Phasing?

[shooter]

 was following the TS discussion n I already learned here phase shifting is not a critical issue, I quit measuring my overall amp phase shift......but, does it come into play when mic'd out is fed back to monitors?  Not being a player.  I'm not talking the 13degree outta phase my last amp had but more the 180 shift, can a player discern his amp vs monitor speaker on stage?

[HotBluePlates]

... does it come into play when mic'd out is fed back to monitors?  ... can a player discern his amp vs monitor speaker on stage?

The player can discern feedback.

You should consider wavelength in air for a moment. I think the velocity of soundwaves in air of median humidity and at sea level is ~343m/sec, or ~1125ft/sec. So that's the speed all soundwaves will propagate in air under those humidity and pressure/density conditions.

What is the wavelength of 1kHz? There are 1,000 cycles per second, so a complete 360-degree cycle occurs over a distance of 1125ft/1000 cycles = 1.125ft (throw away the inch-and-a-half and call it a foot). So if the mic and speaker are in such a position that they're feeding back, and you move the mic 6" you'll have opposite-phase (or 180-degree difference) and no feedback. Move it 6" further beyond, and you have in-phase signal again and feedback.

Lower frequencies have slower cycles, though the wave is propagating just as fast, and so have longer wavelengths. 100Hz would be 11.25ft, so you move either the speaker or mic ~5.5ft and you have 180-degree phase-difference.

So it's a pointless question in a way, because all frequencies have some distance-change whereby you can get opposite-phase or same-phase. The real answer is use a cardioid mic with a null in its response at the back of the capsule, and keep that null pointed towards the speaker. Because even if you could move the mic to some critical distance where you had opposite-phase and no feedback, sound will bounce around the space and find some reflected path that gives in-phase and feedback howl.

Open the May 1959 Electronics World to page 69, and you'll get the real answer for "phase issues" in an amplifier circuit: maximum negative feedback which can be applied. As PRR already said, "phase is absolutely related to the slope of the amplitude response. If the response is flat, phase is zero. If the response tilts, phase is shifted, in that area, in proportion to the amount of tilt." Frequency-shaping, or parasitic reactances in output transformers in the extremes of their response band, shifts phase because they attenuate response. Go far enough along that attenuation curve, and phase shifts far enough to turn negative feedback into positive feedback (if there are enough attenuation points, and therefore phase shifts, in the circuit).

Which is why Futterman wanted to eliminate OT's (he wanted to jack up feedback to clean up the amplifier and lower output impedance/improve damping factor). Direct-coupling also eliminates a capacitor and another source of phase-shift.

A single cap or inductor can only theoretically shift phase 90-degrees, but that assumes zero resistance, so 2 caps or cap-inductor can only approach 180-degrees. Three coupling caps or R-C networks in series assures 180-degree phase shift at some frequency, which is why tremolo oscillators use 3 caps.

But if you have no point of reference (like a feedback signal to compare with the non-feedback signal), then absolute phase is irrelevant. Some will argue that shifting the phase of harmonics of a tone will change its sound quality, like turning a sine wave into a saw-tooth. But that ignores that relative amplitudes of each of the shifted harmonics is important to get that different wave-shape just as much as the degree of phase-shift of the harmonics, and that all those harmonics equate to distortion which we already expect to not sound like clean signal.

[jjasilli]

Shooter, I'm not sure what you're asking; or what the stage set-up(s) is (are). 


Often the el guitarist in a band listens to his own amp, and uses the monitor to hear the vocals.  He's probably not listening to his el guitar in the monitor.  If he is listening to the monitor, then I guess he should listen to that & not his amp.  Anyway. . .


. . .even if there are phase issues on stage for the performers it may be near impossible to address them.  The physical placement of the guitar speaker vs. the monitor speaker and their respective distances from where the performer (or audience member) happens to be standing will affect the phase of soundwaves in the air as heard by the listener.  This is true regardless of, and in addition, to the internal phasing issues of the electronic equipment.

[shooter]

Thanks guys, it was a question with no specific purpose other than gaining a better understanding.  HBP, I was only 2 in '59, don't think I could read, but I will look at the pics :)

thanks again
 

[HotBluePlates]

Shooter, I'm not sure what you're asking; or what the stage set-up(s) is (are).

The thing being overlooked is while a discrete signal in a guitar amp can have a single phase per frequency, which is easily determined, sound in air is a different matter.

