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Chap 4 problems REVIEWED
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Working notes by Phil dated 3.26.05, checking the derivation in section 4.5 of the transmission line chapter. He follows the Lorenz gauge condition and the potential and field relations in the frequency domain, defines V(z) and W(z), and finds the current equation lacks the conductance term (G = 0). He suspects neglected transverse current and transverse A from leakage, and notes the King wave equation includes loss. The text ends mid-thought.
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Chapter 4 Problems PhL 3.26.05
I found a problem of some sort as noted below.
Here I am just tracing through section 4.5 looking for trouble"
There are two equations from Chapter 1 which we now wish to press into service:
E = - grad φ - ∂A/∂t div A = - (4.5.2)
If we regard B = curl A as the definition of A, then the first equation above can be regarded as the definition of φ. The second equation above is our Lorenz gauge condition. In the frequency domain these equations become
E = - grad φ - jωA div A = - j (β2/ωφ (4.5.3)
Since A has only component Az, these equations become,
Ez = - ∂φ/∂z - jωAz
∂Az/∂z = - j (β2/ωφ (4.5.4)
Letting φ be φ12(x) = φ1(x) - φ2(x) and similarly for A and E, we find
E12,z(x) = -∂z φ12(x) - jω A12,z(x)
∂z A12,z(x) = - j (β2/ω φ12(x)
Evaluating at x1 on C1 and x2 on C2 (same z plane) and subtracting gives
E12,z(x1) - E12,z(x2) = -∂z [φ12(x1) - φ12(x2)] - jω [A12,z(x1) - A12,z(x2) ]
∂z [A12,z(x1) - A12,z(x2) ] = - j (β2/ω [φ12(x1) - φ12(x2)]
or
E12,z(x1) - E12,z(x2) = -∂z V(z) - jω W(z)
∂z W(z) = - j (β2/ω V(z)
Now we found that
W(z) = Le i(z) (4.4.7)
so the second equation says
∂z i(z) = - j (β2/ωLe V(z)
STOP! Already this is a problem because we are supposed to get
= - y V(z) y = G +jωC
but we have G = 0. So forget the first equation for now, something is wrong with the second!
It is just the gauge condition. Go backwards,
div A = - j (β2/ωφ
∂Az/∂z = - j (β2/ωφ
Here I have thrown out transverse derivatives in divA. But if there is leakage current, then there is transverse current which I have ignored, and it will produce transverse A !
At(x) = ∫ Jt(x') dx'dy'dz' R = |x - x'| (4.1.1)
At(x) = ∫ Et(x') dx'dy'dz' R = |x - x'| (4.1.1)
Now maybe approximate Et = -tφ12 and you have
At(x) = - ∫ -t'φ12(x') dx'dy'dz' R = |x - x'| (4.1.1)
Then I don't have a plan! Somehow that King gauge just takes care of all this stuff! I would just have to add the conductance in by hand.
Why does the King method get it in there automatically? Because the King wave equation accounts for loss, that's why.
So OK, let's go back to