Notes 9 Transmission Lines (Frequency Domain)

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ECE 3317 Applied Electromagnetic Waves

Prof. David R. Jackson Fall 2021

Frequency Domain

Why is the frequency domain important?

Most communication systems use a sinusoidal signal (which may be modulated).

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(Some systems, like Ethernet, communicate in “baseband”, meaning that there is no carrier.)

Examples: 100BASET, 10GBASET

Why is the frequency domain important?

A solution in the frequency-domain allows to solve for an arbitrary time-varying signal on a lossy line (by using the Fourier transform method).

( ) ( )

1( ) ( )2

j t

j t

v v t e dt

v t v e d

ω

ω

ω

ω ωπ

∞−

−∞

∞

−∞

=

=

∫

∫

3

0

1( ) Re ( ) j tv t v e dωω ωπ

∞

= ∫

For a physically-realizable (real-valued) signal, we can also write

Fourier-transform pair

Jean-Baptiste-Joseph Fourier

Frequency Domain (cont.)

4

0

( )) (Re 1 j tv t ev d ωω ωπ

∞ =

∫

A pulse is resolved into a collection (spectrum) of

infinite sinusoidal waves with different frequencies,

amplitudes, and phases.

A collection of phasor-domain signals!

Frequency Domain (cont.)

Phasor ( )v t

1.0

W

t

Time t

[ ]V

5

( )( ) sinc / 2v W Wω ω=

( ) ( ) j tv v t e dtωω∞

−

−∞

= ∫

Example: rectangular pulse

( ) ( )sinsinc

xx

x≡

Frequency Domain (cont.)

( )v t1.0

t

W

Frequency Domain (cont.)

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Phasor domain:

In the frequency domain, the system has a transfer function H(ω):

( ) 0

1( ) Re ( ) j tout inv t H v e dωω ω ω

π

∞

= ∫

The time-domain response of the system to an input signal is:

( ) ( ) ( )out inV H Vω ω ω=

System

( )H ω( )inV ω ( )outV ω

Time domain: ( )H ω( )inv t ( )outv t

System

Frequency Domain (cont.)

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( )H ω( )inv t ( )outv t

System

( ) 0

1( ) Re ( ) j tout inv t H v e dωω ω ω

π

∞

= ∫

If we can solve the system in the phasor domain (i.e., get the transfer function H(ω)), we can get the output for any time-varying input signal.

This applies for transmission lines also!

This is one reason why the phasor domain is so important!

Summary

( ) ( )( )

out

in

VH

Vω

ωω

≡

Phasor domain :

Telegrapher’s Equations

8

( )2 2

2 2( ) 0v v vRG v RC LG LCz t t

∂ ∂ ∂− − + − = ∂ ∂ ∂

R z∆ L z∆

G z∆ C z∆

z∆

z

( ),v z t

+

-

Frequency Domain

( )2 2

2 2( ) 0v v vRG v RC LG LCz t t

∂ ∂ ∂− − + − = ∂ ∂ ∂

jt

ω∂→

∂To convert to the phasor domain, we use:

( ) ( )2

22 ( ) 0V RG V j RC LG V j LCV

zω ω∂

− − + − =∂

( ) ( )2

22 ( )d V RG V j RC LG V LC V

dzω ω= + + −

or

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Frequency Domain (cont.)

( ) ( )2

22 ( )d V RG j RC LG LC V

dzω ω = + + −

Note that

2( ) ( ) ( )RG j RC LG LC R j L G j Cω ω ω ω+ + − = + +

Z R j LY G j C

ωω

= +=

==+

series impedance / length

parallel admittance / length

We can therefore write: 2

2 ( )d V ZY Vdz

=

10

Telegrapher’s Equations

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2

2 ( )d V ZY Vdz

=

R z∆ L z∆

G z∆ C z∆

z∆

z

( )V z+

-

Z R j Lω= +Y G j Cω= +

Frequency Domain (cont.)

