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1
Transmission Lines
Dr. Sandra Cruz-Pol ECE Dept. UPRM
Introduction to Transmission Lines (T.L.) n Hi frequency or hi power requires T.L. n TEM waves propagate thru T.L. n We will develop T.L. theory to see how
waves propagate thru them
Exercise 11.3 n A 40-m long TL has Vg=15 Vrms, Zo=30+j60 Ω, 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 γ.
n Answers:
ZL Zg
Vg
+ Vin
-
Iin
+ VL
-
Zo=30+j60 γ=α +j β
40 m
Zin = 30+ j60Ω, Iin = 0.112∠− 63.4o A,
Vin = 7.5∠0oVrms, γ = 0.0101+ j0.02094 m−1
7.45∠−111oe−γ 40 = 0.112∠− 63o
Seem to be in parallel, but these V are not equal!
Don’t worry about the details, I’ll teach you
about solving this type of problems pretty soon.
Transmission Lines I. TL parameters (R’,L’,C’, G’) II. TL Equations (for V and I) III. 3 Concepts:
I. Input Impedance, II. Reflection Coefficient and III. Characteristic Impedance
IV. SWR, Power V. Smith Chart VI. Applications - Quarter-wave
transformer, Slotted line, Single stub VI. Microstrips
Transmission Lines (TL) n TL have 2
conductors in parallel with a dielectric separating them
n They transmit TEM waves inside the lines for insulation
outer conductor
Common Transmission Lines Two-wire (ribbon)
Coaxial
Microstrip
Stripline (Triplate)
twinlead
2
Other TL (higher order) [Chapter 12] Fields inside the TL
n V proportional to E, n I proportional to H
∫∫⋅=
⋅−=
dlHI
dlEV
Distributed parameters The parameters that characterize the TL are
given in terms of per length. n R = ohms/meter n L = Henries/ m n C = Farads/m n G = mhos/m
cmckmc
GHz
Hz
15000,000,2000/000,560/
2
60
==
==
λ
λ
At high frequencies we’re dealing with wavelengths comparable to the size of
the circuits.
Common Transmission Lines R’, 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’
€
1πaσ condδ
€
12πσ condδ
1a
+1b
⎛
⎝ ⎜
⎞
⎠ ⎟
€
2wσ condδ
€
Ω /m
€
µπacosh D
2a⎛
⎝ ⎜
⎞
⎠ ⎟
€
µ2πln b
a⎛
⎝ ⎜ ⎞
⎠ ⎟
€
µdw
€
H /m
€
πσ diel
acosh D / 2a( )( )
€
2πσ diel
ln b /a( )
€
σ dielwd
€
S /m
€
πεacosh D / 2a( )( )
€
2πεln b /a( )
€
εwd
€
F /m
TL representation Distributed line parameters Using KVL:
3
Distributed parameters n Taking the limit as Δz tends to 0 leads to
n Similarly, applying KCL to the main node gives
ttzILtzRI
ztzV
∂
∂+=
∂
∂−
),(),(),(
ttzVCtzGV
ztzI
∂
∂+=
∂
∂−
),(),(),(
−∂V∂z
= RI + L ∂I∂t
−∂I∂z=GV +C ∂V
∂t0 2
2
2
=−∂
∂s
s VzV
γ
Wave equation
n Using phasors
n The two expressions reduce to
Wave Equation for voltage
−∂Vs∂z
= R+ jωL( ) Is
−∂Is∂z
= G + jωC( )Vs
])(Re[),(
])(Re[),(tj
s
tjs
ezItzIezVtzVω
ω
=
=
( )( )CjGLjR ωωγ ++=2
Transmission Lines I. TL parameters (R’,L’,C’, G’) II. TL Equations (for V and I) III. 3 Concepts:
I. Input Impedance, II. Reflection Coefficient and III. Characteristic Impedance
IV. SWR, Power V. Smith Chart VI. Applications - Quarter-wave
transformer, Slotted line, Single stub VI. Microstrips
TL Equations n Note that these are the wave eq. for
voltage and current inside the lines.
