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Voltage Standing Wave Ratio Analysis in Transmission Line Circuits

This lecture covers standing wave analysis in transmission line circuits, focusing on voltage maximums, minimums, and special cases. Learn about VSWR, impedance, and examples in circuit design. Explore standing wave ratio calculations and practical applications.

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Voltage Standing Wave Ratio Analysis in Transmission Line Circuits

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  1. + V0 with  = - V0 + - V(z) = V0 + V0 - + + - i(z) = V0 V0 V0 jz jr Z0 Z0 Z0 e e jz jz -jz -jz -jz -jz jz -jz (e e e e e e e e + 1/2 = |V0| [1+ | |² + 2||cos(2z + r)] 16.360 Lecture 6 • Standing wave jz + e V(z) = V0() +  - i(z) = )  + |V(z)| = |V0| || + ||

  2. + V0 with  = - V0 + - V(z) = V0 + V0 - + + - i(z) = V0 V0 V0 jz jr Z0 Z0 Z0 e e jz jz -jz -jz -jz -jz jz -jz (e e e e e e e e + 1/2 = |V0| [1+ | |² + 2||cos(2z + r)] 16.360 Lecture 6 • Standing wave jz + e V(z) = V0() +  - i(z) = )  + |V(z)| = |V0| || + ||

  3. jz + e +  - || + V0 jz jr Z0 e e jz -jz -jz -jz (e e e e + 1/2 = |V0| [1+ | |² + 2||cos(2z + r)] 16.360 Lecture 6 • Standing wave V(z) = V0() - i(z) = )  + |i(z)| = |V0| /|Z0||| + 1/2 = |V0|/|Z0| [1+ | |² - 2||cos(2z + r)] |V(z)|

  4. + |V(z)| |V0| [1+ | |], = max |V(z)| + 1/2 = |V0| [1+ | |² + 2||cos(2z + r)] 16.360 Lecture 6 • Voltage maximum when 2z + r = 2n. –z = r/4+n/2 n = 1, 2, 3, …, if r <0 n = 0, 1, 2, 3, …, if r >= 0

  5. + |V(z)| |V0| [1 - | |], = min |V(z)| + 1/2 = |V0| [1+ | |² + 2||cos(2z + r)] 16.360 Lecture 6 • Voltage minimum when 2z + r = (2n+1). –z = r/4+n/2 + /4 Note: voltage minimums occur /4 away from voltage maximum, because of the 2z, the special frequency doubled.

  6. S  |V(z)| |V(z)| min max 1 + | | = 1 - | | 16.360 Lecture 6 • Voltage standing-wave ratio VSWR or SWR S = 1, when  = 0, S = , when || = 1,

  7. B A Z0 VL Vg(t) Vi ZL l z = - l z = 0 16.360 Lecture 6 • An example Voltage probe S = 3, Z0 = 50, lmin = 30cm, lmin = 12cm, ZL=? Solution: lmin = 30cm,  = 0.6m, S = 3,  || = 0.5, -2lmin + r = -,  r = -36º,  , and ZL.

  8. jz jz + + e e + -   ( ( ) ) V0 V0 V(z) I(z) j2z -j2l -j2l j2z e e e e - + - + (1 (1 (1 (1     ) ) ) ) Z0 Z0 -jz -jz e e 16.360 Lecture 6 • Input impudence B Ii A Zg Vg(t) Z0 VL Vi ZL l z = - l z = 0 Zin(z) = Z0 = = Zin(-l) =

  9. Input Impedance At input, d = l:

  10. -j2l -j2l e e - + (1 (1   ) ) Z0 16.360 Lecture 6 An example A 1.05-GHz generator circuit with series impedance Zg = 10- and voltage source given by Vg(t) = 10 sin(t +30º) is connected to a load ZL = 100 +j5- through a 50-, 67-cm long lossless transmission line. The phase velocity is 0.7c. Find V(z,t) and i(z,t) on the line. Solution: Since, Vp = ƒ,  = Vp/f = 0.7c/1.05GHz = 0.2m.  = 2/,  = 10 .  = (ZL-Z0)/(ZL+Z0),  = 0.45exp(j26.6º) Zin(-l) = = 21.9 + j17.4  Zin(-l) + V0[exp(-jl)+ exp(jl)] Vg = Zin(-l) + Zg

  11. |V(z)| + V0 + |V0| - -3/4 -/2 -/4 |V(z)| + - 2|V0| V0 |V(z)| |V(z)| + 2|V0| - -3/4 -/2 -/4 + 1/2 - -3/4 -/2 -/4 = |V0| [1+ | |² + 2||cos(2z + r)] 16.360 Lecture 6 Standing Wave Special cases • ZL= Z0, = 0 + |V(z)| = |V0| - ZL Z0   = + ZL Z0 2. ZL= 0,short circuit, = -1 + 1/2 |V(z)| = |V0| [2 + 2cos(2z + )] 3. ZL= ,open circuit, = 1 + 1/2 |V(z)| = |V0| [2 + 2cos(2z )]

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