6-1 Chapter 6 Transmission Lines

Transcription

1 6-1 Chapter 6 Transmission ines ECE 3317 Dr. Stuart A. ong

2 6-2

3 General Definitions p Voltage V( z) = α E ds ( C z) 1 C t t ( a) Current I( z) = α H ds ( C0 closed) 2 C 0 ( b)

4 General Definitions 6-4 p.133 Power 1 * 1 * Re V( z) I ( z) = Re ˆ da 2 E H z 2 A () c Characteristic Impedance Z 0 = V( z) I ( z) infinite line with with no reflection

5 Parallel Plate Waveguide (TEM mode) x w 6-5 The Transverse Electromagnetic Fields in a d C t Parallel Waveguide are approximately as follows: C o y E = xˆ Ee 0 - jkz 6.1a H = yˆ E0 e - η jkz 6.1b Using the d 1 E xˆ α V( z) = α dx = E d e - jkz 6.2a General Definitions w Ew 0 I( z) = α2 H yˆ dy = α 0 2 e η - jkz 6.2b

6 Parallel Plate Waveguide (TEM mode) x w 6-6 The time-average power transmitted d C t is given by y C o

7 Parallel Plate Waveguide (TEM mode) x w 6-7 d C t V( z) = E de 0 jkz 6.3a C o y I( z) = Ew 0 η e jkz 6.3b Z d = η w 0 [ Ω] 6.4

8 Coaxial ine The fields inside a coaxial line for the TEM b a C o 6-8 mode are given by C t 6.5a 6.5b Using the 6.6a General Definitions 6.6b

9 Coaxial ine 6-9 The time-average power transmitted is given by b a C o C t

10 Coaxial ine Coaxial ine 6-10 b a C o C t 6.7 Example: What is the ratio b/a for an air-filled coax and = ln( b/ a) 2π b = 2.3 a [ Ω] line?

11 Rectangular Waveguide (Dominant TE 10 mode) y a 6-11 The Electromagnetic fields in a Rectangular Waveguide for the are TE 10 mode are b C t x C o 6.8a 6.8b 6.8c

12 Rectangular Waveguide (Dominant TE 10 mode) y a 6-12 Using the General b C t Definitions x C o b V( z) = α E 1 0 = α E be 1 0 y x= a 2 - jk z z dy 6.9a I( z) = α x ˆ = α 2 C H dx E0akz e - jkz z πωµ 6.9b

13 Rectangular Waveguide (Dominant TE 10 mode) The time-average power transmitted is given by b y C t a 6-13 x C o

14 Rectangular Waveguide (Dominant TE 10 mode) y a 6-14 b C t V( z) = akz E0 be 2bωµ - jk z z C o x I( z) = 2 π bωµ 2E ak 8ak z 0 πωµ z e - jk z z Z = 1 0 [ Ω]

15 Transmission ine Equations (Parallel plate waveguide) 6-15 E = - jωµ H H = jωε E z E = -jωμh - H = jωεe z x y y x jkz Ew 0 jkz V( z) = E0de = d Ex I( z) = e = η wh y V z = jωi I z = jωcv µ d H εw F = C = w m d m

16 Transmission ine Equations (Parallel plate waveguide) z V ω CV = 0 wave equation jkz + e - V= V + Ve + jkz jkz I= V+ e + V-e Z 0 + jkz 6.17 k = ω C Z 0 C = 6.18

17 Standing Waves on Terminated Transmission ines jkz - V( z) = V e + V e + jkz jkz V V I( z) = e e Z Z jkz 6.23 Impedance Z(z) = V( z) I( z) + -jkz - Z( z) V e + V e Z ( z) = = + jkz n Z + -jkz - + jkz 0 V e V e 6.24 (note Z(0) = Z )

18 Standing Waves on Terminated Transmission ines 6-18 Reflection coefficient Γ = V V - + With the reflection coefficient now defined, we can rewrite 6.24 as Z() z n = e- e jkz -jkz + Γ Γ e e + + jkz jkz

19 Standing Waves on Terminated Transmission ines Generator Transmission ine oad 6-19 Z G V G Z Evaluating Z at the oad ( z = 0) n Z( z = 0) 1+Γ Z 1 Z ( z = 0) or n n = = Z 1 Γ Γ = Z n 6.25 Evaluating Z at a point along line ( z = l) n Z( z = l) Z + jz tan kl Z ( z = l) = = 0 n Z Z + jz tan kl

20 Standing Waves on Terminated Transmission ines 6-20 V = V ( z) Γ e j2kz with Γ = Γ e jφ V = V + 1+ ( z) Γ e j( φ + 2 kz) V when max V when min φ + 2kz = 0, - 2π,... φ + 2kz = π, 3 π,...

21 Standing Wave Pattern 6-21 V = V ( z) Γ j( φ + 2 kz) e V( z) V + V ( z ) j( φ + 2 kz) = 1+ Γ e V + 1+ Γ 1 1- Γ z Fig. 6.6

22 Crank Diagram 6-22 V = V + 1+ ( z) Γ j( φ + 2 kz) e Im V ( z ) j( φ + 2 kz) = 1+ Γ e V + 1 Γ Re j( φ + 2 kz) 1+ Γ e Fig. 6.6

23 Crank Diagram 6-23 Im 1+ Γ 1 1- Γ 1+ Γ 1 Γ Re 1- Γ z V ( z ) ( 2 ) 1 Γ j φ + = + e kz V +

24 Standing Wave Ratio (SWR or VSWR) 6-24 V I VSWR = max = max V I min min [ ] To determine ( Γ ) or Z also need position of V min 1+ Γ VSWR = 1- Γ [ ] 6.29 Note : Power ~ V 2 P P + = Γ 2 Γ = [ ] [ ] VSWR -1 VSWR Power to load ~ 1- Γ 2

25 Example 6-25 Ex. 6.6 [ Ω] and [ Ω] Z = 17.4 j30 Z = 50 0 Z 0 + V = V Z Z 1 Z Z Γ = n = 0 = 0.24 j.55 Z + 1 Z + Z n 0 or j1. 99 Γ = 0.6e V(z) V where 0.4 Γ = 0.6 φ = Γ = 1.99 V when max V when min λ φ + 2kz = 0, - 2π,... φ + 2kz = π, 3 π, λ λ z

26 Example 6-26 Ex. 6.6 V where φ+ 2kz min m = π z m π φ ( π ) λ = = =.092λ 2k 2(2 π ) V 8.4 V max 2.1 V min λ λ λ z

27 Example Ex Reverse problem VSWR Vmax 8.4 = = = 4.0 Vmin 2.1 VSWR Γ = = = 0.6 VSWR To determine ( Γ ) or Z also need position of V min 2π Γ = φ = π 2kz = π 2 (.092λ) = 1.99 m λ 1+ Γ Z = Z = 17.4 j Γ Ω

28 Open Circuit 6-28 Z + jz tan kl 1 Z( l) = Z = Z Z jz tan kl j tan kl O/C Z =+ l inductive X pure imaginary jx = jz cot kl 0 capacitive π 2π 3π kl If the impedance is purely imaginary, Z = R+jX becomes Z = jx, where X is called the reactance.

29 Open Circuit 6-29 O/C Z =+ l X pure imaginary jx = jz cot kl 0 inductive capacitive π kl Note: π λ at kl = l = 2 4 ( λ ) Z = 0 4 near kl (short circuit) π Z is inductive near kl 0 Z is capacitive

30 Smith Chart 6-30 Philip Smith Shows the entire universe of complex impedances in one convenient circle. Invented at Bell abs by Philip Smith in By 1975 about 9 million copies of his chart sold to microwave engineers all over the world. Its usefulness continues to this day as a method of displaying measured and calculated data produced by computer software and modern measurement instruments.

