I would just add to what has been said the thought that if there are in fact mechanisms by which warmup beyond the point of thermal stabilization can produce audibly significant changes, I see no reason to expect those changes to necessarily be for the better. And I would certainly expect both the magnitude and the character of the changes to be dependent on the specific design.
Regards, -- Al |
09-12-11: Rower30 What there is, is an increase in capacitance as the leads get longer (say bye bye to higher frquencies at 3 dB per octave at the filter fo frequency which drops as the leads get longer). So all the gold and silver on earth won't do anthing except take your money. What WILL help, is short LENGTH and low capacitance.
Balance leads have HALF the capacitance as single ended RCA leads. The two wires are about TWICE as far apart, leading to LOW capacitance (about 8 Pf ft verses 17 PF/ ft). So the roll off is half the single ended RCA's roll-off. And, you get NOISE rejection added in for a bonus. Some minor factual corrections to your post: "3db per octave" should be "6db per octave" (at frequencies above the 3db point). Balanced cables don't necessarily have half the capacitance of single-ended cables. Besides the capacitance between the two balanced conductors, each conductor will have a significant capacitance to ground, via the shield and/or separate return conductor. In conjunction with the source impedance, that capacitance will also result in a low pass filter effect, reducing the voltage differential between the two signal conductors at high frequencies. See for example this datasheet on Mogami 2534, which in the popular balanced quad configuration has a higher capacitance from conductor to shield than from conductor to conductor. And the two wires are certainly not necessarily "twice as far apart." The idea is that noise pickup should be common mode (equal between the two conductors) to the greatest extent possible. Moving the conductors apart would work in the opposite direction. Also, as you probably realize, the 8pf and 17pf figures you cited will vary very widely depending on the particular cable design. Regards, -- Al |
Lev, your amp's power consumption is specified as 650W in "on" mode (presumably under no signal conditions), 130W in "standby," and 10W in "sleep" mode.
If it is in "on" mode all the time, that computes to 0.65 kilowatts x 24 hours per day x 30 days per month (approx.) = 468 kilowatt-hours per 30 days (yikes!). Multiply that by your kilowatt-hour rate to get the approximate cost per month.
For the other modes, use 0.13 and 0.01, respectively, in place of the 0.65 factor.
Best regards, -- Al |
Paperw8, what Rower30 is referring to is the fact that the capacitance of a line level interconnect will interact with the output impedance of the component which drives the cable, forming an RC low pass filter. The "R" corresponds to that output impedance, not to the resistance of the cable.
The bandwidth of that low pass filter can certainly be low enough under some circumstances to have audible consequences, if cable length is long, cable capacitance per unit length is high, and the output impedance of the component driving the cable is high.
For speaker cables, on the other hand, that effect will almost always be insignificant, because of the very low output impedance of the power amplifier. Inductance and resistance may be significant considerations in the case of speaker cables, though, because of their interaction with speaker impedance.
Regards, -- Al |
Paperw8, on the 3db vs. 6db per octave question, see this. Regards, -- Al |
09-13-11: Paperw8 the equation for dB measurement is: q_dB=10*log(q1/q2) where q1 and q2 are measured quantities and q_dB is the dB measurement for those quantities. when q1=q2/2, q_dB is -3dB. if you read closely the reference that you cited, you can see that what the writer is saying is clearly wrong. 6dB/octave is the falloff for a *second* order filter; the filter shown in the cited reference is a first order filter. Paperw8, we had discussed the definition of db for electrical signals in another thread a while back, and as I indicated then, with all due respect you are simply wrong. Please do some further research, and I think you will see that: db = 10log(P1/P2) = 20log(V1/V2) where P1/P2 is the ratio of two powers, and V1/V2 is the ratio of two voltages. 6db is 6db, regardless of whether it is derived on the basis of power (where it represents a factor of 4) or voltage (where it represents a factor of 2). I should qualify my statement about 6db/octave rolloff for a first order filter, though, by adding that at frequencies that are just a small amount higher than the 3db point (Fc) the rolloff will be somewhat less than 6db/octave. As frequency increases further the rate of rolloff will become progressively closer to 6db/octave. The actual equation is: Vout/Vin = 1/(square root(1 + (f/Fc)^2)) During the first octave above Fc, a rolloff of about 4db will occur. As f increases further, the "1" in the denominator becomes progressively less significant, bringing the rolloff rate progressively closer to 6db/octave. Regards, -- Al |
Paperw8, I repeat. With all due respect your continued insistence that db = 10log(V1/V2) is completely wrong. Please do some further research.
I could perhaps further support the credibility of my statements by describing my academic and professional background, but we've already diverted the topic of this thread far enough.
BTW, in case it's not clear, my statement that db = 10log(P1/P2) = 20log(V1/V2) obviously assumes that the load impedance is the same for all of the measurements.
