Prof. David R. Jackson

Notes 9 Transmission Lines(Frequency Domain)

ECE 3317

1

Spring 2013

Frequency Domain

Why is the frequency domain important?

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

2

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

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

3

0

1( ) Re ( ) j tV t V e d

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

Fourier-transform pair

Jean-Baptiste-Joseph Fourier

Frequency Domain (cont.)

4

0

1( ) Re ( ) j tV t V d e

A pulse is resolved into a collection (spectrum) of infinite sine waves.

A collection of phasor-domain signals!

Frequency Domain (cont.)

t

V t

1.0

W

5

( ) sinc / 2V W W

( ) ( ) j tV V t e dt

Example: rectangular pulse

sinsinc

xx

x

Frequency Domain (cont.)

t

V t

1.0

W

Frequency Domain (cont.)

6

Phasor domain:

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

0

1( ) Re ( ) j t

out inV t H V e d

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

out inV H V

System

H Vin Vout

Time domain: H inV t outV t

System

Frequency Domain (cont.)

7

H inV t outV t

System

0

1( ) Re ( ) j t

out inV t H V e d

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

This applies for transmission lines

Telegrapher’s Equations

RDz LDz

GDz CDz

zDz

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]

8

Telegrapher’s Equations (cont.)

V IRI L

z tI V

GV Cz t

2 2

2 2( ) 0

V V VRG V RC LG LC

z t t

Take the derivative of the first TE with respect to z.

Substitute in from the second TE.

Recall: There is no exact solution in the time domain, for the lossy case.

9

Frequency Domain

2 2

2 2( ) 0

V V VRG V RC LG LC

z t t

jt

To convert to the phasor domain, we use:

2

2

2

VV ( )V V 0RG j RC LG j LC

z

2

2

2

VV ( )V V

dRG j RC LG LC

dz

or

10

Frequency Domain (cont.)

2

2

2

VV ( )V V

dRG j RC LG LC

dz

Note that

= series impedance/length

2( ) ( )( )RG j RC LG LC R j L G j C

Z R j L

Y G j C

= parallel admittance/length

Then we can write:2

2

V( )V

dZY

dz

11

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

V( )V

d

dzThen

Define

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

12

Propagation Constant

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

( )( )R j L G j C (lossy case)

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

13

1/2

/2

j

j

c c e

c e

jc c e

Principle branch of square root:

Re 0c

Note:

/ 2 / 2 / 2

14

j

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

Denote:

( )( )R j L G j C

Choosing the principle branch means that

Re 0 0

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 LC (lossless case)

15

( ) ( ) 1R j L G j C LC

Hence

Propagation Constant (cont.)

Hence we have that 0

LC

Note: = 0 for a lossless line.

Physical interpretation of waves:

V ( ) zz Ae

(This interpretation will be shown shortly.)

V ( ) az j zz Ae e

(forward traveling wave)

(backward traveling wave)V ( ) zz B e

Forward traveling wave:

Propagation Constant (cont.)

16

Propagation Wavenumber

Alternative notations:

j

zk j j

(propagation constant)

(propagation wavenumber)

V ( ) zjk zzz Ae Ae

17

Forward Wave

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

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

DenotejA A e

V ( ) j az j zz A e e e

Hence we have

In the time domain we have:

( , ) Re V j tV z t z e

Then

18

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

Snapshot of Waveform:

0t

z

zA e

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

= wavelength

Forward Wave (cont.)

19

Wavelength

2

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

Hence:

20

Attenuation Constant

The attenuation constant controls how fast the wave decays.

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

zA e envelope

21

0t

z

zA e

Phase Velocity

The 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 22

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

t z constantSet

We thus have pv

Take the derivative with respect to time: 0dz

dt

dz

dt

Hence

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

Phase Velocity (cont.)

23

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.)

24

Backward Traveling Wave

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

( , ) cos( )V z t A t z

z [m]

v = velocity ,V z t t = t1

t = t2

25

Attenuation in dB/m

10

lnlog

ln10

xx Use the following logarithm identity:

ln ( )dB 20 20

ln10 ln10

ze z

Therefore

Attenuation in dB:+

10 10+

V ( )dB 20log 20log ( )

V (0)zz

e

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

20attenuation [dB/m]

ln10

Hence we have:

26

20attenuation [dB/m]

ln10

attenuation 8.6859 [dB/m]

Final attenuation formulas:

Attenuation in dB/m (cont.)

27

Example: Coaxial Cable

a

bz

r

0

0

2F/m

ln

ln H/m2

2S/m

ln

r

d

Cba

bL

a

Gba

1 1/m

2 2ma ma mb mb

Ra b

2m

m

(skin depth)

a = 0.5 [mm]b = 3.2 [mm]r = 2.2tand = 0.001ma = mb = 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 )

28

d = effective conductivity of the dielectric material

a

bz

r

02F/m

ln

2S/m

ln

r

d

Cba

Gba

0

tan d dd

r

Relation between G and C

29

0

d

r

G C

0

d

r

G

C

tanG

C

Hence

Note: The loss tangent of practical insulating materials (e.g., Teflon) is approximately constant over a wide range of frequencies.

Example: Coaxial Cable (cont.)

00 ln

2 r

L bZ

C a

0 75 [ ]Z

2m

m

62.955 10 [m]m

Example: Coaxial Cable (cont.)

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

30

a

bz

r

Ignore loss for calculating Z0:

a = 0.5 [mm]b = 3.2 [mm]r = 2.2tand = 0.001ma = mb = 5.8107 [S/m]

f = 500 [MHz] (UHF)

R = 2.147 [/m]L = 3.713 E-07 [H/m]G = 2.071 E-04 [S/m]C = 6.593 E-11 [F/m] = 0.022 +j (15.543) [1/m]

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

attenuation = 0.191 [dB/m]

= 0.404 [m]

Example: Coaxial Cable (cont.)

( )( )R j L G j C

31

a

bz

r

a = 0.5 [mm]b = 3.2 [mm]r = 2.2tand = 0.001ma = mb = 5.8107 [S/m]

f = 500 [MHz] (UHF)

Current

Use the first Telegrapher equation:

V IRI L

z t

VI IR j L

z

jt

V( ) z zz Ae Be Next, use

V( ) z zzAe Be

z

so

32

Current (cont.)

Hence we have

I Iz zAe Be R j L

Solving for the phasor current I, we have

I z z

z z

z z

Ae B eR j L

R j L G j CAe B e

R j L

G j CAe B e

R j L

33

Characteristic Impedance

Define the (complex) characteristic impedance Z0:

0

R j LZ

G j C

V( ) z zz Ae Be

Then we have

0

1I( ) z zz Ae B e

Z

34

0

LZ

C

Lossless: