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Transmission Lines – Basic Theories
1 Introduction
At high frequencies, the wavelength is much smaller than
the circuit size, resulting in different phases at different
locations in the circuit.
Quasi-static circuit theory cannot be applied. We need to
use transmission line theory.
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Transmission Lines – Basic Theories
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0
z
A transmission line is a two-port network connecting
a generator circuit at the sending end to a load at the
receiving end.
Unlike in circuit theory, the length of a transmission line
is of utmost importance in transmission line analysis.
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2 Common Types of Transmission Lines
metal walls
dielectric spacing
(d) Microstrip line
(e) Waveguide
We focus on studying the coaxial and the two-wire
transmission lines.
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3 AC Steady-State Analysis
3.1
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Distributed parameter representation
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We use the following distributed parameters to
characterize the circuit properties of a transmission line.
R’ = resistance per unit length, (Ω/m)
L’ = inductance per unit length, (H/m)
G’ = conductance per unit length, (S/m)
C’ = capacitance per unit length, (F/m)
Δz = increment of length, (m)
These parameters are related to the physical properties of
the material filling the space between the two wires.
G' σ
=
C' ε
L'C ' = με
(See Text Book No.3,
pp. 432-433)
where µ, ε, σ = permittivity, permeability, conductivity
of the surrounding medium.
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For the coaxial and two-wire transmission lines, the
distributed parameters are related to the physical
properties and geometrical dimensions as follows:
Surface
resistivity of
the conductors
(See Text
Book No.3,
pp. 445-447)
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3.2
Equations and solutions
Consider a short section Δz of a transmission line
(dropping the primes on R’, L’, G’, C’ hereafter) :
Load
Generator
Using KVL and KCL circuit theorems, we can
derive the following differential equations for this
section of transmission line.
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∂i ( z , t )
v ( z , t ) − RΔzi ( z , t ) − LΔz
− v ( z + Δz , t ) = 0
∂t
∂v( z + Δz , t )
i ( z , t ) − GΔzv( z + Δz , t ) − C Δz
− i ( z + Δz , t ) = 0
∂t
By letting Δz→0, these lead to coupled equations:
∂v( z , t )
∂i ( z , t )
−
= Ri ( z , t ) + L
∂z
∂t
∂i ( z , t )
∂v( z , t )
−
= Gv( z , t ) + C
∂z
∂t
General Transmission Line Equations – Coupled!
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For sinusoidal varying voltages and currents, we can use
phasor forms.
v ( z , t ) = Re{V (z )e jωt }
i (z , t ) = Re{I (z )e jωt }
V(z) and I(z) are called phasors of v(z,t) and i(z,t). In
terms of phasors, the coupled equations can be written as:
dV ( z )
−
= ( R + jω L) I ( z )
dz
dI ( z )
−
= (G + jωC )V ( z )
dz
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After decoupling,
2
d V ( z)
2
V (z)
=
γ
2
dz
d 2 I ( z)
2
=
γ
I (z)
2
dz
γ = α + jβ =
( R + jω L )( G + jωC )
γ is the complex propagation constant whose real part α is
the attenuation constant (Np/m) and whose imaginary
part β is the phase constant (rad/m). Generally, these
quantities are functions of ω.
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Solutions to transmission line equations:
Forward
travelling
wave.
V ( z) = V + ( z) + V − ( z)
+ −γ z
0
=V e
+
− γz
0
+V e
−
I ( z) = I ( z) + I ( z)
Backward
travelling
wave.
= I 0+ e −γ z + I 0− eγ z
V0+ , V0− , I 0+ , I 0− = wave amplitudes in the forward and
backward directions at z = 0. (They
are complex numbers in general.)
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4 Transmission Line Parameters
From the solutions to the transmission line equations, it
can be shown (using the coupled transmission line
equations) that:
V0+
V0− R + jω L
=− − =
+
γ
I0
I0
This ratio is called characteristic impedance Z0.
