Transmission Line Theory

Transcription

1 S. R. Zinka School of Electronics Engineering Vellore Institute of Technology April 26, 2013

2 Outline 1 Free Space as a TX Line 2 TX Line Connected to a Load 3 Some Special Cases 4 Smith Chart 5 Impedance Matching 6 Z 0 &β 7 Problems

3 Outline 1 Free Space as a TX Line 2 TX Line Connected to a Load 3 Some Special Cases 4 Smith Chart 5 Impedance Matching 6 Z 0 &β 7 Problems

4 First of All, You Know How to Represent a Wave Right! The wave shown in the above diagram can be represented as F (x, t) = sin (βx βvt) = sin (βx ωt) (1) where, ω = 2πf = βv. (2)

5 Now, What is Electrical Length? The wave shown in the above diagram can be represented as λ 2π l?? (3) Electrical length θ is given as θ = ( ) 2π l = βl. (4) λ

6 Remember these Equations from Wave Propagation in an Isotropic Medium? v = ω β ε s = ε ( 1 j ωε σ ) ; tan θlt = ωε σ γ 2 = γx 2 + γy 2 + γz 2 = ω 2 µε µε γ = α + jβ = ω 2 [ 1 + ( ] ) 2 σωε µε 1 + jω 2 [ 1 + ( ] ) 2 σωε + 1 P = E H η = µ εs = jωµ σ+jωε

7 Reflection of Plane Wave at Normal Incidence Electric fields on both sides are given as, Ei = E i e γ z,1 z ˆx, Et = E t e γ z,1 z ˆx, and Er = E r e γ z,1 z ˆx. Similarly, magnetic fields on both sides are given as, Hi = H i e γ z,1 z ŷ = E i η 1 e γ z,1 z ŷ, x Ht = H t e γ z,2 z ŷ = E t η 2 e γ z,2 z ŷ, and Hr = H r e γ z,1 z ˆx = E r η 1 e γ z,1 z ŷ. From the boundary conditions (at z = 0), E i + E r = E t, and E i η 1 E r η 1 = E t η 2. (5) From (5), Γ = E r E i = η 2 η 1 η 2 + η 1, (6) τ = E t E i = 2η 2 η 2 + η 1, and (7) 1 + