Chapter2_Notes

# 2 14 figure 15 input impedance of transmission line

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Unformatted text preview: CE331, PSU. (117) Electromagnetics I: Transmission lines 62 • Clearly, the two results must give one and the same voltage so we can equate the two, ˜ Vg Zin = V0+ [ejβl + Γe−jβl ] Zg + Zin (118) And, solve for the unknown V0+ , V0+ = ˜ Vg Zin Zg + Zin 1 ejβl + Γe−jβl ⇒ and we have our solution! Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (119) Electromagnetics I: Transmission lines 63 1.8. Special Cases Here we’ll have a closer look at some special transmission lines: ShortCircuited, Open Circuited, Matched Lines, λ/2 lines and λ/4 lines. • Short-circuited line Now that we know how to calculate Zin , let’s have a look at the short-circuited case. That corresponds to ZL = 0 so we can easily calculate Γ: −Z0 ZL − Z0 = = −1 (120) Γ= ZL + Z0 Z0 The standing wave ratio is given by, S= 1 + |Γ| =∞ 1 − |Γ| Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (121) Electromagnetics I: Transmission lines 64 • Voltage and current variation with z can easily be obtained using ˜ V (d) = V0+ [ejβd + Γe−jβd ] (122) V+ ˜ I (d) = 0 [ejβd − Γe−jβd ] Z0 (123) ˜ Vsc (d) = V0+ [ejβd − e−jβd ] = 2jV0+ sin βd (124) ˜ Isc (d) = V0+ 2V0+ [ejβd + e−jβd ] = cos βd (125) Z0 Z0 and is shown in ﬁg. 16. Note the locations of maxima and minima. • Voltage is zero at the load (d = 0) as it should be for short circuit. • What about input impedance? For a line of length l, sc Zin = ˜ Vsc (l) = jZ0 tan βl ˜ Isc (l) Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (126) Electromagnetics I: Transmission lines 65 which is shown if ﬁg. 16. • Very interesting: Short circuit transmission line has an impedance that is purely reactive (no real part), when (using now length of line l), tan βl &gt; 0 line appears inductive; opposite case ⇒ capacitive. • Say we set up an inductor with the same inductance (tan βl &gt; 0), i.e. jωLeq = jZ0 tan βl → Leq = Z0 tan βl (H) ω (127) • If Z0 is ﬁxed, then the only variable is length l, and the minimum length to obtain inductance Leq is l= 1 tan−1 β ω Leq Z0 (m) Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (128) Electromagnetics I: Transmission lines sc 66 short circuit Z0 Zin (a) ~ Vsc(z) 2jV0+ 1 Voltage -λ -3λ 4 -λ 2 z -λ 4 -1 (b) ~ Isc(z)Z0 2V0+ 1 Current -λ -3λ 4 z -λ 4 -λ 2 -1 (c) sc Impedance -λ -3λ 4 -λ 2 -λ 4 Zin jZ0 z (d) Figure 2-15 Figure 16: Input impedance of a S-C transmission line. Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 67 • What if tan βl &lt; 0? Then, the short-circuit transmission line appears capacitive, and the equivalent capacitance is 1 = jZ0 tan βl (F) jωCeq (129) 1 (F) Z0 ω tan βl (130) Ceq = − • Since βl is a positive number, the ﬁrst place we can get a negative num...
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