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Z0 =
R + jωL
γ
R + jωL
=
=
G + jωC
G + jωC
γ = α + jβ =
γ
(R + jωL)(G + jωC )
Z0 and γ are the two most important parameters of
a transmission line.
They depend on the
distributed parameters (RLGC) of the line itself
and ω but not the length of the line.
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Parameters for lossless transmission lines
For lossless transmission lines, R = G = 0.
α =0
β = ω LC = ω με
ω
u p = phase velocity = =
β
1
=
LC
1
με
γ = complex propagation constant
= jβ = jω με = j 2πf με = j
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2π
λ
= jk
Transmission Lines – Basic Theories
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λ = wavelength along the transmission line
ω 2π
1
=
=
=
=
=
f
fβ
β
f με
f LC
up
1
Z 0 = characteristic impedance
R + jωL
=
G + jωC
L
=
C
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Voltage and current along the line:
V ( z ) = V0+ e − jkz + V0− e jkz
I ( z ) = I 0+ e − jkz + I 0− e jkz
Define a reflection coefficient at z = 0 as ΓL:
reflected voltage at z = 0
ΓL =
incident voltage at z = 0
V0− e jk ×0 V0−
= + − jk ×0 = + = Γ L e jθ L
V0 e
V0
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In terms of the reflection coefficient ΓL, the total
voltage and current can be written as:
V (z ) = V e
+ − jkz
0
−
0
+V e
+
0
jkz
−
0
V − jkz V jkz
I (z ) =
e −
e
−
Z0
Z0
⎛
⎞
V
+ − jkz
2 jkz
0
⎜
= V0 e ⎜1 + + e ⎟⎟
+
−
⎛
V
V
V
− jkz
2 jkz ⎞
0
0
0
⎝
⎠
e ⎜⎜1 − + e ⎟⎟
=
Z0
⎠
⎝ V0
= V + e − jkz (1 + Γ e2 jkz )
0
L
+ − jkz
0
=I e
(1 − Γ e )
2 jkz
L
In subsequent analyses, we will consider only lossless
transmission lines.
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5 Infinitely Long Transmission Line
For an infinitely long transmission line, there can be no
reflected wave (backward travelling wave). So for an
infinite long transmission line, there is only a forward
travelling wave.
V ( z ) = V + ( z ) = V0+ e − jkz
I ( z ) = I + ( z ) = I 0+ e − jkz
V ( z ) V0+ ( z )
Z ( z) =
= +
= Z0
I (z ) I 0 (z )
ΓL = 0
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6 Terminated Transmission Line
Γi
ΓL
z
ℓ
z = -d
ℓ=d
Γ(ℓ)
Z(ℓ)
z=0
ℓ=0
load
source
Note the two coordinate systems and their relation:
z = measuring from the left to the right
ℓ= -z
ℓ = measuring from the right to the left
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In the z coordinate system,
V0+ e − jkz + V0− e jkz = V ( z )
+ − jkz
0
I e
−
0
+I e
jkz
= I (z )
In the ℓ (ℓ = -z) coordinate system,
+
0
V e
jkA
− − jkA
0
+V e
= V (A )
I 0+ e jkA + I 0− e − jkA = I (A )
We will use the ℓ coordinate system in subsequent
analyses.
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The characteristic impedance in the ℓ coordinate system is:
V0+
= Z0
+
I0
The reflection coefficient at ℓ = 0 in the ℓ coordinate
system is:
V0− e − jk ×0
Γ (A = 0 ) = + jk ×0 = Γ L
V0 e
As ΓL is obtained at ℓ = 0 (the load position), it is
called the reflection coefficient at the load.