n The propagation constant is γ and the wavelength and velocity are
d2Vsdz2
−γ 2Vs = 0d 2Isdz2
−γ 2Is = 0
γ =α +β j = (R+ jωL)(G + jωC)
λ = 2π β u =ω β = f λ
Waves move through line n The general solution is
n In time domain is
Vs =V+e−γz + V −eγz
z
V (z, t) = Re[Vs (z)ejω t ]
=V +e−αz cos(ωt −βz)+V −e+αz cos(ωt +βz)
Waves move through line n For Current
n Similarly for time-domain, I
Is = I+e−γz + I −eγz
z
I(z, t) = I +e−αz cos(ωt −βz)+ I −eαz cos(ωt +βz)
4
Transmission Lines I. TL parameters (R’,L’,C’, G’) II. TL Equations (for V and I) III. 3 Concepts:
I. Input Impedance, II. Reflection Coefficient and III. Characteristic Impedance
IV. SWR, Power V. Smith Chart VI. Applications - Quarter-wave
transformer, Slotted line, Single stub VI. Microstrips
We will define 3 concepts:
n Characteristic impedance, Zo
n Reflection Coefficient, ΓL n Input Impedance, Zin
We will define 3 concepts:
n Characteristic impedance, Zo
n Reflection Coefficient, ΓL n Input Impedance, Zin
Characteristic Impedance of a Line, Zo n Is the ratio of positively traveling voltage wave to
current wave at any point on the line
−dV (z)dz
= (R+ jωL)I(z)
z
Zo =V +
I +=R+ jωL
γ=
R+ jωLG + jωC
= Ro + jXo = −V −
I −
substitutingV (z) =V +e−γz
I(z) = I +e−γz
−(−γV +e−γz ) = (R+ jωL)I +e−γz
Characteristic Impedance, Zo
Zo =V +
I += −
V −
I −=
R+ jωLG + jωC
Is = I+e−γz + I −eγz
Is =V +
Zoe−γz +
V −
−Zoeγz
Different cases of TL
n Lossless
n Distortionless
n Lossy
Transmission line
Transmission line
Transmission line
Find the characteristic impedance and propagation constant for each:
5
Lossless Lines (R=0=G) Have perfect conductors and a perfect dielectric
medium between them.
n Propagation:
n Wavelength & Velocity:
n Impedance
α = 0, γ = jβ, β =ω LC
u = ωβ
= 1LC
= f λ, λ =2πβ
Zo = Ro =LC
Xo = 0
Distortionless line (R/L = G/C) Is one in which the attenuation is independent on
frequency.
