California Institute of Technology Physics 77 Transmission Lines and Electronic Signal Handling May 24, 2019 1 Background A great many modern physics experiments involve sending and receiving electronic signals, sometimes over long distances. An electronic signal can be analog, which is simply a voltage as a function of time (e.g. the basis of cable television), or digital, which consists of a series of voltage pulses. For low-speed applications over short distances, for example audio signals between a CD player and a set of speakers, one doesn't need to worry much about signal transmission - a cable can be thought of transmitting a signal instantaneously, so the voltage at one end is equal to the voltage a the other. However, electronic signals cannot propagate faster than the sped of light, roughly a meter in three nanoseconds, and at GHz frequencies we have to worry about signal transmission even over laboratory distances. For these high-speed and/or long-distance applications one must take into account the fact that the voltage on one end of a cable is not equal to the voltage at the other end, simply because it takes some time for any voltage changes (such as a pulse) to propagate along the cable. Thus we need to think of a cable as a transmission line. It is worth noting at the beginning that essentially everything we'll talk about in this lab relating to electronic cables carries over pretty nicely to optical cables, a.k.a. fiber optics. Both transmit electromagnetic signals, and in both cases we will have to worry about how the signals are transmitted and whether pulses are reflected at the cable ends. 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The rate of change of charge is the current, so @(q∆z) ∆I = − @t and thus @I @q = − @z @t @V = −C @t The voltage drop along the cable comes from the cable inductance, giving @V @I = −L @z @t Differentiating both these expressions to eliminate I then yields @2V 1 @2V = @t2 LC @z2 which should be recognizable as a standard wave equation. The reader can verify that the same expression is obtained for I. Thus the cable supports electromagnetic waves that propagate along the cable, at a propagation speed 1 vprop = p LC The complete solution of the wave equation yields forward and backward propagating waves z z V (z; t) = F t − + F t + + v − v 3 lv sxuho| uhvlvwlyh +l1h1 lw*v d uhdo qxpehu,1 Iljxuh 5 vkrzv d furvv vhfwlrq ri d frd{ldo fdeoh dorqj zlwk wkh hohfwulf dqg pdjqhwlf hogv suhvhqw zkhq d vlqh zdyh vljqdo lv sursdjdwlqj grzq wkh fdeoh1 Iljxuh 51 D vhfwlrq ri d frd{ldo fdeoh zlwk wkh lqvwdqwdqhrxv hohfwulf dqg pdjqhwlf hogv wkdw duh suhvhqwFigure zkhq 2: d A vlqxvrlgdo section vljqdo of lv a sursdjdwlqj coaxial cable grzq wkh with fdeoh1 the Wkhvh instantaneous hogv pryh grzq electric wkh fdeoh and dwmagnetic wkh sursdjdwlrq fields vshhg1 that are present when a sinusoidal signal is propagating down the cable. These fields move down the cable at the propagation speed. Wkh uhdghu fdq dovr vkrz +uhphpehu Sk4B, wkdw wkh fdsdflwdqfh dqg lqgxfwdqfh ri wkh fdeoh duh 5% F @ +I2p, oq+e@d, where F+ and F− are arbitrary functions. The reader can show that with O @ oq+e@d, +K2p, this V (x; t) the corresponding current5 is zkhuh d dqg e duh vkrzq lq Iljxuh 51 Wkxv wkh yhorflw| ri wudqvplvvlrq/ dw ohdvw iru wklv lghdo +orvvohvv, fdeoh/ lv htxdo wr V (z; t) I(z; t) = 4 ysurs @ s Zc OF q 4 @ s where Zc = L=C is called the characteristic% impedance of the cable, and Qrwhfor lia wkhuh lossless lv qr cablepdwhuldoZ ehwzhhqc is purely wkh +orvvohvv resistive, frqgxfwruv/ (i.e. wkhqit's a@ real3>% number).@ %3> dqg ysurs Figure@ 2 s 4shows@ 3%3 @ af> crosswkh vshhg section ri oljkw1 of a coaxial cable along with the electric and magnetic fieldsWkh Qrw0vr0L presentghdo when Frd{ldo a sine Fdeoh1 waveIru signal wkh fdvh iswkdw propagatingU dqg J duh qrw down qhjoljleoh the wkh cable. htxdwlrqv duh dThe elw pruh reader frpsoh{/ can dqg also wkh zdyh show htxdwlrq (remember iru wkh vljqdo Ph1?) yrowdjh ehfrphv that the capacitance and C5Y 4 C5Y CY @ . +OJ . UF, . UJY +4, inductance of the cableCw5 areOF C}5 Cw zklfk lv wkh prvw jhqhudo zdyh htxdwlrq iru dq|2π wudqvplvvlrq olqh kdylqj wkh htxlydohqw flufxlw vkrzq lq Iljxuh 41 Li zh orrn iru zdyh Cvroxwlrqv= wr wklv htxdwlrq/(F/m) vr wkdw Y +}> w, @ Y +},hl$w> wkhq wkh jhqhudo vroxwlrq lv ln(b=a) } l+$w.n}, } l+$wn}, Y +}>w, @ Y4h h µ . Y5h h +5, zkhuh L = ln(b=a) (H/m) 2π @Uh+, where a and b are shown in Figure 2. Thus the velocity of transmission, at least for this ideal (lossless) cable, is6 equal to 1 vprop = p LC 4 1 = p µ Note if there is no material between the (lossless) conductors, then µ = µ , p 0 = 0, and vprop = 1= µ00 = c, the speed of light. 1.2 The Not-so-Ideal Coaxial Cable For the case that R and G are not negligible the equations are a bit more complex, and the wave equation for the signal voltage becomes @2V @2V @V = LC + (LG + RC) + RGV (1) @z2 @t2 @t which is the most general wave equation for any transmission line having the equivalent circuit shown in Figure 1. If we look for wave solutions to this equation, so that V (z; t) = V (z)ei!t, then the general solution is αz i(!t+kz) −αz i(!t−kz) V (z; t) = V1e e + V2e e (2) where α = <(γ) k = =(γ) with q γ = (R + i!L)(G + i!C) We see the solution is a traveling wave with velocity v = !=k that is sub- stantially attenuated over distances large than ∼ 1/α. The solution for the current is still V (z; t) I(z; t) = Zc where now the characteristic impedance is given by s R + i!L Z = c G + i!C s L G R ≈ 1 + i − C 2!C 2!L where the latter holds for small R, G. 5 Problem 1. Derive Eqn. 1 and the parameters for the wave-like solution Eqn. 2. A major contributor to cable losses is the skin effect, which causes the current in the cable conductors to confine itself to a thinner and thinner layer near the conductor surface as the frequency is increased.