Transmission Lines in Frequency Domain

頻域傳輸線

Mathematical Toolbox Limit Complex variable ( )

Phasor Hyper-trigonometry

0

( , ) ( , ) ( , )limz

v z t v z z t v z tz z∆ →

∂ + ∆ −=

∂ ∆

( , ) ( ) j tv z t V z e ω= Re

( , ) ( ) ( )j tv z t j V z e jt t

ωω ω∂ ∂= →

∂ ∂Re

(cos sin )jAe A jθ θ θ= +( )| || | | || | [cos( ) sin( )]jAB A B e A B jθ φ θ φ θ φ+= = + + +

| | , | |j jA A e B B eθ φ= ∈ = ∈

j z j zj z zj z j z

β ββ ββ β

===

sinh sincosh costanh tan

Mathematical Toolbox (Con’d) Ordinary second-order differential equation

characteristic equations:

22

2 0∂ Ω− Ω =

∂uγ

1 2( ) u uu c e c eγ γ−Ω = +

2 2 0, =− = ±r rγ γ

Sequence of Study

Intro - why transmission lines?

General transmission line equations

Infinite transmission line

Terminated transmission line

Transmission line circuit Circuit elements

Reflection and Standing wave

Input impedance and power

Quantity Symbol Unit Dimensions

Characteristic impedance Z0 Ohm(Ω) ML2T-1Q-2

Input impedance Zin Ohm(Ω) ML2T-1Q-2

Propagation constant γ Meter -1(m -1) L-1

Attenuation constant α Neper/meter(Np/m) L-1

Phase constant β Radian/meter(rad/m) L-1

Phase velocity Meter/second(m/s) LT-1

Wavelength Meter(m) L

Time-average power Watt(W) ML2T-3

pu

λ

avP

List of Notation and Symbols

Quantity Symbol Unit Dimensions

Reflection coefficient NA NA

Generator reflection coefficient NA NA

Load reflection coefficient NA NA

Standing wave ratio SWR NA NA

Series resistance per unit length R Ohm/meter (Ω/m) MLT-1Q-2

Series inductance per unit length L Henry/meter (H/m) MLQ-2

Shunt conductance per unit length G Siemens/meter (S/m) M-1L-3TQ2

Shunt capacitance per unit length C Farad/meter (F/m) M-1L-3T2Q2

Γ

Γs

LΓ

List of Notation and Symbols (Con’d)

Outline

Introduction to Transmission Lines

Why transmission lines? (1.1)

Sort of Transmission line (1.2)

General Transmission Line Equations

Terminated Lossless Transmission Line

Transmission Lines as Circuit Elements

Transmission-line Circuits

Why transmission Lines ?!

很醜 有點美感好嘛！？

Do we use it in our daily life?

Why transmission Lines ?! Wave propagation over space – power inefficiency. For efficient point-to-point transmission

- the source energy must be guided.

This leads to transmission lines (txlines).

Wireless transmission versus Point-to-point transmission

Why transmission Lines ?! What is the difference between a usual connecting line and a

transmission line? The key fact is The relative length compared to the operating wavelength. Propagation results in time delay.

Distributed circuit versus Lumped circuit

Animation module 2.1 (select “incident wave” only with f = 0.01, 0.1, 1, and 10 GHz) http://140.110.31.72/em/2-1.html

Sort of Transmission Lines Most transmission lines are operated in TEM mode.

Many of the characteristics of TEM-mode transmission lines are

- the same as those of a uniform plane wave.

- similar governing equations.

- same propagation properties.

Additionally - Uniquely defined voltage and current Reduced the problem from EM fields to circuit theory.

Sort of Transmission Lines TEM transmission lines - No field components in the direction of propagation. - Very similar to plane wave propagation. - Uniquely defined voltage and current! E-field

Direction of Propagation

Twin wire line Parallel plate

Strip line

Coaxial line

Sort of Transmission Lines Quasi-TEM Lines (small non-transverse electromagnetic fields) - Small field components in the direction of propagation. - Approximate to plane wave propagation. - Approximately uniquely defined voltage and current. - Planar structure suitable for printed circuit design

CPW (coplanar waveguide)

Slotline

Microstrip Line

Outline Introduction to Transmission Lines

General Transmission Line Equations

Lump-Element Circuit Model and Wave equations (2.1)

Wave on an Infinite Transmission Lines (2.2)

Three Special Cases (2.3)

Points of Interest (2.4)

Examples (2.5)

Terminated Lossless Transmission Line

Transmission Lines as Circuit Elements

Transmission-line Circuits

Lump-Element Circuit Model A bridge between field analysis and basic circuit theory.

The key difference is the electrical size.

A lumped circuit,

- Voltages and currents do not vary spatially over the elements.

A distributed-parameter network,

- Voltages(currents) vary in magnitude and phase over its length.

- Standing wave occurs.

Consider an infinitesimal segment of the line…….

Lump-Element Circuit Model A TEM line is schematically represented as a two-wire line. The short piece of line of length Δz can be modeled as - a lumped-element circuit. By considering its physical properties. Current flow – there is an inductance. Charge Accumulation - there is an capacitance in-between. Power dissipation – R & G.

Lump-Element Circuit Model

R, L, G, C are per unit length quantities defined as follows: R = series resistance per unit length, for both conductors, in Ω/m. L = series inductance per unit length, for both conductors, in H/m. G = shunt conductance per unit length, in S/m. C = shunt capacitance per unit length, in F/m.

Characteristics of the line L represents the total self-inductance of the two conductors

C is due to the close proximity of the two conductors.

Loss of the line R represents the resistance due to the finite conductivity of the conductors

G is due to dielectric loss in the material between the conductors.

Wave Equations Kirchhoff’s voltage law (KVL)

Taking the limit as Δz→0 gives the following differential equation:

Similarly, applying Kirchhoff's current law (KCL)

i(z, t)( , ) ( , ) z ( , ) 0 .v z t zi z t v z z tt

∂− ∆ − ∆ − + ∆ =

∂R L

( , ) ( , ) ( , )( , ) .z

+ ∆ − ∂− = +

∆ ∂v z z t v z t i z ti z t

tR L

( , ) ( , )( , ) .∂ ∂− = +

∂ ∂v z t i z ti z t

z tR L

( , )( , ) ( , ) ( , ) 0.∂ + ∆− ∆ + ∆ − ∆ − + ∆ =

∂v z z ti z t zv z z t z i z z t

tG C

( , ) ( , )( , ) .∂ ∂− = +

∂ ∂i z t v z tv z t

z tG C

Wave Equations General transmission-line equations For time harmonics using phasor notations ( )

These equations are referred to as the telegrapher equations. A pair of first-order differential equations in v(z) and i(z).

- We deal with I/V instead of E/H in transmission line theory. NOTE: The variables in the following slides are all in phasor form, otherwise they will be specified as (z,t).

Phasor form

t jω∂ ∂ =

( , ) ( , )( , ) .∂ ∂− = +

∂ ∂v z t i z ti z t

z tR L

( , ) ( , )( , ) .∂ ∂− = +

∂ ∂i z t v z tv z t

z tG C

( , ) e[ ( ) ],( , ) e[ ( ) ].

j t

j t

v z t V z ei z t I z e

ω

ω

= ℜ

= ℜ

( ) ( ) ( ),

( ) ( ) ( ).

− = +

− = +

dV z j I zdz

dI z j V zdz

ω

ω

R L

G C

Wave on an Infinite Transmission Line Solve for V(z) and I(z):

where

γ : the (complex) propagation constant. α >0 : attenuation constant (Neper/m). β >0 : phase constant or wavenumber (rad/m). Similar to that for plane-wave propagation in lossy media.

-1( )( ) (m ).j j jγ α β ω ω= + = + +R L G C

( ) ( ) ( ),

( ) ( ) ( ).

= − +

= − +

dV z j I zdz

dI z j V zdz

ω

ω

R L

G C

22

2

( ) ( ) 0,d V z V zdz

γ− =2

22

( ) ( ) 0,d I z I zdz

γ− =

Wave on an Infinite Transmission Line Traveling wave solutions

the term represents wave propagation in the +z direction; the term represents wave propagation in the -z direction.

ze γ−

ze γ+

Four unknowns

0 0

0 0

( ) ( ) ( )

,

( ) ( ) ( )

,

z z

z z

V z V z V z

V e V eI z I z I z

I e I e

γ γ

γ γ

+ −

+ −−

+ −

+ −−

= +

= +

= +

= +

22

2

( ) ( ) 0,d V z V zdz

γ− =

22

2

( ) ( ) 0,d I z I zdz

γ− =

Wave on an Infinite Transmission Line

Substituting into

Comparing with

The characteristic impedance, Z0,

0 0( ) z zV z V e V eγ γ+ −−= + ( ) ( ) ( ),dV z j I zdz

ω= − +R L

0 0( ) ,z zI z I e I eγ γ+ −−= +

0 0( ) .z zI z V e V ej

γ γγω

+ −− = − +R L

0 0

0 0

.V V jI I

ωγ

+ −

+ −

+= − =

R L

0 ( ).+ += = = Ω

+ +j jZ

j jω γ ω

γ ω ωR L R L

G C G C

Wave on an Infinite Transmission Line Rewritten in the following form:

Time-domain expression

Two unknowns solved!

0 0+

0 0

0 0

( ) ,

( ) .

z z

z z

V z V e V e

V VI z e eZ Z

γ γ

γ γ

+ −−

−−

= +

= −

,pu fω λβ

= =2 .πλβ

=

0

0

( , ) cos( )

cos( ) ,

z

z

z t V t z e

V t z e

α

α

υ ω β φ

ω β φ

+ + −

− −

= − +

+ + +

Three special cases Lossless Line (R = G = 0 )

Propagation constant

Phase velocity

Characteristic impedance

The general solutions :

, where

L, C can be derived with static fields.

( a real number).

; 0,

(a linear function of ).

= + ==

=

j jγ α β ωα

β ω ω

LC

LC

1 (constant). = =pu ωβ LC

0Z =LC

1 ,pu ωβ

= =LC

2 2 .π πλβ ω

= =LC

0 0+

0 0

0 0

( ) ,

( ) .

j z j z

j z j z

V z V e V e

V VI z e eZ Z

β β

β β

+ −−

−−

= +

= −

Almost the same as a

lossless line except for a small loss

Three special cases Low-loss line (R << ωL, G << ωC)

Propagation constant

Phase velocity

Characteristic impedance.

1/ 2 1/ 2

1 1 1 1 .2 2

1 , 2

(approximately a linear function of ).

j j jj j j j

γ α β ω ωω ω ω ω

α

β ω ω

= + = + + ≅ + +

≅ +

≅

G GR RLC LC

L C L C

C LR G

L C

LC

1 (approximately constant). pu ωβ

= ≅LC

1/ 2 1/ 2

0 0 0 1 1

1 1 .2

Z R jXj j

j

ω ω

ω

−

= + = + +

≅ + − ≅

GL RC L C

GL R LC L C C

Three special cases Distortionless line (R/L = G/C)

Propagation constant

Phase velocity

Characteristic impedance.

( )

( );

j j j

j

γ α β ω ω

ω

= + = + +

= +

RCR L C

L

CR L

L

,

.

α

β ω

=

=

CR

LLC

1 (constant). pu ωβ

= =LC

0

0

,

0.

