1. 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

2. 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.

3. 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,

4. 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 )]

5. 16.360 Lecture 7 e - + e short circuit line Ii A B Zg Vg(t) Zin Z0 VLsc Z0 VL
ZL = 0 l
z = - l
z = 0
ZL= 0,  = -1, S =  (e
-jz e
jz +
V(z) = V ) -
= -2jV0sin(z) (e
-jz + V0 Z0 e
jz +
i(z) = + )
= 2V0cos(z)/Z0
V(-l)
Zin =
= jZ0tan(l)
i(-l)

6. 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

7. 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 = (rad/m)
tan (l) = - 1/CeqZ0 = , -1
l = tan (-0.354) + n,
= n,

8. 16.360 Lecture 7 e + - e open circuit line Ii A B Zg Vg(t) Zin Z0 VLoc Z0 VL
ZL =  l
z = - l
z = 0
ZL = ,  = 1, S =  (e
-jz e
jz
= 2V0cos(z) +
V(z) = V ) + (e
-jz + V0 Z0 e
jz +
i(z) = - )
= 2jV0sin(z)/Z0
V(-l)
Zin oc =
= -jZ0cot(l)
i(-l)

9. Short-Circuit/Open-Circuit Method
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

10. 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.

11. 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

12. 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 
Z0 = (ZinZL) = (50*100)

13. 16.360 Lecture 7 Matched transmission line: ZL = Z0  = 0
 = 0
All incident power is delivered to the load.