n Propagation:
n Velocity:
n Impedance
γ =α + jβ
α = RG β =ω LC
u = ωβ
= 1LC
= f λ
Xo = 0 Zo = Ro =LC=
RG
Summary γ = α + jβ
Propagation ConstantZo
Characteristic Impedance General (Lossy)
Lossless
Distortionless RC = GL LCjRG ωγ +=
LCjωγ += 0
CjGLjRZo ω
ω+
+=
GR
CLRZ oo ===
))(( CjGLjR ωωγ ++=
CLRZ oo ==
u = ωβ= 707 km / s
P.E. 11.2 A telephone line has R=30 Ω/km, L=100 mH/km, G=0, and C= 20µF/km. At 1kHz, FIND: the characteristic impedance of the line, the propagation constant, the phase velocity. n Is it distortionless? Solution: Zo =
R+ jωLG + jωC
=30+ j2π (1k)(100m)0+ j2π (1k)20 ⋅10−6
= 70.75∠−1.37oΩ
γ = R+ jωL( ) G + jωC( )
= 30+ j2π (100)( ) 0+ j2π20 ⋅10−3( )= 0.21+ j8.88 / km
No
We will define 3 concepts:
n Characteristic impedance, Zo
n Reflection Coefficient, ΓL n Input Impedance, Zin
We will define 3 concepts:
n Characteristic impedance, Zo
n Reflection Coefficient, ΓL n Input Impedance, Zin
6
Reflection coefficient at the load, ΓL
Vs (z) =V+e−γz +V −e+γz
Load is usually taken at z=0
and generator at z= -l
ΓL =V −
V +
( )zLz
s eeVzV γγ +−+ Γ+=)(
z=0 z= -l
For a Lossless TL terminated with a load
( )zLz
s eeVzV γγ +−+ Γ+=)(
( )zLz
os ee
ZV
zI γγ +−+
Γ−=)(
( )( )zL
z
zL
z
os
s
eeeeZ
zIzVzZ γγ
γγ
+−
+−
Γ−
Γ+==)()()(
Then,
Similarly,
The impedance anywhere along the line is given by
The impedance at the load end, ZL, is given by
Z(l = 0) = ZL = Zo1+ΓL( )1−ΓL( )
Terminated, Lossless TL
LCjj ωβγ =+= 0
Solving for ΓL
Conclusion: The reflection coefficient is a function of the load impedance and the characteristic impedance.
Recall for the lossless case,
Then ( )zjLzj
s eeVzV ββ +−+ Γ+=)(
( )zjLzj
os ee
ZV
zI ββ +−+
Γ−=)(
oL
oLL ZZ
ZZ+
−=Γ
Definition: Matched line
n Means that Zo=ZL
n Therefore there are no reflection!
ΓL=0 oL
oLL ZZ
ZZ+
−=Γ
What happens when you connect the wrong TL to a speaker?
Terminated, Lossless TL
( )
( )djLdjdj
o
djL
djdj
eeeZV
dI
eeeVdV
βββ
βββ
−+++
−++
Γ−=
Γ+=
)(
)(
Using convention the coordinate system, z = - l , at input.
Rewriting the expressions for voltage and current, we have
Rearranging,
( )
( )ljLlj
o
ljL
lj
eeZV
lI
eeVlV
ββ
ββ
−++
−+
Γ−=−
Γ+=−
)(
)(
( )
( )ljLlj
o
ljL
lj
eeZV
lI
eeVlV
ββ
ββ
2
2
1)(
1)(
−++
−++
Γ−=−
Γ+=−
-z
z = - l
7
Voltage anywhere on the Line
( )
( )djLdjdj
o
djL
djdj
eeeZV
dI
eeeVdV
βββ
βββ
−+++
−++
Γ−=
Γ+=
)(
)(
Recall,-z
z = - l
Vin =V (−l)VL =V (0)
V (l = λ / 4) =V + e jπ /2 + (0)e− jπ /2( )
Input voltage= sending end,
Load Voltage = receiving end
V (l) =V + e jβ l +ΓLe− jβ l( )
Voltage quaterwave from matchedload
λ / 4
We will define 3 concepts:
n Characteristic impedance, Zo
n Reflection Coefficient, ΓL n Input Impedance, Zin
The impedance anywhere along the line is given by
Impedance (Lossless line) ( )( )ljL
ljL
o eeZ
lIlVlZ β
β
2
2
11
)()()(
−
−
Γ−
Γ+==
( )( ))(1
)(1)(llZlZ o Γ−
Γ+=
( )( )ljZZ
ljZZZZlZLo
oLoin β
βtantan)(
+
+==
ljjL
ljL eeel βθβ 22)( −− ΓΓ=Γ=Γ
The reflection coefficient at any point along the line:
Conclusion: The load impedance is “transformed” as we move away from the load.
After some algebra, an alternative expression for the impedance is given by
Then, the impedance can be written as.
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: in lossy lines, we end up with the hyperbolic tangent.