31 Smith Chart 6-31 Im( Γ) Re( Γ)

32 Z Chart Constant R Circles 6-32 on Chart Rn=0 Rn=0.5 Rn=1 Rn=2 Rn=50 Rn: normalized resistance Z = R + jx n n n [ Ω] impedance = resistance + j reactance

33 Z Chart Xn=1 Xn=2 Constant X Arcs 6-33 on Chart Xn=0.3 Xn=0 Xn>0 Xn=-0.3 Xn=-2 Xn<0 Xn=-1 Z = R + jx n n n [ Ω] impedance = resistance + j reactance

34 Z n = R + n jx n [ Ω] 6-34 Z n = 0.2 j0.3 [ Ω] Rn=0.2 circle Xn=-0.3 arc x

35 Z n = R + n jx n [ Ω] 6-35 Z n = 0.2 j0.3 [ Ω] Using your compass draw the 0.2-j0.3 circle with center at z n =1. 1 X circle + j (NOTE: this circle is not on the chart until you draw it. x

36 6-36 Find the value of reflection coefficient Reflection Coeff PHASE Reflection Coeff magnitude MAGNITUDE

37 Smith Chart Γ = Z = 1+ j2 n 6-37 R=0 circle ( Γ = 1) 1 X circle + j 45 X X X S/C Perfect Match ( Γ = 0) O/C fig 6.13

38 Smith Chart 6-38 jkz + jkz + 2 jkz e +Γ e 1+Γe 0 jkz + jkz jkz e Γe Γe Z( z) = Z = Z 1 movement in negative zˆ direction ( toward generator) clockwise motion on circle of constant Im( Γ) Γ Generator Transmission ine oad Z G V G Z Γ Re( Γ) To generator z V min z = -l z = 0

39 Smith Chart Im( Γ) 6-39 V ( z ) j2kz = 1+ Γ e V + Γ Re( Γ) Note: λ complete circle ( 360 =2 π rad ) = 2 1 Y n( z)= can just replace Γ by (-Γ ) Z ( z) n

40 Smith Chart 6-40 V =1+ Γ max (RH Real axis) V =1- Γ min (H Real axis) V( z) = 1+Γ V + e 2 jkz V( z) V min Γ V max R = VSWR

41 Smith Chart Z = 1+ j2 n Γ= V(z) + V V oad V oad 45 zv ( min ) zv ( max ) z Z n Vmin Γ Vmax V =1+ Γ = =1.707 max V =1- Γ min = =

42 Smith Chart Z = 1+ j2 n Γ= λ 360 on Smith Chart= 2 λ 180 on Smith Chart= 4 V(z) + V V oad λ 1 45 on Smith Chart= = λ 4 zv ( min ) zv ( max ) z V oad Z n 45 λ 16 Vmin Γ Vmax

43 Smith Chart Z = 1+ j2 n Γ= V(z) V V oad 5λ 16 λ z V oad 45 Z n Vmin Γ Vmax

44 Smith Chart 0.25λ λ = λ λ 6-44 Γ = Z = 1+ j2 n 45 X Y =.2 j.4 n V max VSWR=5.8

45 Impedance Matching 6-45 Usually want power to be absorbed by load (minimize ). To do so one adds pure reactances (or susceptances) to tune or match the network. Γ 2 i Z = R + jx [ Ω] mpedance = resistance + j reactance Y = G + jb [S] admittance = conductance + j susceptance Note: It is physically easier to add a shunt susceptance than series reactance.

46 Impedance Matching 6-46 Example: Given Z = 2 + j2 Γ = n 2 2 Γ = = 38% power reflected change from Z Y =G + jb n n n n 1 1 Yn = = = 0.25 j0.25 Z 2+ j2 n could add + j.25 at load Y = 0.25 Γ=0.6 0 n 2 2 Γ = 0.6 = 36% power reflected instead rotate toward generator to 1+ jb circle and add a -jb there.