Regards, -- Al |
Paperw8, let me supplement the reference I previously provided with this one, supporting my contention that db = 20log(V1/V2), rather than 10log(V1/V2) as you have claimed. It was written by someone possessing technical credentials that are utterly impeccable. See his biographical information here. With respect to your related erroneous contention, that first order filters roll off at 3db/octave, I suggest that you Google the terms "first order filter" and "6 db per octave," placing both phrases within the search term field, with the quotes. You will find countless references supporting the 6db figure. If after viewing all of that, as well as the reference I previously provided, you find yourself continuing to insist that db = 10log(V1/V2) and that first order filters roll off at 3db per octave, I will have no further comment. Regards, -- Al |
Dave, I'll go with the beers and the music. Further explanation seems pointless. JeffreyBehr, thanks for your comment, which of course is 100% correct.
Regards, -- Al |
Kijanki, no problem with the beer. Thanks for your comment, which as I'd expect is totally correct, with P representing power and U representing voltage.
Ralph (Atmasphere), just saw your post. Thanks also.
Best regards, -- Al |
What the equation means is that when voltage is reduced to half it's original value, power is reduced to one quarter it's original value. Yes, of course. I stated that in my post here. Hold on, this goes to my previous comment: you can't just cite equations without understanding what the equations mean. first, the equation 20log(v2/v1) is a dB relationship in *power* not in voltage; a relationship which is true under certain conditions (which, btw, are articulated in the previously cited henry ort reference as well as in my previous comments). Paperw8, the fundamental misconception you have, which ultimately leads you to incorrectly assert that first order filters roll off at 3db/octave, is the notion that the numerical db value describing the ratio of two signals applied to a given resistive load will be different depending on whether voltage or power is being considered. I'll mention, btw, that early in my career as an electrical design engineer, a great many years ago, I had the exact same misconception, until my boss enlightened me. The value of that number is one and same, whether voltage or power is being considered. For a given resistive load, reducing the applied voltage by a factor of 2 reduces power by a factor of 4 (as you agree), and the change in signal level is 6db. Period. The db change in that situation if voltage is being considered is 6db; the db change if power is being considered is 6db; the db change is 6db, period. That is why the formula for db as computed from voltage levels includes the constant "20," while the formula for db as computed from power levels includes the constant "10." Otherwise the two numerical values wouldn't work out to be the same, as you'll agree. Kijanki's post provides an elegant mathematical proof of the equivalency of the two formulas, 10log(P1/P2) and 20log(V1/V2), as does the Ott paper, in a different way. But until you recognize that a db is a db, regardless of whether power or voltage is being considered, the other differences in our positions, including the issue of filter rolloff which started the discussion, will remain unreconcilable. Regards, -- Al |
09-14-11: Paperw8 v_dB=10*log(v_out(f)/v_in(f)) v_dB=10*log(1/2)/octave v_dB=10*(-0.3)/octave v_dB=-3dB/octave
that's the result; it's not subject to your opinion or my opinion, that's just what it is. As I said, you have a fundamental misconception which underlies all of the differences in our positions. The number "10" in all of your equations that I've quoted above should be "20," since you are computing the number of db based on a voltage ratio. I've explained it, Kijanki provided a proof of it, and Ott provided a proof of it. I've recommended multiple times that you do further research to convince yourself of it. Until you understand that the number 10 is the wrong number to use in converting voltage ratios to db, we will get nowhere. Regards, -- Al |
if power is reduced by half is that a 3dB reduction or a 6dB reduction. I'm sure Kijanki will answer the part of your question that was directed to him, but in the meantime I'll state that if power is reduced by half that is a 3db reduction. That same 3db reduction also reflects voltage being reduced to 0.707 of its original value. ort isn't saying that 20*log(v2/v1) is a voltage dB, what he is saying is that it is a power dB expressed as a ratio of voltages. I have tried to explain in my previous posts that it is incorrect to distinguish between a "voltage db" and a "power db." A db is a db. Regards, -- Al |
09-14-11: Paperw8 09-14-11: Kijanki Voltage ratio in electronics is, and always has been 20log(v2/v1). Pretty much anything other than power is always 20dB(k2/k1) including sound pressure, sound level etc. -3dB of voltage means 0.708 of a value. i see; so in your mind a 3dB reduction in power "has always" meant that power if reduced to 0.708 of it's original value and not 0.5 of it's original value. ok...i can see that when you drill down, there is much inconsistency among the dB equation citing crowd. you're not even consistent with almarg; at least he realizes (i think) that a 3dB reduction in power means that power has been reduced to 0.5 of it's original value.
Paperw8, Kijanki said that a reduction in VOLTAGE to 0.708 corresponds to a 3db reduction, which is the same thing that I said (although I rounded off differently, to 0.707). Obviously, for a given load a reduction in voltage to 0.708 corresponds to a 50% reduction in power, which in turn corresponds to the same 3db. There is no inconsistency among what I, Kijanki, Atmasphere, and JeffreyBehr have said on the db and filter rolloff issues. when you say that a first order filter falls off by 6dB/octave, that is a statement of how the power levels change. As I have tried to explain multiple times in different ways, it is a statement, for a given load, about how both the power levels AND the voltage levels change. It means that in one octave the voltage level has been reduced by a factor of 2, and the power level has been reduced by a factor of 4. Is that not clear? Regards, -- Al |