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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

Showing 5 responses by almarg

almarg's avatar
almarg
9,670 posts
 
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
almarg's avatar
almarg
9,670 posts
 
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
almarg's avatar
almarg
9,670 posts
 
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
almarg's avatar
almarg
9,670 posts
 
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
almarg's avatar
almarg
9,670 posts
 
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
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