Standing Wave Phase
On Dec 12, 8:44 pm, "AI4QJ" wrote:
"Cecil Moore" wrote in message
. net...
wrote:
Although we know that the 200 ohm line is longer, there is no
indication that the length of the 600 ohm line must or must not
change.
The phase shift at the impedance discontinuity depends
upon the *ratio of Z0High/Z0Low*. The following two
examples have the same phase shift at the impedance
discontinuity.
Z0High Z0Low
---43.4 deg 600 ohm line---+---10 deg 100 ohm line---open
---43.4 deg 300 ohm line---+---10 deg 50 ohm line---open
So how long does the 600 ohm line have to be in the following
example for the stub to exhibit 1/4WL of electrical length?
---??? deg 600 ohm line---+---10 deg 50 ohm line---open
Based on what you say above, Zo high/Zo low = 6 resul;ts in 37 degrees.
Then for phase shift, (Zo high/Zo low)*37 deg = 6*D where D = phase shift.
If the above ratio is literally true, then in the above example, 12*37deg/6
= 74 degrees and length of the 600 ohm line = 90-74-10 = 6 degrees.
However, I would like to find a reference for the math showing the
characteristic impedance ratio relationship with phase shift. I am reluctant
to accepting formulas without seeing them derived at least once.
In another thread, "Calculating a (fictitious) phase shift;
was : Loading Colis", David Ryeburn has provided the
arithmetic for the general case which includes the
following expression:
-j*(Z_2)*cot(alpha + beta) = -j*(Z_1)*cot(alpha)
"alpha" is the length of the open line
"beta" is the "phase shift" at the joint
"Z_2" is the impedance of the open line
"Z_1" is the impedance of the driven line
It can be seen, as noted by Cecil, that beta
will be the same if the impedances of the two
lines are scaled proportionally.
But beta is dependant not only on the two
impedances, but also on the length of the
open line.
There are no simple relationships here.
It does not seem to be a concept that is
particularly useful for the solving of problems.
....Keith
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