Fields and Waves I. Lecture 3

Transcription

1 Fields and Waves I ecture 3 Input Impedance n Transmissin ines K. A. Cnnr Electrical, Cmputer, and Systems Engineering Department Rensselaer Plytechnic Institute, Try, NY

2 These Slides Were Prepared by Prf. Kenneth A. Cnnr Using Original Materials Written Mstly by the Fllwing: Kenneth A. Cnnr ECSE Department, Rensselaer Plytechnic Institute, Try, NY J. Darryl Michael GE Glbal Research Center, Niskayuna, NY Thmas P. Crwley Natinal Institute f Standards and Technlgy, Bulder, CO Sheppard J. Saln ECSE Department, Rensselaer Plytechnic Institute, Try, NY ale Ergene ITU Infrmatics Institute, Istanbul, Turkey Jeffrey Braunstein Chung-Ang University, Seul, Krea Materials frm ther surces are referenced where they are used. Thse listed as Ulaby are figures frm Ulaby s textbk. 10 September 2006 Fields and Waves I 2

3 10 September 2006 Fields and Waves I 3

4 Overview Review Henry Farny Sng f the Talking Wire ltages and Currents n Transmissin ines Standing Waves Input Impedance ssy Transmissin ines w ss Transmissin ines 10 September 2006 Fields and Waves I 4

5 What d we knw s far? Slutins lk like ω 2π β ω lc ω με u λ 2π ω 2πf T 2π u λ β f ( m β ) Acs ωt z u 1 1 lc με ε ε r ε μ μ μ r Figure frm 10 September 2006 Fields and Waves I 5

6 Phase elcity Phase cnst A simple way t find the phase velcity Identify sme feature f the sine wave. Fr this we chse cnstant phase ωt m βz cnst Determine it s velcity. Since the phase is a cnstant, we knw that ( ) ω t m t β z 0 ω β z m z ω u β t t 0 10 September 2006 Fields and Waves I 6

7 Phasr Ntatin Fr ease f analysis (changes secnd rder partial differential equatin int a secnd rder rdinary differential equatin), we use phasr ntatin. ( m ) { m jβz} The term in the brackets is the phasr. ( jωt) f (,) z t Acsωt βz Re Ae e f () z Ae m j β z 10 September 2006 Fields and Waves I 7

8 Phasr Ntatin T cnvert t space-time frm frm the phasr frm, multiply by and take the real part. e j t ω m j β z jωt ( ω m β ) f (,) z t Re( Ae e ) Acs t z If A is cmplex A Ae j θ A ( ω m β θ ) jθa m jβz jωt f (,) z t Re( Ae e e ) Acs t z A 10 September 2006 Fields and Waves I 8

9 Wrkspace What is the phasr f the time derivative? ([ jβz] jωt) vzt (,) Re e e ([ β ] ω ) t vzt j z j t (,) Re e e? t 10 September 2006 Fields and Waves I 9

10 Transmissin ines Incident Wave Mismatched lad Reflected Wave Standing wave due t interference 10 September 2006 Fields and Waves I 10

11 10 September 2006 Fields and Waves I 11 Transmissin ines Phasr ltage Slutin Phasr Frm f the Wave Equatin: c l z t c l z ω where: z j e ± β m General Slutin: z j z j e e β β

12 Transmissin ines - Standing Wave Derivatin e jβ z e jβ z Frward Wave cs( ω t β z ) TIME DOMAIN Backward Wave cs( ω t β z ) max ccurs when Frward and Backward Waves are in Phase CONSTRUCTIE INTERFERENCE min ccurs when Frward and Backward Waves are ut f Phase DESTRUCTIE INTERFERENCE 10 September 2006 Fields and Waves I 12

13 Transmissin ine ltages and Currents General Slutin vz e j β () z e jβz iz () e e jβz jβz e jβz e jβz The latter expressin is derived frm vz () jωli() z z Can we easily explain the minus sign? 10 September 2006 Fields and Waves I 13

14 Wrkspace 10 September 2006 Fields and Waves I 14

15 Reflectin Cefficient Derivatin Define the Reflectin Cefficient: m Γ m Maximum Amplitude when in Phase: max m m max m (1 Γ ) Similarly: (1 Γ ) min m Standing Wave Rati (SWR) max min 1 1 Γ Γ 10 September 2006 Fields and Waves I 15

16 Anther General Frm fr the Slutin Using the reflectin cefficient Nte that we will be rewriting the slutin in different frms v z e j β () z Γ e jβz iz e jβz () Γ e jβz 10 September 2006 Fields and Waves I 16

17 Transmissin ines Bth frm Ulaby 10 September 2006 Fields and Waves I 17

18 10 September 2006 Fields and Waves I 18 Reflectin Cefficient et z0 at the OAD ) (1 0 0 j j lad e e Γ β β ( ) I lad Γ Γ 1

19 Reflectin Cefficient This is a key relatinship At the lad I lad lad ( 1 Γ ) ( 1 Γ ) Γ 10 September 2006 Fields and Waves I 19

