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16.360 Lecture 6

+. V 0. with  =. -. V 0. +. -. V(z) = V 0. +. V 0. -. +. +. -. i (z) =. V 0. V 0. V 0. j  z. j  r. Z 0. Z 0. Z 0. 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.

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16.360 Lecture 6

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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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