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Roger wrote:
Keith Dysart wrote: On Dec 30, 5:30 pm, Roger wrote: I don't recall any examples using perfect CURRENT sources. I think a perfect current source would supply a signal that could respond to changing impedances correctly. It should solve the dilemma caused by the rise in voltage which occurs when when a traveling wave doubles voltage upon encountering an open circuit, or reversing at the source. What do you think? A perfect current source has an output impedance of infinity, just like an open circuit. The reflection coefficient is 1. Similar to the reflected voltage for the perfect voltage source, the reflected current cancels leaving just the current from the perfect current source. ...Keith This disagrees with Roy, who assigns a -1 reflection coefficient when reflecting from a perfect voltage source. It appears you're confusing perfect voltage and current sources. A perfect voltage source has a zero impedance, so if it's connected to a transmission line with no series resistance, it presents a reflection coefficient of -1. A perfect current source has an infinite impedance, so if it's connected to a transmission line with no parallel resistance, the reflection coefficient is +1, as Keith says. The Norton or Thévenin equivalent circuits seem capable of positive reflection coefficients. That is all that I am looking for. Reflection coefficients are complex numbers, so they can't properly be described as "positive" or "negative" except in the special cases of +1 and -1. In all other cases, the can only be described by their magnitude and angle, or real and imaginary component. Under normal circumstances, reflection coefficients can have any magnitude from zero to one, and any angle. A Thevein equivalent, like the circuit it's replacing, can present any possible reflection coefficient. For example, a Thevenin equivalent circuit having an impedance of 19 - j172 (that is, the equivalent consists of a perfect voltage source in series with a 19 ohm resistance and a capacitance of 172 ohms reactance) will present a reflection coefficient of about 0.8 - j0.5 to a 50 ohm transmission line. This is also true of a Norton equivalent consisting of a perfect current source shunted by an impedance of 19 - j172 ohms. Of course, a voltage source and series impedance not acting as an equivalent for any other circuit can also be used to drive the line, and the source impedance will equal the impedance that's in series with the perfect source. Likewise a current source with a parallel impedance. . . . I must have missed something, because I can't understand why there is an insistence that a negative reflection coefficient must exist at the source for the 1/2 or 1 wavelength long transmission line fed at one end. No one has insisted on that at all. As I've said, any reflection coefficient can exist at the source. It depends solely on the impedances of the transmission line and the source. Both Keith and I have given the equation describing the simple relationship, and you can find it also in many references. It was -1 for my example only because I used a perfect voltage source with no series impedance. Roy Lewallen, W7EL |
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