20 Wrkspace in? ut? 10 September 2006 Fields and Waves I 20

21 Wrkspace 10 September 2006 Fields and Waves I 21

22 Wrkspace Shrt Circuit ad 10 September 2006 Fields and Waves I 22

23 Wrkspace Open Circuit ad 10 September 2006 Fields and Waves I 23

24 Shrt Circuit ad Fr 0, we have Γ ( ) j z v() z e Γ e e e β jβz jβz jβz jβz e csβz jsin βz jβz e csβz jsin βz ( ) vz () j2sinβz 10 September 2006 Fields and Waves I 24

25 Shrt Circuit ad Cnvert t space-time frm ( j ω t ) ( ( ) jωt 2 β ) vzt (,) Re vze () Re j sin ze (( ) jωt j2 βz e ) 2 βz( j βz βz) This is a standing wave ( ) Re sin Re sin cs sin vzt (,) sin zsin t 2 β ω 10 September 2006 Fields and Waves I 25

26 Shrt Circuit ad What are the vltage maxima and minima? vzt (,) sin zsin t 2 β ω Where are they? The standing wave pattern is the envelpe f this functin. 10 September 2006 Fields and Waves I 26

27 Reflectin 50 Ω 80 m 2 pp 1.5 MHz ~ RG Ω Nte that we are free t chse either the lad end r the surce end as z0 z0 z80 z'80 z'0 10 September 2006 Fields and Waves I 27

28 Reflectin Cefficient Γ Determine the reflectin cefficient at the lad, and the standing wave rati, SWR. Start with a shrt circuit lad and then cnsider a 25 Ohm lad. Then d an pen circuit and 93 Ohm lad. Assume that the frward traveling wave has an amplitude f 10 lts. Sketch the standing wave pattern fr vltage and current fr the shrt circuit lad. Include numbers fr amplitudes and distances. Under what cnditins d yu get a vltage maximum at the lad? a minimum? Can yu answer this in general? If the lad is a 3.3 nf capacitr, what is the reflectin cefficient at the lad? Where is the lcatin f the first minimum? T answer this, we need a bit mre develpment. 10 September 2006 Fields and Waves I 28

29 First Sketch the Standing Wave Patterns by Hand The reflectin cefficient 0 Γ Γ Γ Γ September 2006 Fields and Waves I 29

30 Using Matlab fr the ltage Standing Wave Patterns 10 September 2006 Fields and Waves I 30

31 10 September 2006 Fields and Waves I 31

32 Current Standing Wave Patterns Can we use what we just displayed t find the current standing wave patterns? Yes, because the reflectin cefficient fr current is always just the negative f the vltage reflectin cefficient 10 September 2006 Fields and Waves I 32

33 Standing Wave Pattern We have just seen that: Minimum ccurs at OAD fr 0 Is it als true that: Or, in general, that: Maximum ccurs at OAD fr Γ 0 > 0 < 0 > < Γ 0 Max at OAD Min at OAD IF is REA 10 September 2006 Fields and Waves I 33

34 10 September 2006 Fields and Waves I 34 Standing Wave Pattern If z at OAD and z0 at SOURCE, ) z ( 2 j e ) z ( β Γ Γ ) ( 2 z j j e e Γ Γ β θ When Phase π, the FIRST MINIMUM ccurs ( ) π β θ Γ z 2 ( ) λ π θ λ Γ 4 4 z Other MINs are displaced by λ/2 Phase f the reflectin cefficient

35 A Repeat f HW2 Experiment (600kHz) What did yu see at the 20 ndes? Time Delay Amplitude Did any f yu try an pen r shrt circuit? 10 September 2006 Fields and Waves I 35

36 Change the Frequency t 1.5MHz 10 September 2006 Fields and Waves I 36

37 Cntinuing 10 September 2006 Fields and Waves I 37

38 Capacitive ad? 10 September 2006 Fields and Waves I 38

39 Capacitive ad Nw we can answer the questin abut where the first minimum is lcated fr a capacitive lad. Γ 1 jωc 1 jωc 1 1 jωc jωc 1exp tan 1 ωc 1 ( ωc ) 2 a a jb jb a a jb jb a a jb jb a j2abb 2 2 a b 2 2 ( a b ) exp tan a b 2 2 2ab a b 2 2 exp tan 1 2ab a b September 2006 Fields and Waves I 39

40 Capacitive ad Fr the specific case here Γ 1 jωc 1 jωc j j j091. 1e j π λ z Γ θπ λ λ. 6362π λ λ. 1591λ 0. 09λ π 4 10 September 2006 Fields and Waves I 40

41 Standing Wave Pattern fr Capacitive ad Reflectin Cefficient Γ 1e j π ad End 10 September 2006 Fields and Waves I 41

42 PSpice Input Impedance Fr the same surce and line, but different lad: R1 50 OFF 0 AMP 10 FREQ 1meg 1 T1 R R2 50 OFF 0 AMP 10 FREQ 1meg 2 T2 R September 2006 Fields and Waves I 42

