Chapter2_Notes

# Looking at magnitudes or amplitudes only what happens

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Unformatted text preview: 0+| -λ 4 -λ -λ -3λ 2 4 4 (b) ZL = 0 (short circuit) λ/2 -λ -λ -λ -3λ 2 4 4 (c) ZL = (open circuit) 0 z ~ |V(z)| 2|V0+| 0 z ~ |V(z)| 2|V0+| 0 z Figure 13: Standing waves for matched load, S-C and O-C. Figure 2-12 Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 52 ˜ • The matched load ⇒ |Γ| = 0 ⇒ |V (z )| = |V0+ |, i.e. without any reﬂected waves there can be no interference ⇒ no standing waves . • S-C and O-C cases have |Γ| = 1, or Γ = −1 for S-C and Γ = 1 for O-C. • S-C and O-C have the same maximum value: 2|V0+ |, and minimum value of zero. Their patterns are shifted by λ/4. • The ﬁrst voltage minimum is at z = 0 for S-C, while O-C has ﬁrst maximum at z = 0; why? Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 53 What about a general expression for the position of the ﬁrst maximum (or minimum)? We’ve already seen that when ˜ 2βdmax − θr = 2nπ ⇒ |V |max = |V0+ |[1 + |Γ|] (98) with n=0 or a positive integer. So, what’s dmax ? dmax = θr λ nλ θr + 2nπ = + 2β 4π 2 (99) where n = 1, 2, . . . if θr < 0, and n = 0, 1, 2 . . . if θr ≥ 0. • θr is bounded by −π and π , • When θr ≥ 0 the ﬁrst dmax = θr λ/4π , otherwise it occurs at dmax = (θr λ/4π ) + λ/2. • Maximum of voltage standing wave is also where current standing wave has a minimum! Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 54 • Derivation of positions for minima is analogous and yields ˜ |V |min = |V0+ |[1 − |Γ|], for (2βdmin − θr ) = (2n + 1)π (100) and the ﬁrst minimum occurs for n = 0. • Spacing between dmax and dmin is λ/4 ⇒ no need to calculate minima separately: dmin = dmax + λ/4, if dmax < λ/4 dmin = dmax − λ/4, if dmax ≥ λ/4 (101) ˜ ˜ • Finally, the ratio |Vmax |/|Vmin | is the voltage standing wave ratio S , aka SWR or VSWR: S= ˜ |Vmax | 1 + |Γ| = ˜min | 1 − |Γ| |V Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (102) Electromagnetics I: Transmission lines 55 Sliding probe To detector Probe tip ~ Vg + Slit Zg - 40 cm 30 cm 20 cm 10 cm ZL Figure 14: Slotted coaxial line. Figure 2-13 1.7. Input Impedance We’ve learned a lot without actually solving our original equations • We understand that the voltage and current magnitudes are oscillatory with position on the line and are out of phase with each other. • Since impedance is the ratio of voltage to current, we can talk Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 56 about the input impedance which is, Z (d) = ˜ V (d) ˜ I (d) (103) ˜ ˜ • Substitute the solution we found for V (z ) and I (z ) Z (d) = ˜ V + ejβd + Γe−jβd V (d) 1 + Γe−j 2βd = 0+...
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## This note was uploaded on 09/25/2013 for the course ECE 331 taught by Professor Martinsiderious during the Fall '12 term at Portland State.

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