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# Transmission Line Equations

Transmission lines take on many forms in order to accommodate particular applications. All rely on the same basic components - two or more conductors separated by a dielectric (insulator). The physical configuration and properties of all the components determines the characteristic impedance, distortion, transmission speed, and loss.

See a discussion on transmission lines and coaxial connectors.

The following formulas are presented in a compact text format that can be copied and pasted into a spreadsheet or other application.

For the following equations, ε is the dielectric constant (ε = 1 for air)

 Two Conductors in Parallel (Unbalanced) Above Ground PlaneFor D << d, hZ0= (69/ε½) log10{(4h/d)[1+(2h/D)2]-½} Single Conductor Above Ground Plane  For d << hZ0= (138/ε½) log10(4h/d) Two Conductors in Parallel (Balanced) Above Ground PlaneFor D << d, h1, h2 Z0= (276/ε½) log10{(2D/d)[1+(D/2h)2]-½} Two Conductors in Parallel (Balanced) Different Heights Above Ground PlaneFor D << d, h1, h2Z0= (276/ε½)log10{(2D/d)[1+(D2/4h1h2)]-½} Single Conductor BetweenParallel Ground PlanesFor d/h << 0.75Z0= (138/ε½) log10(4h/πd) Two Conductors in Parallel (Balanced) Between Parallel Ground PlanesFor d << D, hZ0= (276/ε½) log10{[4h tanh(πD/2h)]/πd} Balanced Conductors BetweenParallel Ground PlanesFor d << hZ0= (276/ε½) log10(2h/πd) Two Conductors in Parallel (Balanced) of Unequal DiametersZ0= (60/ε½) cosh-1 (N)N = ½[(4D2/d1d2) - (d1/d2) - (d2/d1)] Balanced 4-Wire ArrayFor d << D1, D2Z0= (138/ε½) log10{(2D2/d)[1+(D2/D1)2]-½} Two Conductorsin Open AirZ0= 276 log10(2D/d) 5-Wire ArrayFor d << DZ0= (173/ε½) log10(D/0.933d) Single Conductor inSquare Conductive EnclosureFor d << DZ0≈ [138 log10(ρ) +6.48-2.34A-0.48B-0.12C]/ε½A = (1+0.405ρ-4)/(1-0.405ρ-4) B = (1+0.163ρ-8)/(1-0.163ρ-8)C = (1+0.067ρ-12)/(1-0.067ρ-12) ρ= D/d Air Coaxial Cable withDielectric Supporting WedgeFor d << DZ0≈ [138 log10(D/d)]/[1+(ε-1)(θ/360)]½) ε = wedge dielectric constantθ= wedge angle in degrees Two Conductors Inside Shield(sheath return)For d << D, hZ0= (69/ε½) log10[(ν/2σ2)(1-σ4)] ν = h/d       σ = h/D Balanced Shielded LineFor D>>d, h>>dZ0= (276/ε½) log10{2ν[(1-σ2)/(1+σ2)]} ν = h/d       σ = h/D Two Conductors in Parallel (Unbalanced)Inside Rectangular EnclosureFor d << D, h, w                          ∞Z0= (276/ε½) {log10[(4h tanh(πD/2h)/πd)- ∑ log10[(1+μm2)/(1-νm2)]}                            m=1μm=sinh(πD/2h)/cosh(mπw/2h) νm=sinh(πD/2h)/sinh(mπw/2h)

Equations appear in "Reference Data for Engineers," Sams Publishing 1993