Impedance (Lossy line) ( )( )lL
lL
o eeZ
lIlVlZ γ
γ
2
2
11
)()()(
−
−
Γ−
Γ+==
( )( ))(1
)(1)(llZlZ o Γ−
Γ+=
( )( )lZZ
lZZZZlZLo
oLoin γ
γtanhtanh)(
+
+==
)()( 222 ljlL
lL eeel βαγ −−− Γ=Γ=Γ
Example: Matched Case A TL has Vg=10 Vrms, Zo=50Ω If the line is matched to the load and the generator, find: the input impedance Zin, the sending-end Voltage Vin
n Answers:
ZL Zg
Vg
+ Vin
-
Iin
+ VL
-
Zo γ=j β
l
Zg = ZL = Z0 ⇒ Zin = 50Ω
ΓL =ZL − ZoZL + Zo
= 0
Zin = ZoZL + jZo tanβl( )Zo + jZL tanβl( )
= Zo
Γ = 0VoltageDivider Vin = 5∠0
o
Example: Matched Case A λ/8 long TL has VL=5 < 30o, Zo=50Ω If the line is matched to the load, find: the input impedance Zin, the sending-end Voltage Vin, the propagation constant γ.
n Answers:
ZL Zg
Vg
+ Vin
-
Iin
+ VL
-
Zo γ=j β
l
ZL = Z0 ⇒ Zin = 50Ω
V (l) =V + e jβ l +ΓLe− jβ l( )
ΓL =ZL − ZoZL + Zo
Zin = ZoZL + jZo tanβl( )Zo + jZL tanβl( )
Γ = 0
VL = 5∠30o =V
l=0=V + e jβ 0 + (0)e− jβ 0( ),
Solve for : V + = 5∠30o
V (l = λ / 8) = 5∠30o e jπ /4 + (0)e− jπ /4( )
Vin = 5∠30o e jπ /4( ) = 5∠75o
8
Exercise 1: Not Matched A 2cm lossless TL has VL=10 ej30o, Zg=60 Ω, ZL=50 Ω and Zo=100Ω, λ=10cm . Find: the input impedance Zin, the sending-end Voltage Vin,
n Use this equation at load and at input, find V+ n Find Vin n Find Zin (at input)
ZL Zg
Vg
+ Vin
-
Iin
+ VL
-
Zo γ=j β
l
V (l) =V + e jβ l +ΓLe− jβ l( ) ΓL =
ZL − ZoZL + Zo
Zin = ZoZL + jZo tanβl( )Zo + jZL tanβl( )
Exercise 2: using formulas n A 2cm lossless TL has Vg=10 Vrms, Zg=60 Ω, ZL=100+j80 Ω and Zo=40Ω,
λ=10cm . find: the input impedance Zin, the sending-end Voltage Vin,
ZL Zg
Vg
+ Vin
-
Iin
+ VL
-
Zo γ=j β
l
( )52tan)80100(4052tan40)80100(
40)2(π
π
jj
jjcmZZin
++
⎟⎠
⎞⎜⎝
⎛ ++==
Ω−= 17.212.12 jZinVoltage Divider:
radZZZV
Vgin
ingin 766.030.3 −∠=
+=
Example 3: Not matched to load A generator with 10Vrms and Rg=50, is connected to a 75Ω load thru a 0.8λ, 50Ω-lossless line. n Find VL
PowerDividerVin =∠
V + = 5.006∠1.26radVL =V (l = 0)
ΓL =ZL − ZoZL + Zo
Zin = ZoZL + jZo tanβl( )Zo + jZL tanβl( )
Zo = 50Ω ZL = 75, l = .8λ
V (l) =V + e jβ l +ΓLe− jβ l( )
ZL Zg
Vg
+ Vin
-
Iin + VL
-
Zo γ=j β
l
ΓL = 0.2Zin = 35.+ j8.75Ω
Exercise 11.3: Matched Case A 40-m long TL has Vg=15 Vrms, Zo=30+j60 Ω, 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 γ.
