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16.360 Lecture 9

|V(z)|. +. V 0. +. | V0|. -. -3/4. -/2. -/4. |V(z)|. +. -. 2| V0|. V 0. |V(z)|. |V(z)|. +. 2| V0|. -. -3/4. -/2. -/4. +. 1/2. -. -3/4. -/2. -/4. = | V 0 | [ 1+ | | ² + 2| |cos(2  z +  r )]. 16.360 Lecture 9. Standing Wave. Special cases.

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16.360 Lecture 9

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  1. |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 9 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 )]

  2. + V0 Z0 jz jz -jz -jz e (e e (e 16.360 Lecture 9 short circuit line B Ii A Zg Vg(t) sc Z0 VL Zin ZL = 0 l z = - l z = 0 ZL= 0, = -1, S =  + V(z) = V0 ) - = -2jV0sin(z) + + i(z) = ) = 2V0cos(z)/Z0 V(-l) Zin = jZ0tan(l) = i(-l)

  3. + = 2V0cos(z) + V0 Z0 jz jz -jz -jz e (e e (e 16.360 Lecture 9 open circuit line B Ii A Zg Vg(t) oc Z0 VL Zin ZL =  l z = - l z = 0 ZL = ,  = 1, S =  + V(z) = V0 ) + - i(z) = ) = 2jV0sin(z)/Z0 V(-l) oc Zin = -jZ0cot(l) = i(-l)

  4. 16.360 Lecture 9 Short-Circuit/Open-Circuit Method For a line of known length l, measurements of its input impedance, one when terminated in a short and another when terminated in an open, can be used to find its characteristic impedance Z0and electrical length

  5. 16.360 Lecture 9 Line of length l = n/2 tan(l) = tan((2/)(n/2)) = 0, Zin = ZL Any multiple of half-wavelength line doesn’t modify the load impedance.

  6. (1 - ) Z0 (1 + ) -j2l -j2l e e - + (1 (1   ) ) Z0 16.360 Lecture 9 Quarter-wave transformer l = /4 + n/2 l = (2/)(/4 + n/2) = /2 , -j  e +  ) (1 Zin(-l) = = Z0 = -j  e -  (1 ) = Z0²/ZL

  7. 16.360 Lecture 9 An example: A 50- lossless tarnsmission is to be matched to a resistive load impedance with ZL = 100  via a quarter-wave section, thereby eliminating reflections along the feed line. Find the characteristic impedance of the quarter-wave tarnsformer. Z01 = 50  ZL = 100  /4 = Z0²/ZL Zin Zin = Z0²/ZL= 50  ½ ½ Z0 = (ZinZL) = (50*100)

  8. 16.360 Lecture 9 Matched transmission line: • ZL = Z0 •  = 0 • All incident power is delivered to the load.

  9. + + - V0 V0 V0 Z0 Z0 Z0 -jz jz -jz (e e e 16.360 Lecture 9 • Instantaneous power • Time-average power jz + e V(z) = V0() +  - i(z) = )  At load z = 0, the incident and reflected voltages and currents: i i + V = V0 i = r - r V = V0 i =

  10. 16.360 Lecture 9 • Instantaneous power i i i P(t) = v(t) i(t) = Re[V exp(jt)] Re[ i exp(jt)] + + + + = Re[|V0|exp(j )exp(jt)] Re[|V0|/Z0 exp(j )exp(jt)] + + = (|V0|²/Z0) cos²(t +  ) r r r P(t) = v(t) i(t) = Re[V exp(jt)] Re[ i exp(jt)] - + - + = Re[|V0|exp(j )exp(jt)] Re[|V0|/Z0 exp(j )exp(jt)] + + = - ||²(|V0|²/Z0) cos²(t +  + r)

  11. 1 T + + (|V0|²/Z0) cos²(t +  )dt 16.360 Lecture 9 • Time-average Time-domain approach: i T  T i Pav = P (t)dt = 2 0 0 + = (|V0|²/2Z0) r + Pav = -||² (|V0|²/2Z0) Net average power: i r Pav + Pav = Pav + = (1-||²) (|V0|²/2Z0)

  12. 16.360 Lecture 9 • Time-average Phasor-domain approach Pav = (½)Re[V i*] i + + + Pav = (1/2) Re[V0 V0*/Z0] = (|V0|²/2Z0) r + Pav = -||² (|V0|²/2Z0) + Pav = (1-||²) (|V0|²/2Z0)

  13. Solution of Wave Equation Wave Equation TL effect Lumped element model TL Equation l/>0.01 Wave (Input) Impedance Reflection coefficient Standing Wave Lossless TL + Complete Solution Solving for V0 Power +

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