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Copyright © 2012 by Wayne Stegall
Updated May 25, 2013.  See Document History at end for details.

Feedback Distortion Analysis

Taming the capricious transfer curve


The process of trying to create a desired distortion result in a feedback amplifier baffled me.  I presumed that the transfer curve would hold about the same shape under feedback only flatter, the distortion as a whole reduced by the feedback factor 1 + AOLβ.  Instead, the transfer curve behaved unpredictably under different open-loop gains and feedback factors even paradoxically flipping the curve for high feedback factors.

First Things

If transfer curve polynomials are to be used to evaluate distortion characteristics, it is first necessary to determine how power terms relate to specific harmonics.  Since input signals are sinusoids the power terms must be interpreted as powers of sinusoids.  The first few powers are well documented.

(1) cos2(x)  = 2 ½ + ½cos(2x) 2

sin2(x)  = 2 ½ – ½cos(2x) 2

cos3(x)  = 2 ¾cos(x) + ¼cos(3x) 2

sin3(x)  = 2 ¾sin(x) – ¼sin(3x) 2

cos4(x)  = 2 ⅜ + ½cos(2x) + ⅛cos(4x) 2

sin4(x)  = 2 ⅜ – ½cos(2x) + ⅛cos(4x) 2

In order to understand these results and extend them to higher powers it is necessary to consider raising a sinusoid to a power to be an intermodulation process.  In this sense all distortion is intermodulation distortion, the case of harmonic distortion being the intermodulation of a single signal with itself.  Since intermodulation produces sum and difference frequencies from the inputs the above power of sinusoidal trend is to have harmonics of some level up to the nth harmonic for an xn term varying according to the bias.

(2) cosn(x + φ)  = 2 a0 + a1cos(x + φ1) + a2cos(2x + φ2) + ... + ancos(nx + φn)2
where φi represents any phase shift that may accrue.


Feedback Hacks Up the Transfer Curve

The gain factors in the negative-feedback equation are valid in the both the time and frequency (s-domain) domains.  In the time domain they can represent the non-linear transfer curves we want to analyze.

Start with the standard negative-feedback equation.
1 + AOLβ

Then solve to separate the linear gain term from the remainder.
1/β + AOL
1 + AOLβ
 – 1/β
1 + AOLβ
 – 1
β + β2AOL

If the undistorted closed loop gain is known to be:

What remains is the error or distortion term:
ε = – 1
β + β2AOL

For a non-linear open-loop transfer curve polynomial
H(x) = h0 + h1x + h2x2 + ... + hnxn 

its non-linear gain AOL is defined as1
AOL(x) = d H(x)
= a0 + a1x + a2x2 + ... + an-1xn-1 

The expanded error term now involves division by a polynomial the general nature of which the constants β and β2 do not alter.
ε = -
β + β2(a0 + a1x + a2x2 + ... + an-1xn-1)

Long division then synthesizes an infinite numerator polynomial of the form:
ε =
b0 + b1x + b2x2 + b3x3 + ...

Feedback will then take any transfer curve, even the simplest second order one producing only second harmonic distortion, and wring an infinite number of harmonics out of it.  It is left to the designer to juxtapose the various amplifier parameters affecting the final transfer curve to reduce any undesirable higher harmonics to vanishing levels or a profile that is suitable to him.

An Analogy

One plausible explanation of this result is the following.  That uncorrected distortion becomes a signal itself which input with the correct signal gets distorted again.  This process would occur repeatedly until an infinite number of harmonics are produced.  This infinite process would be conceived to occur instantly.  This is an analogy only, one representing the long division that created the closed-loop transfer curve polynomial from the open-loop one.


I deduct that obtaining a desired euphonic distortion outcome is an art requiring experience and experimentation.  Designing to an objective specification only is perhaps like throwing dice in comparison, a process not guaranteeing a desirable result.  Also, under certain feedback conditions the BJT transistor with its already infinite transfer polynomial may compare more favorably with square-law devices than at least my design prejudices normally judge.  (I still prefer the square-law devices regardless!)

Example of a Long Division by Polynomial

Here a representative long division created by a first order gain polynomial representing a second order transfer characteristic.  In each case the numerator is manipulated into the sum of a divisible term and a remainder then the possible division completed.  The process proceeds infinitely.

10 + x
1 + 0.1x
10 + x
 – 0.1x
10 + x
= 0.1 – 0.1x
10 + x

0.1 – 0.1x
10 + x
 = 0.1 – 0.1x + 0.01x2
10 + x
10 + x
 = 0.1 – 0.01x + 
10 + x

0.1 - 0.01x +  0.01x2
10 + x
= 0.1 – 0.01x +  0.01x2 + 0.001x3
10 + x
 – 0.001x3
10 + x
= 0.1 – 0.01x + 0.001x2 0.001x3
10 + x

0.1 – 0.01x + 0.001x2 0.001x3
10 + x
 = 0.1 – 0.01x + 0.001x2 0.001x3 + 0.0001x4
10 + x
10 + x
 = 0.1 – 0.01x + 0.001x2 – 0.0001x3 + 0.0001x4
10 + x

What a mess!  Division by a larger polynomial is even more complicated.

Happy amplifiers ;-)

1The gain polynomials in this analysis are derivatives of the transfer functions polynomials they represent and are therefore one order lower than them.  This does not affect the validity of the analysis.

Document History
August 11, 2012  Created.
August 11, 2012  Added footnote about polynomial representation, improved some wording, added derivation of gain polynomial from transfer curve polynomial, and added an example of long division.
May 25, 2013  Corrected + to – in equation 4.