n Answers:
ZL Zg
Vg
+ Vin
-
Iin
+ VL
-
Zo=30+j60 γ=α +j β
40 m
oorms
oin
oinin
emjVV
AIjZ
63112.011145.7
02094.00101.0,05.7
,4.63112.0,6030
40
1
−∠=−∠
+=∠=
−∠=Ω+=
−
−
γ
γ
Transmission Lines I. TL parameters (R’,L’,C’, G’) II. TL Equations (for V and I) III. 3 Concepts:
I. Input Impedance, II. Reflection Coefficient and III. Characteristic Impedance
IV. SWR, Power V. Smith Chart VI. Applications - Quarter-wave
transformer, Slotted line, Single stub VI. Microstrips
Power n The average input power at a distance l from the load is given
by
n which can be reduced to
n The first term is the incident power and the second is the reflected power. Maximum power is delivered to load if Γ=0
[ ])()(Re21 * lIlVPave =
( )22
12
Γ−=+
o
oave Z
VP
9
SWR or VSWR or s Whenever 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
II
VV
SWRs
Γ−
Γ+=
===
11
min
max
min
max
Summary
n Input Impedance
n Reflection Coef
n SWR
Zin = ZoZL + jZo tanβl( )Zo + jZL tanβl( )
s =1+ ΓL1− ΓL
ΓL =ZL − ZoZL + Zo
Three (3) common Cases of line-load combinations: n Shorted Line (ZL=0)
n Open-circuited Line (ZL=∞)
n Matched Line (ZL = Zo)
ljZZ oin βcot−=
oin ZZ =
Zin = 0+ jZo tanβl = jb
∞==Γ sL ,1
1,0 ==Γ sL
ΓL = −1, s =∞
Standing Waves –Short (ZL=0)
n So substituting in V(z)
∞=−=Γ= sljZZ Loin ,1 , tanβ
-z -λ/4 -λ/2 -λ
|V(z)|
])1([)( ljlj eeVzV ββ −+ −+=
)sin2()( ljVzV β+=
( )lVzV βsin2)( +=
⎟⎠
⎞⎜⎝
⎛= + lVzVλπ2sin2)(
Voltage maxima
*Voltage minima occurs at same place that impedance has a
minimum on the line
Standing Waves –Open(ZL=∞)
n So substituting in V(z)
|V(z)|
Zin = − jZo cotβ l , ΓL = +1, s =∞
-z -λ/4 -λ/2 -λ
Voltage minima
V (z) =V +[e jβl + (+1)e− jβl ]V (z) =V +(2cosβl)
V (z) = V + 2cos βl( )
V (z) = V + 2cos 2πλl
⎛
⎝⎜
⎞
⎠⎟
Standing Waves –Matched (ZL = Zo)
n So substituting in V(z)
|V(z)| V (z) =V +[e jβl + (0)e− jβl ]V (z) =V +e jβl
V (z) = V + e jβl
V (z) = V +-z -λ/4 -λ/2 -λ
Zin = Zo , ΓL = 0, s =1
10
Java applets
n http://www.amanogawa.com/transmission.html
n http://physics.usask.ca/~hirose/ep225/ n http://www.educatorscorner.com/
index.cgi?CONTENT_ID=2483
Transmission Lines I. TL parameters (R’,L’,C’, G’) II. TL Equations (for V and I) III. 3 Concepts:
I. Input Impedance, II. Reflection Coefficient and III. Characteristic Impedance
IV. SWR, Power V. Smith Chart VI. Applications - Quarter-wave
transformer, Slotted line, Single stub VI. Microstrips
The Smith Chart
Smith Chart
n Commonly used as graphical representation of a TL.
n Used in hi-tech equipment for design and testing of microwave circuits
n One turn (360o) around the SC = to λ/2
What can be seen on the screen?
