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UNIVERSITI MALAYSIA PERLIS EKT 241/4: ELECTROMAGNETIC THEORY CHAPTER 5 – TRANSMISSION LINES PREPARED BY: NORDIANA MOHAMAD SAAID [email protected] Chapter Outline General Considerations Lumped-Element Model Transmission-Line Equations Wave Propagation on a Transmission Line The Lossless Transmission Line Input Impedance of the Lossless Line Special Cases of the Lossless Line Power Flow on a Lossless Transmission Line The Smith Chart Impedance Matching Transients on Transmission Lines General Considerations • Transmission line – a two-port network connecting a generator circuit to a load. The role of wavelength • The impact of a transmission line on the current and voltage in the circuit depends on the length of line, l and the frequency, f of the signal provided by generator. • At low frequency, the impact is negligible • At high frequency, the impact is very significant Propagation modes • Transmission lines may be classified into two types: a) Transverse electromagnetic (TEM) transmission lines – waves propagating along these lines having electric and magnetic field that are entirely transverse to the direction of propagation b) Higher order transmission lines – waves propagating along these lines have at least one significant field component in the direction of propagation Propagation modes Lumped- element model Lumped- element model • A transmission line is represented by a parallelwire configuration regardless of the specific shape of the line, i.e coaxial line, two-wire line or any TEM line. • Lumped element circuit model consists of four basic elements called ‘the transmission line parameters’ : R’ , L’ , G’ , C’ . Lumped- element model • Lumped-element transmission line parameters: – R’ : combined resistance of both conductors per unit length, in Ω/m – L’ : the combined inductance of both conductors per unit length, in H/m – G’ : the conductance of the insulation medium per unit length, in S/m – C’ : the capacitance of the two conductors per unit length, in F/m Lumped- element model Note: µ, σ, ε pertain to the insulating material between conductors Lumped- element model • All TEM transmission lines share the following relation: L'C ' G' C' Note: µ, σ, ε pertain to the insulating material between conductors Transmission line equations • Complex propagation constant, γ R' jL'G' jC ' j • α – the real part of γ - attenuation constant, unit: Np/m • β – the imaginary part of γ - phase constant, unit: rad/m Transmission line equations • The characteristic impedance of the line, Z0 : Z0 R ' jL' G ' jC ' • Phase velocity of propagating waves: u p f where f = frequency (Hz) λ = wavelength (m) Example 1 An air line is a transmission line for which air is the dielectric material present between the two conductors, which renders G’ = 0. In addition, the conductors are made of a material with high conductivity so that R’ ≈0. For an air line with characteristic impedance of 50Ω and phase constant of 20 rad/m at 700MHz, find the inductance per meter and the capacitance per meter of the line. Solution to Example 1 • The following quantities are given: Z 0 50, 20 rad/m, f 700 MHz 7 108 Hz • With R’ = G’ = 0, Im jL' jC ' L' C ' and Z 0 jL' L' jC ' C' • The ratio is given by 20 C' 90.9 pF/m 8 Z 0 2 7 10 50 • We get L’ from Z0 Z0 L' C' L' 50 90.9 1012 227 nH/m 2 Lossless transmission line • Lossless transmission line - Very small values of R’ and G’. • We set R’=0 and G’=0, hence: 0 (lossless line) L' C ' (lossless line) Z0 R ' jL' G ' jC ' since R' 0 and G' 0, Z0 L' C' (lossless line) Lossless transmission line • Using the relation properties between μ, σ, ε : (rad/m) up 1 (m/s) • Wavelength, λ 0 c 1 f f r r up Where εr = relative permittivity of the insulating material between conductors Voltage reflection coefficient ~ VL ZL ~ IL • The load impedance, ZL Where; ~ VL V0 V0 V0 ~ V0 IL Z0 Z0 ~ V L = total voltage at the load V0- = amplitude of reflected voltage wave V0+ = amplitude of the incident voltage wave ~ I L = total current at the load Z0 = characteristic impedance of the line Voltage reflection coefficient • Hence, load impedance, ZL: V0 V0 Z L V V 0 0 Z 0 • Solving in terms of V0- : V0 Z L Z0 V0 Z L Z0 Voltage reflection coefficient • Voltage reflection coefficient, Γ – the ratio of the amplitude of the reflected voltage wave, V0- to the amplitude of the incident voltage wave, V0+ at the load. • Hence, V0 Z L Z 0 Z L Z 0 1 (dimension less) Z L Z0 Z L Z0 1 V0 Voltage reflection coefficient • Z0 for lossless line is a real number while ZL in general is a complex number. Hence, e jr Where |Γ| = magnitude of Γ θr = phase angle of Γ • A load is matched to the line if ZL = Z0 because there will be no reflection by the load (Γ = 0 and V0−= 0. Voltage reflection coefficient • When the load is an open circuit, (ZL=∞), Γ = 1 and V0- = V0+. • When the load is a short circuit (ZL=0), Γ = -1 and V0- = V0+. Example 2 • A 100-Ω transmission line is connected to a load consisting of a 50-Ω resistor in series with a 10pF capacitor. Find the reflection coefficient at the load for a 100-MHz signal. Solution to Example 2 • The following quantities are given RL 50, CL 10 11 F, Z 0 100, f 100 MHz 108 Hz • The load impedance is Z L RL j / CL 1 50 j 50 j159 8 11 2 10 10 • Voltage reflection coefficient is Z L / Z 0 1 0.5 j1.59 1 0.76 60.7 Z L / Z 0 1 0.5 j1.59 1 Standing Waves • Interference of the reflected wave and the incident wave along a transmission line creates a standing wave. • Constructive interference gives maximum value for standing wave pattern, while destructive interference gives minimum value. • The repetition period is λ for incident and reflected wave individually. • But, the repetition period for standing wave pattern is λ/2. Standing Waves • For a matched line, ZL = Z0, Γ = 0 and ~ V z = |V0+| for all values of z. Standing Waves • For a short-circuited load, (ZL=0), Γ = -1. Standing Waves • For an open-circuited load, (ZL=∞), Γ = 1. The wave is shifted by λ/4 from short-circuit case. Standing Waves • First voltage maximum occurs at: r n Where θr = phase l max where n 0 angle of Γ 4 2 • First voltage minimum occurs at: lmin lmax / 4 if lmax / 4 lmax / 4 if lmax / 4 r n l • For next voltage maximum: max 4 2 where n 0 n 1, 2, 3, ...... if r 0 n 0, 1, 2, ...... if r 0 Voltage standing wave ratio • VSWR is the ratio of the maximum voltage amplitude to the minimum voltage amplitude: ~ V 1 max VSWR ~ (dimension less) 1 V min • VSWR provides a measure of mismatch between the load and the transmission line. • For a matched load with Γ = 0, VSWR = 1 and for a line with |Γ| - 1, VSWR = ∞. Example 3 A 50- transmission line is terminated in a load with ZL = (100 + j50)Ω . Find the voltage reflection coefficient and the voltage standingwave ratio (VSWR). Solution to Example 3 • We have, Z L / Z 0 1 100 j50 50 0.45e j 26.6 Z L / Z 0 1 100 j50 50 • VSWR is given by: 1 1 0.45 VSWR 2.6 1 1 0.45 Input impedance of a lossless line • The input impedance, Zin is the ratio of the total voltage (incident and reflected voltages) to the total current at any point z on the line. ~ V ( z) Z in ( z ) ~ I ( z) • or 1 e j 2 z Z0 j 2 z 1 e Z L cos l jZ0 sin l Z L jZ0 tan l Z in l Z 0 Z 0 Z 0 cos l jZL sin l Z 0 jZL tan l Special cases of the lossless line • For a line terminated in a short-circuit, ZL = 0: ~ Vsc l sc Z in ~ jZ 0 tan l I sc l • For a line terminated in an open circuit, ZL = ∞: Z inoc Voc l ~ jZ0 cot l I oc l Application of short-circuit and open-circuit measurements • The measurements of short-circuit input impedance, Z insc and open-circuit input impedance, Z inoc can be used to measure the characteristic impedance of the line: Z o Z insc Z inoc • and tan l Z insc Z inoc Length of line • If the transmission line has length where n is an integer, l n / 2 , tan l tan 2 / n / 2 tan n 0 • Hence, the input impedance becomes: Zin ZL for l n / 2 Quarter wave transformer • If the transmission line is a quarter wavelength, with l / 4 n / 2 , where n 0 or any positive integer , we have l 2 , then the input 4 2 impedance becomes: 2 Z0 Z in ZL for l / 4 n / 2 Example 4 A 50-Ω lossless transmission line is to be matched to a resistive load impedance with ZL=100Ω via a quarter-wave section as shown, thereby eliminating reflections along the feedline. Find the characteristic impedance of the quarter-wave transformer. Solution to Example 4 • To eliminate reflections at terminal AA’, the input impedance Zin looking into the quarter-wave line should be equal to Z01 (the characteristic impedance of the feedline). Thus, Zin = 50Ω . 