2

2 ( )d V ZY Vdz

=

Solution:

2 ( ) ( )ZY R j L G j Cγ ω ω≡ = + +

( ) z zV z Ae B eγ γ− += +

22

2 ( )d V Vdz

γ=Then

Define

Note: We have an exact solution, even for a lossy line, in the phasor domain! 12

Propagation Constant

Convention: We choose the (complex) square root to be the principle branch:

( )( )R j L G j Cγ ω ω≡ + + (lossy case)

γ is called the propagation constant, with units of [1/m]

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( )/2jc c e φ =jc c e φ=

π φ π− < ≤

Principle branch of square root:

Re 0c ≥

Note:

/ 2 / 2 / 2π φ π− < ≤

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jγ α β= +

γ = propagation constant [1/m] α = attenuation constant [nepers/m] β = phase constant [radians/m]

Denote:

( )( )R j L G j Cγ ω ω= + +

Choosing the principle branch means that

Re 0γ ≥ 0α ≥

Propagation Constant (cont.)

For a lossless line, we consider this as the limit of a lossy line, in the limit as the loss tends to zero:

j LCγ ω= (lossless case)

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( )( ) ( ) 1R j L G j C LCγ ω ω ω= + + = −

Hence

Propagation Constant (cont.)

Hence we have that

0

LC

α

β ω

=

=

Note: α = 0 for a lossless line.

Physical interpretation of waves:

( ) zV z Ae γ+ −=

( ) z j zV z Ae eα β+ − −=

(forward traveling wave)

(backward traveling wave) ( ) zV z B e γ− +=

Forward traveling wave:

Propagation Constant (cont.)

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jγ α β= +

( ) z j zV z B e eα β− + +=Backward traveling wave:

Note: The waves must decay in the

direction of propagation.

Propagation Wavenumber

Alternative notations:

jγ α β= +

zk j jγ β α= − = −

(propagation constant)

(propagation wavenumber)

( ) zjk zz z j zV z Ae Ae Ae eγ α β−+ − − −= = =

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zjkγ =Note :

Forward Wave

( ) z j zV z Ae eα β+ − −=Forward traveling wave:

( , ) cos( )zv z t A e t zα ω β φ+ −= − +

Denote jA A e φ=

( ) j z j zV z A e e eφ α β+ − −=

Hence we have

In the time domain we have:

( ) ( , ) Re j tv z t V z e ω+ +=

Then

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( , ) cos( )zv z t A e t zα ω β φ+ −= − +

Snapshot of Waveform:

λ0t =

z

zA e α−

The distance λ is the distance it takes for the waveform to “repeat” itself in meters.

λ = wavelength

Forward Wave (cont.)

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Wavelength

2πβλ

=

2βλ π=The wave “repeats” (except for the amplitude decay) when:

Hence:

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Note: This equation can be used to find λ if we already know β :

( ) ( )2 2 2

Im Im ( )( )R j L G j Cπ π πλβ γ ω ω

= = =+ +

Wavelength (cont.)

Lossless case:

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02 2 2 1 1 12

dd

r r r r

c cf fLC f LC f LC f

π π π λλ λβ ω π µε µ ε µ ε

= = = = = = = = =

dλ λ=

0d

r r

λλµ ε

=

0cf

λ = [ ]82.99792458 10 m/sc ≡ ×

Summary for lossless case:

Attenuation Constant

The attenuation constant controls how fast the wave decays.

( , ) cos( )zv z t A e t zα ω β φ+ −= − +zA e α−=envelope

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λ0t =

z

zA e α−

( ) ( )Re Re ( )( )R j L G j Cα γ ω ω= = + +

Phase Velocity

The forward-traveling wave is moving in the positive z direction.

( , ) cos( )zv z t A e t zα ω β φ+ −= − +

Consider a sinusoidal wave moving on a transmission line (shown below for a lossless line (α = 0) for simplicity):

Crest of wave: 0t zω β φ− + =23

( ),v z t+

[ ]mz

pv = velocity1t t=

2t t=

The phase velocity vp is the velocity of a point on the wave, such as the crest.

t zω β φ− = − = constantSet

We thus have

pv ωβ

=

Take the derivative with respect to time: 0dzdt

ω β− =

dzdt

ωβ

=Hence

Note: This result holds for a

general lossy line.

