Transmission Line Theory
EE142 Dr. Ray Kwok Transmission Line Dr. Ray Kwok
RF Spectrum
km mm m Å
Advanced Light Source Berkeley Lab Transmission Line Dr. Ray Kwok
RF / Microwave Circuit
1 – port 2 wires network GND
2 – port input output source network load Transmission Line Dr. Ray Kwok
Series connection
low f A B
RF
A B Transmission Line Dr. Ray Kwok
Parallel connection
low f A
B
RF
A
B Transmission Line Dr. Ray Kwok Common transmission lines
Zo l most correct schematic
twisted pair microstrip (line) VLF no distortion lossy & noisy wide freq range lowest cost
co-planar waveguide low cost flip chip access paralllel wire complex design LF HF noisy & lossy waveguide lowest loss coaxial cable freq bands no distortion wide freq range Transmission Line Dr. Ray Kwok
Equivalent circuit
i(z,t) i(z+ z,t) Ideal transmission line V(z,t) L z V(z+ z,t) C z
Taylor ∂ )t,z(i ∂V )t,z( Kirchhoff”s law: V )t,z( − L( )z = V z( + )t,z ≈ V )t,z( + z ∂t ∂z ∂ )t,z(i ∂V )t,z( − L = ∂t ∂z
Q=CV ∂V )t,z( ∂ )t,z(i dQ/dt=i=C dV/dt Junction rule: z(i + )t,z − )t,z(i = −(C )z ≈ z ∂t ∂z
∂V )t,z( ∂ )t,z(i − C = ∂t ∂z Transmission Line Dr. Ray Kwok
Coupled equations (V – i)
∂ )t,z(i ∂V )t,z( ∂V )t,z( ∂ )t,z(i − L = − C = ∂t ∂z ∂t ∂z
∂2i ∂2V ∂ 1 ∂i 1 ∂2i − L = = − = − ∂t 2 ∂t∂z ∂z C ∂z C ∂z2 ∂2i ∂2i = LC current wave ∂z2 ∂t 2
∂ 2V ∂ 2i ∂ 1 ∂V 1 ∂ 2V similarly − C = = − = − ∂t 2 ∂t∂z ∂z L ∂z L ∂z2 ∂ 2V ∂ 2V = LC voltage wave ∂z2 ∂t 2 Transmission Line Dr. Ray Kwok
Wave equation x(f ± vt) ≡ )u(f reverse / forward traveling wave ∂f ∂u = )u('f = )u('f ∂x ∂x ∂2f ∂u note: 2 = )u("f = )u("f 1 2π 2π ∂x ∂x x ± vt = x ± vt k λ λ ∂f ∂u 1 = )u('f = ±vf )u(' = ()kx ± 2πft ∂t ∂t k ∂ 2f ∂u 1 = ±vf )u(" = v2 )u("f = ± ()ωt ± kx ∂t 2 ∂t k 2 2 x(f ± vt) = f ()ωt ± kx ∂ f 1 ∂ f r r = wave equation f ()ωt − k ⋅ r ∂x 2 v2 ∂t 2 (3D) Transmission Line Dr. Ray Kwok
Voltage & Current Waves 2π + (j ωt−β )z − (j ωt+β )z where β = V )t,z( = Vo e + Vo e λ + (j ωt−β )z − (j ωt+β )z λ ω 1 )t,z(i = Io e − Io e v = fλ = 2( π )f = = 2π β LC why “ ” ? ∂i = −jβI+e (j ωt−β )z − jβI−e (j ωt+β )z ∂z o o ∂i ∂V = −C = −C[]jωV+e (j ωt−β )z + jωV−e (j ωt+β )z ∂z ∂t o o + + βIo = CωVo 1 − − v = βIo = CωVo LC L ± β ± LC ± L ± ± Z = V = I = I = I ≡ Z I o C o Cω o C o C o o o Transmission Line Dr. Ray Kwok
Fields and circuits r r E ∂ 2 E ∇2 r = ε r 1 2 v = H ∂t H ε r r r r r r r (j ωt−ki ⋅ )r (j ωt−k r ⋅ )r )t,z(E = Eoie + Eore η = r r r r r r r ε (j ωt−ki ⋅ )r (j ωt−k r ⋅ )r H )t,z( = Hoie − Hore
∂2 V ∂2 V = LC 1 ∂z2 i ∂t 2 i v = LC + (j ωt−β )z − (j ωt+β )z V )t,z( = V e + V e L o o Z = + (j ωt−β )z − (j ωt+β )z o C )t,z(i = Io e − Io e Transmission Line Dr. Ray Kwok
What is Zo?
