Transmission Lines Dr. Sandra Cruz-Pol ECE Dept. UPRM.
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Transcript of Transmission Lines Dr. Sandra Cruz-Pol ECE Dept. UPRM.
Transmission Transmission LinesLines
Dr. Sandra Cruz-PolECE Dept. UPRM
Exercise 11.3 A 40-m long TL has Vg=15 Vrms, Zo=30+j60 , and VL=5e-j48o Vrms. If the
line is matched to the load and the generator, find: the input impedance Zin, the sending-end current Iin and Voltage Vin, the propagation constant .
Answers:
ZL
Zg
Vg
+
Vin
-
Iin
+
VL
-
Zo=30+j60+j
40 m
oo
rmso
in
oinin
e
mjVV
AIjZ
63112.011145.7
02094.00101.0,05.7
,4.63112.0,6030
40
1
Transmission LinesI. TL parameters
II. TL Equations
III. Input Impedance, SWR, power
IV. Smith Chart
V. Applications Quarter-wave transformer Slotted line Single stub
VI. Microstrips
Transmission Lines (TL)
TL have two conductors in parallel with a dielectric separating them
They transmit TEM waves inside the lines
Common Transmission LinesTwo-wire (ribbon)
Coaxial
Microstrip
Stripline (Triplate)
Other TL (higher order)[Chapter 11]
Fields inside the TL V proportional to E, I proportional to H
dlHI
dlEV
Distributed parameters
The parameters that characterize the TL are given in terms of per length.
R = ohms/meter L = Henries/ m C = Farads/m G = mhos/m
cmc
kmc
GHz
Hz
15000,000,2000/
000,560/
2
60
Common Transmission LinesR, L, G, and C depend on the particular transmission line structure and the material properties. R, L, G, and C can be calculated using fundamental EMAG techniques.
Parameter Two-Wire Line Coaxial Line Parallel-Plate Line
Unit
R
L
G
C
1
a cond
1
2 cond1
a
1
b
2
w cond
/m
acoshD
2a
2
lnb
a
d
w
H /m
diel
acosh D / 2a
2 diel
ln b /a
diel
w
d
S /m
acosh D / 2a
2ln b /a
w
d
F /m
TL representation
Distributed line parametersUsing KVL:
Distributed parameters Taking the limit as z tends to 0 leads to
Similarly, applying KCL to the main node gives
t
tzILtzRI
z
tzV
),(),(
),(
t
tzVCtzGV
z
tzI
),(),(
),(
ss
ss
VCjGz
I
ILjRz
V
0 22
2
ss V
z
V
Wave equation
Using phasors
The two expressions reduce to
Wave Equation for voltage
sss
sss
CVjGVz
I
LIjRIz
V
])(Re[),(
])(Re[),(tj
s
tjs
ezItzI
ezVtzV
CjGLjR 2
TL Equations Note that these are the wave eq. for
voltage and current inside the lines.
The propagation constant is and the wavelength and velocity are
fu
CjGLjRj
Idz
IdV
dz
Vds
ss
s
2
))((
00 22
22
2
2
Waves moves through line The general solution is
In time domain is
Similarly for current, I)cos()cos(),(
)cos()cos(
])(Re[),(
zteIzteItzI
zteVzteV
ezVtzV
eVeVV
zz
zz
tjs
zzs
z
Characteristic Impedance of a Line, Zo
Is the ratio of positively traveling voltage wave to current wave at any point on the line
I
VjXR
CjG
LjRLjR
I
VZ
eILjReV
eIzI
eVzV
zILjRdz
zdV
ooo
zz
z
z
)()(
)(
)(
)()()(
z
Example: An air filled planar line with w=30cm, d=1.2cm,
t=3mm, c=7x107 S/m. Find R, L, C, G for 500MHz Answer
0
/2213.0
/24.50
/036.02
d
wG
mpFd
w
d
wC
mnHw
dL
mw
R
o
c
w
d
Exercise 11.1
A transmission line operating at 500MHz has Zo=80 , =0.04Np/m,=1.5rad/m. Find the line parameters R,L,G, and C.
Answer: 3.2 /m, 38.2nH/m, 0.0005 S/m, 5.97 pF/m
03.00005.
1202.3
jCjGZ
jLjRZ
o
o
Different cases of TL
Lossless
Distortionless
Lossy
Transmission line
Transmission line
Transmission line
Lossless Lines (R=0=G)
Has perfect conductors and perfect dielectric medium between them.
Propagation:
Velocity:
Impedance C
LRZX
fLC
u
LCj
ooo
0
2,
1
,,0
Distortionless line (R/L = G/C)
Is one in which the attenuation is independent on frequency.
Propagation:
Velocity:
Impedance G
R
C
LRZX
fLC
u
LCRG
j
ooo
0
1
Summaryj Zo
=General
Lossless
Distortionless
RC = GLLCjRG
LCj 0
CjG
LjRZo
G
R
C
LRZ oo
))(( CjGLjR
C
LRZ oo
Excersice 11.2
A telephone line has R=30 /km, L=100 mH/km, G=0, and C= 20F/km. At 1kHz, obtain: the characteristic impedance of the line, the propagation constant, the phase velocity.
Solution:
skmu
kmjjjCjGLjR
kj
mkj
CjG
LjRZ o
o
/707
/88.821.0102020)100(230
37.175.701020)1(20
)100)(1(230
3
6
Define reflection coefficient at the load, L
V
V
eVeVzV
L
zzs
)(
Terminated, Lossless TL
zL
zs eeVzV )(
zL
z
os ee
Z
VzI
)(
z
Lz
zL
z
os
s
ee
eeZ
zI
zVzZ
)(
)()(
Then,
Similarly,
The impedance anywhere along the line is given by
The impedance at the load end, ZL, is given by
L
LoL ZZZ
1
1)0(
Terminated, Lossless TL
oL
oLL ZZ
ZZ
LCjj 0
Then,
Conclusion: The reflection coefficient is a function of the load impedance and the characteristic impedance.
Recall for the lossless case,
Then zjL
zjs eeVzV )(
zjL
zj
os ee
Z
VzI
)(
Example A generator with 10Vrms and Rg=50, is connected to
a 75load thru a 0.8 50-lossless line. Find VL
oL
in
in
L
V
radV
V
jZ
26.1006.5
75.8.35
2.0
Terminated, Lossless TL
djL
djdj
o
djL
djdj
eeeZ
VdI
eeeVdV
)(
)(
It is customary to change to a new coordinate system, z = - l , at this point.
Rewriting the expressions for voltage and current, we have
Rearranging,
ljL
lj
o
ljL
lj
eeZ
VlI
eeVlV
)(
)(
ljL
lj
o
ljL
lj
eeZ
VlI
eeVlV
2
2
1)(
1)(
-z
z = - l
The impedance anywhere along the line is given by
The reflection coefficient can be modified as follows
Then, the impedance can be written as
After some algebra, an alternative expression for the impedance is given by
Conclusion: The load impedance is “transformed” as we move away from the load.
Impedance (Lossy line)
lL
lL
o e
eZ
lI
lVlZ
2
2
1
1
)(
)()(
)(1
)(1)(
l
lZlZ o
lZZ
lZZZZlZ
Lo
oLoin
tanh
tanh)(
)()( 222 ljlL
lL eeel
The impedance anywhere along the line is given by
The reflection coefficient can be modified as follows
Then, the impedance can be written as
After some algebra, an alternative expression for the impedance is given by
Conclusion: The load impedance is “transformed” as we move away from the load.
Impedance (Lossless line)
lj
L
ljL
o e
eZ
lI
lVlZ
2
2
1
1
)(
)()(
)(1
)(1)(
l
lZlZ o
ljZZ
ljZZZZlZ
Lo
oLoin
tan
tan)(
ljjL
ljL eeel 22)(
Exercise : using formulas A 2cm lossless TL has Vg=10 Vrms, Zg=60 , ZL=100+j80 and Zo=40,
=10cm . find: the input impedance Zin, the sending-end Voltage Vin,
ZL
Zg
Vg
+
Vin
-
Iin
+
VL
-
Zo
j
2 cm
52tan)80100(40
52
tan40)80100(40)2(
jj
jjcmZZ in
17.212.12 jZ in
Voltage Divider: rad
ZZ
ZVV
gin
ingin 766.030.3
o
oL
oLL ZZ
ZZ4.2362.0
SWR or VSWR or sWhenever there is a reflected wave, a standing wave will form out of the combination of incident and reflected waves.