47 Smith Chart Solution: Add + j1.57 at λ Or - j1. 57 at 0.219λ λ λ = 0.219λ λ λ = 0.363λ 0.178λ j1.57 Z = 2 + j2 n 0.041λ 1 j1.57 Y = 0.25 j0. 25 n 0.322λ

48 Smith Chart Solution: Add + j1.57 at λ Or - j1. 57 at 0.219λ λ λ = 0.219λ λ λ = 0.363λ 0.178λ j λ Z = 2 + j2 n 0.041λ 0.363λ Y = 0.25 j0. 25 n 1 j λ

49 Smith Chart Applet

50 Standing Wave Pattern 6-50 Voltages: Unmatched ine Z 0.219λ 0 Vload V V V max min = V ( 0) = 1.62 = 0.38 = ( 0.219λ ) 1.55 z = Z n 0.219λ 0.042λ

51 Standing Wave Pattern 6-51 Currents: Unmatched ine Z 0.219λ 0 z 1.45 Z n λ

52 Standing Wave Pattern 6-52 Voltages: Matched ine -jb Z 0.219λ 0 V load V V V max min = 1.55 = 1.62 = 0.38 ( 0.219λ ) z = Z n 0.219λ

53 Standing Wave Pattern UNMATCHED z 0 Z V I λ 0.219λ 0.042λ z 6-53 fig 6.19a 1+ Γ V V V V I 1- Γ I I I load max min ( 0.219λ ) Γ load max min = 1.55 = 1.62 = 0.38 = 0.55 = 1.55 = 0.38 ( 0.219λ ) = 0.78 = 1.45 MATCHED I fig 6.19b 1+ Γ -jb Z V Γ Γ 0.219λ 0 z 0.219λ 0.042λ z

54 How to Add Shunt Susceptance 6-54 For a transmission line usually add a short circuit section of a line placed perpendicular to the main line. l Y n (l) S/C Y n (l)= -j cotkl 6.36 for kl varying from 0 to π λ 0 l 2 l Z all possible values -j jb + j S/C d fig 6.20

55 How to Add Shunt Susceptance 6-55 For a rectangular waveguide usually insert a metal iris inside the waveguide. d b d a Inductive Iris -λg 2 π d B= 0 cot a 2a 6.37 Capacitive Iris 4b π d B 0= ln csc λ 2b g 6.38

56 S/C Stub Smith Chart Solution 6-56 Rotate clockwise from S/C to desired jb Im( Γ) 0+j0.5 0+j1 0+j2 0+j0 Re( Γ) S/C Y = 0± j n 0-j0.5 0-j2 0-j1

57 S/C Stub Smith Chart Solution 6-57 Example: To add jb = - j1. 57 analytically Yn = j cot kl j1.57 = j cot kl 1 cot kl = 1.57 ; tan kl = 1.57 O/C 0 j1.57 S/C 0.09λ kl 2π = l = λ [ radian s] l = λ

58 S/C Stub Smith Chart Solution 6-58 Example: Given Z =0.5-j2 find everything n Smith chart Analytically Γ = Γ = Y = j0.47 n Y = j n SWR = 10.5 SWR = d = λ B = 2.9 l = = λ B = λ l = λ d Γ d 2 = 0.227λ B l 2 2 = = 0.447λ Γ Γ λ 0.179λ z λ λ

59 Smith Chart d2 = 0.227λ d1 = 0.131λ λ Y = j0. 47 n 0.202λ 1+ j2.9 1 j2.9 Γ= Z = 0.5 j2 n 0.298λ fig 6.18

60 Smith Chart 6-60 l l 1 1 = ( ) λ = 0.053λ Note: l 0.197λ + l = 0.5λ 1 2 j2.9 S/C j2.9 l 1 l 2 l l 2 2 = ( ) λ = 0.447λ 0.303λ