R

X

=

=

LC0 0 0 ,

( / )

jZ R jXj

ωω

+= + = =

+R L L

RC L C C

Point of interest- Distributed Parameters of TxLines

Derived from electrostatics/magnetostatics

σε

=GC

=µεLC

-1

-1

ln cosh 2 22 ' '' '

ln cosh ( 2 )1 1

2s

b D da a w

wb a dD a

Ra b

µ µ µπ ππε πε ε

π

+

L

C

R

-1

2

2 ' '' '' ln cosh ( 2 )

s sR Ra w

wb a dD a

ππωε πωε ωε

G

TWO-WIRE COAX PARALLEL PLATE

′ ′′= −′ ′′= −

jj

ε ε εµ µ µ

Point of interest- Analogous to plane wave in lossy medium

Maxwell equations

A plane wave with Ex(z,t)

Propagation constant and characteristic impedance

' ''

' ''

( )( )

j jj j

ω µ µ

ω ε ε

∇× = − −

∇× = −

Ε HH E

'' '( ) ( ) ,xy

dE z j Hdz

ωµ ωµ− = +

'' '( )( ) .y

x

dH zj E

dzωε ωε− = +

22

2

( ) ( ), xx

d E z E zdz

γ=

22

2

( )( ).y

y

d H zH z

dzγ=

'' ' '' '( )( ), j j jγ α β ωµ ωµ ωε ωε= + = + +

'' '

'' ' .cjj

µ µηε ε

+=

+

There is a current on a transmission line given as i(t) = 1.5cos( 1.885 ∙ 1010 t－78.54 z ) A. Find a) The frequency, b) The wavelength, c) The phase velocity, d) The phasor representation of this current,

Example I

101.885 10 3 ( ).2 2

f ωπ π

⋅= = =

GHz

2 2 0.8 ( ).78.54

π πλβ

= = = m

1081.885 10 2.4 10 ( ).

78.54pu ωβ

⋅= = = ×

m/s

1.5 78.54 ( ).I z= ∠ − A

Known parameters

101.885 10 ( ), 78.54 .ω β= ⋅ = rad/s (rad/m)

Example II The attenuation on a 50 (Ω) distortionless txline is 0.02(dB/m). The line has a capacitance of 0.1 (nF/m). Find a) R, L, G per meter of the line,

Known parameters

,=GR

L C3 0.02 ( ) 2.3 10 ( ).α −= = = ⋅dB/m Np/mC

RL

0 50 ( ),R = = ΩLC

30

2 10 20

2 20

(2.3 10 ) 50 0.115 ( );

10 50 0.25 ( );0.115 46 ( ).50

R

R

R

α

µ

µ

−

−

= = ⋅ ⋅ = Ω

= = ⋅ =

= = = =

/m

H/m

S/m

R

L CRC R

GL

The attenuation on a 50 (Ω) distortionless txline is 0.02(dB/m). The line has a capacitance of 0.1 (nF/m). Find b) Phase velocity, c) The percentage to which the amplitude of a voltage traveling wave

decreases in 1 (km) and in 4 (km).

8

6 10

1 1 2 10 ( ).(0.25 10 10 )

pu− −

= = = ×⋅ ⋅

m/sLC

2

1

.zV eV

α−=

1000 2.32 1

4000 9.22 1

After 1 ( ), ( / ) 0.1, or 10%.After 4 ( ), ( / ) 0.0001, or 0.01%.

V V e eV V e e

α

α

− −

− −

= = =

= = =

kmkm

Example II

In this figure, using the distributed parameters of transmission line to solve the following problems. Find a) A coaxial cable of Z0 = 75 Ω if εr = 2.5 ε0 , find the ratio b/a.

Example III

00 ln ,

2 r

bZa

ηπ ε

=

12075 ln .2 2.5

=ba

ππ

ln 1.976 ,=ba

1.976 7.21 .= =b ea

ln ,22 .ln

r

ba

b a

µππε

=

=

L

C0

0 ln .2 r

bZa

ηπ ε

=

0 .Z =LC

COAXIAL

- Commonly used in cable TVs.

In this figure , using the distributed parameters of transmission line to solve the following problems. Find

b) The parallel-wire line of Z0 = 300 Ω if ε = 2ε0, find the ratio D/2a.

Example III

100 cosh ,

2r

DZa

ηπ ε

−=

-1

-1

cosh ,2

'' .cosh ( 2 )

Da

D a

µπ

πε

=

=

L

C10

0 cosh .2r

DZa

ηπ ε

−=

0 .Z =LC

1120300 cosh .22

−=Da

ππ

1cosh 3.535,2

− =Da

cosh 3.535 17.16 .2Da

==

TWO-WIRE

- Commonly used in TV reception with a Yagi-Uda antenna.

Outline Introduction to Transmission Line

General Transmission Line Equations

Terminated (Finite) Lossless Transmission Line

Terminated Lossless Transmission Line (3.1)

Standing Wave Patterns (3.2)

Power Flow on a Terminated Line (3.3)

Input Impedances (3.4)

Points of Interest (3.5)

Examples (3.6)

Transmission Lines as Circuit Elements

Transmission-line Circuits

A lossless line terminated in an arbitrary load ZL

An incident wave and , from an unknown source.

The ratio of incident voltage to incident current is Z0.

However, at the load termination (ZL≠Z0).

the ratio of voltage to current at the load must be ZL.

A reflected wave must be excited, satisfy the boundary condition.

Terminated Lossless Transmission Line

0j zV e β+ −

0j zI e β+ −

LL

L

.V ZI

=

unknown source (shadow region)

unknown source (shadow region)

A lossless line terminated in an arbitrary load ZL

The total voltage on the line, as a sum of incident and reflected waves:

Terminated Lossless Transmission Line

V0: The unknown !!! 0 0

0 0

0 0

( ) .

( ) .

j z j z

j z j z

V z V e V e

V VI z e eZ Z

β β

β β

+ −−

+ −−

= +

= −0 0

00 0

.V V ZI I

+ −

+ −= − =

Terminated Lossless Transmission Line

The total voltage and current at the load are

The voltage to current ratio is confined by the load termination

The voltage reflection coefficient, Γ

Solved

0 0

0 0

0 0

( 0) ,

( 0) .

V z V V

V VI zZ Z

+ −

+ −

= = +

= = −

0 0

0 0

0 0

( ) .

( ) .

j z j z

j z j z

V z V e V e

V VI z e eZ Z

β β

β β

+ −−

+ −−

= +

= −

0 00

0 0

(0) .(0)L

V VVZ ZI V V

+ −

+ −

+= =

−0

0 00

.L

L

Z ZV VZ Z

− +−=

+

0 0

00

.L

L

V Z ZZ ZV

−

+

−Γ = =

+

Terminated Lossless Transmission Line The total voltage and current waves on the line can then be written as

A superposition of an incident and reflected wave; such waves are called standing waves.

Only when Γ = 0 is there no reflected wave.

Γ= 0 => ZL = Z0 (matched to the line)

In general, Γ is a complex number | | ,je θΓΓ = Γ( )

0( 2 )

0

( ) ( | | )

(1 | | ).

j zj z

j zj z

V z V e e

V e e

β θβ

θ ββ

Γ

Γ

++ −

++ −

= + Γ

= + Γ

0

( ) [ ],

( ) [ ].

j z j zo

j z j zo

V z V e e

VI z e eZ

β β

β β

+ −

+−

= + Γ

= − Γ

Standing Wave Patterns

If , : a constant

When the load is mismatched, the standing wave occurs on the line:

( ) ( 2 )0 0( ) ( | | ) (1 | | ).j z j zj z j zV z V e e V e eβ θ θ ββ βΓ Γ+ ++ − + −= + Γ = + Γ

( ) oV z V +=0Γ =

2 2

( 2 )

( ) [ ],

( ) 1 1

1 .

j z j zo

j z j lo o

j lo

V z V e e

V z l V e V e

V e

β β

β β

θ βΓ

+ −

+ + −

−+

= + Γ

= − = + Γ = + Γ

= + Γ

Animation module 2.2 http://140.110.31.72/em/2-2.html

Animation module 2.1 (time domain) http://140.110.31.72/em/2-1.html

Standing Wave Patterns

The maximum voltage: The minimum value:

The voltage standing wave ratio (VSWR), can be defined as

SWR is a real number.

1 ≦ SWR ≦ ∞, SWR = 1 implies a matched load. (SWR= ∞, when |Γ|=1)

( 2 )0| ( ) | | ||1 | | | .j lV z V e θ βΓ −+= + Γ

( 2 ) 1,j le θ βΓ − =

( 2 ) 1,j le θ βΓ − = −

max

min

1SWR .

1VV

+ Γ= =

− Γ

max (1 ).oV V += + Γ

min (1 ).oV V += − Γ

Standing Wave Patterns the distance between two successive voltage maxima (or minima) is

the distance between a maximum and a minimum is

is the wavelength on the transmission line.

As an illustration,

2 / 2 / 2 / 2.l π β πλ π λ= = =

/ 2 / 4.l π β λ= =

λ

|V(z)|

|I(z)| Animation module 2.2 http://140.110.31.72/em/2-2.html

( 2 )0| ( ) | | ||1 | | | .j lV z V e θ βΓ −+= + Γ

Power Flow on a Terminated Line Time-average power flow along the line at a point z:

or

If Γ = 0, maximum power is delivered to the load. If |Γ| = 1 , no power is delivered.

Reflection loss

Reactance power

0

( ) [ ],

( ) [ ].

j z j zo

j z j zo

V z V e e

VI z e eZ

β β

β β

+ −

+−

= + Γ

= − Γ

[ ]2

0 22 2av

0

1 1( ) ( )* 1 * ,2 2

j z j zV

P e V z I z e e eZ

β β+

− = ℜ = ℜ − Γ + Γ − Γ 2

0 2av

0

1 (1 ),2

VP

Z

+

= − Γ

Power Flow on a Terminated Line Time-average power flow along the line at a point z:

This "loss" is called return loss (RL),

Γ = 0 => RL= ∞ dB (no reflected power), |Γ| = 1 => RL= 0 dB (all incident power is reflected).

( a positive value),

2

0 2av

0

1 (1 ),2

VP

Z

+

= − Γ

max

min

1SWR .

1VV

+ Γ= =

− Γ

RL 20log dB,= − Γ

Input impedance Input impedance on the line: Total voltages and currents

The input impedance at z = - l seen looking toward the load is

0

( ) [ ],

( ) [ ].

j z j zo

j z j zo

V z V e e

VI z e eZ

β β

β β

+ −

+−

= + Γ

= − Γ

0 0in 0

0 0

00

0

00

0

( ) ( )( ) ( )

cos sin cos sin

tan ( ).tan

j l j lL L

j l j lL L

L

L

L

L

Z Z e Z Z eZ ZZ Z e Z Z e

Z l jZ lZZ l jZ lZ jZ lZZ jZ l

β β

β β

β ββ β

ββ

−

−

+ + −=

+ − −+

=+

+= Ω

+

-2

in 0 0-2[ ]( ) 1= = ,

I( ) [ ] 1

j l j l j lo

j l j l j lo

V e eV z l eZ Z Zz l V e e e

β β β

β β β

+ −

+ −

+ Γ= − + Γ=

= − − Γ − Γ

Animation module 2.3 http://140.110.31.72/em/2-3.html

Input impedance Equivalent lumped circuit for the finite transmission line

Assume a lossless line Input power = power along the line = power deliver to the load

2* 2 * 2

0

| |1 1 1 1( ) Re( ) (1 | | ) Re( ) | | .2 2 2 2

oav in i i L L L L

VP V I V I I RZ

+

= = − Γ = =

0in 0

0

tan .tan

L

L

Z jZ lZ ZZ jZ l

ββ

+=

+

in

in

in

,

.

i

i

ZV VZ Z

VIZ Z

=+

=+

ss

s

s

Point of interest – input impedance In a general lossy line

Total voltage and current

Input impedance

( )

( )

0

00

( ) cosh sinh ,2

( ) sinh cosh .

LL

LL

IV z l Z l Z l

II z l Z l Z lZ

γ γ

γ γ

= − = +

= − = +

00

0

cosh sinh( )( ) .( ) sinh cosh

L

L

Z l Z lV z lZ z l ZI z l Z l Z l

γ γγ γ

+= −= − = =

= − +

0in 0

0

tanh( ) ( ).tanh

L

L

Z Z lZ Z z l ZZ Z l

γγ

+= = − = Ω

+

0 0+

0 0

0 0

( ) ,

( ) .

z z

z z

V z V e V e

V VI z e eZ Z

γ γ

γ γ

+ −−

−−

= +

= −

Point of interest – input reflection coefficient

The reflection coefficient at any point on the line : The total voltage at z = - l (l: arbitrary),

By definition, the ratio of the reflection to the incident wave at z=0 is

Thus,

Transforming the effect of a load mismatch down the line.

By only a phase shift!

( ) .j l j lo oV z l V e V eβ β+ + − −= − = +

( 0) (0)zΓ = Γ

( ) ( )z l lΓ = − Γ

Incident Reflected

0 0

00

.L

L

V Z ZZ ZV

−

+

−= =

+

2= (0) .j l

j loj l

o

V e eV e

ββ

β

− −−

+= Γ

Animation module 2.5 http://140.110.31.72/em/2-5.html

Example IV A 50 Ω transmission line, with a length of 0.2 λ, is terminated a load impedance of 100+j80 Ω.

Find The reflection coefficient at the load,

The reflection coefficient at the input,

SWR on the line,

( )( )

L 0

L 0

100 80 500.555 29.9 .

100 80 50jZ Z

Z Z j+ −−

Γ = = = ∠ °+ + +

1 1 0.555SWR 3.494 .1 1 0.555

+ Γ += = =

− Γ −

2 2 (0.4 )( ) (0) (0.555 29.9 ) 0.555 114.1j l je eβ π− −Γ = Γ = ∠ ° = ∠ − °

Example IV A 50 Ω transmission line, with a length of 0.2 λ, is terminated a load impedance of 100+j80 Ω.

Find The return loss,

The impedance at the input of the line,

20log 20log(0.555) 5.11 dB.RL = − Γ = − =

0in 0

0

tantan

(100 80) 50 tan(72 ) 5050 (100 80) tan(72 )

19.65 28.77 .

L

L

Z jZ lZ ZZ jZ l

j jj jj

ββ

+=

++ + °

=+ + °

= − Ω

( )( )

L 0

L 0

100 80 500.555 29.9 ,

100 80 50jZ Z

Z Z j+ −−

Γ = = = ∠ °+ + +

Example V A 50 (Ω) lossless transmission line is connected with a signal generator having an internal resistance 5 (Ω) and an open-circuit voltage vg(t) = 0.5 cos π x l08t (V). The line is 5 (m) long, and the velocity of wave propagation on the line is 2 x 108 (m/s). Now, assume the termination is a matched load (50 Ω), find :

Instantaneous voltage and current at an arbitrary location on the line

Given parameters : Propagation constant is

0 08

8

0.5 0 (V), 5 ( ),50 ( ),

10 (rad/s),2 10 (m/s),

5 (m).p

VZ RZ R

uω π

= ∠ °

= = Ω

= = Ω

= ×

= ×

=

g

g g

0j l

iI I e β+=

0j l

iV V e β+=

gZ

LZ

( ), ( )V z I z

0Z , β

0z l= − z

gV0V +

0I +

0j zV e β+ −

0j zI e β+ −

8

8

10 0.5 (rad/m).2 10pu

ω πβ π×= = =

×50 ( )0

= ΩΓ =L

ZL

Example V

V(z) and I(z) at an arbitrary location on the line, Input voltage/current,

Incident voltage/current,

0

0

50( ) 0.5 0 0.45 0 (V),5 50

0.5 0( ) 0.0091 0 (A).5 50

j l ini s

s in

j l si

s in

ZV V z V e VZ Z

VI I z I eZ Z

β

β

+

+

= = − = = = ∠ ° = ∠ °+ +

∠ °= = − = = = = ∠ °

+ +

2.5 0.50 (0.5 5)

2.5 0.50 (0.5 5)

1 10.45 0 0.45 =0.45 (V),

1 10.0091 0 0.0091 =0.0091 (A).

j jins j l j

s in

j jsj l j

s in

ZV V e eZ Z e e

VI e eZ Z e e

π πβ π

π πβ π

+ − −⋅

+ − −⋅

= = ∠ °⋅ =+

= = ∠ °⋅ =+

( ) [ ],

( ) [ ]. ( 0)

j z j zo

j z j zo

V z V e e

I z I e ematch

β β

β β

+ −

+ −

= + Γ

= − Γ⇒ Γ =

Example V

The phasor form of V(z) and I(z) are, Instantaneous expressions are ,

(0.5 0.5 )

(0.5 0.5 )

( ) 0.45 (V),( ) 0.0091 (A).

j z

j z

V z eI z e

π π

π π

− +

− +

=

=

8( 10 0.5 0.5 )

8

( , ) 0.45

0.45cos( 10 0.5 0.5 ) (V),

j t zv z t e e

t z

π π π

π π π

⋅ − − = ℜ = ⋅ − −

8( 10 0.5 0.5 )

8

( , ) 0.0091

0.0091cos( 10 0.5 0.5 ) (A).

j t zi z t e e

t z

π π π

π π π

⋅ − − = ℜ = ⋅ − −

Example V

Instantaneous voltage and current at the load ( z = 0 ).

The average power delivered to the load ( Γ = 0 ).

8

8

(0, ) 0.45cos( 10 0.5 ) (V),(0, ) 0.0091cos( 10 0.5 ) (A).

v t ti t t

π π

π π

= ⋅ −

= ⋅ −

8

8

( , ) 0.45cos( 10 0.5 0.5 ) (V),( , ) 0.0091cos( 10 0.5 0.5 ) (A).

v z t t zi z t t z

π π π

π π π

= ⋅ − −

= ⋅ − −

[ ]av L av

3

1( ) ( ) ( ) *(z)2

1 (0.45 0.0091) 2.05 10 (W) 2.05 (mW).2

iP P e V z I

−

= = ℜ

= ⋅ = ⋅ =

Example VI A 50 (Ω) lossless transmission line is connected with a signal generator having an internal resistance 50 (Ω), an open-circuit voltage vs(t) = 0.5 cos π ∙ l08t (V), and terminated by loading impedance 5(Ω). The line is 5 (m) long, and the velocity of wave propagation on the line is 2 ∙ 108 (m/s).

For a mismatched load (5 Ω) , Find

V(z) and I(z) at an arbitrary location on the line expressed in phasor form , Given parameters : Reflection coefficient at the load phase constant

0 08

8

0.5 0 (V), 50 ( ),50 ( ),

10 (rad/s),2 10 (m/s),

5 (m).p

VZ RZ R

uω π

= ∠ °= = Ω= = Ω

= ⋅

= ⋅

=

s

s s

0

0

5 50 0.82 .5 50

LL

L

Z ZZ Z

− −Γ = = = −

+ +

8

8

10 0.5 (rad/m).2 10pu

ω πβ π⋅= = =

⋅

Incident voltage/current at , Incident voltage/current at , The phasor form of V+(z) and I+(z) are ,

Example VI

2.5 0.500 0.5 5

0

2.5 0.50 0.5 5

0

1 50 10.5 0 0.25 0.25 (V),50 50

1 1 1 10.5 0 5 5 (mA).50 50

j js j l j

s

j js j l j

s

ZV V e eZ Z e e

I V e eZ Z e e

π πβ π

π πβ π

+ ° − −⋅

+ ° − −⋅

= = ∠ ⋅ = =+ +

= = ∠ ⋅ = =+ +

(0.5 0.5 )0

(0.5 0.5 )0

( ) 0.25 (V),

( ) 5 (mA).

j z j z

j z j z

V z V e e

I z I e e

β π π

β π π

++ − − +

++ − − +

= =

= =

z = −

0z =

00

0

00

(z ) ,

( ) .

js

s

j s

s

ZV V e VZ Z

VI z I eZ Z

β

β

++

++

= − = =+

= − = =+

Reflect voltage/current at , The phasor form of V-(z) and I-( (z) are , The phasor form of V(z) and I(z) are ,

Example VI

(0.5 0.5 )0

(0.5 0.5 )0

( ) 0.205 (V),

( ) 4.1 (mA).

j z j z

j z j z

V z V e e

I z I e e

β π π

β π π

−− −

−− −

= = −

= = −

2.5 2.5 0.50

2.5 2.5 0.50

5 50( 0) 0.25 0.205 0.205 (V),5 50

5 50( 0) 5 4.1 4.1 (mA).5 50

j j jL

j j jL

V z V e e e

I z I e e e

π π π

π π π

− + − − −

− + − − −

−= = Γ = = − = −

+−

= = Γ = = − = −+

(0.5 0.5 ) (0.5 0.5 )

(0.5 0.5 ) (0.5 0.5 )

( ) ( ) ( ) 0.25 0.205 (V),( ) ( ) ( ) 5 4.1 (mA).

j z j z

j z j z

V z V z V z e eI z I z I z e e

π π π π

π π π π

+ − − + −

+ − − + −

= + = −

= − = +

0z =

Example VI

The average power delivered to the load ( Γ≠0 ),

[ ]

0.5 0.5 0.5 0.5

0.5 0.5

4

1( ) ( 0) *( 0)21 (0.25 0.205 ) (0.005 0.0041 )21 (0.045 ) (0.0091 )2

2.048 10 (W) 0.2048 (mW).

av L

j j j j

j j

P e V z I z

e e e e e

e e e

π π π π

π π

− −

−

−

= ℜ = =

= ℜ − ⋅ +

= ℜ ⋅

= ⋅ =

Point of interest – (optional) More discussion on Standing wave As a reminder : Total voltage as

Voltage maximums occur Voltage minimums occur

For resistive loading (XL =0), The minimum (RL<Z0) or maximum (RL > Z0) occurs at the load

(pp.41).

( 2 )0| ( ) | | ||1 | | | .j zV z V e θ βΓ −+= + Γ

2 ' 2 , 0,1, 2,....Mz n nθ β πΓ − = − =

2 ' (2 1) , 0,1, 2,....mz n nθ β πΓ − = − + =

Point of interest – (optional) More discussion on Standing wave For a line with arbitrary termination The load ZL can be determined experimentally by measuring SWR

and the distance (Vmin).

1) Find |Γ| from SWR, 2) Find θΓ from

3) Find ZL from

,mz ′

mz ′

NOTE 0 ,m

ZRSWR

= 2 .m ml zλ ′= −The equivalent impedance:

1= .1

SWRSWR

−Γ

+

2 ' for 0 .mz nθ β πΓ = − =

L L L 0

1.

1

j

j

eZ R jX Z

e

θ

θ

Γ

Γ

+ Γ= + =

− Γ0

0

| | .Γ−

Γ = Γ =+

j L

L

Z Ze

Z Zθ

Example VII (optional) The standing-wave ratio is found to be 7.0 on a lossless 50 (Ω) TxLines terminated in an unknown load impedance. The distance between successive voltage minima is 50 (cm), and the first minimum is located at 37.5 (cm) from the load. Find a) The reflection coefficient Γ,

2 22 0.5 1 (m), = 2 ( rad m ).1

π πλ β πλ

= × = = =

1 7 1 0.75.1 7 1

SWRSWR

− −Γ = = =

+ +

( )2 π 2 2π 0.375 π 0.5π rad ,mzθ βΓ ′= − = ⋅ ⋅ − =

0.5π0.75 0.75.j je e jθΓΓ = Γ = =

b) The load impedance ZL ,

c) The equivalent length and terminating resistance of a line such that the input impedance is equal to ZL ,

Example VII (optional)

L L L 0

11

j

j

eZ R jX Z

e

θ

θ

Γ

Γ

+ Γ= + =

− Γ

( ) ( )1 0.7550 50 0.28 0.96 14 48 .1 0.75

j j jj

+= = + = + Ω −

( )

( )

= = 0.5 - 0.375 0.125 m ,250 7.14 .7

m m

m

l z

R

λ ′− =

= = Ω

50 tan50 14 48 .

50 tanm m

Lm m

R j l j ZjR l

ββ

+= + = +

Outline Introduction to Transmission Line

General Transmission Line Equations

Terminated (Finite) Lossless Transmission Line

Transmission Lines as Circuit Elements

Open Circuit (4.1)

Short Circuit (4.2)

Half- and Quarter-wavelength Lines (4.2)

Point of Interest and Example(4.3)

Transmission-line Circuits

Transmission Lines as Circuit Elements Not only as wave-guiding structures but function as circuit components at microwave and

millimeter-wave frequencies.

Provide an inductive or capacitive impedance. For matching purpose.

