Transmission Lines

Chapter 2 Transmission Lines ECE 130a ZLocosbl + j Z sin bl ZLo+ j Z tan bl ZlZinaf-= o = Zo ZljZloLcosbb+ sin ZjZloL+ tan b Examples of Transmission Lines: (Chapter 2) Parallel Strip Line: metal conductors d w Coaxial Line: metal radius b metal radius a dielectric Parallel Wire Line: Air Line Dielectric Line D ε r conductor radius a Microstrip Line: w Dielectric Metal d There is a simple way to view the guided wave on a transmission line. z Iztaf, Vztaf, Iztaf, The potential difference (voltage) between the metal conductors with equal and opposite current flowing in them are circuit concepts, except they depend not only on time, but also on the distance z. So we describe the wave as voltage and current waves. itza , f + generator vtza , f load - Length L >>2πβ/ = λ Other guiding structures: 1. Waveguides -- consist of a single hollow metal tube of various cross- sectional geometry. An EM wave propagates longitudinally inside the hollow structure. The wave propagation in waveguides is not transverse (not TEM). That is, it has longitudinal field component(s). Transverse spatial dependence is fairly complicated. The propagation constant β of a waveguide wave is not equal to that of plane waves, and velocity of propagation thus is not the same as light. 2. Optical Fibers -- are used at optical frequencies, at infrared, and at visi- ble wavelengths. An optical fiber consists of a very thin (50-300 mm) dielectric circular cross section cylinder. The material is usually glass or plastic. Inner and outer portions have different dielectric constants (index of refraction), as shown below. core radius a, index n1 cladding, index n2 Optical fibers do not support TEM waves, like the hollow metallic guides. Their propagation constant and mode structures are even more complicated than for hollow metallic guides. Derivation of Transmission Line Equations (1-3) Let us consider a length of a transmission line at location z . The circuit model is clearly a series inductance and resistance (since the whole line with a load at the end forms a loop) and a shunt capacitance and shunt leak- age conductance (between the good conductors, across the dielectric which also has a small conductance). With the above circuit parameters being LRC,,, and G , the model is iztaf, L R iza +D zt, f + + vzta , f C G vza +D zt, f − − A straight forward application of Kirchoff’s Loop Law gives ∂iztaf, vza +-DDD zt,,f vzta f =-LR z -zia zt, f , ∂t and Kirchoff’s Current Law at the upper node gives ∂ vza +D zt, f izafafaf+-DDDD zt,, izt =-+-GC zvz zt , z . ∂ t Dividing through by ∆z and taking the limit ∆z→0 , we get: ∂v ∂i =-LR - i (1-5) ∂z ∂t ∂i ∂v =-CG - v (1-6) ∂z ∂t Sinusoidal Analysis of Transmission Lines (using phasors) ∂v =-aRL +jw fi (1) ∂z ∂ i =-afGC +jw v (2) ∂ z Differentiate (1) with respect to z : ∂ 2 vid =-afRL +jw ∂ z 2 dz Substitute from (2) ∂∂i/ z : ∂ 2v =+afafGCRLjjww + v ∂ z 2 Similarly, we get: ∂ 2i =+afafGCRLjjww + i ∂ z 2 The Lossless Transmission Line R = 0 (perfect conductor) G = 0 (lossless dielectric) This is not only instructive and simple, but also a good approximation of many real lines which are made of very good conductors (typically, copper) and nearly lossless dielectrics (e.g., teflon). For example, a commercial coaxial line is a nearly lossless line. ∂ 2 v = jjwwCLv ∂z 2 ◊ 2 ∂ 2 i ∂ v 2 2LC =-chw LC v Also, 2 =-chw i ∂ z 2 ∂ z ∂ 2 v ∂ 2 i =-b 2 v =-b 2 i ∂z 2 ∂z 2 ∂V ∂ I ∂ I ∂V =-L =-C ∂z ∂t ∂z ∂t The solutions are travelling waves: e ± jzβ (withe jtω already assumed) jtafwb- z jtafwb+ z e ore or sum of both. βω==LC ωµε It will be shown later, when we consider transmission lines from the Electro- magnetic Field’s point of view, that these voltage and current waves corre- spond to EM waves, also. It will be shown that: 1 LC ==µε , v where v is velocity of light in the medium between the conductors. We can get the relationship of V and I by substituting back into the original transmission line equations. jtafwb- z jtafwb- z v = Ve i = Ie (i.e., only +z going wave) dv di =−jωLi =−jω Cv dz dz −=−jjβωviL −=−jjβωivC v ωL β == βω==LC ωµε, i β ω C v L + + ==Zo This is for z going wave only. i + C Zo is called the characteristic impedance of the line. For this lossless line, Zo is real. It means that V and I are in phase. For example, power is propagated. None is absorbed. (Contrast this with ordinary circuits, where real impedance means power is absorbed and