• Characteristic Impedance. • 50 ohms for most communications system, • 75 ohms for TV cable. • Measure 75 ohms with a ohmmeter? • Two 75 cables together (in series) makes a 150 cable? • 75 + 75 = 75 !!!!
• What does Zo represent? Transmission Line Dr. Ray Kwok
Reflection at Load + − jβx − jβx V )x( = Vo e + Vo e + − jβx − jβx Zo l Z L )x(i = Io e − Io e
+ − V )0( ≡ VL = Vo + Vo at the load x = − l x = 0 + − 1 + − )0(i ≡ IL = Io − Io = ()Vo − Vo Zo V V+ + V− L ≡ Z = Z o o L o + − Define normalized impedance IL Vo − Vo Z Z ()()V+ − V− = Z V+ + V− Z ≡ L o o o o o + − Zo Vo ()()ZL − Zo = Vo ZL + Zo Z −1 − L Vo ZL − Zo ΓL = ≡ Γ = Z ≠ Z Z +1 + L L o L Vo ZL + Zo reflection Transmission Line Dr. Ray Kwok
Example
does it work?
75
50
75 Transmission Line Dr. Ray Kwok Impedance at Input l + jβl − − jβl Vin V(− ) Vo e + Vo e Zin ≡ = = l 1 l l Iin (i − ) + jβ − − jβ ()Vo e − Vo e Z Z l Zo in o ZL e jβl + Γ e− jβl Z = Z L in o jβl − jβl e − ΓLe x = − l x = 0 l 1+ jtanβl Z −1 l e− jβ + L e− jβ l 1− jtanβ ZL +1 Zin = l 1+ jtanβl Z −1 l V )x( = V+e− jβx + V−e jβx e− jβ − L e− jβ o o l 1− jtanβ ZL +1 + − jβx − jβx )x(i = Io e − Io e ()Z +1 ()1+ jtanβl + ()Z −1 ()1− jtanβl Z = L L in l l ()ZL +1 ()1+ jtanβ − ()ZL −1 ()1− jtanβ 2()Z + jtanβl Z = L in l 2()1+ Zj L tanβ Z + jtanβl Z = L in l 1+ Zj L tanβ Z + jZ tanβl Z = Z L o in o l Zo + jZL tanβ Transmission Line Dr. Ray Kwok
Exercise
Zo = 50 Zin Zo l Z L ZL = 100 Zin = ?
x = − l x = 0 For length = λ/8? λ/4? λ/2?
What if Z = Z = 50 ? Z + jtanβl o L Z = L Would the length make any difference? in l 1+ Zj L tanβ Z + jZ tanβl Z = Z L o in o l Zo + jZL tanβ 50 ( 37 o) 25 100
Zin =Zo=Z L Transmission Line Dr. Ray Kwok
Transmission Line Impedance
case 1: βl= 0, or l = 0 Zin Z l o ZL tan β = 0
Zin = Z L
x = − l x = 0 case 2: βl= π, or l = λ/2 tan βl = 0 Z + jtanβl Z = Z Z = L in L in l 1+ Zj L tanβ case 3: βl= π/2 , or l = λ/4 Z + jZ tanβl Z = Z L o tan β l → ∞ in o l 2 Zo + jZL tanβ Zin = Z o / Z L Quarter wave transformer (impedance), real to real, complex to complex .
