Re: Transmission Line Reflections
- From: Roy Lewallen <w7el@xxxxxxxxx>
- Date: Wed, 04 Jan 2006 13:45:52 -0800
W. Watson wrote:
It seems reasonable that if I have an open ended transmission line that the current would reflect back and change phase. After all, the electrons at the end have nowhere to go but back down the line. However, the voltage is a different matter. There is no phase reversal (or polarization change) There doesn't seem to be an intuitive reason for this. However, if one examines the equations for current and capacitive voltage, then it falls out of the math. Still, where is the non-math that indicates this is true?
Now suppose instead the line is short circuited. The voltage returns down the line, and the current does not.
That's not true. Both are totally reflected.
In this case, there doesn't seem to be any non-math or intuitive feel for why the short should cause this--either for voltage or current. Can one clue me in on what's really happening above (voltage) and here (voltage and current)? Inquiring minds want to know.
The math is dictated by the boundary conditions, and those can be used to gain an intuitive feel for what's happening. When both forward and reflected waves are present, the voltage and current at any point on the line are the sum of the forward and reverse waves. When the line is open circuited, the current at the end of the line is of course zero. So the sum of the forward and reverse current wave is zero at the end, and the only way that can be is for the forward and reverse waves to be equal in magnitude and out of phase. Likewise, when the line is shorted, the voltage is zero, and the only way that can happen is for the reverse voltage wave to be equal in magnitude and opposite in phase to the forward voltage wave.
Some confusion can result when discussing the reverse wave. The reverse current wave can be, and usually is, defined as positive in the same direction as the forward wave. If you use this definition, the current changes phase at a shorted end, since the sum of the two current waves has to equal zero when both are defined as being positive in the same direction. However, you can also say that the current simply reverses direction and remains the same in phase. Those two viewpoints are equivalent. I've been using, and will continue to use, the convention that the direction of positive current is toward the load for both waves, for this discussion.
At an open end, the forward and reverse voltages are equal and in phase, and at a shorted end, the forward and reverse currents are equal and in phase. This can be seen by realizing that the voltage/current ratio of the forward wave is the same as for the reverse wave -- both equal the line Z0, but because of the current convention, the sign of the current reverses for the reflected wave. So we know that Vf/If = -Vr/Ir, where If and Ir are defined as positive when flowing toward the load. Now let's apply what we know about shorted ends to find what happens at an open end. At an open end, we know that If = -Ir. Since Vf/If = -Vr/Ir, it follows that Vr = Vf. Likewise, at an open end, we know that Vf = -Vr, so from the same equation If = Ir.
Roy Lewallen, W7EL .
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