power cables - not in the signal path?


According to popular wisdom the AC power is not in the signal path and therefore a power cord, AC conditioner or similar should have zero audible effect.

In a don quixotesque attempt I'd like to turn this perception around: the AC is 100% in the signal path - more so that the actual low-level signal that gets amplified, and I think I found the simple words to clarify this.

The low-level signal is actually only modulating the high-voltage (high-intensity) signal produced by the transformer. Those electrons from the transformer are the actual electrons we "hear". The low-level signal is simply lost in translation. In a simple example, a 0.1V peak-to-peak sine signal gets amplified (say) 10x by a 10 V continuous (transformed) DC. The output is (say) a 1V sinusoid oscillating back and forth in time. If the 10V continuous is NOT actually exactly 10V (but is actually has noise) - then the noise will directly reappear "riding" on the 1V output.

Hence the need to keep the AC noise-free.

(Of course I purposefully neglected for simplicity the other effects (need for instantaneous delivery of power, etc..) for which I did not find a simple enough description (without reference to I/V curves and impedance / capacitance details, that is).

Does is make sense?

Thanks
C.
cbozdog
In a don quixotesque attempt I'd like to turn this perception around: the AC is 100% in the signal path - more so that the actual low-level signal that gets amplified, and I think I found the simple words to clarify this.

The low-level signal is actually only modulating the high-voltage (high-intensity) signal produced by the transformer. Those electrons from the transformer are the actual electrons we "hear". The low-level signal is simply lost in translation. In a simple example, a 0.1V peak-to-peak sine signal gets amplified (say) 10x by a 10 V continuous (transformed) DC. The output is (say) a 1V sinusoid oscillating back and forth in time. If the 10V continuous is NOT actually exactly 10V (but is actually has noise) - then the noise will directly reappear "riding" on the 1V output.
No, that is not correct. The main thing you are not realizing is that the circuit stages which provide amplification have a property known as Power Supply Rejection Ratio (PSRR), which (assuming reasonably competent design) results in the amount of noise appearing at the outputs of those stages being GREATLY less than the amount of noise which may be present on the DC power being supplied to them.

Also, of course, in the process of converting AC to DC the component's power supply itself will greatly reduce noise that is present on the incoming AC, as will filtering and "decoupling" capacitors that are normally provided at numerous locations elsewhere in the design.

IMO it is a misleading oversimplification to consider the AC power provided to a component as being in the signal path, while at the same time it would be incorrect to view AC power as not being able to affect the signal path.

Regards,
-- Al
C, R U trippin? Can you share info with substance abuser like I am on substance to use? It must be darn great!
EVERYTHING IS IN THE SIGNAL PATH
Would this be a good way to visualize it?

Think of a car as an audio system and the road represents your power/electricity. The car/audio system has a level of maximum performance it can achieve. But you can only get the best performance if you drive on a very good road. If you drive on a road that is in bad condition, you can't drive the car to its fullest potential. So even though the electricity going through the power cables in not part of the signal path, it can still effect it.
Eh - sure the original post is oversimplification. Not doubting the other effects that are meant to improve the (former) AC signal, i neglect them in order to keep it easy to follow. All conversion, filtering and capacitive storage aside - the low-level signal does nothing more than opening and closing (I e modulating) the passage of a huge amount of electrons (obtained from the AC originally) through the device. It is not the low-level signal electrons that eventually get heard. if anything, it is the latter electrons that passed through the floodgate that eventually reach the audio transducer, etc.

I do stand corrected though regarding complexity - surely all amp designers make provisions to have a time-invariant pool (DC) to draw from. Maintaining such pool is hard though (is probably what separates boys from men).