Phase Velocity (cont.)

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( )Im ( )( )R j L G j Cβ ω ω= + +

Let’s calculate the phase velocity for a lossless line:

1pv

LC LCω ωβ ω

= = =

Also, we know that 2

1

d

LCc

µε= =

Hence p dv c= (lossless line)

Phase Velocity (cont.)

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0 0

1 1 1d

r r r r

ccµε µ ε µ ε µ ε

= = =Recall :

Backward Traveling Wave

Let’s now consider the backward-traveling wave (lossless, for simplicity):

( , ) cos( )zv z t A e t zα ω β φ− += + +

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( ),v z t−

[ ]mz

1t t=

2t t=pv = velocity

This wave has the same phase velocity, but it travels backwards.

The group velocity vg is the velocity of a pulse.

We have (derivation omitted):

gdvd

ωβ

= Note: This result holds for a

general lossy line.

Group Velocity

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zgv

Note: for a lossless line we have: 1 / 1 /g p dv v LC cµε= = = =

( )LCβ ω=Lossless line :

Attenuation in dB/m

10lnlogln10

xx =Use the following logarithm identity:

( )ln ( ) 20dB 20 20ln10 ln10 ln10

ze z zα

α α−

− = = = −

Therefore, the “gain” is:

Gain in dB: 10 10( )dB 20log 20log ( )(0)

zV z eV

α+

−+= =

20Attenuation [dB/m]ln10

α =

Hence we have:

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( ) zjk zz z j zV z Ae Ae Ae eγ α β−+ − − −= = =

20Attenuation [dB/m]ln10

α =

( )Attenuation 8.686 [dB/m]α≈

Final attenuation formulas:

Attenuation in dB/m (cont.)

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Example: Coaxial Cable

[ ]

[ ]

[ ]

0

0

2 F/mln

ln H/m22 S/mln

r

d

Cba

bLa

Gba

πε ε

µππσ

=

=

=

[ ]1 1 /m2 2ma ma mb mb

Ra bπ σ δ π σ δ

= + Ω

2 2ma mb

ma ma mb mb

δ δωµ σ ωµ σ

= =

(skin depth of the two conductors)

Copper conductors (nonmagnetic: µm = µ0 )

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[ ][ ]

[ ][ ] ( )

7

0.5 mm

3.2 mm2.2

tan 0.001

5.8 10 S/m

500 MHz

r

d

ma mb

a

b

f

εδ

σ σ

=

=

==

= = ×

= UHF

rε a

bz

Note: The “loss tangent” of the dielectric is called tanδd.

TV coax

Example: Coaxial Cable (cont.)

0

tan d dd

r

σ σδωε ωε ε

≡ =

Dielectric conductivity is often specified in terms of the loss tangent:

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σd = effective conductivity of the dielectric material*

Note: The loss tangent of practical insulating materials (e.g., Teflon) is approximately constant

over a wide range of frequencies. (For example, tanδd ≈ 0.001 for Teflon.)

*The effective conductivity accounts for the actual conductivity as well as molecular friction and other effects.

[ ]

[ ]

02 F/mln

2 S/mln

r

d

Cba

Gba

πε ε

πσ

=

=

0

tan d dd

r

σ σδωε ωε ε

≡ =

Relation between G and C

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0

d

r

G C σε ε

=

0

d

r

GC

σω ωε ε

⇒ =

tan dGC

δω

=

Hence

Example: Coaxial Cable (cont.)

rε a

bz

This relationship holds for any type of transmission line.

Recall

00 ln

2 r

L bZC a

ηπ ε

= =

0 75 [ ]Z = Ω

2m

m

δωµσ

=

62.955 10 [m]mδ −= ×

Example: Coaxial Cable (cont.)

