EELE 3332 – Electromagnetic II Chapter 11 Transmission Lines Islamic University of Gaza Electrical Engineering Department Dr. Talal Skaik 2012 1 11.5 The Smith Chart The Smith Chart is a graphical tool for analyzing transmission lines. The Smith Chart is made up of circles and arcs of circles. The circles are called “constant resistance circles” The arcs are “constant reactance circles” The combination of intersecting circles and arcs inside the chart allow us to locate the normalized impedance and then to find the impedance anywhere on the line. We will assume lossless transmission lines. (Zo=R0) 2 The Smith Chart The Smith Chart is constructed within a circle of unit radius (|Γ|≤1). || ri j Where Γr and Γi are the real and imaginary parts of Γ. 3 The Smith Chart Smith Chart is normalized so that all impedances are normalized to the characteristic impedance Z0. (Hence, it can be used for any line) For the load impedance ZL , the normalized impedance zL is: ZL zL r jx Z0 Resistance Reactance Circles Circles r-circles x-circles 4 The Smith Chart Plot normalized impedance zL=1+j2 5 The Smith Chart 6 The Smith Chart • Impedance: – Z1 = 100 + j50 – Z2 = 75 -j100 – Z3 = j200 – Z4 = 150 – Z5 = infinity (an open circuit) – Z6 = 0 (a short circuit) – Z7 = 50 – Z8 = 184 -j900 • Normalized impedance (line impedance (50 Ohms) : – z1 = 2 + j – z2 = 1.5 -j2 – z3 = j4 – z4 = 3 – z5 = infinity – z6 = 0 – z7 = 1 – z8 = 3.68 -j18 7 The Smith Chart 8 The Smith Chart •A complete revolution (3600) around Smith Chart represents a distance of (λ/2) on the line. λ distance on the line corresponds to a 7200 movement on the chart. λ→ 7200 •Clockwise movement represents moving toward the Generator. •Counter-clockwise movement represents moving toward the Load. 9 The Smith Chart Three scales: •Outermost scale: distance on line in terms of wavelength, moving toward Generator. •Middle scale: distance on line in terms of wavelength, moving toward Load. •Innermost scale : to determine θΓ (in degrees) Given Z0, ZL, λ and length of line, We can determine Zin, Yin, s, and Γ. 10 Example 11.4 •A lossless transmission line with Z0=50 Ω is 30 m long and operates at 2 MHz. The line is terminated with a load ZL=60+j40 Ω. If u=0.6c on the line, find •(a) The reflection coefficient Γ •(b) The standing wave ratio s •(c) The input impedance. Solution Method 1 (Without the Smith Chart) Z Z 60 jj 40 50 10 40 (a) =L0 0.3523 560 ZL0 +Z 60jj 40 50 110 40 1 | | 1 0.3523 (b ) s= 2.088 1 | | 1 0.3523 11 Example 11.4- Solution continued 2 (2 106 ) 2 (c) ll= (30) 1200 (electrical length) u 0.6(3 108 ) 3 ZL jZ0 tan l ZZin 0 Z0 jZL tan l 60jj 40 50tan1200 500 50jj (60 40) tan120 23.97 j 1.35 24.01 3.220 12 Example 11.4 ZL (a) Normalized: zL Z0 60 j 40 =1.2+j0.8 50 Locate zL at point P 0P z | | 0.3516 L L 0Q is between 0SP and 0 : 0 angle PS0 56 0 L 0.3516 56 _____________________ (b ) To find standing wave (c) to find Zin , express l interms of or in degrees. ratio s, draw s-circle u 0.6(3 1080 ) 30 720 90 m, l=30 m= 2400 f 2 106 90 3 3 (radius 0P and center 0) 0 Move clockwise from load toward Generator 240 on the s-circle meets r -axis s-circle from point PG to point . At G, We obtain: zin 0.47 j 0.03 at s 2.1 Hence Zin Z0 z in 50(0.47 j 0.03) 23.5 j 1.5 13 Example 11.5 A load of 100+j150 Ω is connected to a 75 Ω lossless line. Find: a) The reflection coefficient Γ b) The standing wave ratio s c) The load admittance YL d) Zin at 0.4λ from the load. e) The locations of Vmax and Vmin with respect to the load if the line is 0.6 λ long. f) Zin at the generator. 14 Example 11.5 Z L 100 j 150 (a) zL Z0 75 =1.33+j2 (point P) 0P | | 0.659 L 0Q 0 angle PS0 40 0 L 0.659 40 ZZL 0 Check : L ZZL 0 100j 150 75 L 100j 150 75 0 L 0.659 40 ____________________ (b ) Draw s-circle passing through P and obtain s=4.82 1 | | check: s= 4.865 1 | | 15 Example 11.5 (c) yL 0.228j 0.35 YYLL 0 y 1 (0.228j 0.35) 75 YL 3.04 j 4.67 mS Check : 11 YL ZjL 100 150 3.07j 4.62 mS (d ) The 0.4 corresponds to 0.4 72000 288 Move 2880 on s-circle from P toward generator (clockwise) to reach point R. zjin 0.3 0.63 Zin Z0 z in 75(0.3 j 0.63) =22.5+j47.2516 Example 11.5 – Solution continued ()d Check : 2 l (0.4 ) 360 (0.4) 1440 0 ZL jZ0 tan l ZZin 0 Z0 jZL tan tan l 100jj 150 75tan1440 =750 75jj (100 150) tan tan144 =54.41 65.250 or Zin 21.9j 47.6 17 Example 11.5 (e) 0.6 0.6 7200 4320 1 revolution 720 Start from P (Load), move alone s-circle 4320 , or one revolution +720 , and reach Generator at point G. We have passed through point T (location of Vmin ) once, and point S (location of Vmax ) twice. From load: 1st Vmax is located at : 400 0.055 7200 2nd Vmax is located at : 0.055 0.555 2 The only Vmin is located at : 0.055 / 4 0.3055 18 Example 11.5 (f) At G (Generator end), zjin 1.8 - 2.2 Zjin 75(1.8 - 2.2) 135-j 165 19 .