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#81
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"dangling entity" wrote in message m... Kevin McMurtrie wrote in message ... It's all irrelevant for audio frequencies and normal lengths of wire. Having the two conductors side by side is perfectly good. Just don't split the wires and route them to the speaker along opposite walls. Just curious, but what *would* happen if you did that? The listener(s) would be inside a 1-turn magnetic loop. Actually, some hearing-assist systems work by exactly that method. They use "receivers" with pickup coils and amps that drive the headphones. And some hearing aids will pick it up directly (from the coils they use to pick up telephone receiver audio.) |
#82
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"dangling entity" wrote in message m... Kevin McMurtrie wrote in message ... It's all irrelevant for audio frequencies and normal lengths of wire. Having the two conductors side by side is perfectly good. Just don't split the wires and route them to the speaker along opposite walls. Just curious, but what *would* happen if you did that? The listener(s) would be inside a 1-turn magnetic loop. Actually, some hearing-assist systems work by exactly that method. They use "receivers" with pickup coils and amps that drive the headphones. And some hearing aids will pick it up directly (from the coils they use to pick up telephone receiver audio.) |
#83
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"Richard Crowley" wrote in news:vvb788bo5ro346
@corp.supernews.com: "Bruce" wrote ... Transmission line theory is useless below about a 1/10 wavelength. At 20 kHz, this is several thousand feet. Why would anyone want to worry about 100kHz, when most people can no longer hear 20kHz? If you have to ask then you'll never get it! :-)) Riiiiggghhhhtt..... |
#84
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"Richard Crowley" wrote in news:vvb788bo5ro346
@corp.supernews.com: "Bruce" wrote ... Transmission line theory is useless below about a 1/10 wavelength. At 20 kHz, this is several thousand feet. Why would anyone want to worry about 100kHz, when most people can no longer hear 20kHz? If you have to ask then you'll never get it! :-)) Riiiiggghhhhtt..... |
#85
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"Richard Crowley" wrote in news:vvb788bo5ro346
@corp.supernews.com: "Bruce" wrote ... Transmission line theory is useless below about a 1/10 wavelength. At 20 kHz, this is several thousand feet. Why would anyone want to worry about 100kHz, when most people can no longer hear 20kHz? If you have to ask then you'll never get it! :-)) Riiiiggghhhhtt..... |
#86
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#87
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#88
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#89
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#91
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#92
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"Rusty Boudreaux" wrote in message
So how do you enter into the computer the load impedance being it is a complex fuction of frequency? There a two ways one could do it. 1) One could use a general lumped constant model of a speaker as the termination of the line. 2) The better way is: measure the input impedance of a given speaker at many frequencies (at 100 frequencies would be good). Mathemetically convert the input impedance into S11 (input reflection cofficient). One can then put in dummy values for S22 (output reflection cofficient), S12 (reverse transmission cofficient) and S21 (forward transmission cofficient). Create a two-port (data file) device from the data. Terminate the cable (in the computer) with the two-port (data) device. The two-port device will present exactly the impedance you measured, at the frequencies you measured. The computer will interpolate values in between the frequencies you measured, to give the correct complex impedance, at all frequencies. Simple, no? :-) Bob Stanton |
#93
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"Rusty Boudreaux" wrote in message
So how do you enter into the computer the load impedance being it is a complex fuction of frequency? There a two ways one could do it. 1) One could use a general lumped constant model of a speaker as the termination of the line. 2) The better way is: measure the input impedance of a given speaker at many frequencies (at 100 frequencies would be good). Mathemetically convert the input impedance into S11 (input reflection cofficient). One can then put in dummy values for S22 (output reflection cofficient), S12 (reverse transmission cofficient) and S21 (forward transmission cofficient). Create a two-port (data file) device from the data. Terminate the cable (in the computer) with the two-port (data) device. The two-port device will present exactly the impedance you measured, at the frequencies you measured. The computer will interpolate values in between the frequencies you measured, to give the correct complex impedance, at all frequencies. Simple, no? :-) Bob Stanton |
#94
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"Rusty Boudreaux" wrote in message
So how do you enter into the computer the load impedance being it is a complex fuction of frequency? There a two ways one could do it. 1) One could use a general lumped constant model of a speaker as the termination of the line. 2) The better way is: measure the input impedance of a given speaker at many frequencies (at 100 frequencies would be good). Mathemetically convert the input impedance into S11 (input reflection cofficient). One can then put in dummy values for S22 (output reflection cofficient), S12 (reverse transmission cofficient) and S21 (forward transmission cofficient). Create a two-port (data file) device from the data. Terminate the cable (in the computer) with the two-port (data) device. The two-port device will present exactly the impedance you measured, at the frequencies you measured. The computer will interpolate values in between the frequencies you measured, to give the correct complex impedance, at all frequencies. Simple, no? :-) Bob Stanton |
#96
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#97
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#98
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#99
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#101
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#102
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#103
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#104
