. Let's call Vn = K1's input voltage From your fig 1, Vn = K3.{ Vi - K2.(V'o - Vn) } Vn = K3.Vi - K3.K2.V'o + K3.K2.Vn Vn.( 1 - K2.K3 ) = K3.Vi - K2.K3.V'o for loop gain we want zero input signal, so make Vi=0. Vn.(1 - K2.K3) = -K2.K3.V'o Vn/V'o = -K2.K3/(1 - K2.K3) Vo is equal to Vn.K1, loop gain = Vo/Vo' = -K1.K2.K3/(1 - K2.K3)
 28th November 2006, 03:04 PM #922 jcx   diyAudio Member   Join Date: Feb 2003 Location: .. my mistake, shouldn't have added the comma http://www.control.lth.se/~kja/modeluncertainty.pdf From K J Astrom “Model Uncertainty and Robust Control” “fair use” excerpt: Be sure to read that last para!
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Join Date: Jan 2002
Location: Earth
Quote:
 Hint: K1 is error + ideal output stage gain.
I am begining to wonder whether this is where the mistakes are being made. A non-linear gain block should not be modelled as a linear gain block PLUS an error term. This is mathmatical folly because if you then manipulate your equations as if this "addition" holds true your equations will be wrong. Again, addition and subtraction do not work in a non-linear function (by definition).

mikeks
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Join Date: Jun 2004
Location: Animal farm
Quote:
 Originally posted by traderbam Mike, Sorry about your head. Where have I gone wrong? We have to make a break in the feedback from the output. Let's call the S1 side of the break Vo' to distinguish it from Vo. Let's call Vn = K1's input voltage From your fig 1, Vn = K3.{ Vi - K2.(V'o - Vn) } Vn = K3.Vi - K3.K2.V'o + K3.K2.Vn Vn.( 1 - K2.K3 ) = K3.Vi - K2.K3.V'o for loop gain we want zero input signal, so make Vi=0. Vn.(1 - K2.K3) = -K2.K3.V'o Vn/V'o = -K2.K3/(1 - K2.K3) Vo is equal to Vn.K1, loop gain = Vo/Vo' = -K1.K2.K3/(1 - K2.K3)

http://www.control.lth.se/~kja/nybok...1/lecture9.pdf

http://encon.fke.utm.my/nikd/Dc_dc_c...r/Blockdia.pdf

http://www.columbia.edu/~ag2363/e3410/mult_sys.pdf

 28th November 2006, 03:16 PM #925 traderbam   diyAudio Member     Join Date: Jan 2002 Location: Earth So what's wrong with my algebra, Mike?
mikeks
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Join Date: Jun 2004
Location: Animal farm
Quote:

What are you doing with figure 1 ; what is the point of looking at figure 1?

Are you satisfied that figure 3 is equivalent to figure 1?

Then, to find loop-gain from figure 3, simply multiply the gains round the loop. Period.

This is abundantly clear from the links i have given.

Then surely it must be apparent that you are wrong?

Finding out why is down to you.

mikeks
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Join Date: Jun 2004
Location: Animal farm
Quote:
 Originally posted by janneman ...we have a system that - with the simple tuning of a passive voltage divider - can actually reduce distortion to zero and even go beyond this null and make the distortion re-appear in opposite phase... Jan Didden
This, consistent with Bob's experience, is what i would expect of an error-cancellation-by-feedback system.

 28th November 2006, 03:45 PM #928 mikeks   Account Disabled   Join Date: Jun 2004 Location: Animal farm Traderbum, More stuff: http://web.mit.edu/6.302/www/Rec3_F05.pdf
 28th November 2006, 03:49 PM #929 mikeks   Account Disabled   Join Date: Jun 2004 Location: Animal farm
 28th November 2006, 03:55 PM #930 traderbam   diyAudio Member     Join Date: Jan 2002 Location: Earth Mike, No I don't think fig 2 or fig 3 are equivalent to fig 1. Reason: K1 is a non-linear function. In fig 2 you have assumed that K1.1/K1 = 1. But K1 is a non-linear function so this does not hold true. Example: suppose K1(x) = x^3. Therefore 1/K1(x) = 1/(x^3). K1.1/K1 = (x^3)/[ (x^3)^3 ] = 1/(x^6). This does not equal x. To make fig 2 work you need to define the inverse of K(x)...let's call this G(x). You define G(x) such that G[K(x)] = x.

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