note: at low freq, β → 0, Zin = Z L regardless of line length or line impedance. Transmission Line Dr. Ray Kwok
Reflection at Input Z + jtanβl Z = L in l 1+ Zj L tanβ Z l Z’ o Zin o ZL l ZL + jZo tanβ Zin = Zo Γin l Zo + jZL tanβ ' ' x = − l x = 0 Zin − Zo Zin −1 Γin = ' = ' Zin + Zo Zin +1
Zin − Zo Zin −1 In general Γin = = Zin + Zo Zin +1
just have to know what Z to use Transmission Line Dr. Ray Kwok
Exercise
Z l Z’ o Zin o ZL Zo = 50 Γin Z’ o = 50 Z = 100 l L x = − x = 0 Length = λ/8
ΓL = ? Γin = ? Z + jtanβl What if Z’ is 75 ? Z = L o in l 1+ Zj L tanβ Z + jZ tanβl Z = Z L o in o l Zo + jZL tanβ ' ' Zin − Zo Zin −1 Γin = ' = ' Zin + Zo Zin +1 1/3 o 1/3 ( 90 ) only change phase !?! 0.388 (235 o) Transmission Line Dr. Ray Kwok
Voltage wave in transmission line
+ − jβx − jβx V )x( = Vo e + Vo e + − jβx j2 βx Z Z l V )x( = V e ()1+ Γ e in o ZL o L + j2 βx V = Vo 1+ ΓLe jθ x = − l x = 0 ΓL ≡ ρe + (j θ+2β )x V = Vo 1+ ρe
+ 2 2 2 V = Vo ()1+ ρcos(θ + 2β )x + ρ sin (θ + 2β )x
+ 2 min when sine = 1 V = Vo 1+ 2ρcos(θ + 2β )x + ρ 2 + 2 + V = V+ ()()1+ ρ − 2ρ 1− cos(θ + 2β )x Vmin = Vo ()1+ ρ − 4ρ = Vo ()1− ρ o
2 θ + 2βx max when sine = 0 V = V+ ()1+ ρ − 4ρsin 2 o 2 + 2 + Vmax = Vo ()1+ ρ = Vo ()1+ ρ Transmission Line Dr. Ray Kwok
Voltage Standing Wave
+ − jβx − jβx V )x( = Vo e + Vo e V V min max ZL standing wave + − Γ ≡ ρ = ±1 If V o = V o , L perfect standing wave with nodes
x = − l x = 0 + 2 2 θ + 2βx V = Vo ()1+ ρ − 4ρsin 2
min when θ + 2βx 2( n + )1 π max when θ + 2βx = ± = ±nπ 2 2 2 λ λ (x < 0) x = −[]θ m 2( n + )1 π x = −[]θ m 2nπ 4π 4π Transmission Line Dr. Ray Kwok
VSWR (Voltage Standing Wave Ratio)
+ 2 2 θ + 2βx V V V = Vo ()1+ ρ − 4ρsin min max ZL 2
+ 2 + Vmin = Vo ()1+ ρ − 4ρ = Vo ()1− ρ
+ 2 + x = − l x = 0 Vmax = Vo ()1+ ρ = Vo ()1+ ρ
V 1+ ρ 1+ Γ VSWR ≡ max = = Vmin 1− ρ 1− Γ
perfect match: ρ = 0, VSWR = 1.0 open / short: ρ = 1, VSWR → ∞ It is an indicator on how well the load matches the line. VSWR is the standing wave pattern INSIDE the line. Only Γ at the reflected junction that counts Transmission Line Dr. Ray Kwok
Exercise
Zo = 50 Z’ Z Zo l o in ZL Z’ o = 75 Γin ZL = 100 Length = λ/8 x = − l x = 0 VSWR = ?
ΓL = 1/3 VSWR = 2 Transmission Line Dr. Ray Kwok Return Loss
RL ≡ − 20 log ρ (dB)
perfect match: ρ → 0, VSWR → 1.0, RL → ∞
open / short: ρ = 1, VSWR → ∞, RL → 0 dB 1+ ρ 1+ Γ VSWR −1 VSWR ≡ = ρ = Γ = 1− ρ 1− Γ VSWR +1
Typical VSWR = 1.1 to 2 ρ = 0.048 to 0.33 RL = 26 dB to 9.5 dB Transmission Line Dr. Ray Kwok
Stub
Transmission line connecting nowhere(?)
Open stub
Short stub (short)
Series stub
Shunt stub
(short) Transmission Line Dr. Ray Kwok
Open Shunt Stub
L Band Transmission Line Dr. Ray Kwok
Short Shunt Stub
20 GHz Interdigital Filter Transmission Line Dr. Ray Kwok
Radial Stub
18 GHz Rat Race Transmission Line Dr. Ray Kwok Tuning stub (open) Transmission Line Dr. Ray Kwok Short Stub Z + jtanβl Z = L in l 1+ Zj L tanβ Z Zo l in ZL (Z → 0) l L Zsh = jtanβ 1 l Ysh ≡ = −jcotβ Zin Zsh
Zcoil = jωL
Zcap = j/ ωC 1 Zsres = jωL1− ω2LC π/2 π 3π /2 2π βl 1 Z = pres 1 jωC1− ω2LC
period of π Transmission Line Dr. Ray Kwok
Open Stub Z + jtanβl Z = L in l Z l 1+ Zj L tanβ Zin o ZL Z = −jcotβl (Z L → ∞) op l Zin Yop = jtanβ
Zcap = j/ ωC
Zcoil = jωL 1 Zsres = jωL1− ω2LC π/2 π 3π /2 2π βl 1 Z = pres 1 jωC1− ω2LC
period of π Transmission Line Dr. Ray Kwok
Exercise 75 , λ /8 Find Zin & Γin .
Zin Zsh = j Zotan( βl) 50 = j 75 tan(45 o) = j 75 λ/2 100 Zop = j Zocot( βl) 0.1 λ = j 100 cot(36 o) = j 138
Y =1/j75 = j 0.0133 Zin 50 λ/2 Y = j/138 = j0.0073
Zin Z = 1/Y = 1/( j0.006) = j166 50 Z + jtanβl λ/2 Z = L = Z in l L Zin = j 166 1+ Zj L tanβ
Zin − Zo j166 − 50 o o o Γin = = =1∠(107 − 73 ) =1∠34 Zin + Zo j166 + 50 Transmission Line Dr. Ray Kwok
Admittance (Y = 1/Z)
Y Y l YL l 1 in Yo ZL + jtan β Zin = l Γin 1+ Zj L tan β l 1 1+ Zj L tan β Yin ≡ = Z Z + jtan βl Zin − Z1 in L Γin = 1+ /1(j Y ) tan βl Zin + Z1 Y = L in l /1 Y − /1 Y /1 YL + jtan β in 1 Γin = Y + jtan βl /1 Yin + /1 Y1 Y = L in 1+ jY tan βl L Y1 − Yin 1− Yin Γin = = Y + jY tan βl Y + Y 1+ Y Y = Y L o 1 in in in o l Yo + jYL tan β useful for shunt circuits Transmission Line Dr. Ray Kwok
Earlier exercise 75 , λ /8 Find Zin & Γin .
Zin Ysh = j Yocot( βl) 50 = j (1/75) cot(45 o) = j 0.0133 λ/2 100 Yop = j Yotan( βl) 0.1 λ = j 0.01 tan(36 o) = j 0.0073
Y = j 0.0133 Zin 50 λ/2 Y = j0.0073
Zin Y = j 0.006 50 Y + jtanβl Y = L = Y YL = j 0.006 λ/2 in l L 1+ jYL tanβ Zin = j 166
Yo − Yin /1 50 + .0j 006 o o o Γin = = =1∠(17 +17 ) =1∠34 Yo + Yin /1 50 − .0j 006 Transmission Line Dr. Ray Kwok
Earlier Exercise – power consideration
Z = 50 Z l o Z’ o Zin o ZL Z’ o = 50 Γin ZL = 100 Length = λ/8 x = − l x = 0 o ΓL = ? 1/3 Γin = ?1/3 ( 90 ) Z + jtanβl only change phase Z = L in l 1+ Zj L tanβ l ZL + jZo tanβ Z = Z 2 in o − 2 Z + jZ tanβl V 2 1 o L Power reflected = ? + = Γ = =11% Z − Z' Z' −1 V 3 Γ = in o = in in ' ' 2 Zin + Zo Zin +1 Power delivered = ? 1− Γ = 89%
Don’t double count reflection…. ΓL & Γin
Return Loss (RL) = 20 log| ρ| = + 9.5 dB Transmission Line Dr. Ray Kwok
High f circuit elements
1 GHz lumped element Band pass filter
CAPCAPCAP IND CAP CAPID=INDID=ID=C2ID=C6IND C3 C4 ID=ID=ID=CAPC1L3L4ID= C= 1L7 pF C=ID=C=C=1 pFC511 pF pF C=L=L=11 1 pFnH nHL= 1 nH ….. …..C= 1 pF
12 GHz lumped element Low pass filter much smaller
IND CAPINDCAP ID=CAPINDINDID=L1 C4 ….. ID=ID=ID=ID=ID=C1L3L4 L7C6 C= 1 pF …..L= L=1L=L=C= nH11 11nH nH nHpF C= 1 pF
A small loop of thin wire is an inductor !! Transmission Line Dr. Ray Kwok
High-Z Line as inductor
l Z Zo Zin Z1 L
Γin
l ZL + jZ1 tanβ Z1 >> Z L Z = Z line length < λ/4 ( π/2) in 1 Z + jZ tanβl 1 L ZL ~ Zo (order of magnitude) a∠ + Ψ Zin = Z1 = Zin ∠ + θ b∠ + ϕ
L z C z Zin has a positive phase → inductor like !!!
small C, large L, series inductor Transmission Line Dr. Ray Kwok
Low-Z Line as capacitor
l Z Zo Zin Z1 L
Γin
l ZL + jZ1 tanβ Z1 << Z L Z = Z line length << λ/4 ( π/2) in 1 Z + jZ tanβl 1 L ZL ~ Zo a∠ + ϕ Zin = Z1 = Zin ∠ − θ b∠ + Ψ Y = Y ∠ + θ L z in in C z
Yin has a positive phase → capacitor like !!! small L, large C, shunt capacitor Transmission Line Dr. Ray Kwok
Low pass filter
5 GHz low pass filter
IND CAPINDCAP ID=CAPINDINDID=L1 C4 ….. ID=ID=ID=ID=ID=C1L3L4 L7C6 C= 1 pF …..L= L=1L=L=C= nH11 11nH nH nHpF C= 1 pF
14 GHz low pass filter high low impedance lines waveguide high power low loss Transmission Line Dr. Ray Kwok
High-Low-Z lines
20 GHz band pass filter high Z lines → inductors Short shunt stubs λ/4 resonators
….. INDCAPIND CAPIND ID=INDCAPINDL2IND ID=ID=ID=ID=INDC6C1C4L3L4 …..L=ID=ID=ID=1 L=L6nHL7L5L11 nH C= L=1 1pF nH C=L=L=L=11 1nH pFnH nHnH
13 GHz coupler Tuning with stubs (shunt open) Think of them as shunt capacitors → low Z lines Transmission Line Dr. Ray Kwok
Homework 1. A 100 tranmission line has an effective dielectric constant of 1.65. Find the shortest open circuited length of this line that appears at its input as a capacitor of 5 pF at 2.5 GHz. Repeat for an inductance of 5 nH.
2. A radio transmitter is connected to an antenna having an impedance 80 + j40 with a 50 coaxial cable. If the 50 transmitter can deliver 30 W when connected to a 50 load, how much power is delivered to the antenna?
3. A 75 coaxial transmission line has a length of 2 cm and is terminated with a load impedance of 37.5 + j75 . If the dielectric constant of the line is 2.56 and the frequency is 3 GHz, find the input impedance to the line, the reflection coefficient a the load, the reflection coefficient at the input, and the SWR on the line.
4. The VSWR on a lossless 300 transmission line terminated in an unknown load impedance is 2.0, and the nearest voltage minimum is at a distance 0.3 λ
from the load. Determine (a) ΓL, (b) Z L. Transmission Line Dr. Ray Kwok
VSWR ρρρ RL (dB) 5. Calculate VSWR, ρ, and return loss 1.00 0.00 ∞ values to complete the entries in the 1.01 following table. 0.01 1.05 32 30 1.10 1.20 0.10 1.50 10 6. Measurements on a 0.6 m lossless 2.00 coaxial cable at 100 kHz show a 2.50 capacitance of 54 pF when the cable is open circuited, and an inductance of 0.30 H when it is short circuited. (a)
Determine Zo and εr of the medium.