Lecture 7 Last lecture Standing wave Input impedance i(z) = V(z) = V 0 ( ) + +  e jzjz - e -j  z (e(e + V0V0 Z0Z0 e jzjz  ) |V(z)| = | V 0 | | | + e -j  z |||| e jzjz + e jrjr = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2 |i(z)| = | V 0 | /|Z0|| | + e -j  z |||| e jzjz - e jrjr = | V 0 |/|Z 0 | [ 1+ |  |² - 2|  |cos(2  z +  r )] + 1/2

Lecture 7 Special cases 1.Z L = Z 0,  = 0 = | V 0 | + |V(z)| 2. Z L = 0, short circuit,  = -1 |V(z)| |V0| /4- /2- /4 = | V 0 | [ 2 + 2cos(2  z +  )] + 1/2 |V(z)| 2|V0| /4- /2- /4 3. Z L = , open circuit,  = 1 = | V 0 | [ 2 + 2cos(2  z )] + 1/2 |V(z)| 2|V0| /4- /2- /4

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

Lecture 7 Input impudence Vg(t) VLVL A z = 0 B l ZLZL z = - l Z0Z0 ViVi Zg IiIi Z in (z) = V(z) I(z) = + +  e jzjz e -j  z V0V0 ( ) + -  e jzjz e V0V0 ( ) Z0Z0 = + (1  e j2  z ) - (1  e j2  z ) Z0Z0 + (1  e -j2  l ) - (1  e -j2  l ) Z0Z0 Z in (-l) =

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

Lecture 7 short circuit line Z L = 0,  = -1, S =  Vg(t) VLVL A z = 0 B l Z L = 0 z = - l Z0Z0 Zg IiIi Z in sc i(z) = V(z) = V 0 ) - e jzjz + (e(e -j  z (e(e + V0V0 Z0Z0 e jzjz ) = -2jV 0 sin(  z) + = 2V 0 cos(  z)/Z 0 + Z in = V(-l) i(-l) = jZ 0 tan(  l)

Lecture 7 short circuit line Z in = V(-l) i(-l) = jZ 0 tan(  l) If tan(  l) >= 0, the line appears inductive,j  L eq = jZ 0 tan(  l), If tan(  l) <= 0, the line appears capacitive,1/j  C eq = jZ 0 tan(  l), l = 1/  [  - tan (1/  C eq Z 0 )], The minimum length results in transmission line as a capacitor:

Lecture 7 An example: tan (  l) = - 1/  C eq Z 0 = , 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)  l = tan (-0.354) + n , = n ,

Lecture 7 open circuit line Z L = 0,  = 1, S =  Vg(t) VLVL A z = 0 B l Z L =  z = - l Z0Z0 Zg IiIi Z in oc i(z) = V(z) = V 0 ) + e jzjz - (e(e -j  z (e(e + V0V0 Z0Z0 e jzjz ) = 2V 0 cos(  z) + = 2jV 0 sin(  z)/Z 0 + Z in = V(-l) i(-l) = -jZ 0 cot(  l) oc

Lecture 7 Application for short-circuit and open-circuit Network analyzer Measure Z in oc Z in sc and Calculate Z 0 Measure S paremeters Z in = -jZ 0 cot(  l) oc Z in = jZ 0tan (  l) sc =Z0Z0 Z in sc Z in oc Calculate  l = -jZ0Z0 Z in sc Z in oc

Lecture 7 Line of length l = n /2 = Z L tan(  l) = tan((2  / )(n /2)) = 0, + (1  e -j2  l ) - (1  e -j2  l ) Z0Z0 Z in (-l) = Any multiple of half-wavelength line doesn’t modify the load impedance.

Lecture 7 Quarter-wave transformer l = /4 + n /2 = Z 0 ²/Z L  l = (2  / )( /4 + n /2) =  /2, + (1  e -j2  l ) - (1  e -j2  l ) Z0Z0 Z in (-l) = + (1  e -j  ) - (1  e -j  ) Z0Z0 = (1 +  ) Z0Z0 (1 -  ) =

Lecture 7 An example: A 50-  lossless tarnsmission is to be matched to a resistive load impedance with Z L = 100  via a quarter-wave section, thereby eliminating reflections along the feed line. Find the characteristic impedance of the quarter-wave tarnsformer. Z L = 100  /4 Z 01 = 50  Z in = Z 0 ²/Z L = 50  Z 0 = (Z in Z L ) = (50*100) ½ ½

Lecture 7 Matched transmission line: 1.Z L = Z 0 2.  = 0 3.All incident power is delivered to the load.

Lecture 7 Next lecture: Power flow on a lossless transmission line