Bob Cordell Interview: Error Correction - Page 93 - diyAudio
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Old 28th November 2006, 02:59 PM   #921
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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)
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Old 28th November 2006, 03:04 PM   #922
jcx is offline jcx  United States
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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:
Click the image to open in full size.

Be sure to read that last para!
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Old 28th November 2006, 03:07 PM   #923
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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).
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Old 28th November 2006, 03:10 PM   #924
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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
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Old 28th November 2006, 03:16 PM   #925
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So what's wrong with my algebra, Mike?
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Old 28th November 2006, 03:30 PM   #926
mikeks is offline mikeks  United Kingdom
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Quote:
Originally posted by traderbam
Mike,

From your fig 1,


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.

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Old 28th November 2006, 03:39 PM   #927
mikeks is offline mikeks  United Kingdom
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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.
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Old 28th November 2006, 03:45 PM   #928
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Traderbum,

More stuff:

http://web.mit.edu/6.302/www/Rec3_F05.pdf
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Old 28th November 2006, 03:49 PM   #929
mikeks is offline mikeks  United Kingdom
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http://www.ece.ubc.ca/~elec360/lectures/Lecture04.pdf
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Old 28th November 2006, 03:55 PM   #930
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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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