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#106
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#108
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"Stewart Pinkerton" wrote in message ... On 3 Jan 2004 04:39:58 -0800, (Bob-Stanton) wrote: "Rusty Boudreaux" wrote in message So how do you enter into the computer the load impedance being it is a complex fuction of frequency? There a two ways one could do it. 1) One could use a general lumped constant model of a speaker as the termination of the line. 2) The better way is: measure the input impedance of a given speaker at many frequencies (at 100 frequencies would be good). Mathemetically convert the input impedance into S11 (input reflection cofficient). One can then put in dummy values for S22 (output reflection cofficient), S12 (reverse transmission cofficient) and S21 (forward transmission cofficient). Create a two-port (data file) device from the data. Terminate the cable (in the computer) with the two-port (data) device. The two-port device will present exactly the impedance you measured, at the frequencies you measured. The computer will interpolate values in between the frequencies you measured, to give the correct complex impedance, at all frequencies. Simple, no? :-) Indeed yes. Now tell us how you optimise a transmission line between the sub-ohm source impedance of the amplifier, and the wildly varying multi-ohm load impedance of the speaker. Sheesh, whatta maroon! -- Stewart Pinkerton | Music is Art - Audio is Engineering Go to the Linear Technology web site, download the (free) Spice modelling program, and try out various cable configurations. Or measure it. Really, if you measure it someone might learn what is significant, and what is not. Regards Ian (Stewart, not getting at you, I agree with your .sig) |
#109
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"Stewart Pinkerton" wrote in message ... On 3 Jan 2004 04:39:58 -0800, (Bob-Stanton) wrote: "Rusty Boudreaux" wrote in message So how do you enter into the computer the load impedance being it is a complex fuction of frequency? There a two ways one could do it. 1) One could use a general lumped constant model of a speaker as the termination of the line. 2) The better way is: measure the input impedance of a given speaker at many frequencies (at 100 frequencies would be good). Mathemetically convert the input impedance into S11 (input reflection cofficient). One can then put in dummy values for S22 (output reflection cofficient), S12 (reverse transmission cofficient) and S21 (forward transmission cofficient). Create a two-port (data file) device from the data. Terminate the cable (in the computer) with the two-port (data) device. The two-port device will present exactly the impedance you measured, at the frequencies you measured. The computer will interpolate values in between the frequencies you measured, to give the correct complex impedance, at all frequencies. Simple, no? :-) Indeed yes. Now tell us how you optimise a transmission line between the sub-ohm source impedance of the amplifier, and the wildly varying multi-ohm load impedance of the speaker. Sheesh, whatta maroon! -- Stewart Pinkerton | Music is Art - Audio is Engineering Go to the Linear Technology web site, download the (free) Spice modelling program, and try out various cable configurations. Or measure it. Really, if you measure it someone might learn what is significant, and what is not. Regards Ian (Stewart, not getting at you, I agree with your .sig) |
#110
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(Bob-Stanton) wrote in message . com...
(Stewart Pinkerton) wrote in message While true, this is irrelevant to the fact that lumped theory is perfectly adequate for audio frequencies in domestic situations. Yes, I agree that lumped constant models are perfectly adaquate for audio frequencies, if one doesn't have the software necessary for modeling a true transmission line. BTW, have *you* ever tried modeling a *real* speaker cable using lumped constants? BTW, have *you* EVER bothered to see if your "theory" results in predictions that you have then compared with ACTUAL measurements? I'm curious to see what you used as a lumped constant model. Please show us a model of 100 ft of standard (Home Depot), 12 gage cable (terminated by an ideal 8 Ohm load). Please show us who is using 100 ft of standard (Home Depot) 12 gauge cable in a typical home listening situation. Please show us ANYONE whose speaker cables are terminated by an ideal 8 ohm load. Mr. Stanton, your model is **** NOT because its a transmission line or any other model, it's **** becuase of your grossly incorrect assumptions and the fact that these assumptions simply don't exist in actual situations. BTW, have you ever tried modelling a *real* loudspeaker as the load? Dick Pierce in the past, has presented a lumped constant loudspeaker model. I'm sure it is adaquate for use as a load. And that is yet more evidence of how far from physical reality your "model" is. I have presented a NUMBER of lumped parameter (not "lumped constant") models, as each and every speaker system presents a significantluy different load. So, let's review your assumptions behind your "model" once again: 1. You assume that people are using 100 feet of cable. But people VERY RARELY use 100 feet of cable, it's more typically 1/10th that distance, making the necessity of a transmission line model even more irrelevant and unnecessary. 2. You assume that the cable is terminated by an ideal 8 ohm load. But NO speaker is anything approaching an ideal 8 ohm load. 3. You have looked at ONE example of a non-ideal load. But, apparently, you have never incorporated such a non- ideal load in ANY of your models. Further, you have apparently ignored the fact that one lumped parameter model simlpy is not representative of the enormous variations in actual speaker loads. And, finally: 4. You have never once presented a single shred of physical evidence in support of your "theory" that demonstrates its superiority or even its very efficacy. You insist your "theory" is right, but are unable or, more likely, simply unwilling to do ANY of the work YOU need to do to support it. YOUR theory, based on your gross missapplication of transmission line principles, your preposterous assumptions of operating conditions, and your long-demonstrated inability to relate it to any real-world performance issues indeed makes YOUR theory useless. |
#111
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(Bob-Stanton) wrote in message . com...
