Prof. D. R. Wilton

Notes 6 Notes 6 Transmission LinesTransmission Lines

(Frequency Domain)(Frequency Domain)

ECE 3317

[Chapter 6]

Frequency Domain

Why is the frequency domain important?

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

A solution in the frequency-domain allows us 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 t V e d

V V t e dt

0

0

cos

cos cos

cos cos

s

s

s

A t

A t t

A t B t

0s

Arbitrary signals can be expressed as a superposition of phasors of different frequencies:

Frequency Domain

( ) ( ) j tV t V e d

0 0

0 0

0 0

( ) ( )0

0 0 0

( )cos cos ( )

( )2

1( ) ( ) ;

2

1( )cos ( ) ( )

2

j t

j t j tj t

j t j t

j t

V t t t V e d

e eV e d

V e V e d

V t t V V e d

( )V

0 0( ) ( )V V

00

Amplitude modulation:

Spectrum of a single square pulse

Video or audio spectrum

0Modulation shifts spectrum to

Telegrapher’s Equations

Rz Lz

Gz Cz

zz

Transmission-line model for a small length of transmission line:

C = capacitance/length [F/m]

L = inductance/length [H/m]

R = resistance/length [/m]

G = conductance/length [S/m]

Telegrapher’s Equations (cont.)

V IRI L

z tI V

GV Cz t

Re V , Re I

Re V , Re I

j t j t

j t j t

V e I eV I

j e j et t

Assume a steady state sinusoidal signal :

VRe Re I I

IRe Re V V

j t

j tj t

j tj t j t

d eR e j L e

dz

d eG e j C e

dz

:Recall

Telegrapher’s Equations (cont.)

, V , Ij t j tj V e I et

d

z dz

Hence to convert to the phasor domain:

3 2 Re V Im V ,0

V VRe Re I I Im Im I I

I IRe Re V V Im Im V

:

V

: At etc.At t jt

d dR j L R j L

dz dzd d

G j C G j Cdz dz

factors canceleverywhere !

j te

VI

IV

dR j L

dzd

G j Cdz

VRe Re I I

IRe Re V V

j tj t j t

j tj t j t

d eR e j L e

dz

d eG e j C e

dz

Must hold for all t !

Telegrapher’s Equations (cont.)

VI

IV

dR j L

dzd

G j Cdz

Z R j L

Y G j C

= series impedance/length

= parallel admittance/length

Define

VI

IV

dZ

dzd

Ydz

Frequency domain telegrapher’s equations

Telegrapher’s Equations (cont.)

VI

IV

dZ

dzd

Ydz

To eliminate I, differentiate the first equation and substitute from the second:

2

2

V IV

d dZ ZY

dz dz

Frequency Domain (cont.)

2

2

V( )V

dZY

dz

Solution:

2 ( )( )ZY R j L G j C

V( ) z zz Ae Be

2

2

2

VV

d

dzThen

Define

Note: We have an exact solution, even for a lossy line!

Propagation Constant

Convention: we choose the (complex) square root to be the principal branch:

( )( )ZY R j L G j C

Choosing the principal branch means that Re 0

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

j LC (lossless case)

(lossy case)

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

Physical interpretation of waves:

V ( ) zz Ae

j

(This will be shown shortly.)

V ( )nepers radians

j zzz Ae e

= propagation constant [ m-1] = attenuation constant [nepers/m] = phase constant [radians/m]

(forward traveling wave)

(backward traveling wave)V ( ) zz B e

Forward traveling wave:

Note: For a lossless line = 0,

Propagation Constant (cont.)

Denote:

LC

Current

Use the first Telegrapher equation:

VI

dZ

dz

V( ) z zz Ae Be Next, use

V( ) z zd zAe Be

dz

so

Iz zAe B e Z

Current (cont.)

Solving for the current I, we have

I

1

1

z z

z z

z z z z

z z

Ae B eZ

YZAe B e

Z

YAe B e Ae B e

Z Z Y

Ae B eR j LG j C

Iz zAe B e Z

Current (cont.)

Define the (complex) characteristic impedance Z0:

0

Z R j LZ

Y G j C

V( ) z zz Ae Be

Then we have

0

1I( ) z zz Ae B e

Z

Special case of a lossless line:

j

Hence we realize that

R j L G j C j LC

Compare with

0

LC

Propagation Constant for Lossless Line

Forward Wave

V ( ) z j zz Ae e Forward traveling wave:

( , ) cos( )zV z t A e t z

DenotejA A e

V ( ) j z j zz A e e e

Hence we have

In the time domain we have:

( , ) Re V j tV z t z e

Then

Note that ( , ) cos

( ) ,

z

z

V z t A e t z v

A e f t z v v

if

( , ) cos( )zV z t A e t z Snapshot of Waveform

• The distance is the distance (in meters) over which the waveform repeats itself.

Forward Wave (cont.)