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At the position of the load (ℓ = 0), the voltage is VL and
the current is IL. Then we have:
V0+ + V0− = VL
V0+ V0−
−
= IL
Z0 Z0
VL
= ZL
IL
Solve these two equations, we have:
1
+
V0 = I L (Z L + Z 0 )
2
1
−
V0 = I L (Z L − Z 0 )
2
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-
Putting the expressions for V0 and V0 into the equations
for the voltage and current, we have:
+
[
]
[
]
1
jkA
jkA
− jkA
− jkA
(
)
V A = I L Z L (e + e ) + Z 0 (e − e )
2
= I L [Z L cos(kA ) + jZ 0 sin (kA )]
1 IL
I (A ) =
Z L (e jkA − e − jkA ) + Z 0 (e jkA + e − jkA )
2 Z0
IL
= [Z 0 cos(kA ) + jZ L sin (kA )]
Z0
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Using V(ℓ) and I(ℓ), we can obtain the impedance Z(ℓ) at
an arbitrary point ℓ on the transmission line as:
V (A )
Z L + jZ 0 tan (kA )
Z (A) =
= Z0
I (A )
Z 0 + jZ L tan (kA )
The reflection coefficient at the load ΓL can be expressed
as:
1
I L (Z L − Z 0 )
−
Z L − Z0
V0
2
=
ΓL = + =
1
+
Z
Z
V0
L
0
I L (Z L + Z 0 )
2
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In fact, we can further define a reflection coefficient Γ(ℓ)
at any point ℓ on the transmission line by:
reflected voltage at point A
Γ (A ) =
incident voltage at point A
V0− e − jkA V0− − j 2 kA
= + jkA = + e
= Γ L e − j 2 kA
V0 e
V0
As we know (by solving the two equations on page 22
with ℓ ≠0):
1
+ jkA
V0 e = I (A )(Z (A ) + Z 0 )
2
1
− − jkA
V0 e
= I (A )(Z (A ) − Z 0 )
2
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Therefore, alternatively we can write,
1
I (A )[Z (A ) − Z 0 ]
Z (A ) − Z 0
2
Γ (A ) =
=
1
(
)
Z
+
Z
A
0
I (A )[Z (A ) + Z 0 ]
2
Then,
1 + Γ (A )
Z (A ) = Z 0
1 − Γ (A )
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At the position of the generator (ℓ = d),
Z L + jZ 0 tan (kd )
Zi = Z ( A = d ) = Z 0
Z 0 + jZ L tan (kd )
Zi − Z 0
Γ (A = d ) = Γi =
= Γ L e − j 2 kd
Zi + Z 0
Vg
Γi
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Example 1
A 100-Ω transmission line is connected to a load consisted
of a 50-Ω resistor in series with a 10-pF capacitor.
(a) Find the reflection coefficient ГL at the load for a 100MHz signal.
(b) Find the impedance Zin at the input end of the
transmission line if its length is 0.125λ.
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Solutions
The following information is given
RL = 50Ω, CL = 10 −11 F, Z 0 = 100Ω, f = 100MHz = 108 Hz
The load impedance is
Z L = RL − j ωCL
1
= 50 − j
= 50 − j159
−11
8
2π × 10 × 10
(Ω)
(a) Voltage reflection coefficient is
Z L / Z 0 − 1 0.5 − j1.59 − 1
ΓL =
=
= 0.76∠ − 60.70°
Z L / Z 0 + 1 0.5 − j1.59 + 1
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(b) d =0.125λ
Zin = Z ( A = 0.125λ )
Z L + jZ0 tan (π 4 )
= Z0
Z0 + jZ L tan (π 4 )
Z L + jZ0
= Z0
Z0 + jZ L
Normalized zin = 0.1437-j 0.2555 Ω
(Ω)
(Ω)
= 14.3717 - j 25.5544
= 29.32∠ − 60.65°
See animation “Transmission Line Impedance Calculation”
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6.1 Voltage/current maxima and minima
V (A ) = V0+ e jkA + V0− e − jkA
−
⎛
V
+ jkA
− j 2 kA ⎞
0
⎟⎟
= V0 e ⎜⎜1 + + e
⎠
⎝ V0
= V0+ e jkA (1 + Γ L e − j 2 kA )
V ( A ) = V 1 + Γ Le
+
0
− j 2kA
= V 1+ Γ L e
+
0
L
|ΓL|≤1
Γ = Γ L e j (θ
L −2kA
)
= a complex number
= V 1+ Γ
+
0
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j (θ L −2 k A )
Γ L = Γ L e jθ
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Im
1+ Γ
1
0
V (A )
θ =θL-2kℓ
Re
θ
1− Γ L
Γ = Γ L e j (θ
L − 2 kz '
)
AM
Am
A=0
1+ Γ L
Complex plane of (1+Γ )
See animation “Transmission Line Voltage Maxima and Minima”
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V ( A ) is maximum when 1 + Γ = (1 + Γ L
)
EE2011
V ( A ) max ⇒ θ = θ L − 2k A = −2nπ
θ L λ nλ
⇒ AM =
+
, n = 0,1, 2,"
4π
2
Note:θL has to be specified in the range [ −π , π ) .
V ( A ) is minimum when 1 − Γ = (1 − Γ L
)
V ( A ) min ⇒ θ = θ L − 2k A = − ( 2n + 1) π
θ L λ ( 2n + 1) λ
⇒ Am =
+
, n = 0,1, 2,"
4π
4
Note:θL has to be specified in the range [ −π , π ) .
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As current is
I ( A ) = I 0+ 1 − Γ L e − j 2 k A
V0+
=
1− Γ
Z0
Current is maximum when voltage is minimum and
minimum when voltage is maximum.
θ L λ (2n + 1)λ
I (A ) max at A M =
+
, n = 0,1,2,", with θ L ≤ π
4π
4
θ L λ nλ
I (A ) min at A m =
+
, n = 0,1,2,", with θ L ≤ π
4π
2
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Define a voltage standing wave ratio (VSWR) as:
S = voltage standing wave ratio (VSWR)
=
V ( A ) max
V ( A ) min
=
V0+ (1 + Γ L
+
0
V
(1 − Γ
L
) = 1+ Γ
) 1− Γ
L
(dimensionless)
L
S −1
ΓL =
S +1
|V(z)|
|I(z)|
|V|max
|I|max
|V|min
|I|min
lmax
load
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lmax
load
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Transmission Lines – Basic Theories
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Special terminations
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ΓL
S
ZL
0
1
ZL= Z0 (matched)
-1
∞
ZL= 0 (short-circuited)
1
∞
ZL= ∞ (open-circuited)
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6.2 Power flow in a transmission line
Power flow at any point z on a transmission line
is given by:
1
Pav ( z ) = Re{V ( z )I * ( z )}
2
Power delivered by the source:
1
Ps = Re{Vg I i* }
2
Power dissipated in the source impedance Zg:
{
}
1
1
1 2
*
*
PZ g = Re VZ g I Z g = Re{Z g I i I i } = I i Re{Z g }
2
2
2
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Power input to the transmission line:
1
Pi = Pav (− d ) = Re{V (− d )I * (− d )}
2
1
1
1 2
*
*
= Re{Vi I i } = Re{Z i I i I i } = I i Re{Z i }
2
2
2
1 ⎧ Vi * ⎫ 1 2 ⎧ 1 ⎫
= Re ⎨Vi * ⎬ = Vi Re ⎨ * ⎬
2 ⎩ Zi ⎭ 2
⎩ Zi ⎭
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Power dissipated in the terminal impedance:
1
PL = Pav (0 ) = Re{V (0 )I * (0 )}
2
1
1
1 2
= Re{VL I L* } = Re{Z L I L I L* } = I L Re{Z L }
2
2
2
1 ⎧ VL* ⎫ 1 2 ⎧ 1 ⎫
= Re ⎨VL * ⎬ = VL Re ⎨ * ⎬
2 ⎩ ZL ⎭ 2
⎩ ZL ⎭
Transmission Lines – Basic Theories
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By the principle of conservation of power:
Ps = PZ g + Pi
Pi = PL
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Example 2
A lossless transmission line with Z0 = 50 Ω and d =1.5 m
connects a voltage Vg source to a terminal load of ZL = (50 +
j50) Ω. If Vg = 60 V, operating frequency f = 100 MHz, and
Zg = 50 Ω, find the distance of the first voltage maximum ℓM
from the load. What is the power delivered to the load PL?
Assume the speed of the wave along the transmission line
equal to speed of light, c.
Zg
d
Ii
A
Vg ~
Vi
Zi
Z0 = 50 Ω
ZL
A’
ℓ
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Solutions
The following information is given:
Z 0 = 50Ω, d = 1.5 m,
Vg = 60 V, Z g = 50Ω, Z L = 50 + j50Ω,
f = 100MHz = 10 Hz
8
c
up = c ⇒ λ = 8 = 3 m
10
The reflection coefficient at the load is:
Z L − Z 0 50 + j50 − 50
ΓL =
=
= 0.2 + j 0.4 = 0.45e j1.11
Z L + Z 0 50 + j50 + 50
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Therefore,
Then,
EE2011
Γ L = 0.45, θ L = 1.11 rad
θ L λ nλ
AM =
+
, when n = 0
4π
2
1.11λ
=
= 0.09λ = 0.27 m (from the load)
4π
The input impedance Zi looking at the input to the
transmission line is:
Z L + jZ 0 tan (kd )
Zi = Z 0
Z 0 + jZ L tan (kd )
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⎞
⎛ 2π
× 1. 5 ⎟
50 + j50 + j50 tan ⎜
3
⎠
⎝
Zi = 50
= 50 + j50Ω
⎞
⎛ 2π
× 1.5 ⎟
50 + j (50 + j50 ) tan ⎜
⎠
⎝ 3
The current at the input to the transmission line is :
EE2011
Vg
60
Ii =
=
= 0.48 − j 0.24 A
Z g + Zi 50 + 50 + j50
As the transmission line is lossless, power delivered to the
load PL is equal to the power input to the transmission line
Pi. Hence,
1
1 2
PL = Pi = I i Re{Z i } = × 0.288 × 50 = 7.2 W
2
2
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6.3 Complete solutions for voltage and current
The voltage and current on the transmission line can
be written as:
(
V (A ) = V0+ e jkA + V0− e − jkA = V0+ e jkA 1 + Γ L e − j 2 kA
)
V0+ jkA V0+ − jkA V0+ jkA
− j 2 kA
e −
e
=
e 1 − ΓLe
I (A ) =
Z0
Z0
Z0
(
)
We still have one unknown V0+ in V(ℓ) and I(ℓ). We
need the knowledge of voltage source Vg to further
determine V0+.
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At ℓ = d, V(d) = Vi and I(d) = Ii.
(
Vi = V0+ e jkd 1 + Γ L e − j 2 kd
V0+ jkd
− j 2 kd
Ii =
e 1 − ΓLe
Z0
(
)
)
Vi and Ii are related to the source voltage Vg as:
Vg = Vi + I i Z g
From the expressions of Vi, Ii, and Vg, we can find V0+.
V0+ =
Γg =
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(Z
Vg Z 0 e − jkd
g
Z g − Z0
Z g + Z0
(
+ Z 0 ) 1 − Γ g Γ L e − j 2 kd
)
= source reflection coefficient
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Putting V0+ into the expressions of V(ℓ) and I(ℓ), we have:
V (A ) =
I (A ) =
(Z
(Z
Vg Z 0 e − jkd
g
(
+ Z0 )1 − Γ g ΓLe
− j 2 kd
Vg e − jkd
g
(
+ Z0 )1 − Γ g ΓLe
− j 2 kd
(
)
)
(
)
)
e jkA 1 + Γ L e − j 2 kA
e jkA 1 − Γ L e − j 2 kA
Now the voltage and current on the transmission line are
expressed in terms of the known parameters of the
transmission line.
