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16.360 Lecture 8

16.360 Lecture 8. Last lecture. short circuit line open circuit line quarter-wave transformer matched transmission line. +. V 0. Z 0. j  z. j  z. -j  z. -j  z. e. ( e. e. ( e. 16.360 Lecture 8. short circuit line. B. I i. A. Zg. Vg(t). sc. Z 0. V L. Z in.

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16.360 Lecture 8

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  1. 16.360 Lecture 8 • Last lecture • short circuit line • open circuit line • quarter-wave transformer • matched transmission line

  2. + V0 Z0 jz jz -jz -jz e (e e (e 16.360 Lecture 8 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. -1 l = 1/[- tan (1/CeqZ0)], 16.360 Lecture 8 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:

  4. + = 2V0cos(z) + V0 Z0 jz jz -jz -jz e (e e (e 16.360 Lecture 8 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)

  5. 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 8 Application for short-circuit and open-circuit • Network analyzer • Measure S paremeters • Calculate Z0 • Calculate l

  6. -j2l -j2l e e + - (1 (1   ) ) Z0 16.360 Lecture 8 Transmission 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.

  7. (1 - ) Z0 (1 + ) -j2l -j2l e e - + (1 (1   ) ) Z0 16.360 Lecture 8 Quarter-wave transformer l = /4 + n/2 l = (2/)(/4 + n/2) = /2 , -j  e +  ) (1 Zin(-l) = = Z0 = -j  e -  (1 ) = Z0²/ZL

  8. 16.360 Lecture 8 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 8 Today: Power flow on a lossless transmission line • 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 8 • 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 8 • 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 8 • 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. 16.360 Lecture 8 Next lecture • The Smith Chart

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