Network Analyzer Smith Chart n Use the reflection coefficient real and imaginary parts .
and define the normalized ZL:
( )( )
ir
irL
L
L
L
L
o
LL
jjjxrz
zz
jxrZZz
Γ−Γ+
Γ+Γ+=+=
+
−=Γ
+=Γ−
Γ+==
11
1111
( )( )oL
oLir ZZ
ZZj+
−=Γ+Γ=∠Γ=Γ Γθ
Γi
Γr
|Γ|
11
Now relating to z =r +jx
n After some algebra, we obtain two eqs.
n Similar to general equation of a circle of radius a, center at (h,k)
Γr −r1+ r
⎛
⎝⎜
⎞
⎠⎟2
+Γi2 =
11+ r⎡
⎣⎢⎤
⎦⎥
2
Γr −1[ ]2 + Γi −1x
⎡
⎣⎢⎤
⎦⎥
2
=1x⎡
⎣⎢⎤
⎦⎥
2
Circles of r
Circles of x
(x − h)2 + (y− k)2 = a2
Examples of circles of r and x
2
2
1Radius 1,1 Center
11Radius 0,
1 Center
⎥⎦
⎤⎢⎣
⎡=⎟⎠
⎞⎜⎝
⎛=
⎥⎦
⎤⎢⎣
⎡+
=⎟⎠
⎞⎜⎝
⎛+
=
xx
rrr
⎥⎦
⎤⎢⎣
⎡+
=⎟⎠
⎞⎜⎝
⎛+
=rr
r1
1Radius 0,1
Center
Circles of r Circles of x
⎥⎦
⎤⎢⎣
⎡=⎟⎠
⎞⎜⎝
⎛=xx1Radius 1,1 Center
Examples
Find z= r+jxgiven Γ
Also Find Γ
Given Z
Examples of circles of r and x
Circles of r Circles of x
rrr
+=⎟
⎠
⎞⎜⎝
⎛+
=1
1Radius 0,1
Center
Find SWR on the SC
n Numerically s = r on the +axis of Γr in the SC Proof:
1s1-s0 but then
11)when (
11
+≡+Γ=Γ
+
−===
+
−=Γ
j
rrrz
zz
r
L
L
Moving on the TL on the SmithC n A lossless TL is represented as
a circle of constant radius, |Γ|, or constant s
n Moving along the line from the load toward the generator, the phase decrease, therefore, in the SC equals to moves clockwisely.
ljjL
ljL eeel βθβ 22)( −− ΓΓ=Γ=Γ To
generator
12
One-turn on the Smith Chart
n One turn (360o) around the SC = to λ/2 because in the formula below, if you substitute length for half-wavelength, the phase changes by 2π, which is one turn.
n Find the point in the SC where Γ=+1,-1, j, -j, 0, 0.5 n What is r and x for each case?
