Home / Math and Physics Files / Physics / Transmission Lines / Notes By Chapter and Appendix / Chapter 4 TL equations
K calculation for symmetric REVIEWED
DOCX · 56.2 KB
Open DOCX file
Short calculation note dated 1.8.14, part of Phil's transmission line notes (Chapter 4). It evaluates the angular integrals of ln(s^2) for two round wires, using the standard integral of ln(A ± B cosθ), and finds all four integrals take the same form. It then inserts axially symmetric charge densities into the KL expression (4.10.9) to see whether KL reduces to K. Equation text is partly garbled.
AI-written summary; may contain errors. This description is approximate.
Extracted text (machine-read; may contain errors)
K calculation for symmetric PhL 1.8.14
This example is installed.
We are supposed to have K = KL in general, but I don't see how that will work in simple symmetric cases.
Here I will try some examples.
1. The widely space unequal round wire transmission line.
Need a new picture, so
s212 = r12 + (b-a1)2 - 2 r1(b-a1) cos(θ1)
s112 = r12 + a12 - 2 r1 a1 cos(θ1)
s222 = r22 + a22 + 2 r2 a2 cos(θ2)
s122 = r22 + (b-a2)2 + 2 r2(b-a2) cos(θ2) (4.5.4)
The general rule is
!Syntax Error, Idθ ln (A ± Bcosθ) = 2π ln[(1/2)(A + )] . (4.5.5)
The θ integrals of interest are:
!Syntax Error, Idθ1 ln(s212) = !Syntax Error, Idθ1ln([r12 + (b-a1)2 - 2 r1(b-a1) cos(θ1)]
A = r12 + (b-a1)2 B = 2 r1(b-a1)
A2-B2 = [r12 + (b-a1)2]2 - 4 r12(b-a1)2 = [r12 - (b-a1)2]2 => = (b-a1)2- r12 > 0 b >> a1
=> !Syntax Error, Idθ1 ln(s212) = 2π ln[(1/2)( r12 + (b-a1)2 + (b-a1)2 - r12 ) = 2π ln[(b-a1)2] .// same
The fourth integral is the same with 1↔ 2, and the different sign of the second term in s122 makes no difference,
!Syntax Error, Idθ2 ln(s122) = 2π ln[(b-a2)2] .
The second integral is the first with b-a1 → a1 so
!Syntax Error, Idθ1 ln(s112) = 2π ln(a12)
!Syntax Error, Idθ2 ln(s222) = 2π ln(a22)
So all four integrals are exactly this same! I was not expecting that result.
Now let's look at the KL calculation
KL ≡ !Syntax Error, Idx1' dy1' b1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' b2(x2',y2') ln(s222/s122) (4.10.9)
In a case where each bi is axially symmetric for its round wire we have
b1(r1,θ1) = b1(r1)
where
!Syntax Error, Idθ1 !Syntax Error, Ir1dr1 b1(r1) = 1 => !Syntax Error, Ir1dr1 b1(r1) = 1/2π
Then,
KL = !Syntax Error, Idθ1 !Syntax Error, Ir1dr1 b1(r1) ln(s212/s112) -!Syntax Error, Idθ2 !Syntax Error, Ir2dr2b2(r2) ln(s222/s122)
= !Syntax Error, Ir1dr1 b1(r1) !Syntax Error, Idθ1 ln(s212/s112) - !Syntax Error, Ir2dr2b2(r2) !Syntax Error, Idθ2 ln(s222/s122)
= 2π !Syntax Error, Ir1dr1 b1(r1) [ln[(b-a1)2]- ln(a12)] - !Syntax Error, Ir2dr2b2(r2) [ ln[(b-a2)2] - ln(a22)]
= 2π [ln[(b-a1)2/a12] !Syntax Error, Ir1dr1 b1(r1) - 2π [ln[(b-a2)2/a22] !Syntax Error, Ir2dr2 b2(r2)
= [ln[(b-a1)2/a12] - [ln[(b-a2)2/a22]
= 2π ln []
= K