dissipated!) Power Propagated on a Lossless Line The power propagated can be calculated either from the electromagnetic wave or from the voltage-current wave. It is usual to use voltage and current for a transmission line. The power, of course, is an oscillating quantity. We are mostly interested, however, in the average power flow. If (real), then from the circuit picture: 2 ==1 1 Vo Pav Re vi* for lossless line 2 2 Zo If we have only a − z going wave, the same kind of derivation we did for the + z going wave gives: v− =− − Zo (z going wave) i− Circuits vs. Transmission Lines 1) Q: When does the “regular” circuit (E17) method hold correct; that is, only e jtω dependence (no propagation) and when does the transmission line method need to be used? A: Simple. When the wavelength is large compared to the dimensions, we can neglect transmission line concepts: π β =→2 zzλ 0 That is, whenever the distance scale (z) is small and/or the wavelength is long; that is, z << λ . = 1 2 2) For circuits: If is real (that is, ZR= ), PIRav 2 . This is power absorbed in the real impedance (resistance). For a lossless transmission line: Zo is real. If we have only + going wave: 2 ==1 v+ 1 2 Pav i + Zo 2 Zo 2 This is not absorbed power. It is power propagating in the +z direction down the lossless line. = RZo ⇒ Power is absorbed in the load to Load where it does “something.” If we have - going wave: 2 ==−=−1 * 1 v− 1 2 Pav Re vi−− i − Zo 2 2 Zo 2 The negative sign means that power is propagating in the negative direction. Unless otherwise stated, we always will mean average power. Transmission Line Circuits Transmission lines are used to carry signals from one location to another. Thus, they usually have a signal generator at one end and a load at another end. Let’s now examine the behavior of a transmission line with load. We use phasors and will assume lossless transmission line. 1) Transmission line terminated in its characteristic impedance. This is called “matched.” = Zo ZZLo Let’s assume we launched a + z going wave on this transmission line of characteristic impedance Zo . Then, at any point z, we know that = = vi++/ Zo . At the load, viLL/ Z L. (This is definition of ZL .) Thus, if = = = = = ZZLo, then vvL + (ZL ) and iiL + (ZL ). Thus, there is no reflected wave (− z going wave). Rather, the wave continues into the load, where it gets dissipated. (In practical situations, the load is a whole circuit, for example, a detector, a radio, an oscilloscope. As long as its input impedance is Zo , the foregoing discussion is true.) 2) Transmission line terminated in other impedance (1-7, text) ≠ ⇒ ZZLo Zo Assume again we launched a signal at some frequency. Then, = = ≠ vi++/ Zo . But at the load, viLL/ Z L. If ZZLo, v L cannot be v+ and iL cannot be i + . The boundary condition at the load can only be satisfied if there is also a reflected wave vi−−, . This wave propagates backward in the − z direction. Then, at any point on the line: jtafwb- z jt af wb+ z vvv=+=+Ve+ Ve- +- V+ jtafwb- z V- jt af wb+ z v− iii=-=+e e +- i− =− Zo Zo Zo Note the − sign on the current of the reflected wave! Now we can satisfy = the condition at the load vi/ ZL . Only one value of V− will satisfy. Let us define: v− =Γ, the reflection coefficient v+ Then: vvv+++Γ = (1) vv++v − Γ = (2) ZZZooL We solve for Γ to get: (Problem: Solve to G ) − Γ= ZZLo + ZZLo Γ Note: is complex, in general, because ZL may be complex. -£11G £ Fraction of the incident power reflected back: 1 1 221 Incident Power PZVZ= ReL vi* O = Re L v //O = iooNM2 ++QP NM2 +QP 2 + 1 1 Reflected Power P = ReL vi***O = Re L GG v iOa-1f r NM2 --QP NM2 - +QP 221 =-Re L GGvi* O =- P NM 2 ++QP i L1 ***O L1 O Total Power PTTT= Revi=+- Re a11GGf v++ej i NM2 QP NM2 QP L1 ***O =+--Re ej1 GG GGvi++ NM2 QP =−Γ 2 ≡ 1 Pi Transmitted power (into Load) 1 * Note: ejGG-=Im Gis purely imaginary. 2 3) Special Cases of Termination = A special case of the general load is ZL 0 , that is, a short circuited transmission line. (A transmission line may be shorted deliberately in order to achieve a certain result, or it may be shorted accidentally, something that has to be repaired.) For this case: Zo Z Γ=−Zo =− 1 ZL = 0 Zo Z = 0 ~ We can say here that ΓΓ==eejjϕπ1 + here. If jtafwb- z, v + = Ae vafzAeAe=-- jzbb jz ce jtw understoodf 1 i afz =+Ae- jzbb Ae jz Zo Note that this time, the z dependence has two different parts.

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