Assume lossless

0in 0

0

tan( ) .tan

L

L

Z jZ lZ Z z l ZZ jZ l

ββ

+= = − =

+

Special Case I – Open Circuit Consider a line is terminated in a open circuit, ZL = ∞,

NOTE: when l = 0, Zin = ∞, but for l = λ/4, Zin = 0 (short circuit!!).

*Re 0.VI =(Reactive power)

0 0

00

1 (SWR = ).L

L

V Z ZZ ZV

−

+

−Γ = = = ∞

+

0in 0

0

tan .tan

L

L

Z jZ lZ ZZ jZ l

ββ

+=

+0

0 cot .tan

= = − = −ioc iocjZ

Z jX jZ ll

ββ

0 0( ) = 2 cos ,j z j zV z V e e V zβ β β+ − + = +

0 0

0 0

2( ) = sin ,j z j zV jVI z e e zZ Z

β β β+ +

− − = −

Special Case I – Open Circuit Periodic in l, repeating for multiples of λ/2.

NOTE

.eqC l= C

00 cot .

tan= = − = −ioc ioc

jZZ jX jZ l

lβ

β

0 1ioc ioc

ZZ jX j j jl llβ ωω

= ≅ − = − = −L C

CLC

tan ,l lβ β≅1lβ <<

Consider a line is terminated in a short circuit, ZL = 0,

Special Case II – Short Circuit

NOTE: when l = 0, Zin = 0, but for l = λ/4, Zin = ∞ (open circuit!!).

*Re 0.VI =(Reactive power)

0 0

00

1 (SWR= ).L

L

V Z ZZ ZV

−

+

−Γ = = = − ∞

+

0in 0

0

tan .tan

L

L

Z jZ lZ ZZ jZ l

ββ

+=

+ 0Z tan .isc iscZ jX j lβ= =

0 0( ) = 2 sin ,j z j zV z V e e jV zβ β β+ − + = − −

0 0

0 0

2( ) = cos ,j z j zV VI z e e zZ Z

β β β+ +

− = +

Special Case II – Short Circuit Periodic in l, repeating for multiples of λ/2.

NOTE

.eqL l= L

tan ,l lβ β≅1lβ <<

0isc iscZ jX jZ l j l j lβ ω ω= ≅ = =L

LC LC

0Z tan .isc iscZ jX j lβ= =

Special Case III – Half- and Quarter-Wavelength Lines

If l = nλ/2 (n=1,2,3,...),

A half-wavelength line (or any multiple of λ/2) does not alter or

transform the load impedance.

If the line is a quarter-wavelength long or, more generally, l = λ/4 + nλ/2, for n = 1,2,3,..., the input impedance is

A quarter-wave transformer has the effect of transforming the

load impedance, in an inverse manner.

0in 0

0

tan .tan

L

L

Z jZ lZ ZZ jZ l

ββ

+=

+in L ,Z Z=

20

inL

Z .ZZ

=

Point of interest – Determination of Z0 and γ

For a general lossy line section under open- and short-circuited conditions:

L 0

L 0

: coth .0 : tanh .

→ ∞ == =

ioc

isc

Open - circuited line, Z Z Z lShort - circuited line, Z Z Z l

γγ

0 ( ).ioc iscZ Z Z= Ω

1 11 tanh (m ).− −= isc

ioc

Zl Z

γ

Point of interest – Application of Open/Short-circuited Stubs Virtual short/open circuit in bias networks

Radial stub

Example VIII The open-circuit and short-circuit impedances at the input terminals of a 2-meter-long lossless transmission line, which is less than a quarter wavelength, are –j25 (Ω) and j100 (Ω), respectively.

Find Find Zo and γ of the line.

Without changing the frequency, find Zin of a short-circuited line with l = 4m

(capacitive! 2 m<λ/4< 4 m)

25 ( ),100 ( ),

2 ( ).

ioc

isc

Z jZ jl m

= − Ω= Ω=

0 25( 100) 50 ( ),Z j j= − = Ω

1 11 100tanh tan 2 0.554 (rad/m).2 25 2

− −= = = =−j jj jj

γ β

4.0 (m),l =0.554 4.0 2.216 (rad).j l j jβ = ⋅ =

50 tanh 2.216 50 tan 127 = 50( 1.33) 66.35 ( ).

= = °− = − Ω

iscZ j jj j

(lossless)

Example VIII How long should the short-circuited line become an open circuit at

the input terminals ? Figure out the wavelength first Hence the required length (odd multiple of λ/4) is

2 2 11.34 (m).0.554

= = =π πλβ

( 1)4 2

= 2.835 5.67( 1) (m), 1, 2,3,....

= + −

+ − =

l n

n n

λ λ

Outline

Introduction to Transmission Line

General Transmission Line Equations

Lump-Element Circuit Model

Terminated (Finite) Lossless Transmission Line

Transmission Lines as Circuit Elements

Transmission-line Circuits

General Concept (5.1)

Formulation (5.2)

Example (5.3)

In general, both generator and load may present mismatched impedances - Zs and ZL , which may be complex values.

General circuit model for most passive and active networks. Multiple reflections can occur.

0( ) ( ), 0,j z j zlV z V e e zβ β+ −= + Γ <

General Concept

(still unknown!)

General Concept Multiple reflections due to both generator and load mismatched

0

0.s

ss

Z ZZ Z

−Γ ≡

+

0

0.

−Γ ≡

+L

lL

Z ZZ Z

Reconsider the general solution on a transmission line

0

0

0

( ) ( ),

( ) ( ),

+ −

+−

= − = = = + Γ+

= − = = = − Γ+

j l j lini s l

in s

j l j lsi l

in s

ZV z l V V V e e

Z Z

V VI z l I e e

Z Z Z

β β

β β

.i s i sV V I Z= −( is the only unknown) 0V +

00 2

0,

(1 )

−+

−=

+ − Γ Γ

j l

s j ls s l

Z eV VZ Z e

β

β

0

0.s

ss

Z ZZ Z

−Γ ≡

+

0 0

00

.−

+

−Γ = =

+L

lL

V Z ZZ ZV

0

0

0

( ) ( ),

( ) ( ).