2 Z 02 Z in ZL Z 02 50 100 70.7 • Since the lines are lossless, all the incident power will end up getting transferred into the load ZL. Matched transmission line • For a matched lossless transmission line, ZL=Z0: 1) The input impedance Zin=Z0 for all locations z on the line, 2) Γ =0, and 3) all the incident power is delivered to the load, regardless of the length of the line, l. Power flow on a lossless transmission line • Two ways to determine the average power of an incident wave and the reflected wave; – Time-domain approach – Phasor domain approach • Average power for incident wave; Pavi V0 2 (W) 2Z 0 • Average power for reflected wave: Pavr 2 V0 2 2Z 0 2 Pavi Power flow on a lossless transmission line • The net average power delivered to the load: Pav Pavi Pavr V0 2 2Z 0 1 2 (W) Smith Chart • Smith chart is used to analyze & design transmission line circuits. • Impedances on Smith chart are represented by normalized value, zL for example: ZL zL Z0 • the normalized load impedance, zL is dimensionless. Smith Chart • Reflection coefficient, Γ : Z L / Z0 1 Z L / Z0 1 ZL • Since z L Z0 , Γ becomes: zL 1 zL 1 1 • Re-arrange in terms of zL: z L rL jxL 1 • Normalized load admittance: y L 1 1 zL 1 Smith Chart • The complex Γ plane. Smith Chart • The families of circle for rL and xL. Smith Chart • Plotting normalized impedance, zL = 2-j1 Input impedance • The input impedance, Zin: 1 e j 2 l Z in Z 0 1 e j 2 l • Γ is the voltage reflection coefficient at the load. • We shift the phase angle of Γ by 2βl, to get ΓL. This will zL to zin. The |Γ| is the same, but the phase is changed by 2βl. • On the Smith chart, this means rotating in a clockwise direction (WTG). Input impedance • Since β = 2π/λ, shifting by 2 βl is equal to phase change of 2π. • Equating: 2l 2 2 l 2 • Hence, for one complete rotation corresponds to l = λ/2. • The objective of shifting Γ to ΓL is to find Zin at an any distance l on the transmission line. Example 5 • A 50-Ω transmission line is terminated with ZL=(100-j50)Ω. Find Zin at a distance l =0.1λ from the load. Solution to Example 5 at B, zin = 0.6 –j0.66 VSWR, Voltage Maxima and Voltage Minima VSWR, Voltage Maxima and Voltage Minima • Point A is the normalized load impedance with zL=2+j1. • VSWR = 2.6 (at Pmax). • The distance between the load and the first voltage maximum is lmax=(0.25-0.213)λ. • The distance between the load and the first voltage minimum is lmin=(0.037+0.25)λ. Impedance to admittance transformations zL=0.6 + j1.4 yL=0.25 - j0.6 Example 6 • Given that the voltage standing-wave ratio, VSWR = 3. On a 50-Ω line, the first voltage minimum occurs at 5 cm from the load, and that the distance between successive minima is 20 cm, find the load impedance. Solution to Example 6 • The distance between successive minima is equal to λ/2. Hence, λ = 40 cm. • First voltage minimum (in wavelength unit) is at 5 l min 0.125 on the WTL scale from point B. 40 • Intersect the line with constant SWR circle = 3. • The normalized load impedance at point C is: z L 0.6 j0.8 • De-normalize (multiplying by Z0) to get ZL: Z L 500.6 j 0.8 30 j 40 Solution to Example 6 Impedance Matching • Transmission line is matched to the load when Z0 = ZL. • This is usually not possible since ZL is used to serve other application. • Alternatively, we can place an impedancematching network between load and transmission line. Single- stub matching • Matching network consists of two sections of transmission lines. • First section of length d, while the second section of length l in paralllel with the first section, hence it is called stub. • The second section is terminated with either short-circuit or open circuit. Single- stub matching Single- stub matching • The distance d is chosen so as to transform the load admittance, YL=1/ZL into an admittance of the form Yd = Y0+jB when looking towards the load at MM’. • The length l of the stub is chosen so that its input admittance, YS at MM’ is equal to –jB. • Hence, the parallel sum of the two admittances at MM’ yields Y0, which is the characteristic admittance of the line. Example 7 50-Ω transmission line is connected to an antenna with load impedance ZL = (25 − j50)Ω. Find the position and length of the shortcircuited stub required to match the line. Solution to Example 7 • The normalized load impedance is Z 25 j50 (located at A). zL L 0.5 j Z0 50 • Value of yL at B is yL 0.4 j 0.8 which locates at position 0.115λ on the WTG scale. • Draw constant SWR circle that goes through points A and B. • There are two possible matching points, C and D where the constant SWR circle intersects with circle rL=1 (now gL =1 circle). Solution to Example 7 Solution to Example 7 First matching points, C. • At C, yd 1 j1.58 is at 0.178λ on WTG scale. • Distance B and C is d 0.178 0.155 0.063 • Normalized input admittance yin ys yd 1 j 0 ys 1 j1.58 at the juncture is: ys j1.58 E is the admittance of short-circuit stub, yL=-j∞. Normalized admittance of −j 1.58 at F and position 0.34λ on the WTG scale gives: l1 0.34 0.25 0.09 Solution to Example 7 Second matching point, D. • At point D, yd 1 j1.58 • Distance B and C is d2 0.322 0.115 0.207 • Normalized input admittance ys j1.58 at G. • Rotating from point E to point G, we get l2 0.25 0.16 0.41 Solution to Example 7