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In Reply to: RE: another commentary... posted by Tre' on April 30, 2012 at 21:52:36
Good idea to go through the calculations on this assertion.
You wrote:
"Let's say we have two hypothetical gain stages. Each with a gain of 10 and having only 2nd order harmonic distortion. Let's say the percentage of 2nd order distortion is 10% (each stage).
If we apply 1 volt of 1kHz to the first stage we get 10 volts of 1kHz and 1 volt of 2khz feeding the second stage.
The second stage amplifier the 10 volts of 1kHz to give us 100 volts of 1kHz.
The second stage amplifies the 1 volt of 2kHz to give us 10 volts of 2kHz.
The second stage distorts the 1Khz and gives us 10 volts MORE of 2khz for a total of 20 volts 2kHz.
The second stage also distorts the 1 volt of 2kHz input signal, to give us 1 volt of 4kHz.
What have I missed?"
I will now edit all those lines to correct the errors, one by one:
"If we apply 1 volt of 1kHz to the first stage we get -10 volts of 1kHz (polarity inversion needs a minus sign) and 1 volt MAGNITUDE of 2khz feeding the second stage, phase not found due to not knowing the distortion transfer function completely. It could be -180 degrees, where it is likely. That means we might assume it is actually -1 volts of 2khz midband.
The second stage amplifier with -10 volts of 1kHz then gives us (1)(-10)(-10) = +100 volts of 1kHz. Note two polarity inversions for inverting common cathode stages.
The second stage amplifies and inverts the -1 volt of 2kHz to give us +10 volts of 2kHz.
The second stage distorts the 1Khz and gives us -10 volts MORE of 2khz for a total of a magnitude of (+10)+(-10) = 0 volts 2kHz!"
It gets complicated quickly when the order is increased to calculate % distortion of a single even order harmonic. But you can see that it's doing it quite well to the 2nd order in terms of cancellation.
Follow Ups:
"(polarity inversion needs a minus sign) and 1 volt MAGNITUDE of 2khz feeding the second stage, phase not found due to not knowing the distortion transfer function completely. It could be -180 degrees, where it is likely. "So you are saying that, at the plate, the 2nd harmonic distortion product of a grounded cathode triode stage is out of phase with the the fundamental?
Why would that be?
BTW The voltage in question is AC. We should not use a - sign. The phase is inverted but the voltage is neither - or +, it's AC.
In the studio we use a circle with a diagonal line through it. Ø
From Wikipedia "In audio engineering, the ø is used to represent a signal whose polarity (sometimes called phase) has been reversed."
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Edits: 05/05/12 05/05/12
In EE circuit analysis texts, there are two components to describe the sinusoidal signal: either use cartesian, or Rectangular on your calculator (Real and Imaginary) or Polar (Magnitude and Phase). They are convertible using a scientific calculator with R-> P and P-> R keys. EE's prefer to use "j" as the imaginary number term instead of "i" since "i" is already taken to be AC current.
Real is on the X-axis and Imaginary is on the Y-axis, something that is graphically representative of how the math works in this realm of "phasors".
If a Magnitude is 1 + j1 in Rectangular, then the Polar Magnitude is the length of the diagonal showing, SQRT(1^2 + 1^2) = SQRT(2), and the phase is +45 degrees, with respect to the input reference. Positive phase is lagging, negative is leading.
Now for the example of +1 Vrms vs. -10 Vrms. First we know that the input to the first tube is 1 Vrms Magnitude and is the reference, 0 degrees. That is a vector (arrow) pointing from 0,0 to 1,0 in Rectangular. It's on the real line, in positive direction.
The output is best seen as a real Magnitude of 10 Vrms, but in the opposite direction, +/-180 degrees lagging or leading (it's the same). In Rectangular form, that shows up as a vector pointing from 0,0 to -10,0. The magnitude is ABS(-10) = 10, but the vectors show an important inversion using vector arithmetic, and is represented as a negative number. The entire problem is one of vector addition, not just magnitude addition, because we have phase changes.
Another example for kicks: If you sum +10, zero phase reference, and a partial inverted 1 Magnitude, Phase = +/-180 degrees, you get a net 10 - 1 = 9 Magnitude, zero phase output.
NEXT QUESTION:
How are the harmonics handled in vector arithmetic, not going to the old simple 'magnitude only' addition?
This is helpful if you have a solid understanding of FFT's. But this time I opt to go by observation instead of polarity signs of +/-Acos(wt)+/-Bcos(2wt)...+/-Ksin(wt)+/-Lsin(2wt)... in an infinite series.
Instead of this equation, we can determine by inspection what "polarity" the B coefficient in Bcos(wt) is.
The transfer function of the 2nd harmonic is determined from a simple graph you can create from measurements of such a common cathode stage for Vpk v. Vgk (it's a different special plate curve). We can think of it like this: At about -15 VDC at Vgk, the tube cuts off and we get all the B+ to the output (high voltage) to say +250 VDC. At +1 VDC Vgk (slightly into class A2 in this example), the plate voltage will go down as the tube drains current away from B+. Let's say to +30 VDC.
And then let's say normal bias offers us +140 VDC on the plate. So the cutoff swings up to +250 VDC, or +110 VDC more than quiescent bias point, and small class A2 pull offers +30 VDC, or -110 VDC less than quiescent bias point. Note the bias is centered.