+ −

+−

= + Γ

= − Γ

j z j zl

j z j zl

V z V e e

VI z e e

Z

β β

β β

z l= −

Formulation – Transmission-line Circuits

Formulation – Transmission-line Circuits Reconsider the general solution on a transmission line

( solved) 0V +00 2

0.

(1 )

−+

−=

+ − Γ Γ

j l

s j ls s l

Z eV VZ Z e

β

β

20

20

2

20

(1 )( ) ,

(1 )

(1 )( ) .

(1 )

−−

−

−−

−

+ Γ=

+ − Γ Γ

− Γ=

+ − Γ Γ

j z j lj z l

s j ls s l

j z j lj zs l

j ls s l

Z e eV z V e

Z Z e

V e eI z e

Z Z e

β ββ

β

β ββ

β

( )( )

( )( )

( ) ( )

12 20

0

2 2 2 2 40

0

( ) 2 ( )0

0

( ) 1 1 ,

1 1 ,

,

−− − −

− − − −

− + − + − − +

+ − −+

= + Γ − Γ Γ+

= + Γ + Γ Γ + Γ Γ + ⋅⋅⋅+

= + Γ + Γ Γ + ⋅⋅⋅ +

= + + +

j z j l j z j ll l

j z j l j z j l j ll l l

j z l j l j z j l j z ll l

Z VV z e e e e

Z ZZ V

e e e e eZ Z

Z Ve e e e e

Z Z

V V V

β β β β

β β β β β

β β β β β

ss

s

ss s

s

ss

s

,−+− + ⋅⋅ ⋅V

represent multiple reflections z l= −

Reconsider the general solution on a transmission line

( )( )

( )( )

( ) ( )

12 20

0

2 2 2 2 40

0

( ) 2 ( )0

0

( ) 1 1 ,

1 1 ,

,

−− − −

− − − −

− + − + − − +

+ − −+

= + Γ − Γ Γ+

= + Γ + Γ Γ + Γ Γ + ⋅⋅⋅+

= + Γ + Γ Γ + ⋅⋅⋅ +

= + + +

j z j l j z j ll l

j z j l j z j l j ll l l

j z l j l j z j l j z ll l

Z VV z e e e e

Z ZZ V

e e e e eZ Z

Z Ve e e e e

Z Z

V V V

β β β β

β β β β β

β β β β β

ss

s

ss s

s

ss

s

,−+− + ⋅⋅ ⋅V

Formulation – Transmission-line Circuits

(the initial wave by voltage division) 0

0.M

Z VVZ Z

=+

s

s

( )( )

( ) ( )0

0

2 ( )

,

,

.

+ − + − +

− − +

−+ − − +

= =+

= Γ

= Γ Γ

j z l j z lM

j l j zl M

j l j z ll M

V ZV e V e

Z Z

V V e e

V V e e

β β

β β

β β

s

s

s

(the initial incident wave)

(the first reflected wave by load)

(the first reflected wave by source)

Reconsider the general solution on a transmission line

( )( )

( )( )

( ) ( )

12 20

0

2 2 2 2 40

0

( ) 2 ( )0

0

( ) 1 1 ,

1 1 ,

,

−− − −

− − − −

− + − + − − +

+ − −+

= + Γ − Γ Γ+

= + Γ + Γ Γ + Γ Γ + ⋅⋅⋅+

= + Γ + Γ Γ + ⋅⋅⋅ +

= + + +

j z j l j z j ll l

j z j l j z j l j ll l l

j z l j l j z j l j z ll l

Z VV z e e e e

Z ZZ V

e e e e eZ Z

Z Ve e e e e

Z Z

V V V

β β β β

β β β β β

β β β β β

ss

s

ss s

s

ss

s

,−+− + ⋅⋅ ⋅V

Formulation – Transmission-line Circuits

A Vs = 5∠0° (V) generator operating at 200 (MHz) with an internal resistance 50 (Ω) is connected to a 3-meter-long lossless 50 (Ω) air line that is terminated in a 30 +j30 (Ω) load. a) V(z) at a location z from the generator,

( ) ( )

( ) 20

04 8(3 ) 0.573 4 /3 (4 /3 0.573)3 3

( ) 1

50 5 = 1+0.422 =2.5 0.422 V .

100

− +

− + − +

= + Γ +

+

j l z j zl

j z j z j j z j z

Z VV z e e

Z Z

e e e e e

β β

π ππ π π

s

s

Example IX

( ) ( ) ( )( ) ( ) ( )

8

0 L

5 0 V , 50 , 2 10 Hz ,

50 , 30 30 42.43 45 , 3 m .

V Z f

R Z j l

= ∠ ° = Ω = ⋅

= Ω = + = ∠ ° Ω =s s

( ) ( )8

82 2 10 4 rad /m , 4 rad ,

33 10cω π πβ β π⋅ ⋅

= = = =⋅

0.Γ =s

Known parameters

( )( )

L 0

L 0

30 30 50 20 30 36.06 123.7 0.422 0.573 ,30 30 50 80 30 85.44 20.6

+ −− − + ∠ °Γ = = = = = ∠

+ + + + ∠ °ljZ Z j

Z Z j jπ

V- (V-+ = 0, why?) V+

b) Vi at the input terminals and VL at the load,

c) the SWR on the line,

Example IX

( ) ( )( ) ( )

4 3.4273 2.5 0.422

2.5 0.904 0.411 2.48 24.44 V .

j jiV V z e e

j

π π−= = − = +

= + = ∠ °

( ) ( )( ) ( )

0 0.573L 0 2.5 0.422

2.5 0.904 0.411 2.48 24.44 V .

j jV V z e e

j

π= = = +

= + = ∠ °

1 1 0.422 2.46.1 1 0.422

+ Γ += = =

− Γ −l

lSWR

Example IX d) the average power delivered to the load.

Comparison – maximum available power in all matched condition

(only the V+ term exists)

( )2 2

Lav L

L

1 1 2.48 30 0.051 W .2 2 42.43

VP RZ

= = ⋅ =

( )2 2

Lav

L

2.5 0.0625 W .2 2 50VPR

= = =⋅

( )L 2.5 V ,2i

VV V= = =s

0 50 ( ),LZ Z= = Ω