43 Changing the ad The vltages and currents at the input change the input impedance changes SE>> (T2:A) (T2:B) us 10.5us 11.0us 11.5us 12.0us 12.5us 13.0us (T1:A) (T1:B) Time 10 September 2006 Fields and Waves I 43

44 Changing the ength and ine Prperties Frm the standing wave patterns r the expressins fr the vltages and the currents n the line, we can see that the rati f the vltage t the current will depend n the length f the line and the line prperties. 10 September 2006 Fields and Waves I 44

45 Wrkspace 10 September 2006 Fields and Waves I 45

46 Input Impedance What des in in, lk like? I in Define: When is cmplex, s is. T address the input impedance, we need t generalize the reflectin cefficient. Γ( z) e e jβ z jβ z Γ e j2β z Γ j 2 z e β if z 0 at OAD 10 September 2006 Fields and Waves I 46

47 Anther Frm fr the General Slutin Using the Generalized Reflectin Cefficient ( 1 Γ ) vz () e jβz () z iz e jβz () () ( 1 Γ z ) 10 September 2006 Fields and Waves I 47

48 10 September 2006 Fields and Waves I 48 Input Impedance Previusly, we have seen: )) ( (1 ) ( ) ( ) ( ˆ z e z z z z j Γ β What abut I? )) ( (1 ) ( ) ( ) ( ˆ z e z z z I z j Γ β Als, ) z ( j 2 e ) z ( β Γ Γ

49 Input Impedance Frm the Rati (the generalized impedance): ˆ( z) Iˆ( z) 1 Γ( z) 1 Γ( z) ( z) We are primarily interested in z0 value treat cnnectin t rest f circuit as 2 prt with, in ( z 0) 1 Γ( z 0) 1 Γ( z 0) 10 September 2006 Fields and Waves I 49

50 10 September 2006 Fields and Waves I 50 Input Impedance After lts f algebra, ne can shw: ) tan( ) tan( 0) ( j j z in β β Special Case example: 0 (shrt circuit) ) tan( ) tan( 0 ) tan( 0 0) ( j j j z in β β β

51 Input Impedance - SHORT CIRCUIT in ( z 0) j tan( β ) Can change in by changing these tw parameters Fix β, vary - different effects ary β, fix - get same effects Nte that is the length f the Transmissin ine 10 September 2006 Fields and Waves I 51

52 Input Impedance Shrt Circuit Fr varying frequency, the input impedance is imaginary and can achieve any value. 10 September 2006 Fields and Waves I 52

53 Input Impedance - T Cnsider sme ther cases 80 Ω l 80 m 1 Peak ~ in RG September 2006 Fields and Waves I 53

54 Input Impedance - T Open Circuit Case small j j tan( β ) tan( β ) small 93Ω - lts f cmplex algebra, but straight frward 10 September 2006 Fields and Waves I 54

55 Using the Input Impedance We knw in (z0) - treat as 2-PORT 80 Ω 1 in in ltage Divider Pwer 1 2 Re { I } in in 1 2 in Re in in 1 2 Re 2 in in 10 September 2006 Fields and Waves I 55

56 10 September 2006 Fields and Waves I 56

57 Using the Input Impedance In a ssless Transmissin ine, P in flws int the Transmissin ine and it is dissipated at the 2 OAD What is the vltage at the lad? ˆ ( z) e jβ z (1 Γ( z)) ˆ jβ z ( z 0) e (1 Γ( z in P in Γ(0) in 0)) Can then plug back and get the full phasr expressin 10 September 2006 Fields and Waves I 57

58 Using the Input Impedance The full frm f the vltage z e j β () z e jβz All infrmatin is nw available t determine the vltage and current everywhere n the line. Yu will be ding this n the prject. 10 September 2006 Fields and Waves I 58

59 Special Cases Recall that the standing wave pattern repeated every half wavelength. Thus, we expect that this will als happen fr in. First, cnsider the trivial case f 0. j tan β in j tan β Nw let the line be a half wavelength lng 2π λ tan β tan tan( π) λ 2 0 in September 2006 Fields and Waves I 59

60 Special Cases Thus, fr a line that is exactly an integer number f half wavelengths lng in Thus, if yu have a transmissin line with the wrng characteristic impedance, yu can match the lad t the surce by selecting a length equal t a half wavelength. 10 September 2006 Fields and Waves I 60

61 Special Cases If the line is an dd multiple f a quarter wavelength, we als get an interesting result. j tan β in j tan β j tan β j tan β 2π λ π tan β tan tan λ 4 2 Thus, such a transmissin line wrks like an impedance transfrmer and has a real input impedance September 2006 Fields and Waves I 61

62 10 September 2006 Fields and Waves I 62 Tday s Majr Result Input Impedance ) tan( ) tan( 0) ( j j z in β β