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16.360 Lecture 7 Last lecture Standing wave Input impedance i(z) = V(z) = V 0 ( ) + + e jzjz - e -j z (e(e + V0V0 Z0Z0 e jzjz ) |V(z)| = | V 0 | | | + e -j z |||| e jzjz + e jrjr = | V 0 | [ 1+ | |² + 2| |cos(2 z + r )] + 1/2 |i(z)| = | V 0 | /|Z0|| | + e -j z |||| e jzjz - e jrjr = | V 0 |/|Z 0 | [ 1+ | |² - 2| |cos(2 z + r )] + 1/2

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

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16.360 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,

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16.360 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 jzjz e -j z V0V0 ( ) + - e jzjz e V0V0 ( ) Z0Z0 = + (1 e j2 z ) - (1 e j2 z ) Z0Z0 + (1 e -j2 l ) - (1 e -j2 l ) Z0Z0 Z in (-l) =

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

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16.360 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 jzjz + (e(e -j z (e(e + V0V0 Z0Z0 e jzjz ) = -2jV 0 sin( z) + = 2V 0 cos( z)/Z 0 + Z in = V(-l) i(-l) = jZ 0 tan( l)

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16.360 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:

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16.360 Lecture 7 An example: tan ( l) = - 1/ C eq Z 0 = -0.354, 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 , = -0.34 + n ,

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16.360 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 jzjz - (e(e -j z (e(e + V0V0 Z0Z0 e jzjz ) = 2V 0 cos( z) + = 2jV 0 sin( z)/Z 0 + Z in = V(-l) i(-l) = -jZ 0 cot( l) oc

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16.360 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

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16.360 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.

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16.360 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 - ) =

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16.360 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

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

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

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