( )0 tand r dσ ωε ε δ=

56.12 10 [S/m]dσ −= ×33

Characteristic impedance (ignore R and G for this):

rε a

bz

[ ][ ]

[ ][ ] ( )

7

0.5 mm

3.2 mm2.2

tan 0.001

5.8 10 S/m

500 MHz

r

d

ma mb

a

b

f

εδ

σ σ

=

=

==

= = ×

= UHF

Skin depth of metal:

Effective conductivity of dielectric:

( )0µ µ=

Example: Coaxial Cable (cont.)

( )( )R j L G j Cγ ω ω= + +

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a

b z

rε

[ ][ ]

[ ][ ] ( )

7

0.5 mm

3.2 mm2.2

tan 0.001

5.8 10 S/m

500 MHz

r

d

ma mb

a

b

f

εδ

σ σ

=

=

==

= = ×

= UHF

( ) [ ]0.022 15.543 1/mjγ = +

[ ][ ]

0.022 nepers/m15.544 rad/m

α

β

=

=

[ ]0.191 dB/m=Attenuation

[ ]0.404 mλ =

[ ][ ][ ][ ]

7

4

11

2.147 /m

3.713 10 H/m

2.071 10 S/m

6.593 10 F/m

R

L

G

C

−

−

−

= Ω

= ×

= ×

= ×

Results for (R, L, G, C):

Current

Use the first Telegrapher equation:

v iRi Lz t

∂ ∂= − −

∂ ∂

V RI j LIz

ω∂= − −

∂

jt

ω∂→

∂

( ) z zV z Ae B eγ γ− += +Next, use

( ) z zV z Ae B ez

γ γγ − +∂ = − − ∂so

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Current (cont.)

Hence, we have

z zAe B e RI j LIγ γγ ω− + − − = − −

Solving for the phasor current I, we have

( ) ( )

z z

z z

z z

I Ae B eR j L

R j L G j CAe B e

R j L

G j C Ae B eR j L

γ γ

γ γ

γ γ

γω

ω ωω

ωω

− +

− +

− +

= − + + + = − +

+ = − +

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Characteristic Impedance

Define the (complex) characteristic impedance Z0 in the frequency domain for a lossy line:

0R j LZG j C

ωω

+≡

+

Then we have:

0

1( ) z zI z Ae B eZ

γ γ− + = −

37

Note: In the time domain, we only define Z0 for a lossless line. In the frequency domain, we can define it for a lossy line.

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( ) 0( )V z Z

I z

+

+ =

Practical note: Even though Z0 is always complex for a practical line (due to loss), we usually neglect this and take it to be real.

0R j L LZG j C C

ωω

+= ≈

+

Characteristic Impedance (cont.)

The characteristic impedance is the ratio of the voltage to the

current, for a wave traveling in the positive z direction.

+ - ( )V z

( )I z

z

Note the reference directions!

( )V z+

Summary of Solution Characteristic Impedance

0R j LZG j C

ωω

+=

+

( ) z zV z Ae B eγ γ− += +

0

1( ) z zI z Ae B eZ

γ γ− + = −

39

0LZC

=

Lossless: Lossy

Voltage and Current

Appendix: Summary of Formulas

0R j LZG j C

ωω

+=

+

( ) z zV z Ae B eγ γ− += +

0

1( ) z zI z Ae B eZ

γ γ− + = −

40

0

tan d dd

r

σ σδωε ωε ε

≡ =

tan dGC

δω

=

( )( )R j L G j Cγ ω ω≡ + +

( )Attenuation 8.686 [dB/m]α=jγ α β= +

2πβλ

=

pv ωβ

=

General Lossy Case

Appendix: Summary of Formulas

( ) j z j zV z Ae B eβ β− += +

0

1( ) j z j zI z Ae B eZ

β β− + = −

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0LZC

=

jγ β=

LCβ ω=

2πβλ

=

p dv c=

Lossless Case

dλ λ=

0d

r r

λλε µ

=

0cf

λ =

dr r

ccε µ

=

[ ]82.99792458 10 m/sc = ×