The (Voltage) Standing Wave Ratio - SWR (or VSWR) is defined as
L
Ls
I
I
V
VSWRs
1
1
min
max
min
max
Power
The average input power at a distance l from the load is given by
which can be reduced to
The first term is the incident power and the second is the reflected power. Maximum power is delivered to load if =0
)()(Re2
1 * lIlVPave
2
2
12
o
o
ave Z
VP
Three common Cases of line-load combinations: Shorted Line (ZL=0)
Open-circuited Line (ZL=∞)
Matched Line (ZL = Zo)
ljZZ oin cot
oin ZZ
sjbljZZ Loin ,1 tan0
sL ,1
1,0 sL
Standing Waves -Short
Shorted Line (ZL=0), we had
So substituting in V(z)
sljZZ Loin ,1 , tan
-z /4/2
|V(z)|
])1([)( ljlj eeVzV
)sin2()( ljVzV
lVzV sin2)(
lVzV2
sin2)(
Voltage maxima
*Voltage minima occurs at same place that impedance has a
minimum on the line
Standing Waves -Open
Open Line (ZL=∞) ,we had
So substituting in V(z) |V(z)|
lVzV
lVzV
lVzV
eeVzV
sljZZ
ljlj
Loin
2cos2)(
cos2)(
)cos2()(
])1([)(
,1 , cot
-z /4/2
Voltage minima
Standing Waves -MatchedMatched Line (ZL = Zo), we had
So substituting in V(z)|V(z)|
VzV
eVzV
eVzV
eeVzV
sZZ
lj
lj
ljlj
Loin
)(
)(
)(
])0([)(
1,0 ,
-z /4/2
Java applets
http://www.amanogawa.com/transmission.html
http://physics.usask.ca/~hirose/ep225/ http://www.educatorscorner.com/index.cgi
?CONTENT_ID=2483
The Smith Chart
Smith Chart
Commonly used as graphical representation of a TL.
Used in hi-tech equipment for design and testing of microwave circuits
One turn (360o) around the SC = to /2
What can be seen on the
screen?
Network Analyzer
Smith Chart Suppose you use as coordinates the reflection
coefficient real and imaginary parts.
and define the normalized ZL:
ir
irL
L
L
L
L
o
LL
j
jjxrz
z
z
jxrZ
Zz
1
1
1
1
1
1
oL
oLir ZZ
ZZj
i
r
Now relating to z=r+jx
After some algebra, we obtain two eqs.
Similar to general equation of a circle of radius a, center at (h,k)
222
222
22
2
)()(
111
1
1
1
akyhx
xx
rr
r
ir
ir
Circles of r
Circles of x
Examples of circles of r and x
2
2
1Radius
1,1 Center
1
1Radius 0,
1 Center
xx
rr
r
rr
r
1
1Radius 0,
1 Center
Circles of r Circles of x
xx
1Radius
1,1 Center
Examples of circles of r and x
Circles of r Circles of x
rr
r
1
1Radius 0,
1 Center
The joy of the SC
Numerically s = r on the +axis of r in the SC
Proof:
1s
1-s0 but then
1
1)when (
1
1
j
r
rrz
z
z
r
L
L
Fun facts about the Smith Chart A lossless TL is represented as
a circle of constant radius, ||, or constant s
Moving along the line from the load toward the generator, the phase decrease, therefore, in the SC equals to moves clockwisely.
ljjL
ljL eeel 22)( To
generator
Fun facts about the Smith Chart
One turn (360o) around the SC = to /2 because in the formula below, if you substitute length for half-wavelength, the phase changes by 2, which is one turn.