With some exceptions/limitations, once sound leaves a sound source (maybe a speaker) and is moving through air, it spreads in all directions. Unless you're listening to the sound in an anechoic chamber, that sound is bouncing off every surface. It is also traveling a great many different distances over a great many different direct and reflected paths between the source and your ears. Therefore, the sound becomes a "phase soup" with the direct sound usually loudest, and with an innumerable number of lower intensity delayed (and therefore phase-shifted) copies.

The end product of this mixture gives your brain information on the size and nature of the space in which you & the sound source exist. Do an internet search for "stereo audio phase meter" to get an idea; these show signal phase of stereo signals on a polar plot; a single-phase signal would be a straight diagonal line from upper-right to bottom-left, and that won't describe the images you'll find in a search.

[2deaf]

Frequency-shaping, or parasitic reactances in output transformers in the extremes of their response band, shifts phase because they attenuate response.

Do you mean that the attenuation is causing the phase shift?

[jjasilli]

Yes; I think this is the explanation. When the signal slopes, say down for attenuation, it must travel along a longer line (the hypotenuse of a triangle) to get from Point A to Point B. Like changing lanes in your car on the highway:  if you maintain a constant speed on your speedometer, your actual forward rate of travel will slow down.  So if you want to maintain forward speed while changing lanes, you need to speed up slightly.  There is probably no way to speed up signal in an amplifier, so if you cause signal to slope, it will take a longer to time to get where it's going, compared to signal which was left flat.  The sloped signal will go out of phase with the flat signal.


This is also like the problem posed by Einstein of bouncing a ball while walking.  The ball is no longer traveling in a vertical line up & down as when the person is standing still.  So the ball has a longer distance to travel to return to your hand to bounce it again, compared to when you're standing still. 

[2deaf]

Izzatso.

I can see that PRR is correct by viewing just two equations.  However, the combination of those two equations doesn't convey any cause and effect relationship that is obvious to me.

[shooter]

Like the visuals by JJasilli, the cause are the limits of RC, LC at the edges,(attenuation?) the effect is phase shifting, (signal?)

[PRR]

If frequency response is rising, the faster parts of the signal get-through before the slower parts.

If frequency response is falling, the faster parts of the signal get-through after the slower parts.

Or see HBP's signature.

[PRR]

There is the large question of microphone-loudspeaker feedback. (Accept that an electric guitar is "a microphone", even if not made for that and not very sensitive.)

For any reasonable mike-speaker distance, and a fairly wide frequency band, speed of sound in air ensures that "all" phase-shifts will happen. At 6,750Hz, moving the mike (or speaker) one inch will add 180 deg of phase shift, another inch another 180 deg. At 560Hz, a one foot move is 180 deg phase.

So the feedback path goes in-and-out of phase over the frequency band. As feedback is increased (as gain approaches the howling-point), some frequencies are boosted and some are reduced. Any change of mike-speaker distance changes all these peak/dip frequencies. Adjacent objects (go-go dancers, other players) also change the secondary paths (reflected sounds). So in general, absolute phase is NOT meaningful in mike-speaker systems. If you flip the phase you may lose one howl-point but will find another. Same if you move the mike or speaker.

One scenario where it *may* be useful to try both phases: amplified harmonica "harp". The bandwith may be quite small, the mike-speaker distance may be set by the player. One phase or the other may reduce a troublesome howl, and the other-phase howl may be less troublesome. But unless the player sets the amp the SAME every time (tie the amp and the mouth to a stick?), it is not likely to be the same phase every time. A DPDT switch on the speaker leads may be useful.

[2deaf]

When the signal slopes, say down for attenuation, it must travel along a longer line (the hypotenuse of a triangle) to get from Point A to Point B.

So . . .  when you say a signal slopes down for attenuation, maybe you mean the amplitude of the signal deceases as a function of frequency?  So then as the signal increases in frequency and decreases in amplitude, it must travel along a longer and longer hypotenuse of a right triangle to get from point A to point B?  If the signal is traveling along the hypotenuse to get from A to B, then A and B must be on this hypotenuse somewhere, so where is this shorter line from A to B?  What is this right triangle, anyways?  Since the signals have different travel quantities, is the hypotenuse time?  Maybe the hypotenuse . . .  ah, forget it.

[2deaf]

Like the visuals by JJasilli, the cause are the limits of RC, LC at the edges,(attenuation?) the effect is phase shifting, (signal?)

The limits of a capacitor at the edges are at or approaching DC and when the wavelength of the electromagnetic field generated by the oscillating signal is compatible with the space between the plates of the capacitor.  Both of these are way out of the range of the audio signals with the noted phase effects.

