Appendix Q orig

Appendix Q: Properties of the function βd'(ω) According to Section 5.3 (a), when losses are included in the transmission line model, one must make the following replacement to the exponent in the general traveling wave factor ej(ωt-βz) : βd = (ω/vd) → βd' = -j= -j . (Q.1) Fact 1: The real and imaginary decomposition of βd' is given by βd' = - j (Q.2) where a ≡ (R2+ω2L2)1/4 (G2+ω2C2)1/4 dim(a) = 1/m a > 0 (Q.3) c ≡ RG - ω2LC dim(c) = 1/m2 c = real, |c| < a2 (Q.4) Proof: Let q ≡ zy = (R+jωL)(G+jωC) = (RG-ω2LC) + jω(LG+RC) = c + jω(LG+RC) = |q| ejθ Then |q|2 = | (R+jωL)(G+jωC) |2 = | (R+jωL)|2|(G+jωC) |2 = (R2+ω2L2) (G2+ω2C2) = a4 |q| = a2 cosθ = c /|q| = c/a2 Now write s ≡ = = = |s| eiφ |s| = = a φ = θ/2 Re(s) = |s| cosφ = a cos(θ/2) = a = (a/) = Im(s) = |s| sinφ = a sin(θ/2) = a = (a/) = s = = + j βd' = -j = -js = - j QED Maple verification, where βd' = bdp and a2_ = a2 : Fact 2: In the high frequency limit, Re(β'd) = ω Im(β'd) = - (Q.5) Proof: At high ω write a2 ≡ (R2+ω2L2)1/2 (G2+ω2C2)1/2 = LC (R2/L2+ω2)1/2(G2/C2+ω2)1/2 = ω2LC ( 1 + R2/[ω2L2])1/2( 1 + G2/[ω2C2])1/2 ≈ ω2LC ( 1 + R2/[2ω2L2] + ) ( 1 + G2/[2ω2C2]) // (1+x)1/2 = 1+x/2 twice ≈ ω2LC [1 + (1/2ω2)(R2/L2+G2/C2) ] a2 + c = ω2LC [1 + (1/2ω2)(R2/L2+G2/C2) ] + RG - ω2LC = ω2LC [(1/2ω2)(R2/L2+G2/C2) ] + RG = (1/2)LC(R2/L2+G2/C2) + RG = (1/2) [LC(R2/L2+G2/C2) + 2RG] = (1/2LC) [L2C2(R2/L2+G2/C2) + 2RGLC] = (1/2LC) [R2C2+G2L2 + 2RGLC] = (1/2) [RC+GL]2/LC a2 - c = ω2LC [1 + (1/2ω2)(R2/L2+G2/C2) ] - RG + ω2LC ≈ 2ω2 LC Therefore Re(β'd) = = ω Im(β'd) = - = - QED Note: The expansion (1+x)1/2 = 1+x/2 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 Fact 3: In the low frequency limit, Re(β'd) = (ω/2) Im(β'd) = - [ 1 + (ω2/8) ()2 ] (Q.6) Proof: 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(β'd) = = (ω/2) Im(β'd) = - = - [ 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: Fact 4: In the low frequency limit with G = 0 , Re(β'd) = Im(β'd) = - Proof: a2 ≡ (R2+ω2L2)1/2 (ω2C2)1/2 = ωRC (1 + ω2L2/R2)1/2 ≈ ωRC(1 + ω2L2/2R2) a2 + c = ωRC(1 + ω2L2/2R2) - ω2LC ≈ ωRC a2 - c = ωRC(1 + ω2L2/2R2) + ω2LC ≈ ωRC Therefore Re(β'd) = = Im(β'd) = - = - QED Summary of results: Summary: βd' and its limits (Q.7) βd' ≡ -j= -j // definition of the wavenumber βd'(ω) βd' = - j = + j a ≡ (R2+ω2L2)1/4 (G2+ω2C2)1/4 Re(β'd) = Im() c ≡ RG - ω2LC Im(β'd) = - Re() For large ω: Re(β'd) = ω Im(β'd) = - ω >> (R/L) and ω >> (G/C) For small ω: Re(β'd) = (ω/2) Im(β'd) = - [ 1 + (ω2/8) ()2 ] ω << (R/L) and ω << (G/C) For small ω: Re(β'd) = Im(β'd) = - G = 0 and ω << (R/L)