16.360 Lecture 7
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16.360 Lecture 7 • Last lecture • Standing wave • Input impedance i(z) = V(z) = V0 ( ) + + e jz - e -jz (e -jz + V0 Z0 e jz ) V(z)| = |V0| | | + e -jz || e jz + e jr = |V0| [1+ | |² + 2||cos(2z + r)] + 1/2 = |V0| /|Z0|| | + e -jz || e jz - e jr = |V0|/|Z0| [1+ | |² - 2||cos(2z + r)] + 1/2
description
-. ||. +. V 0. j r. j z. j z. j r. Z 0. e. e. e. e. j z. -j z. -j z. -j z. -j z. e. ( e. e. e. e. +. 1/2. = | V 0 | [ 1+ | | ² + 2| |cos(2 z + r )]. 16.360 Lecture 7. Last lecture. Standing wave Input impedance. j z. +. e. - PowerPoint PPT Presentation
Transcript of 16.360 Lecture 7
16.360 Lecture 7
• Last lecture
• Standing wave • Input impedance
i(z) =
V(z) = V0 ( )++e jz
-
e-jz
(e-jz+
V0
Z0 ejz )
|V(z)| = |V0| | |+ e-jz || e jz+ e jr
= |V0| [1+ | |² + 2||cos(2z + r)] + 1/2
|i(z)| = |V0| /|Z0|| |+ e-jz || e jz- e jr
= |V0|/|Z0| [1+ | |² - 2||cos(2z + r)] +
1/2
16.360 Lecture 7
Special cases
1. ZL= Z0, = 0
= |V0| +|V(z)|
2. ZL= 0, short circuit, = -1
|V(z)|
|V0| +
- -3/4 -/2 -/4
= |V0| [2 + 2cos(2z + )] + 1/2
|V(z)||V(z)|
2|V0| +
- -3/4 -/2 -/4
3. ZL= , open circuit, = 1
= |V0| [2 + 2cos(2z )] + 1/2
|V(z)||V(z)|
2|V0| +
- -3/4 -/2 -/4
16.360 Lecture 7
• Voltage maximum and minimum
|V0| [1+ | |],+|V(z)|max = when 2z + r = 2n.
|V0| [1 - | |],+|V(z)|min = when 2z + 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,
16.360 Lecture 7
• Input impudence
Vg(t) VL
A
z = 0
B
l
ZL
z = - l
Z0Vi
ZgIi
Zin(z) = V(z)
I(z)
=
++e jz
e-jzV0( )
+- e jz
e-jzV0( )
Z0 =+(1 e j2z
)
-(1 e j2z )
Z0
+(1 e-j2l )
-(1 e-j2l )
Z0Zin(-l) =
16.360 Lecture 7
• Today: special cases
• short circuit line• open circuit line• quarter-wave transformer• matched transmission line
16.360 Lecture 7
short circuit line
ZL= 0, = -1, S =
Vg(t) VL
A
z = 0
B
l
ZL = 0
z = - l
Z0
ZgIi
Zinsc
i(z) =
V(z) = V0 )- ejz
+
(e-jz
(e-jz+
V0
Z0 ejz )
= -2jV0sin(z)+
= 2V0cos(z)/Z0+
Zin =V(-l)
i(-l)= jZ0tan(l)
16.360 Lecture 7
short circuit line
Zin =V(-l)
i(-l)= jZ0tan(l)
• If tan(l) >= 0, the line appears inductive, jLeq = jZ0tan(l),
• If tan(l) <= 0, the line appears capacitive, 1/jCeq = jZ0tan(l),
l = 1/[- tan (1/CeqZ0)], -1
• The minimum length results in transmission line as a capacitor:
16.360 Lecture 7
An example:
tan (l) = - 1/CeqZ0 = -0.354,
Choose the length of a shorted 50- lossless line such that its input impedance at2.25 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/ = 2ƒ/Vp = 62.8 (rad/m)
l = tan (-0.354) + n, = -0.34 + n,
-1
16.360 Lecture 7
open circuit line
ZL= 0, = 1, S =
Vg(t) VL
A
z = 0
B
l
ZL =
z = - l
Z0
ZgIi
Zinoc
i(z) =
V(z) = V0 )+ ejz
-
(e-jz
(e-jz+
V0
Z0 ejz )
= 2V0cos(z)+
= 2jV0sin(z)/Z0+
Zin =V(-l)
i(-l)= -jZ0cot(l)
oc
16.360 Lecture 7
Application for short-circuit and open-circuit
• Network analyzer
• Measure ZinocZin
scand
• Calculate Z0
• Measure S paremeters
Zin = -jZ0cot(l)ocZin = jZ0tan(l)sc
=Z0 ZinscZin
oc
• Calculate l
= -jZ0Zin
sc
Zinoc
16.360 Lecture 7
Line of length l = n/2
= ZL
tan(l) = tan((2/)(n/2)) = 0,
+(1 e-j2l )
-(1 e-j2l )
Z0Zin(-l) =
Any multiple of half-wavelength line doesn’t modify the load impedance.
16.360 Lecture 7
Quarter-wave transformer l = /4 + n/2
= Z0²/ZL
l = (2/)(/4 + n/2) = /2 ,
+(1 e-j2l )
-(1 e-j2l )
Z0Zin(-l) =+(1 e
-j )
-(1 e-j
)Z0=
(1 + ) Z0(1 - )
=
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.
ZL = 100
/4
Z01 = 50
Zin = Z0²/ZL= 50
Z0 = (ZinZL) = (50*100)½ ½
16.360 Lecture 7
Matched transmission line:
1. ZL = Z0
2. = 03. All incident power is delivered to the load.
16.360 Lecture 7
Next lecture:
• Power flow on a lossless transmission line