[2deaf]

Once the characteristics of a resister, capacitor, and inductor are defined, everything else is purely mathematical from there on in and the effects are only mathematical relationships and are not cause and effect.

[HotBluePlates]

Frequency-shaping, or parasitic reactances in output transformers in the extremes of their response band, shifts phase because they attenuate response.

Do you mean that the attenuation is causing the phase shift?

No, attenuation does not cause phase shift; they are 2 results of the interaction of reactance and resistance which happen to occur simultaneously, and have a set, repeatable relationship.

Imagine a 2-resistor voltage divider, with no reactance anywhere in the circuit. Regardless of applied frequency the attenuation of the signal is the same, and the phase of the attenuated signal is exactly the same as that of the original signal.

Now imagine a voltage divider made of a resistor and a reactive part (could be a cap or an inductor; we'll talk about a cap here). Because the reactance changes with frequency (i.e., looks like a different number of ohms, and is frequency-dependent) there is changing attenuation with frequency. If you want to know the phase of the resulting voltage output, you do vector algebra to combine the resistive and reactance portions of impedance and determine the resulting phase.

In a purely resistive circuit, the phase angle of voltage and current is 0-degrees; voltage and current waveforms are in lock-step with each other.

In a purely capacitive circuit, voltage lags current by 90 degrees (or the applied voltage appears to have been phase-shifted by -90 degrees). But you never absolutely zero-resistance in any circuit, so the phase shift of voltage across the total impedance is something between 0- and 90-degrees. When capacitive reactance equals the circuit resistance, the phase shift is 45-degrees. This is also the point where the circuit response is -3dB, and a the handy memory tool. When the attenuation is less than -3dB, phase shift is between 0-45 degrees, and when the attenuation is greater than -3dB, the phase shift is between 45-90 degrees.

To determine the amount of phase shift at a single frequency, perform vector algebra. See Chapter 4 of NEETS Module 2 for explanation and examples (but do read the whole thing, as well as Modules 1, and maybe 3-8 or 9, plus 16, 19 & 21).

Scroll down to the Interactive RLC Graph on this page. You only get to use resistance/reactance up to 10Ω, but you can mentally scale that up by adding zeros. Slide the Blue Dot (Inductive Reactance) to zero. Leave the Red Dot (Resistance) on 10Ω. Slide the Green Dot (Capacitive Reactance) down to 10Ω. The diagonal Black Line is the resulting Impedance (in this case 14.15Ω), and the angle between the horizontal resistance axis and the Black Line is the resulting phase angle (in this case, minus 45-degrees because resistance and reactance are equal).

Trigonometry comes back to haunt you: You could have used the Pythagorean Theorem to find the total impedance. Impedance = √(X_c² + R²)  ->  √(10Ω² + 10Ω²) = √(200) = 14.1421356... (Looks like the graphing tool rounds up) You could find the phase angle by calculating tan^-1 -X_c/R  -> tan^-1 -10Ω/10Ω = tan^-1 -1 = -45 degrees.

Keystrokes on your computer's calculator will be: [-] [X_c] [/] [R], then [Inv], then [tan]  ([tan] may have changed to [tan^-1] after hitting [Inv]).

Since the total impedance calculation and the phase-angle calculation are based on X_c (capacitive reactance in ohms), and that reactance changes depending on the applied frequency, then both the total impedance (and attenuation if this is a voltage divider) and phase angle will change accordingly. Which is why I said attenuation does not cause phase shift, but it correlates to phase shift; they're both happening simultaneously.

[HotBluePlates]

... There is probably no way to speed up signal in an amplifier ...

My last example talked about capacitive reactance, where the phase on the voltage lags the current in the circuit. But in inductive reactance, the phase of the voltage leads the current in the circuit, and so the voltage appears advanced in phase by 90-degrees.

The memory aid for this is "ELI the ICE man." In an L (inductive) circuit, Voltage (E) leads Current (I); in a C (capacitive) circuit, I (current) leads Voltage (E). Or for the latter you could say "voltage lags current".

Since we are generally more concerned about voltage in the parts of amplifiers we're likely to care about phase issues, you could say that if the circuit contains only R and C, the voltage will be delayed in phase compared to the original applied signal. If the circuit contains only L and R, voltage will be advanced in phase (opposite of delayed, or "sped up" in Jjasili's mental picture) compared to the original signal.

If L and C are present (with or without R), you find the total as X_L-X_C, then plug into the equations I showed to find total impedance and/or phase angle. Or you could plot the values graphically; X_L is always the positive vertical axis, X_C is always the negative vertical axis, R is always the positive horizontal axis. To find the total resulting reactance, subtract the smaller of X_L or X_C from the larger.