Find the point in the SC where =+1,-1, j, -j, 0, 0.5 What is r and x for each case?
ljLel 2)(
Fun facts : Admittance in the SC The admittance, y=YL/Yo where Yo=1/Zo, can be
found by moving ½ turn (/4) on the TL circle
1
1
1
1
1
11)0(
1
1
)1(1
)1(1
1
1)4/(
1
1
1
1
1
1
1
11)0(
4
222
0
0
2
2
02
02
2
2
2
2
j
j
lj
lj
o
o
j
j
j
j
lj
lj
lj
o
lj
oo
L
e
e
eV
eZV
Yly
e
elz
e
e
e
e
eZV
eV
ZZ
Zlz
l
Fun facts about the Smith Chart
The r +axis, where r > 0 corresponds to Vmax
The r -axis, where r < 0 corresponds to Vmin
Vmax (Maximum
impedance)
Vmin
Exercise: using S.C. A 2cm lossless TL has Vg=10 Vrms, Zg=60 , ZL=100+j80 and Zo=40, =10cm . find: the input impedance Zin, the sending-end Voltage Vin,
Load is at .2179 @ S.C. Move .2 and arrive to .4179 Read
ZL
Zg
Vg
+
Vin
-
Iin
+
VL
-
Zo
j
2 cm
55.3. jzin
Voltage Divider:
radZZ
ZVV
gin
ingin 775.032.3
25.240
80100j
jzL
2.l
2212 jZ in
oL 5.23622.0
ocm 120622.0)2(
4.176. jyin
zL
zin 0.2
Exercise: cont….using S.C. A 2cm lossless TL has Vg=10 Vrms, Zg=60 , ZL=100+j80 and Zo=40, =10cm . find: the input impedance Zin, the sending-end Voltage Vin,
Distance from the load (.2179to the nearest minimum & max Move to horizontal axis toward the generator and arrive to .5Vmax and to .25 for the Vmin.
Distance to min=.5-.2179=.282 Distance to 2st voltage maximum is .282See drawing
ZL
Zg
Vg
+
Vin
-
Iin
+
VL
-
Zo
j
2 cm
25.2 jzL 2.lo
L 5.23622.0
Exercise : using formulas A 2cm lossless TL has Vg=10 Vrms, Zg=60 , ZL=100+j80 and Zo=40,
=10cm . find: the input impedance Zin, the sending-end Voltage Vin,
ZL
Zg
Vg
+
Vin
-
Iin
+
VL
-
Zo
j
2 cm
52tan)80100(40
52
tan40)80100(40)2(
jj
jjcmZZ in
17.212.12 jZ inVoltage Divider:
radZZ
ZVV
gin
ingin 766.030.3
o
oL
oLL ZZ
ZZ4.2362.0
ooL
ljLecm
6.12062.0144
)2( 2
Another example: A 26cm lossless TL is connected to load ZL=36-j44 and Zo=100, =10cm . find: the input impedance Zin
Load is at .427 @ S.C. Move .1 and arrive to .527 Read
ZL
Zg
Vg
+
Vin
-
Iin
+
VL
-
Zo
j
26cm
16.31. jzin
Distance to first Vmax:
44.36. jzL 1.)5(.56.2 l
1631 jZ in
oL 12754.0
028.0427.5.0min l
278.25.028.0max l
Exercise 11.4
A 70 lossless line has s =1.6 and =300o. If the line is 0.6 long, obtain , ZL, Zin and the distance of the first minimum voltage from the load.
Answer
The load is located at: Move to .4338and draw line from center to this place, then read where it crosses you TL circle. Distance to Vmin in this case, lmin =.5-.3338=
oL 30023.0
1s
1-s
6.335.80
48.15.1
jzZZ
jz
LoL
L
6/
5.176.47
25.68.0
jZ
jz
in
in3338.
Java Applet : Smith Chart
http://education.tm.agilent.com/index.cgi?CONTENT_ID=5
Applications Slotted line as a frequency meter Impedance Matching
If ZL is Real: Quarter-wave Transformer (/4 Xmer)
If ZL is complex: Single-stub tuning (use admittance Y)
Microstrip lines
Slotted Line
Used to measure frequency and load impedance
HP Network Analyzer in Standing Wave Display http://www.ee.olemiss.edu/software/naswave/Stdwave.pdf
Slotted line exampleGiven s, the distance between adjacent minima, and lmin for an
“air” 100 transmission line, Find f and ZL
s=2.4, lmin=1.5 cm, lmin-min=1.75 cm
Solution: =8.6GHz
Draw a circle on r=2.4, that’s your T.L.