(Stewart Pinkerton) wrote in message While true, this is irrelevant to the fact that lumped theory is perfectly adequate for audio frequencies in domestic situations. Yes, I agree that lumped constant models are perfectly adaquate for audio frequencies, if one doesn't have the software necessary for modeling a true transmission line. BTW, have *you* ever tried modeling a *real* speaker cable using lumped constants? BTW, have *you* EVER bothered to see if your "theory" results in predictions that you have then compared with ACTUAL measurements? I'm curious to see what you used as a lumped constant model. Please show us a model of 100 ft of standard (Home Depot), 12 gage cable (terminated by an ideal 8 Ohm load). Please show us who is using 100 ft of standard (Home Depot) 12 gauge cable in a typical home listening situation. Please show us ANYONE whose speaker cables are terminated by an ideal 8 ohm load. Mr. Stanton, your model is **** NOT because its a transmission line or any other model, it's **** becuase of your grossly incorrect assumptions and the fact that these assumptions simply don't exist in actual situations. BTW, have you ever tried modelling a *real* loudspeaker as the load? Dick Pierce in the past, has presented a lumped constant loudspeaker model. I'm sure it is adaquate for use as a load. And that is yet more evidence of how far from physical reality your "model" is. I have presented a NUMBER of lumped parameter (not "lumped constant") models, as each and every speaker system presents a significantluy different load. So, let's review your assumptions behind your "model" once again: 1. You assume that people are using 100 feet of cable. But people VERY RARELY use 100 feet of cable, it's more typically 1/10th that distance, making the necessity of a transmission line model even more irrelevant and unnecessary. 2. You assume that the cable is terminated by an ideal 8 ohm load. But NO speaker is anything approaching an ideal 8 ohm load. 3. You have looked at ONE example of a non-ideal load. But, apparently, you have never incorporated such a non- ideal load in ANY of your models. Further, you have apparently ignored the fact that one lumped parameter model simlpy is not representative of the enormous variations in actual speaker loads. And, finally: 4. You have never once presented a single shred of physical evidence in support of your "theory" that demonstrates its superiority or even its very efficacy. You insist your "theory" is right, but are unable or, more likely, simply unwilling to do ANY of the work YOU need to do to support it. YOUR theory, based on your gross missapplication of transmission line principles, your preposterous assumptions of operating conditions, and your long-demonstrated inability to relate it to any real-world performance issues indeed makes YOUR theory useless. |
#112
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(Bob-Stanton) wrote in message . com...
(Stewart Pinkerton) wrote in message While true, this is irrelevant to the fact that lumped theory is perfectly adequate for audio frequencies in domestic situations. Yes, I agree that lumped constant models are perfectly adaquate for audio frequencies, if one doesn't have the software necessary for modeling a true transmission line. BTW, have *you* ever tried modeling a *real* speaker cable using lumped constants? BTW, have *you* EVER bothered to see if your "theory" results in predictions that you have then compared with ACTUAL measurements? I'm curious to see what you used as a lumped constant model. Please show us a model of 100 ft of standard (Home Depot), 12 gage cable (terminated by an ideal 8 Ohm load). Please show us who is using 100 ft of standard (Home Depot) 12 gauge cable in a typical home listening situation. Please show us ANYONE whose speaker cables are terminated by an ideal 8 ohm load. Mr. Stanton, your model is **** NOT because its a transmission line or any other model, it's **** becuase of your grossly incorrect assumptions and the fact that these assumptions simply don't exist in actual situations. BTW, have you ever tried modelling a *real* loudspeaker as the load? Dick Pierce in the past, has presented a lumped constant loudspeaker model. I'm sure it is adaquate for use as a load. And that is yet more evidence of how far from physical reality your "model" is. I have presented a NUMBER of lumped parameter (not "lumped constant") models, as each and every speaker system presents a significantluy different load. So, let's review your assumptions behind your "model" once again: 1. You assume that people are using 100 feet of cable. But people VERY RARELY use 100 feet of cable, it's more typically 1/10th that distance, making the necessity of a transmission line model even more irrelevant and unnecessary. 2. You assume that the cable is terminated by an ideal 8 ohm load. But NO speaker is anything approaching an ideal 8 ohm load. 3. You have looked at ONE example of a non-ideal load. But, apparently, you have never incorporated such a non- ideal load in ANY of your models. Further, you have apparently ignored the fact that one lumped parameter model simlpy is not representative of the enormous variations in actual speaker loads. And, finally: 4. You have never once presented a single shred of physical evidence in support of your "theory" that demonstrates its superiority or even its very efficacy. You insist your "theory" is right, but are unable or, more likely, simply unwilling to do ANY of the work YOU need to do to support it. YOUR theory, based on your gross missapplication of transmission line principles, your preposterous assumptions of operating conditions, and your long-demonstrated inability to relate it to any real-world performance issues indeed makes YOUR theory useless. |
#113
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#114
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#116
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#117
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#118
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#119
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