But plot the graph of the output. It will have that 10% 2nd order distortion of asymmetry that is the feature of 2nd order distortion. Now observe the orientation of where the bend is. Since the tube can work slightly into A2, it is still pretty linear over in that part. BUT! The plate curves show that remote cutoffs and all triodes with imperfect linearity towards cutoff really distort more over on this side when in the signal operating range. The upshot for this side of distortion is an elongation of the sinusoidal negative going curve. For odd order distortion, it's about each end with symmetry.
On the other side, it just hit hard clipping at the end of the power supply voltage. Let's ignore the part about overload.
So by observation, we have the polarity of the 2nd harmonic set to distort on the low voltage side every time when not clipping. But let's say the input was pre-distorted where the low voltage output would amplify that distorted end linearly, from a previous stage. What you can picture is that one signal is distorted on one side once, with one stage; but another signal is distorted by a previous stage on one side once, and then let go of linearly the second time.
What we have now is distortion on both ends, moving the distortion to a more odd order problem of distortion. If one end distorted: even order; if two ends distorted: odd order. That was with no math to see this happening.
But the math using vectors and polarity changed to plus and minus Rectangular coordinates offers a way to get a calculation. The harmonics are being played out in different mathematical phases through different polarity orientations.
That's the best I can do to try to explain this. I hope you got something from this.
"But plot the graph of the output. It will have that 10% 2nd order distortion of asymmetry that is the feature of 2nd order distortion. Now observe the orientation of where the bend is. Since the tube can work slightly into A2, it is still pretty linear over in that part. BUT! The plate curves show that remote cutoffs and all triodes with imperfect linearity towards cutoff really distort more over on this side when in the signal operating range. The upshot for this side of distortion is an elongation of the sinusoidal negative going curve. For odd order distortion, it's about each end with symmetry."
Should read:
But plot the graph of the output. It will have that 10% 2nd order distortion of asymmetry that is the feature of 2nd order distortion. Now observe the orientation of where the bend is. Since the tube can work slightly into A2, it is still pretty linear over in that part. BUT! The plate curves show that remote cutoffs and all triodes with imperfect linearity towards cutoff really distort more over on this side when in the signal operating range. The upshot for this side of distortion is a compression at the positive end of the output swing and an elongation of the sinusoidal negative going swing.
For odd order distortion, this will be symmetrical on each swing sides, which is what will happen in cascaded identical stages. Cancellation of evens plus creation of odds is happening here, I believe.
"But plot the graph of the output. It will have that 10% 2nd order distortion of asymmetry that is the feature of 2nd order distortion. Now observe the orientation of where the bend is. Since the tube can work slightly into A2, it is still pretty linear over in that part. BUT! The plate curves show that remote cutoffs and all triodes with imperfect linearity towards cutoff really distort more over on this side when in the signal operating range. The upshot for this side of distortion is a compression of the sinusoidal positive going curve AND and elongation. For odd order distortion, it's about each end with symmetry."
The fundamental passes with inversion on each stage, -10 being the gain of each stage.
The 2nd harmonic passes with DOUBLE INVERSION on each stage, thereby it is not inverted in effect through the stage generating that distortion. Let me explain. The first inversion is the polarity flip of the circuit stage itself, the next one comes from the polarities of what constitutes "positive 2nd HD" and "negative 2nd HD."
The triode common cathode circuit produces an inversion of a "negative 2nd HD." To illustrate this, a normal expected non-inverted 2nd harmonic distortion product has the positive cosine peaks of the fundamental coinciding with the positive cosine peaks of the 2nd harmonic. An inversion in the triode stage's distortion curve in the transfer function causes the opposite inverted distortion within it: the positive cosine peaks coincide with an inverted 2nd harmonic signal where the peaks of the positive fundamentals coincide with the peaks of the most negative 2nd harmonic signal output.
When you start with the fundamental of the first stage, the fundamental proceeds obviously as: +1 V => -10 V => +100 V.
The harmonic generated from the 2 triodes is thus: 0 => (1)(0.1)(-10)(-10) = +10 and an addditional (10)(0.1)(-10) just generated by the fundamental on the last triode stage alone = -10. Add the two and you get a complete cancellation.
It can be done by graphical calculations to show this to you better. The graph of Vpk vs. Vgk shows negative gain with the flat top of the bending curve for a straight line to appear flattened on the positive signal end and elongated at the negative going end. This one curve is mainly a 2nd order distortion.
Do this again a second time, doing fundamental gain and curvature of distortion from 2nd order, and this time the anode flattens the opposite polarity of the signal on the positive side. This is converting the 2nd to an odd order "square wave" type distortion. It takes away some harmonic richness for people who like it.
The effect can be taken to extreme with a strong signal generator and see it with a scope. Take a look at each stage. The input will be a nice sinewave, we hope. The input to the second stage will be a cut-off on the positive end of the scope (highest voltage) while an elongated sinewave feature will show on the negative going half-cycle. Finally backing off on amplitude, the output of the second stage will show cutoffs at a certain signal level input to have only one cutoff still, the final tube. Small signal distortion is less noticeable in normal use, but it counted to me.
The most problematic amp area with even order cancellation will probably be between driver and final output triodes. The signals start to get large and both these tubes can generate a great deal about the overall tone.
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