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The voltage of the reflected wave at a short circuit must cancel the voltage of the incident wave so that zero potential exists across the short circuit (equal to and 180? out of phase with). In other words, the voltage reflection coefficient must be -1 : |1| at an angle of 180?. Angles are measured from the positive real axis, which is horizontal. The pure short circuit location is at the left. (Only applies to voltage, not current.) So the measured response of the reflected wave at the plane of the VNA must be 180? out of phase with transmission.

The reflected voltage of an open circuit is equal to and /in phase/ with the incident voltage (reflection coefficient of +1) so that the open circuit location is on the right.

The ideal is not realized in real life. If you use the NanoVNA calibrated, you can watch the Smith chart marker move from 50 + 0i to something moving left and up to something moving right and down, as you change from a 50 ohm load > short > open.

~R~

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On the banks of the Piscataqua
Rich NE1EE