0t

zzA e

1/1A e

= wavelength

• The distance 1/α is the distance (in meters) over which the (envelope) amplitude

drops by a factor of e

A

Wavelength

2

2

The oscillation repeats i.e.,

Hence:

(ignoring the amplitude decay) when:

cos cost z t z

Attenuation Constant

The attenuation constant controls how fast the wave decays.

( , ) cos( )zV z t A e t z

VzA e z envelope

0t

z

zA e

0t

z

zA e

V ( ) j j z j zzz A e e e A e e

( ,0) cos( )zV z A e z Snapshot of Waveform

Spatial vs. Time-Varying Signals

• The distance is the distance (in meters) over which the (envelope) amplitude drops by

a factor of e

0t

zzA e

1/1A eA T

ttA e

1A eA

( ) cos( )tf t A e t Oscilloscope signal trace

2Im [rad/m]

pv

is the period in [m]spatial

22 [rad/s]f

TT

is the period in [s]temporal

• The distance is the time (in seconds) over which the (envelope)

amplitude drops by a factor of e

1 / 1 / Re

Phase Velocity

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

( , ) cos( )V z t A t z Consider a lossless transmission line for simplicity ( = 0):

z [m]

v = velocity ,V z t t = t1

t = t2

crest of wave: 0t z

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

t z constantSet

We thus havep

v

Take the derivative with respect to time: 0dz

dt

pv

dz

dt

(velocity of a constant phase point)

Hence

Note: this result holds also for a lossy line.)

Phase Velocity (cont.)

pt z t z v Hence note that

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

1p

vLC LC

Also, we know that2

1

d

LCc

Hencep d

v c (lossless line)

Phase Velocity (cont.)

Backward Traveling Wave

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

( , ) cos( )V z t B t z

z [m]

v = velocity ,V z t t = t1

t = t2

V ( ) z j j zz B e B e e

Attenuation in dB/m

Attenuation in dB:+

10 10+

10

0.4343

V ( )dB 20log 20log ( )

V (0)

20 log ( )

8.686

zze

z e

z

( , ) cos( )zV z t A e t z

Hence we have: attenuation 8.686 [dB/m]

Recall: 3 dB attenuation signal reduced to 1 2 power

10 dB attenuation signal reduced to 1 10 power

6 dB attenuation signal reduced to 1 4 power

20 dB attenuation signal reduced to 1 100 power

Attenuation (cont’d)

Attenuation in dB:

2 1

+2

10 10+1

2 1 10

2 1

0.4343

V ( )dB 20log 20log

V ( )

20 log ( )

8.686

z zze

z

z z e

z z

Forward traveling wave measured at two points on line:

1 2

+

+ +1 2

+2

+1

V ( )

V ( ) , V ( )

V ( )

V ( )

z

z z

z Ae

z Ae z Ae

z A

z

2ze

A

2 1 2 1 2 1

1

z z z z j z z

z e e ee

1z 2z

Example: Coaxial Cable

Example (coaxial cable)

r a

bz

0

0

2F/m

ln

ln H/m2

rCba

bL

a

2S/m

ln

1 1 1/m

2 2

d

m m

Gba

Ra b

2m

m

(skin depth)

a = 0.5 [mm]b = 3.2 [mm]r = 2.2tan d = 0.001m = 5.8107 [S/m] f = 500 [MHz] (UHF)

0

tan d dd

r

Dielectric conductivity is specified in terms of the loss tangent:

copper conductors (nonmagnetic: = 0 )

r a

bz

2m

m

62.955 10 [m]m

Example: Coaxial Cable (cont.)

0 tand r d 56.12 10 [S/m]d

7

9

0.5[mm]

3.2[mm]

2.2

tan 0.001

5.8 10 [S/m]

500 [MHz] (UHF)

3.14 10 [rad/s]

r

d

m

a

b

f

r a

bz

R = 2.147 [/m]L = 3.713 E-07 [H/m]G = 2.071 E-04 [S/m]C = 6.593 E-11 [S/m]

= 0.022 +j (15.543) [1/m]

= 0.022 [nepers/m] = 15.543 [rad/m]

attenuation = 8.686 x 0.022 = 0.191 [dB/m]

= 0.404 [m]

Example: Coaxial Cable (cont.)

( )( )R j L G j C

7

9

0.5[mm]

3.2[mm]

2.2

tan 0.001

5.8 10 [S/m]

500 [MHz] (UHF)

3.14 10 [rad/s]

r

d

m

a

b

f

1

0

0

00

0

1166

2.0171 10

ln 752

377

r

R L

G C

R j L LZ

G j C C

b

a

0,R G Z

Ques : Why do we neglect in calculating , but not in calculating ?

Hint : How much atten. for 100[m] cable?

Propagation Wavenumber

Alternative notations:

j

zk j j

(propagation constant)

(propagation wavenumber)

V ( ) zjk zzz Ae Ae