prof. d. r. wilton notes 6 transmission lines (frequency domain) ece 3317 [chapter 6]
TRANSCRIPT
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?