Standing Wave 16.360 Lecture 7 Voltage maximum = |V0| [1+ | |² + 2||cos(2z + r)] + 1/2 |V(z)| |V0| [1+ | |], + |V(z)| max = when 2z + r = 2n. –z = r/4 + n/2 n = 1, 2, 3, …, if r <0 n = 0, 1, 2, 3, …, if r >= 0

16.360 Lecture 7 Voltage minimum |V(z)| = |V0| [1+ | |² + 2||cos(2z + r)] + 1/2 |V(z)| |V0| [1 - | |], + |V(z)| min = when 2z + r = (2n+1). –z = r/4 + n/2 + /4 Note: voltage minimums occur /4 away from voltage maximum, because of the 2z, the special frequency doubled.

16.360 Lecture 7 = Voltage standing-wave ratio VSWR or SWR S  |V(z)| 1 - | | |V(z)| min S  max = 1 + | | S = 1, when  = 0, S = , when || = 1,

16.360 Lecture 7 = |V0| [1+ | |² + 2||cos(2z + r)] |V(z)| 1/2 |V(z)| Standing Wave |V(z)| |V0| + - -3/4 -/2 -/4 Special cases ZL= Z0,  = 0 + |V(z)| = |V0| - V0 + ZL - Z0   = ZL + Z0 |V(z)| 2|V0| + - -3/4 -/2 -/4 2. ZL= 0, short circuit,  = -1 + 1/2 |V(z)| = |V0| [2 + 2cos(2z + )] |V(z)| 2|V0| + - -3/4 -/2 -/4 3. ZL= , open circuit,  = 1 + 1/2 |V(z)| = |V0| [2 + 2cos(2z )]

16.360 Lecture 7 e - + e short circuit line Ii A B Zg Vg(t) Zin Z0 VL sc Z0 VL ZL = 0 l z = - l z = 0 ZL= 0,  = -1, S =  (e -jz e jz + V(z) = V0 ) - = -2jV0sin(z) (e -jz + V0 Z0 e jz + i(z) = + ) = 2V0cos(z)/Z0 V(-l) Zin = = jZ0tan(l) i(-l)

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: l = 1/[- tan (1/CeqZ0)], -1

16.360 Lecture 7 An example: Solution: l = 1/[- tan (1/CeqZ0)], 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: l = 1/[- tan (1/CeqZ0)], -1 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,

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

Short-Circuit/Open-Circuit Method 16.360 Lecture 7 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 Z0 and electrical length

16.360 Lecture 7 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.

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

16.360 Lecture 7 An example: Z01 = 50  ZL = 100  /4 Zin = Z0²/ZL 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 Zin = Z0²/ZL= 50  ½ ½ Z0 = (ZinZL) = (50*100)

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