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Example 3
A 1.05-GHz generator circuit with a series impedance Zg = 10Ω
and voltage source given by:
vg (t ) = 10 sin (ωt + 30°) (V )
is connected to a load ZL = (100 + j50) through a 50-Ω, 67-cmlong lossless transmission line. The phase velocity of the line is
0.7c, where c is the velocity of light in a vacuum. Find the
instantaneous voltage and current v(ℓ,t) and i(ℓ,t) on the line and
the average power delivered to the load.
Zg
d
Ii
A
vg
Vi
Zi
Z0 = 50 Ω
ZL
A’
ℓ
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Transmission Lines – Basic Theories
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Solutions
EE2011
up
0.7 × 3 × 108
λ= =
= 0.2 m
9
1.05 × 10
f
0.67
d = 67cm =
= 3.35λ
0.2
source reflection coefficient Γ g
Z g − Z0
10 − 50
2
=
=
=−
Z g + Z 0 10 + 50
3
load reflection coefficient Γ L
Z L − Z 0 100 + j 50 − 50
=
=
= 0.45e j 0.46
Z L + Z 0 100 + j 50 + 50
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Transmission Lines – Basic Theories
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vg (t ) = 10 sin (ωt + 30°)
{
= 10 cos(ωt − 60°) = Re 10e − j 60° e jωt
V (A ) =
=
(Z
− jkd
+ Z 0 )(1 − Γ g Γ L e − j 2 kd )
10e
− jπ / 3
50e
−j
e jkA (1 + Γ L e − j 2 kA )
2π
λ
(3.35λ )
4π
×
− j (3.35 λ ) ⎤
⎡
j 0.46
(10 + 50)⎢1 − (− 2 / 3)(0.45e )e λ
⎥
⎣
⎦
e jkA + (0.45e j 0.46 )e − jkA = 10.18e j 2.77 e jkA + 0.45e − j (kA −0.46 )
[
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Vg Z 0 e
g
− jπ / 3
Vg = 10e
Phasor form:
} (V )
]
[
49
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Transmission Lines – Basic Theories
NUS/ECE
EE2011
I (A ) =
(Z
Vg e − jkd
g
+ Z 0 )(1 − Γ g Γ L e
[
− j 2 kd
)
e jkA (1 − Γ L e − j 2 kA )
= 0.20e j 2.77 e jkA − 0.45e − j (kA −0.46 )
]
Therefore instantaneous forms are:
v (A, t ) = Re{V (A )e jωt }
[
]
[
]
= Re{10.18e j 2.77 e jkA + 0.45e − j (kA −0.46 ) e jωt }
= 10.18 cos(ωt + kA + 2.77 ) + 4.58 cos(ωt − kA + 3.23)
i (A, t ) = Re{I (A )e jωt }
= Re{0.20e j 2.77 e jkA − 0.45e − j (kA −0.46 ) e jωt }
= 0.20 cos(ωt + kA + 2.77 ) − 0.09 cos(ωt − kA + 3.23)
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Transmission Lines – Basic Theories
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Z L + jZ 0 tan (kd )
Zi = Z0
Z 0 + jZ L tan (kd )
⎛ 2π
(100 + j50) + j50 tan⎜ × 3.35λ ⎞⎟
λ
⎠
⎝
= 50
⎛ 2π
⎞
50 + j (100 + j 50 ) tan⎜
× 3.35λ ⎟
⎝ λ
⎠
= 21.9 + j17.4 Ω
10e − jπ / 3
Ii =
=
= 0.28e − j1.55
Z g + Z i 10 + 21.9 + j17.4
Vg
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Transmission Lines – Basic Theories
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Power delivered to the load
= power input to the transmission line at AA’
1
= Re{Vi I i* }
2
1
= Re{I i Z i I i* }
2
1 2
= I i Re{Z i }
2
1
= 0.282 Re{21.9 + j17.4}
2
= 0.86 Watt
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Transmission Lines – Basic Theories
NUS/ECE
EE2011
7 Special Cases of Terminations in a Transmission Line
7.1 Matched line
For a matched line, ZL = Z0. Then,
Z 0 + jZ 0 tan (kA )
⎫
Z (A ) = Z 0
= Z0 ⎪
Z 0 + jZ 0 tan (kA )
⎪
⎬for any length A of the line
Z (A ) − Z 0
⎪
(
)
A
=
=
0
Γ
Note ℓ =-z
⎪⎭
Z (A ) + Z 0
Thus, there is no reflection on a matched line. There is
only an incident voltage. It is same as the case of an
infinitely long line.
