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Author Topic: Plotting VSWR and Phase On A Smith Chart  (Read 1944 times)
TANAKASAN
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Posts: 933




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« on: February 19, 2013, 03:21:37 AM »

If I have a VSWR of 2:1 and a phase difference of 30 degrees between the forward and reflected signals how do I plot this on a Smith chart? For VSWR I use the circles radiating out from the center. Assuming an inductive load (reflected power leading), I have two alternatives for the phase line on the top of the chart, pointing towards the short circuit side or pointing towards the open circuit side.

Tanakasan
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KA4POL
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« Reply #1 on: February 19, 2013, 03:34:23 AM »

Check out: http://www.youtube.com/watch?v=j7UsQqqEY9E
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WB6BYU
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Posts: 13578




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« Reply #2 on: February 19, 2013, 07:21:58 AM »

There should only be one point on the SWR = 2 curve with a +30 degree
phase shift.

Phase shift is given by the azimuth angle from the center of the chart to
impedance, and is marked around the perimeter of the chart with 0 degrees
at the bottom, +90 degrees in the middle of the right side, and +/- 180
degrees at the top.  +30 degrees is in the lower right-hand corner.
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W5DXP
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« Reply #3 on: February 19, 2013, 01:18:32 PM »

There should only be one point on the SWR = 2 curve with a +30 degree
phase shift.

Would it be 1.65+j0.6 ?
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TANAKASAN
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« Reply #4 on: February 20, 2013, 03:31:44 AM »

Sorry folks but I'm still stuck here. Each of the following mixed impedances gives a line from the center of a Smith chart with a thirty degree angle to the horizontal:

R = 25.5 ohms
jX = 10 ohms
VSWR = 2.06:1

R = 82.9 ohms
jX = 31.0 ohms
VSWR = 1.99:1

So, if my scope shows that the reflected signal is 30 degrees ahead of the forward signal then which point do I pick on the Smith chart so that I can match the load to 50 ohms?

Tanakasan
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W5DXP
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« Reply #5 on: February 20, 2013, 05:07:44 AM »

R = 25.5 ohms. jX = 10 ohms. VSWR = 2.06:1

R = 82.9 ohms, jX = 31.0 ohms, VSWR = 1.99:1

Note: When a resistive load impedance is less than Z0, the reflected voltage undergoes a 180 degree phase shift. When a resistive load impedance is greater than Z0, the reflected current undergoes a 180 degree phase shift.

The first impedance has a reflection coefficient of 0.3333 at 150 degrees. That would put the two waves 150 degrees out of phase at the load, not 30 degrees out of phase.

The second impedance has a reflection coefficient of 0.3333 at 30 degrees. Seems that is the only load impedance for Z0=50 ohms that will cause an SWR=2:1 with Vref leading Vfor by 30 degrees. Note this is the 1.65+j0.6 impedance that I mentioned in my previous posting. When you un-normalize that impedance to 50 ohms you get the second impedance above.
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WB6BYU
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« Reply #6 on: February 20, 2013, 07:41:32 AM »

Quote from: TANAKASAN
...Each of the following mixed impedances gives a line from the center of a Smith chart with a thirty degree angle to the horizontal...


On my Smith Chart at least, the phase shift is measured starting with zero
at the bottom of the chart.  +30 degrees is in the lower right quadrant.
30 degrees from vertical at the top of the chart is a phase shift of
+150 degrees, not +30 degrees.  A horizontal line should have a phase shift
of +/- 90 degrees.
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