Home |
Search |
Today's Posts |
#1
![]() |
|||
|
|||
![]()
I've been told that this discussion continues in Letters
to the Editor in QEX magazine, to which I don't subscribe any more. Without knowing the context of those present discussions, I composed a letter to the QEX editors. Here it is: “Where Does the Power Go?” was answered in my magazine article, “An Energy Analysis at an Impedance Discontinuity in a Transmission Line”, published by Worldradio magazine and available on my web page at: http://www.w5dxp.com/energy.htm Here is a thought experiment that should lay the subject to rest. Assume a one-second long lossless 50 ohm open-circuit transmission line being driven by a 141.4 volt, 50 ohm Thevenin source. At time, t = 0, source power is supplied to the transmission line. After one second, 100 joules will have been delivered to the line’s forward wave and the forward wave will have just reached the open-circuit at the other end of the transmission line. What happens next simply follows the laws of physics. The conservation of energy principle tells us that the 100 joules in the forward wave must be conserved. The conservation of momentum principle tells us that the momentum in the forward wave must be conserved. The open-circuit prohibits the forward wave from continuing in the forward direction and the forward wave ceases to exist beyond the open-circuit point. The forward wave ceases to exist so what happens to the energy and momentum in the forward wave? Physically, there is only one possibility. The energy and momentum in the forward electromagnetic wave is moving at the speed of light and is indeed conserved by transferring its energy and momentum to the reflected wave at the open-circuit point in accordance with the rules of the distributed-network/wave-reflection model. This starts to happen at t = 1 second at which time 100 joules of electromagnetic energy exists in the forward wave in the transmission line. (One can actually calculate the force being exerted on the open- circuit by the forward wave. It is akin to the pressure of sunlight.) The transfer of energy and momentum from the forward wave to the reflected wave continues throughout the t =1 to t = 2 second time frame. At the end of two seconds, the forward wave contains 100 joules and the reflected wave contains 100 joules. Thus 200 joules are "stored" in the transmission line during the first two seconds. After two seconds, steady-state has been reached and the Thevenin source ceases to supply any energy. No additional energy is needed since the system is lossless. But please note that 200 joules exist in the transmission line all during steady-state, exactly enough energy to support the 100 joules/second forward watts and the 100 joules/second reflected watts. Any theory of where the energy goes must account for the 200 joules of energy in the transmission line. Those 200 joules cannot be destroyed by sweeping them under the steady-state rug. An ideal (lossless) directional wattmeter will tell us that forward watts = 100 watts and that reflected watts = 100 watts. 200 joules is required to support that number of forward and reflected watts. Is it just a coincidence that the 200 joules existing in the transmission line during steady-state is *exactly* the magnitude of energy required to support 100 watts forward and 100 watts reflected? The number of joules in a transmission line is *always* exactly the number required to support the forward watts and reflected watts. Since EM waves cannot stand still, it seems logical to leave the energy right where it is at the beginning of steady-state - 200 joules per second being exchanged between the forward wave and the reflected wave. The energy exchanges at the two ends of the transmission line are balanced and equal so one might argue that there is no net energy exchange between the forward and reflected waves during steady-state. But since the two energy transfers are occurring thousands of miles apart, that would seem to be a moot point. The transmission line is charged with this energy during the power on transient phase before steady-state is reached. The transmission line is discharged (energy dissipated) during the power down transient phase after steady-state. It is simply conservation of energy at work. It is interesting to note that if the transmission is discharged through a 50 ohm resistor at the far open-end, it will be discharged at a rate of 100 watts for two seconds. -- 73, Cecil http://www.w5dxp.com |
Thread Tools | Search this Thread |
Display Modes | |
|
|
![]() |
||||
Thread | Forum | |||
FS: Collins 32V-3 HF Transmitter NICE!!! | Boatanchors | |||
FCC: Broadband Power Line Systems | Policy | |||
Wanted: Power Supply for TR-4C | Boatanchors | |||
Wanted: Power Supply for TR-4C | Homebrew | |||
Mobile Power Fluctuations | Equipment |