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Transmission Lines – Basic Theories
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Normalized voltage magnitude
1
z
z
0
ℓ
-1
0
Normalized current magnitude
1
Z0
Z0
z
-1
Zin
Normalized impedance (Zin/Z0)
1
z
Note:
Normalized voltage = voltage/max. |voltage|
Normalized current = current/max. |current|
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Transmission Lines – Basic Theories
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7.2 Short-circuited line
For a short circuit, ZL = 0. Then
Normalized voltage magnitude
Z insc = jZ 0 tan (kA ) = − jZ 0 tan (kz )
Normalized current magnitude
Normalized impedance (=-tan(kz))
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Transmission Lines – Basic Theories
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7.3 Open-circuited line
For an open circuit, ZL = ∞. Then
Z inoc = − jZ 0 cot (kA ) = jZ 0 cot (kz )
Normalized voltage magnitude
Note that:
Z insc Z inoc = [ jZ 0 tan (kA )][− jZ 0 cot (kA )]
Normalized current magnitude
= Z 02
Z
sc
in
Z
oc
in
= [ jZ 0 tan (kA )] [− jZ 0 cot (kA )]
Normalized impedance (=cot(kz))
= − tan 2 (kA )
Given Z insc , Z inoc , and A, compute Z 0 and k.
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Transmission Lines – Basic Theories
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7.4 λ/4 transmission line terminated in ZL
Z L + jZ 0 tan (π 2 ) Z 02
Zin = Z ( A = λ 4) = Z 0
=
Z 0 + jZ L tan (π 2 ) Z L
ℓ
Z0
ZL
Zin
7.5 λ/2 transmission line terminated in ZL
Z L + jZ 0 tan (π )
Zin = Z ( A = λ 2) = Z 0
= ZL
Z 0 + jZ L tan (π )
ℓ
Z0
ZL
Zin
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Transmission Lines – Basic Theories
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Example 4
The open-circuit and short-circuit impedances measured at
the input terminals of a lossless transmission line of length
1.5 m (which is less than a quarter wavelength) are −j54.6 Ω
and j103 Ω, respectively.
(a) Find Z0 and k of the line.
(b) Without changing the operating frequency, find the
input impedance of a short-circuited line that is twice
the given length.
(c) How long should the short-circuited line be in order
for it to appear as an open circuit at the input
terminals?
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Transmission Lines – Basic Theories
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Solution
The given quantities are
Z inoc = − j54.6 Ω
Z insc = j103 Ω
A = 1.5m
(a) Z 0 = Z inoc Z insc = 75 Ω
1 −1
k = tan
− Z insc Z inoc = 0.628 rad m
A
2π
λ=
= 10m
k
(b) For a line twice as long, ℓ = 3 m and k ℓ =1.884 rad,
Z insc = jZ 0 tan kA = − j 232 Ω
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Transmission Lines – Basic Theories
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(c) Short circuit input impedance
= Z = jZ 0 tan (kA )
sc
in
For Z insc = ∞, ⇒ kA = π 2 + nπ , n = 0,1,2,"
A=
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π 2 + nπ
k
2n + 1
=
λ
4
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Transmission Lines – Basic Theories