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16.360 Lecture 7

-. ||. +. V 0. j  r. j  z. j  z. j  r. Z 0. e. e. e. e. j  z. -j  z. -j  z. -j  z. -j  z. e. ( e. e. e. e. +. 1/2. = | V 0 | [ 1+ | | ² + 2| |cos(2  z +  r )]. 16.360 Lecture 7. Last lecture. Standing wave Input impedance. j  z. +. e.

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16.360 Lecture 7

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  1. - || + V0 jr jz jz jr Z0 e e e e jz -jz -jz -jz -jz e (e e e e + 1/2 = |V0| [1+ | |² + 2||cos(2z + r)] 16.360 Lecture 7 • Last lecture • Standing wave • Input impedance jz + e V(z) = V0() +  - i(z) = )  + |V(z)| = |V0| || + || + |i(z)| = |V0| /|Z0||| 1/2 = |V0|/|Z0| [1+ | |² - 2||cos(2z + r)] +

  2. |V(z)| + |V0| - -3/4 -/2 -/4 |V(z)| + 2|V0| |V(z)| + 2|V0| - -3/4 -/2 -/4 - -3/4 -/2 -/4 16.360 Lecture 7 Special cases • ZL= Z0, = 0 + |V(z)| = |V0| 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 )]

  3. S  |V(z)| |V(z)| min max 1 + | | = + + 1 - | | |V(z)| |V(z)| |V0| [1+ | |], |V0| [1 - | |], = = max min 16.360 Lecture 7 • Voltage maximum and minimum when 2z + r = 2n. when 2z + r = (2n+1). • Voltage standing-wave ratio VSWR or SWR S = 1, when  = 0, S = , when || = 1,

  4. 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 7 • Input impudence B Ii A Zg Vg(t) Z0 VL Vi ZL l z = - l z = 0 Zin(z) = Z0 = = Zin(-l) =

  5. 16.360 Lecture 7 • Today: special cases • short circuit line • open circuit line • quarter-wave transformer • matched transmission line

  6. + V0 Z0 jz jz -jz -jz e (e e (e 16.360 Lecture 7 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)

  7. -1 l = 1/[- tan (1/CeqZ0)], 16.360 Lecture 7 short circuit line V(-l) Zin = jZ0tan(l) = i(-l) • If tan(l) >= 0, the line appears inductive, jLeq= jZ0tan(l), • If tan(l) <= 0, the line appears capacitive, 1/jCeq= jZ0tan(l), • The minimum length results in transmission line as a capacitor:

  8. 16.360 Lecture 7 An example: Choose the length of a shorted 50- lossless line such that its input impedance at 2.25 GHz is equivalent to the reactance of a capacitor with capacitance Ceq = 4pF. The wave phase velocity on the line is 0.75c. Solution: Vp= ƒ,  = 2/ = 2ƒ/Vp = 62.8 (rad/m) tan (l)= - 1/CeqZ0 = -0.354, -1 l= tan (-0.354) + n, = -0.34 + n,

  9. + = 2V0cos(z) + V0 Z0 jz jz -jz -jz e (e e (e 16.360 Lecture 7 open circuit line B Ii A Zg Vg(t) oc Z0 VL Zin ZL =  l z = - l z = 0 ZL= 0, = 1, S =  + V(z) = V0 ) + - i(z) = ) = 2jV0sin(z)/Z0 V(-l) oc Zin = -jZ0cot(l) = i(-l)

  10. oc sc Zin Zin • Measure and sc oc = -jZ0cot(l) = jZ0tan(l) Zin Zin oc sc sc oc Zin Zin Zin Zin Z0 = Z0 = -j 16.360 Lecture 7 Application for short-circuit and open-circuit • Network analyzer • Measure S paremeters • Calculate Z0 • Calculate l

  11. -j2l -j2l e e + - (1 (1   ) ) Z0 16.360 Lecture 7 Line of length l = n/2 tan(l) = tan((2/)(n/2)) = 0, Zin(-l) = = ZL Any multiple of half-wavelength line doesn’t modify the load impedance.

  12. (1 - ) Z0 (1 + ) -j2l -j2l e e - + (1 (1   ) ) Z0 16.360 Lecture 7 Quarter-wave transformer l = /4 + n/2 l = (2/)(/4 + n/2) = /2 , -j  e +  ) (1 Zin(-l) = = Z0 = -j  e -  (1 ) = Z0²/ZL

  13. 16.360 Lecture 7 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 Zin = Z0²/ZL= 50  ½ ½ Z0 = (ZinZL) = (50*100)

  14. 16.360 Lecture 7 Matched transmission line: • ZL = Z0 •  = 0 • All incident power is delivered to the load.

  15. 16.360 Lecture 7 Next lecture: • Power flow on a lossless transmission line

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