No substance abuse here - sorry C.
I've found power supply mods to make major audible improvements in whatever audio device applied(tubed, SS, analog, digital, all of my own and quite a few happy customers), for a few decades. The new breed of PCs and fuses have seemed to me, the icing on the cake. My take on the function of power supplies and gain stages/output devices was reinforced by Viktor Khomenko, in an interview with Stereophile, some years back(made me feel quite good). Couldn't find that interview(online), but it was partially echoed here: (http://www.theabsolutesound.com/articles/balanced-audio-technology-vk-600m-se-monoblock-power-amplifier-1/)
C, It's entertaining indeed! I should read it again and than few more times when I get home and get stoned.
Cbozdog writes:
It is not the low-level signal electrons that eventually get heard. if anything, it is the latter electrons that passed through the floodgate that eventually reach the audio transducer, etc.
Fascinating. You are attributing properties to electrons that they simply do not have.

Regards,
Please keep in mind that it's the much much faster electric field and not the electron flow. Sure, those electrons do make it down the wire eventually, but it's more like a train with a lot of cars. You move the 1st one, and the Caboose moves. That's the electric field. It's not like a gun firing electrons down the wire.
You may find this treatise interesting: (http://amasci.com/miscon/whatis.html)
One assumes the electromagnetic field, the signal, in the conductor is made up of photons, no?
Czarivey, I expect great philosophical insight from you on the topic. Please, do give it a stoned thought.
Signal path or not, clean AC is key to great sound. My personal experience is that providing exceptionally clean AC to your equipment via the best conditioners and cables can result in improvements to SQ that are absolutely stunning, often greater than a major component upgrade.
C, Beleive me it's very haaard to insite some theory or even philosophy or at least simply to relate neither your first original post or second, but I'll try to make sense on my next post when I'm really stoned and if I find even THAT hard, I'll try to get even more stoned.
For simplicity in general:
AC -> Power Cord -> Power Transformer -> Modified AC voltage(s) increased or decreased -> Rectifier(s) -> DC voltage(s) provides proper offset(s) for active circuit elements such as transistors, tubes, diodes etc.
Did I mention anything about signal path?
Don't go towards bunch of electrons, because it's even more confusing.
Czarivey - you said "proper offsets"? Please, do continue.
06-10-14: Stevecham
EVERYTHING IS IN THE SIGNAL PATH
"Everything," so to speak, can AFFECT the signal path, to one degree or another. For example, turning on a dimmer switch in a different part of the house from the listening room can introduce sufficient distortion on the AC line to have audible consequences, in some situations. But I don't think that most of us would consider that dimmer switch to be "in the signal path," i.e., to be part of the path that is traveled by the audio signal.

Regards,
-- Al
Geoffkait - no. EM radiation is produced when electrons move (and radiates away from the source that produced it). Such EM radiation decays rapidly with distance but can influence (make them move a bit) other charge carriers (electrons) nearby (for example, those in another cable). Such EM radiation is carried by photons with energies in the RF range, I think.
Cbozdog, thanks for the response. I suspect we are pretty close, I am pretty sure ALL electromagnetic waves are actually photons, including Gamma Rays, RF, x-rays, and music signals. RF waves! to take an example do not attenuate much over distance, traveling at the speed of light or close to it.
Almarg - sure, the signal path is that which carries the information. Nothing is in the signal path until it is. The signal changes many hands (digital, analog, sometimes back and forth) - each time is actually physically converted into something else. In that sense, even amplification is a conversion (to a higher-intensity signal). For the purpose of this example, is might be viewed as taking clay from a jar (AC, or a rectified version thereof) and molding a faithful reproduction (or encryption) of the original.

Blah, this came out too poetic - sorry.
C, offset determines point of quiescent operation or stand-by or zero signal.
Czar - if that's what you were gonna say next... you probably stopped in time in the first place. My bad.
06-11-14: Cbozdog
... EM radiation decays rapidly with distance...

06-11-14: Geoffkait
RF waves! to take an example do not attenuate much over distance, traveling at the speed of light or close to it.
You're both right, sort of. I believe that the main reason RF attenuates as distance increases is that it "spreads out," and therefore while the total amount of energy that is present at a given distance does not decrease a great deal as distance increases (assuming propagation through a medium that is conductive to it, such as air or a vacuum), the amount of energy reaching a given cross-sectional area at which it may be received will decrease considerably as distance increases.