move from Vmin to zL
3850
38.5.
jzZZ
jz
LoL
L
cm
cf
5.3
103 8
429./5.3
5.1min l
Quarter-wave transformer …for impedance matching
2/tan
tan
42
tan
ljZZ
jZZ
ZZLo
oL
oin
ZLZin Zo ,
l= /4
L
oin Z
ZZ
2
Conclusion: **A piece of line of /4 can be used to change the impedance to a desired value (e.g. for impedance matching)
Single Stub Tuning …for impedance matching A stub is connected in parallel to sum the
admittances Use a reactance from a short-circuited
stub or open-circuited stub to cancel reactive part
Zin=Zo therefore z =1 or y=1 (this is our goal!)
Single Stub Basics
We work with Y, because in parallel connections they add.
YL (=1/ZL) is to be matched to a TL having characteristic admittance Yo by means of a "stub" consisting of a shorted (or open) section of line having the same characteristic admittance Yo
http://web.mit.edu/6.013_book/www/chapter14/14.6.html
Single Stub Steps
First, the length l is adjusted so that the real part of the admittance at the position where the stub is attached is equal to Yo or yline = 1+jb
Then the length of the shorted stub is adjusted so that it's susceptance cancels that of the line, or ystub= -jb
Example: Single StubA 75 lossless line is to be
matched to a 100-j80 load with a shorted stub. Calculate the distance from the load, the stub length, and the necessary stub admittance.
Answer:
Change to:
.4338- .3393=0.094j or next intersection:0.272,
Short stub:.25-.124=0.126
With ystub= -j.96/75 j mhos
067.133.1 jzL 366.457. jyL
Example: Single StubA 50 lossless T.L. is 20 m long and terminated into a 120+j220 load. To perfectly match ,
what should be the length and location of a short-circuited stub line. Assume an operating frequency of 10MHz.
Answer: =30m, zL= 2.4+j4.4 (circulito rojo).
Trazo raya por el centro y leo yL al otro lado (circulito azul).
La yL está en la posición 0.472. Trazo círculo amarillo, esa es mi T.L. donde está la carga. Busco donde interseca el
circulo de r=1(circulo turquesa). Lo interseca en 1+j3.2 Ese es mi 1+jb. Y está en la posicion 0.214. Por tanto la distancia desde la carga al segmento(stub) = .5-.472(debido a cambio de
escala) +.214= .242 . Ahora miro el círculo del segmento= circulo grande en el SC (donde Gamma =1).y busco
donde jb= -j3.2 (abajo ver flecha roja). Para buscar su posición, trazo línea desde el centro hasta afuera y leo posición está
en .2958. El segmento empieza en carga de corto circuito (ver palabra roja que dice short) está en .25. Por lo tanto el largo del segmento es
.25-.202= .0485
Microstrips
Microstripsanalysisequations &Pattern of EM fields
1/8
ln60
hw
h
w
w
hZ
eff
o
]1/for[)444.1/ln(667.0393.1
1201
hw
hwhw
Zeff
o
MicrostripDesign Equations
Falta un radical en eff
eff
cu
2/2
8/
2
hw
e
ehw
A
A
2/61.039.)1ln(
2
1
)12ln(12
/
hwB
BB
hw
rr
r
rr
rroZA
11.0
23.01
1
2
1
60
roZB
160 2
Microstrip Design Curves
ExampleA microstrip with fused quartz (r=3.8) as a
substrate, and ratio of line width to substrate thickness is w/h=0.8, find:
Effective relative permittivity of substrate Characteristic impedance of line Wavelength of the line at 10GHz
Answer:eff=2.75, Zo=86.03 , =18.09 mm
75.28.0/1212
8.2
2
8.4/1212
1
2
1
wh
rreff
03.868.8.
8ln
75.2
60
]1/for[8
ln60
hw
h
w
w
hZ
eff
o
mmf
c
eff
1.1875.21010
1039
8
Diseño de microcinta:
Dado (r=4) para el substrato, y h=1mm halla w para Zo=50 y cuánto es eff?
Solución: Suponga que como Z es pequena w/h>2
