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16.360 Lecture 7
Standing Wave
•
Voltage maximum
+
1/2
|V(z)| = |V0| [1+ | |² + 2||cos(2z + r)]
+
|V(z)|max = |V0| [1+ | |], 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
+
1/2
|V(z)| = |V0| [1+ | |² + 2||cos(2z + r)]
+
|V(z)|min = |V0| [1 - | |], 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)| max
|V(z)| min
=
1 + | |
1 - | |
S = 1, when  = 0,
S = , when || = 1,
16.360 Lecture 7
+
Standing Wave
1/2
|V(z)| = |V0| [1+ | |² + 2||cos(2z + r)]
Special cases
1.
|V(z)|
+
|V0|
ZL= Z0,  = 0
+
ZL - Z0
V0
 + =
ZL + Z0
V0
2.
|V(z)| = |V0|
-
-3/4
-/2
ZL= 0, short circuit,  = -1
+
-/4
|V(z)|
+
2|V0|
1/2
|V(z)| = |V0| [2 + 2cos(2z + )]
-
3.
-3/4
-/2
ZL= , open circuit,  = 1
-/4
|V(z)|
+
2|V0|
1/2
+
|V(z)| = |V0| [2 + 2cos(2z )]
-
-3/4
-/2
-/4
16.360 Lecture 7
short circuit line
Zg
Vg(t)
Ii
B
A
Zsc
in
Z0
VL
l
z=-l
z=0
ZL= 0,  = -1, S = 
V(z) = V0(e-jz - e jz ) = -2jV+0sin(z)
+
V0 -jz jz
) = 2V+
i(z) =
(
e
0cos(z)/Z0
+
e
Z0
Zin =
V(-l)
= jZ0tan(l)
i(-l)
ZL = 0
16.360 Lecture 7
short circuit line
Zin =
V(-l)
= 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:
-1
l = 1/[- tan (1/CeqZ0)],
16.360 Lecture 7
An example:
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.
-1
Solution:
l = 1/[- tan (1/CeqZ0)],
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
open circuit line
Zg
Vg(t)
Ii
B
A
Zoc
in
Z0
VL
l
z=-l
z=0
ZL = ,  = 1, S = 
-jz
jz
e
V(z) = V0 (e
) = 2V+0cos(z)
+
+
V0 -jz jz
) = 2jV+0sin(z)/Z0
i(z) =
(
e
e
Z0
oc
Zin =
V(-l)
= -jZ0cot(l)
i(-l)
ZL = 
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
Quarter-wave transformer l = /4 + n/2
l = (2/)(/4 + n/2) = /2 ,
Zin(-l) =
(1 + e
-j2l
(1 + e
-j 
(1 - )
=
Z
0
Z
0
Z0
=
-j 
-j2l
(1
+
)
)
(1 - e
(1 - e
)
= Z0²/ZL
)
)
16.360 Lecture 7
An example:
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 =
½
(ZinZL) =
½
(50*100)
16.360 Lecture 7
Matched transmission line:
1.
2.
3.
ZL = Z0
 =0
All incident power is delivered to the load.