The degree to which that happens will of course vary greatly depending on the directionality of how the energy is launched, e.g., omni-directionally, or as a beam that is focused with some particular degree of sharpness.

Regards,
-- Al
Well, not sure I agree with your detective work, unless by "spreads out" you mean absorption and scattering. Low level signals for cell phones work pretty well even though they are "spread out." RF is not necessarily line of sight. Cheers, Geoff
Geoffkait and Almarg - agree with both, with the small exception that attenuation of radiation takes place in any medium that is not vacuum to various degrees depending on medium and frequency (with the most striking example being metals - which are very effective in absorbing said radiation for practically all frequencies relevant here). The reason is (again grossly simplifying) that metals have those free carriers ready to move when they get pushed around by EM, so the EM energy reaching them is converted to electron motion.
The EM discussion touches on another topic (which I just realized). Kinda fun actually. Why do the "magic stones" work? (not that I have ever tried one, but I think I might).

The EM radiation at these frequencies is fairly delocalized - according to Heisenberg - hehe (the wavelengths can be on the order of yards to miles, compared to the wavelength of visible light which is on the order of fractions of micron). As such - any given quantum of radiation can be anywhere within that radius (with a certain probability). However, as soon as the wavefunction of said radiation overlaps with a strong absorber - there is a probability that it will get absorbed into it - and thus disappear for all other possible absorbers (such as unsuspecting cables lying around).

With that in mind - having such "magic stones" placed around the source of EM (but not too close so that their action as secondary emitters would not have least impact) would effectively suck RF from the environment.

Ha! If true then here goes the "magic stone" business out of business (now anyone can dump their spare iron and wrenches and such around their electronics and get crystal clear audio).

(incidentally - delocalization of wavefunction can be viewed as a reason for diminished RF influence with distance - the probability that the wave/quantum will be absorbed does decrease with distance - the more localized the wf, the quicker the probability that it will affect nearby cables diminishes).
06-12-14: Cbozdog
Geoffkait and Almarg - agree with both, with the small exception that attenuation of radiation takes place in any medium that is not vacuum to various degrees depending on medium and frequency (with the most striking example being metals - which are very effective in absorbing said radiation for practically all frequencies relevant here). The reason is (again grossly simplifying) that metals have those free carriers ready to move when they get pushed around by EM, so the EM energy reaching them is converted to electron motion.
Cbozdog, agreed. As I indicated, I was "assuming propagation through a medium that is conductive to it, such as air or a vacuum," air being a bit less conductive than a vacuum, consistent with your comment.

It should perhaps also be stated that the earlier comments pertain to propagation of electromagnetic energy through free space, as opposed to through coaxial cables, waveguides, fiberoptic cables, etc., for which the factors affecting attenuation as a function of distance are of course completely different than the "spreading out" effect I referred to.

Regards,
-- Al
Geoff, no, I'm not referring to absorption and scattering. I'm referring simply to a matter of geometry. A radio antenna, a radar antenna, a flashlight, etc., will not radiate energy in a beam that remains at precisely/exactly/100.00000000% the same diameter is it travels out over an extended distance. The cross-sectional area of the beam will increase in size as the distance from the source increases. That will happen to the greatest degree, of course, if there is no beam to begin with, i.e., if the radiating antenna or other source is omnidirectional. The surface area of a sphere increases exponentially as its radius increases (or more precisely, in proportion to the square of the radius).

Therefore the fraction of the originally radiated energy which intersects a GIVEN cross-sectional area will decrease as the distance of that cross-sectional area from the source becomes greater.

That is the main reason why although a radio station's transmitter may send out many thousands of watts, the signal seen by the input circuit of a radio receiver may be only a few microvolts, corresponding as a rough order of magnitude to perhaps a trillionth of a watt.
Low level signals for cell phones work pretty well even though they are "spread out." RF is not necessarily line of sight.
True, but not relevant to my point.

Regards,
-- Al
"...not relevant..."

Hey a new advertising slogan for many MD products!!! %^)
Hey, Mapman, see you finally woke up out of your stupor, welcome back!

:-)