ljLel β2)( −Γ=Γ
Fun facts : Admittance in the SC n The admittance, y=YL/Yo where Yo=1/Zo, can be
found by moving ½ turn (λ/4) on the TL circle
( )( )
( )( ) Γ+
Γ−=
Γ+
Γ−=
Γ+
Γ−⎟⎠⎞
⎜⎝⎛==
Γ+
Γ−=
−Γ−
−Γ+=
Γ−
Γ+==
Γ−
Γ+=
Γ−
Γ+=
Γ−
Γ+=
Γ−
Γ+===
=⎟⎠
⎞⎜⎝
⎛=
+
+
+
+
11
11
1
11)0(
11
)1(1)1(1
11)4/(
11
11
11
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
ee
eV
eZV
Yly
eelz
ee
ee
eZV
eVZZ
Zlz
l
β
β
π
π
β
β
β
β
λ
πλ
λπ
β
Vmax and Vmin on the SmithC n The Γr +axis, where r > 0 corresponds to Vmax
n The Γr -axis, where r < 0 corresponds to Vmin
Vmax (Maximum impedance)
Vmin
Exercise: using S.C. n A 2cm lossless TL has Vg=10 Vrms, Zg=60 Ω, ZL=100+j80 Ω and Zo=40Ω,
λ=10cm . find: the input impedance Zin, the sending-end Voltage Vin,
n Load is at .2179λ @ S.C. n Move .2λ and arrive to .4179λn Read
ZL Zg
Vg
+ Vin
-
Iin
+ VL
-
Zo γ=j β
2 cm
55.3. jzin −=
Voltage Divider:
radZZZV
Vgin
ingin 775.032.3 −∠=
+=
25.24080100 jjzL +=
+= λλ 2.=l
Ω−= 2212 jZin
oL 5.23622.0 ∠=Γ
ocm 120622.0)2( −∠=Γ
Ω−= 4.176. jyin
zL
zin 0.2λ
Exercise: cont….using S.C. n A 2cm lossless TL has Vg=10 Vrms, Zg=60 Ω, ZL=100+j80 Ω and Zo=40Ω,
λ=10cm . find: the input impedance Zin, the sending-end Voltage Vin,
n Distance from the load (.2179λ) to the nearest minimum & max n Move to horizontal axis toward the generator and arrive to .5λ
(Vmax) and to .25λ for the Vmin.
n Distance to min=.5-.2179=.282λ n Distance to 2st voltage maximum is .282λ +.25λ=.482 See drawing
ZL Zg
Vg
+ Vin
-
Iin
+ VL
-
Zo γ=j β
2 cm
25.2 jzL += λλ 2.=l oL 5.23622.0 ∠=Γ
13
Exercise : using formulas n A 2cm lossless TL has Vg=10 Vrms, Zg=60 Ω, ZL=100+j80 Ω and Zo=40Ω,
λ=10cm . find: the input impedance Zin, the sending-end Voltage Vin,
ZL Zg
Vg
+ Vin
-
Iin
+ VL
-
Zo γ=j β
2 cm
( )52tan)80100(4052tan40)80100(
40)2(π
π
jj
jjcmZZin
++
⎟⎠
⎞⎜⎝
⎛ ++==
Ω−= 17.212.12 jZinVoltage Divider:
radZZZV
Vgin
ingin 766.030.3 −∠=
+=
o
oL
oLL ZZ
ZZ 4.2362.0 ∠=+
−=Γ
ooL
ljLecm
6.12062.0144
)2( 2
−∠=−∠Γ=
Γ=Γ − β
Another example: n A 26cm lossless TL is connected to load ZL=36-j44 Ω and Zo=100Ω,
λ=10cm . find: the input impedance Zin
n Load is at .427λ @ S.C. n Move .1λ and arrive to .527λ (=.027λ)
n Read
ZL Zg
Vg
+ Vin
-
Iin
+ VL
-
Zo γ=j β
26cm
16.31. jzin +=
Distance to first Vmax:
44.36. jzL −= λλλλ 1.)5(.56.2 +==l
Ω+= 1631 jZin
ΓL = 0.54∠−127o
λλλ 028.0427.5.0min =−=lλλλ 278.25.028.0max =+=l
Exercise 11.4
n A 70 Ω lossless line has s =1.6 and at the load θΓ =300o. If the line is 0.6λ long, obtain Γ, ZL, Zin and the distance of the first minimum voltage from the load.
n Answer
n The load is located at: n Move to .4338 λ and draw line from n center to this place, then read where it crosses you TL circle. n Distance to Vmin in this case, lmin =.5λ-.3338λ= ~
oL 30023.0 ∠=Γ
1s1-s+
=Γ
Ω−==
−=
6.335.8048.15.1
jzZZjz
LoL
L
6/
5.176.4725.68.0
λ
Ω−=
−=
jZjz
in
inλ3338.
Java Applet : Smith Chart n http://education.tm.agilent.com/index.cgi?CONTENT_ID=5