1//5 Th Transmission Lin Wav Equation.doc 1/6 Th Transmission Lin Wav Equation Q: So, what functions I (z) and V (z) do satisfy both tlgraphr s quations?? A: To mak this asir, w will combin th tlgraphr quations to form on diffrntial quation for V (z) and anothr for I(z). First, tak th drivativ with rspct to z of th first tlgraphr quation: V ( z) = ( R jωl) I ( z) V ( z) I ( z) = = ( R jωl) Not that th scond tlgraphr quation xprsss th drivativ of I(z) in trms of V(z): I ( z) = ( G jωc ) V ( z) Combining ths two quations, w gt an quation involving V (z) only:
1//5 Th Transmission Lin Wav Equation.doc /6 V ( z) = ( R jωl)( G jωc ) V ( z) = γ V ( z) whr it is apparnt that: ( R j L)(G j C ) γ ω ω In a similar mannr (i.., bgin by taking th drivativ of th scond tlgraphr quation), w can driv th diffrntial quation: I(z) = γ I(z) W hav dcoupld th tlgraphr s quations, such that w now hav two quations involving on function only: V(z) = I(z) = γ γ V(z) I(z) Not only spcial functions satisfy ths quations: if w tak th doubl drivativ of th function, th rsult is th original function (to within a constant)!
1//5 Th Transmission Lin Wav Equation.doc 3/6 Q: Yah right! Evry function that I know is changd aftr a doubl diffrntiation. What kind of magical function could possibly satisfy this diffrntial quation? A: Such functions do xist! For xampl, th functions V ( z) = γ z and V ( z) γ z = ach satisfy this transmission lin wav quation (insrt ths into th diffrntial quation and s for yourslf!). Likwis, sinc th transmission lin wav quation is a linar diffrntial quation, a wightd suprposition of th two solutions is also a solution (again, insrt this solution to and s for yourslf!): γ z V ( z) = V V γ z In fact, it turns out that any and all possibl solutions to th diffrntial quations can b xprssd in this simpl form!
1//5 Th Transmission Lin Wav Equation.doc 4/6 Thrfor, th gnral solution to ths wav quations (and thus th tlgraphr quations) ar: γ z γ z V (z) = V V γ z γ z I(z) = I I whr V,V, and γ ar complx constants. It is unfathomably important that you undrstand what this rsult mans! It mans that th functions V(z) and I(z), dscribing th currnt and voltag at all points z along a transmission lin, can always b compltly spcifid with just four complx constants ( V,V )!! W can altrnativly writ ths solutions as: V ( z) = V ( z) V ( z) whr: I ( z) = I ( z) I ( z) γ z V ( z) V V ( z) V γ z I ( z) I I ( z) I γ z γ z
1//5 Th Transmission Lin Wav Equation.doc 5/6 Th two trms in ach solution dscrib two wavs propagating in th transmission lin, on wav (V (z) or I (z) ) propagating in on dirction (z) and th othr wav (V - (z) or I - (z) ) propagating in th opposit dirction (-z). V ( z ) = V γ z V ( z ) = V γ z z Thrfor, w call th diffrntial quations introducd in this handout th transmission lin wav quations. Q: So just what ar th complx valus V,V? A: Considr th wav solutions at on spcific point on th transmission lin th point z =. For xampl, w find that: ( ) V z = = V γ ( z = ) = V = V = V () 1 In othr words, V is simply th complx valu of th wav function V (z) at th point z = on th transmission lin! ( )
1//5 Th Transmission Lin Wav Equation.doc 6/6 Likwis, w find: ( ) V = V z = ( ) I = I z = ( ) I = I z = Again, th four complx valus V,V ar all that is ndd to dtrmin th voltag and currnt at any and all points on th transmission lin. Mor spcifically, ach of ths four complx constants compltly spcifis on of th four transmission lin wav V z I z V z I z. functions ( ), ( ), ( ), ( ) Q: But what dtrmins ths wav functions? How do w find th valus of constants V,V? A: As you might xpct, th voltag and currnt on a transmission lin is dtrmind by th dvics attachd to it on ithr nd (.g., activ sourcs and/or passiv loads)! Th prcis valus of V,V ar thrfor dtrmind by satisfying th boundary conditions applid at ach nd of th transmission lin much mor on this latr!