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App Q hand calc results
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Word-processed working notes dated 3.26.05, signed PhL, from the Transmission Lines appendix on k and Z0. They expand the quantities a^2 plus or minus c, built from R, L, G, C and frequency ω, to get Re(k) and Im(k) for ω much larger than R/L and G/C, and again for small ω. Validity conditions of the square-root expansions are noted, with Maple checks; equations are partly garbled in the text.
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This is the Title PhL 3.26.05
Algebra done in Appendix G, now replaced by Maple calculations:
a2 ≡ (R2+ω2L2)1/2 (G2+ω2C2)1/2 = LC (R2/L2+ω2)1/2(G2/C2+ω2)1/2
= ω2LC ( 1 + R2/(ωL)2)1/2( 1 + G2/(ωC)2)1/2
= ω2LC ( 1 + a/ω2)1/2( 1 + a'/ω2)1/2 a = (1/2)(R2/L2) a' = (1/2)
= ω2LC [ 1 + a/ω2 + b/ω4] [ 1 + a'/ω2 +b'/ω4] b = - (1/8)(R4/L4) b = - (1/8)(G4/C4)
= ω2LC [ 1 + (a+a')/ω2 +(b+b'+aa')/ω4] aa' = (1/4) (R2/L2)(G2/C2)
= ω2LC [ 1 + (1/2ω2)(R2/L2 + G2/C2) - (1/8ω4) [R4/L4 + G4/C4 - 2(R2/L2)(G2/C2) ]
= ω2LC [ 1 + (1/2ω2)(R2/L2 + G2/C2) - (1/8ω4) (R2/L2 - G2/C2)2 ]
a2 + c = ω2LC [ 1 + (1/2ω2)(R2/L2 + G2/C2) - (1/8ω4) ( R2/L2 - G2/C2)2 ] + RG - ω2LC
= LC [ (1/2)(R2/L2 + G2/C2) - (1/8ω2) ( R2/L2 - G2/C2)2 ] + RG
= LC [ (1/2)(R2/L2 + G2/C2 + 2R/L*G/C) - (1/8ω4) ( R2/L2 - G2/C2)2 ]
= LC [ (1/2)(R/L + G/C)2 - (1/8ω2) ( R2/L2 - G2/C2)2 ]
= LC (1/2)(R/L + G/C)2 { 1 - (1/8ω2) ( R2/L2 - G2/C2)2 2 /(R/L + G/C)2 }
= LC (1/2)(R/L + G/C)2 { 1 - (1/4ω2) (R/L - G/C)2 }
= (R/L + G/C) { 1 - (1/8ω2) (R/L - G/C)2 }
= 1/() (RC + GL) { 1 - (1/8ω2) (RC - GL)2 (LC)-2 }
= 1/(2) (RC + GL) { 1 - (1/8ω2) (RC - GL)2 (LC)-2 }
= 1/(2) (RC + GL) - (1/8ω2) (RC - GL)2 (LC)-2 * 1/(2) (RC + GL)
= 1/(2) (RC + GL) - (1/16ω2) (RC - GL)2 (RC + GL) (LC)-5/2
a2 - c = a2 + c - 2c = LC (1/2)(R/L + G/C)2 { 1 - (1/4ω2) (R/L - G/C)2 } - 2RG + 2 ω2LC
= 2 ω2LC + LC (1/2)(R/L + G/C)2 - 2RG + O(1/ω2)
= 2 ω2LC + LC (1/2)(R/L + G/C)2 - 2RG
= 2 ω2LC + (1/2)(RC-LG)2/LC
= 2 ω2LC { 1 + (1/2)(RC-LG)2/LC * 1/( 2 ω2LC) }
= 2 ω2LC { 1 + (1/4ω2)(RC-LG)2/L2C2 }
= ω { 1 + (1/8ω2)(RC-LG)2/L2C2 }
= ω + (1/8ω2)(RC-LG)2/L2C2 * ω
= ω + (1/8ω)(RC-LG)2/(LC)3/2
Therefore,
Re(k) = = ω +
Im(k) = - = - +
Maple verification:
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Therefore
dim(R/L) = ohm/henry = ohm * mho*sec-1 = sec-1
Re(k) = ω + (1/8ω)(RC-LG)2/(LC)3/2
dim(G/C) = mho/far = mho*ohm/sec = sec-1
Re(k) = = ω dim(LC) = sec2/m2 dim(RC) = dim(GL) = sec/m2
Im(k) = - = - 1/(2) (RC + GL) { 1 - (1/8ω2) (RC - GL)2 (LC)-2 }
= - [ 1 + ]
= - 1/(2) (RC + GL) { 1 - (1/8ω2) (RC - GL)2 (LC)-2 }
= - 1/(2) (RC + GL) + 1/(2) (RC + GL) (1/8ω2) (RC - GL)2 (LC)-2
= - 1/(2) (RC + GL) + (1/16ω2) (RC + GL) (RC - GL)2 (LC)-5/2
} QED
Note: The expansion (1+x)1/2 = 1+x/2-x2/8 used in the proof is valid only if
R2/ω2L2 << 1 and G2/ω2C2 << 1
which requires that
ω >> (R/L) and ω >> (G/C)
For any values of R,L,C,G we can thus find ω large enough to make our high ω limit valid.
Maple verification
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a2 ≡ (R2+ω2L2)1/2 (G2+ω2C2)1/2 = RG(1+ω2L2/R2)1/2(1+ω2C2/G2)1/2
≈ RG (1+ω2L2/2R2) (1+ω2C2/2G2) ≈ RG [1+(1/2)ω2(L2/R2+C2/G2)]
a2 + c = RG [1+(1/2)ω2(L2/R2+C2/G2)] + RG - ω2LC
= [RG +(1/2)ω2(GL2/R+RC2/G)] + RG - ω2LC
= 2RG +(1/2)ω2(GL2/R+RC2/G -2LC)
= 2RG +(1/2)ω2(G2L2+R2C2 - 2LCRG)/RG
= 2RG +(1/2)ω2(RC-GL)2/RG
= RG +(ω/2)2 (RC-GL)2/RG = RG [ 1 + (ω/2)2 ()2 ]
= [ 1 + (ω/2)2 ()2 ]1/2 = [ 1 + (ω2/8) ()2 ]
a2 - c = same as a2 + c but the main RG's cancel and the cross term changes sign, so
= (1/2)ω2(RC+GL)2/RG
Therefore for low ω we find,
Re(k) = = (ω/2)
Im(k) = - = - [ 1 + (ω2/8) ()2 ] QED
Note: The expansion (1+x)1/2 = 1+x/2 used in the proof is valid only if
ω2L2/R2 << 1 and ω2C2/G2 << 1
which requires that
ω << (R/L) and ω << (G/C)
If R = 0 and/or G = 0, our proof fails since at least one of these inequalities cannot be met.
Maple verification, continuing the code shown above:
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