[HotBluePlates]

Like the visuals by JJasilli, the cause are the limits of RC, LC at the edges,(attenuation?) the effect is phase shifting, (signal?)

The limits of a capacitor ...

I think Shooter was talking about the upper and lower limits ("edges") of the audio pass-band, where lows and highs are rolled off. Then his statement is accurate.

[shooter]

I was indeed and I forgot about eli but then I remembered,(MFC), Mary's fuzzy, um, well I think that's a different thread.
Thanks HBP!  time for a 3rd cup-o-coffee

[2deaf]

NEETS defines the vertical axis as X_C-X_L on page 4-14 and again on 4-37.  In their example they came up with positive 10 ohms, but drew it in a negative direction.  In college physics, they taught us that the vertical vector was X_L-X_C and showed us the math that lead to that definition.  Of course NEETS never says that the vertical axis is a vector and it obviously is not.  The physics version is mathematically correct and it also eliminates the possibilities for error that the NEETS version has.   

[HotBluePlates]

NEETS defines the vertical axis as X_C-X_L on page 4-14 and again on 4-37. 

No, they don't. Look at Figure 4-7 on page 4-14; the X_L portion of the vertical axis is in the positive direction, matching the information on the website I provided. Likewise, X_C is in a negative direction, matching the website. Page 4-14 shows X_C-X_L but X_C is the larger quantity, and both are treated as positive numbers. However, on page 4-16 X_L is the larger quantity, so the formula becomes X_L-X_C.

The worthwhile correction is "the smaller quantity is always subtracted from the bigger quantity." I've changed that in the post above.

... In college physics, they taught us ...

Absolute Book-Correct Answer: The form is always X_L+X_C however the result always has an absolute-value smaller then the figure with the greatest magnitude. X_C is always a negative number because it is in a negative direction. If X_C is the larger value,it  gets squared to find the resulting impedance, so the minus sign doesn't matter.

If you carry-over the minus sign to the calculation of phase angle, you get a semi-useful indication that the resulting negative angle indicates a capacitive circuit. However, it is still up to the user to apply their knowledge to determine if it is current or voltage which is lagging in phase.

The NEETS process, as amended above, always gives the correct answer. NEETS  places the same burden on the user to have understanding of the type of circuit they're looking at. It does, however, avoid the possibility of the user seeing "addition" and simply adding the numbers. The potential danger is when the resulting figure is bigger than each of the 2 values used, the user may not see this as an obvious error.

I don't believe the question or this explanation really contributed meaningfully to an understanding of impedance and phase angle.

[2deaf]

The NEETS process, as amended above, always gives the correct answer. It is up to the individual to know that if X_C is the greater quantity, the resulting phase angle will be negative. . .

Yeah, it's up to the individual to know and that is where the errors start.  If you set the triangle up like they taught me, you can ascertain whether the phase angle is positive or negative by which way the vertical vector is pointing.  The only trick now is to define that vector properly and that is where all those gimmicks posted above come in. 

[HotBluePlates]

Yeah, it's up to the individual to know and that is where the errors start.  If you set the triangle up like they taught me ...

Yes, draw the triangle. NEETS assumes you draw the triangle.

And when I was in the Navy and being taught this stuff, we spent days drawing triangles and doing the calculations. I said "NEETS assumes ..." because after drawing the triangle, you know the stuff which NEETS assumes you know.

I said before this wasn't contributing to the thread because I think there is disagreement over the fact that we really agree and are saying the same thing! 

[2deaf]

NEETS defines the vertical axis as X_C-X_L on page 4-14 and again on 4-37. 

No, they don't.

What?  I'm looking at it right now and there is a vertical vector that forms a non-hypotenuse leg that is labeled "X" and it says right there that X=X_C-X_L.  It's right there in black and white.  The worthwhile correction would be to define it as X_L-X_C.

The equation on pg. 4-13 that he squared both sides of to get a Pythagorean equation has X_L-X_C in it, but when he drew the triangle he reversed it.  It's completely ridiculous to draw something that requires a rather questionable, convoluted interpretation in order to get the correct answer.

I don't believe the question or this explanation really contributed meaningfully to an understanding of impedance and phase angle.

A correctly drawn vector triangle is immensely useful when it comes to LCR circuits.

And when I was in the Navy and being taught this stuff, we spent days drawing triangles and doing the calculations. I said "NEETS assumes ..." because after drawing the triangle, you know the stuff which NEETS assumes you know.

