Frequency-domain interferometry diagnostic system for the detection of relativistic plasma waves
Notes 9 Transmission Lines (Frequency Domain)
Transcript of Notes 9 Transmission Lines (Frequency Domain)
Notes 9 Transmission Lines (Frequency Domain)
1
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).
2
(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.)
6
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.)
7
( )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
9
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
11
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]
13
( )/2jc c e φ =jc c e φ=
π φ π− < ≤
Principle branch of square root:
Re 0c ≥
Note:
/ 2 / 2 / 2π φ π− < ≤
14
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)
15
( )( ) ( ) 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.)
16
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−+ − − −= = =
17
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
18
( , ) 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.)
19
Wavelength
2πβλ
=
2βλ π=The wave “repeats” (except for the amplitude decay) when:
Hence:
20
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:
21
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
22
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.)
24
( )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.)
25
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− += + +
26
( ),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
27
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:
28
( ) 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.)
29
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 )
30
[ ][ ]
[ ][ ] ( )
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:
31
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
32
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= + +
34
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
35
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
− +
− +
− +
= − + + + = − +
+ = − +
36
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.
38
( ) 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− + = −
41
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 = ×