Chapter2_Notes

0 13 014 036 80 037 038 byo j 0 07 wave lengt hs

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Unformatted text preview: TI CI PA CA 0. 0. 1.4 ) B/Yo (+j CE TAN EP SC SU 0.7 0.14 0.36 80 0.37 0.38 0.9 0.8 110 0. 07 0. 43 0. 0 13 0.4 , 0. 3 0 3. 6 0. 2 0. 40 3 0 1. 0.2 8 0. 6 10 0.1 20 10 20 50 5.0 Pmax 4.0 3.0 2.0 1.8 1.6 1.4 1.2 1.0 0.9 0.8 0.7 Pmin 0.6 0.5 0.4 0.3 0.2 0.2 0.1 A 0.4 0.0 180 0.037 λ 0. 0.0 0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.28 0.23 0.27 LECTION COEFFICIENT IN 0.22 DEGR F REF 0.22 EES LE O ANG 0.28 -20 20 5.0 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) 0.2 50 20 0.4 0.1 10 6 0. 8 0. 1. 0 0.2 5.0 1. 0 0 4. 0. 29 0. 21 -3 0 8 0. 3 0. 0. 6 3. 0 0 0.4 0.39 0.4 0.11 0.1 -100 0. 41 -110 0. 09 0. CAPA 42 CIT IVE -1 0.0 20 8 REA 0. CT AN 43 CE 0. CO -1 07 MP 30 ON EN T (jX / .0 2 8 6 1. 1.4 0.35 0.15 1.2 -70 0.36 0.14 -80 1.0 16 34 0.9 0. 0. -90 0.13 0.37 0.12 0.38 0.8 33 0. 7 1 0 0. -6 0.7 1. 0.2 6 32 5 18 0. 0 -5 0. 0. 31 0. 19 0. 0. 3 -4 0. 4 0. 2 0. 0.49 0.48 AD <— RD LO 0.47 S TOWA -170 TH ENG 46 VEL -160 0. WA Yo) <— -jB/ 5 E( 4 50 ANC 0. 5 -1 EPT 0 SC 0. SU E IV 44 CT 0. 06 140 DU IN 0. OR , ) Zo 0.213 λ 0 4. 0 1. 50 0. 4 0. 8 0. 21 0. 29 0. 30 —> WAVE LENGT HS T 0.49 OWA RD 0.48 GEN 170 0.47 ERAT OR 160 —> 0. INDUC 46 TIVE 0. REA 15 0 CTA 0 0. 5 NC 45 E CO 0. MP ON 06 14 0 EN 0 .4 T 4 (+ jX /Z o) 41 0.12 0.11 0.39 100 0.1 0.4 09 0. 0. 08 0. 2 4 0 0. 12 103 0.287 λ Figure 2-24 Figure 27: Position of voltage maxima and minima on transmission line. Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 104 • Impedance to Admittance • Sometimes it is easier to work with admittance rather than impedance (as we will see later). An impedance Z can be written in terms of resistance and reactance, Z = R + jX . • Admittance Y (siemens S), is the inverse of impedance and can be written in terms of conductance and susceptance, Y= 1 R − jX 1 = =2 Z R + jX R + X2 (188) • We can group in terms of real and imaginary terms, Y = G + jB (189) • Where G is conductance and B is susceptance. • Comparing the above equations we can see, G= R R2 + X 2 Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (190) Electromagnetics I: Transmission lines B= −X R2 + X 2 105 (191) • We can also normalize the admittance just like the impedance, y= Y G B = +j = g + jb Y0 Y0 Y0 (192) • Where Y0 is called the characteristic admittance of the transmission line. • The normalized load admittance yL is, yL = 1−Γ 1 = zL 1+Γ (193) • Now consider the wave impedance at a distance z (d) where d = λ/4 from the load, z (d) = 1 + Γd 1 + Γe−j 2βd = 1 − Γd 1 − Γej 2βd Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (194) Electromagnetics I: Transmission lines 106 • Putting in 2βd = 4πd/λ = π , z (d = λ/4) = 1−Γ 1 + Γe−jπ = = yL jπ 1 − Γe 1+Γ (195) • We notice that the impedance at a distance d = λ/4 is the same as yL . • Therefore, rotating by λ/4 on the sta...
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