It's certainly not my intent to insult the U.S. Navy and this book was taken from civilian work, anyways.

You can draw triangles with an error until the cows come home and it still won't make it right.  You shouldn't have to do that, anyways.

Let's try NEETS drawing.  Let's say X_C is greater than X_L.  They show a vector (right there in black and white) that is defined as X_C-X_L.  X_C is greater, so the vector is positive.  Now let's calculate theta using the tangent anti-function.  We get a positive number, but we know it should be negative from that Eli the Ice detective gimmick.  We got the wrong answer.  It doesn't matter what NEETS assumes and assumptions have no place in mathematics.  We got the wrong answer.   

EDIT: Expletive removed.

[HotBluePlates]

Let's try again.

NEETS defines the vertical axis as X_C-X_L on page 4-14 ...

No, the vertical axis is reactance.

In Fig 4-7, Resistance is along the horizontal axis; take the arrowhead of that vector to be pointing at the origin of the graph. Upward from the origin is X_L, and is labeled "X_L=10Ω". The distance from the arrowhead of R's vector to the top of the X_L=10Ω vector is obviously 10Ω. Downward from the R vector arrowhead is X_C which is labeled for this example "X_C=20Ω". This is consistent with what you're saying regarding direction of X_C. The length of this vector is twice and long from the origin as the vector of X_L, because 20Ω is twice 10Ω.

Because X_C is larger than X_L the text example is overlaying an additional resultant vector of  X_C-X_L which is 20Ω-10Ω=10Ω. That's why there are 2 arrowheads moving downward from the origin. The equation next to the resultant vector is not the "definition of the axis" but the math operation that determined the length of the resultant vector.

  Now let's calculate theta using the tangent anti-function.  We get a positive number, but we know it should be negative from that Eli the Ice___ gimmick.  We got the wrong answer.  It doesn't matter what NEETS assumes and assumptions have no place in mathematics.  We got the wrong answer.     

Do you agree the triangle in Fig 4-17 is drawn correctly? That is, for the values of reactance given, the resulting reactance is 10Ω in the capacitive direction, so the hypotenuse is below the horizontal axis? If so, we're in agreement, at least up to this point.

There are different ways used by different books to attack the phase angle issue, though the overall math is the same:

• Always start by getting a resultant reactance using X_L-X_C. As you note, this avoids later confusion about the direction of the impedance phase angle, especially if you don't draw the triangle. Positive angles mean the circuit is inductive, negative angles mean the circuit is capacitive.
• Always draw the triangle; subtract the smaller reactance from the larger, regardless of which that places first. Calculate the phase angle, then look back at the triangle diagram to determine if the circuit is net-inductive or net-capacitive.

Neither method is right or wrong, though I would prefer #1 personally. You are correct that it is less prone to error.
The drawback of #1 is one might assume since the phase angle is negative, that indicates a delay of phase. This is incorrect in that voltage would be delayed, but current is not. You still have to know whether it is voltage or current which is delayed in phase, depending on whether you're analyzing an inductive or capacitive circuit.

[2deaf]

Let's try again.

You don't need to write out all this explanation.  I understood it perfectly 42 years ago and I still understand it today.

Quote from: 2deaf on January 23, 2015, 10:07:07 am
NEETS defines the vertical axis as XC-XL on page 4-14 ...

No, the vertical axis is reactance.

You got me there.  It's the vector that is defined and that's what I've been talking about the whole time.

That's why there are 2 arrowheads moving downward from the origin.

I understood that immediately upon viewing it.  I also read what the man in NEETS said about it being customary to draw X_L upwards and X_C downwards.  Is there a gimmick to remember that?

The equation next to the resultant vector is not the "definition of the axis"

I only made that mistake once (I hope).

the math operation that determined the length of the resultant vector.

Vectors have a direction and a magnitude.  The magnitude is the only place where you and I disagree.  Because the endpoint is placed on the line with positive numbers above it and negative numbers below it, it makes a difference if the magnitude is positive or negative.  The equation in NEETS yields a vector in the wrong direction.

Do you agree the triangle in Fig 4-17 is drawn correctly? That is, for the values of reactance given, the resulting reactance is 10Ω in the capacitive direction, so the hypotenuse is below the horizontal axis? If so, we're in agreement, at least up to this point.

Of course.

The drawback of #1 is one might assume since the phase angle is negative, that indicates a delay of phase

People assume all kinds of goofy things.  It never occurred to me to interpret a negative phase angle in that fashion.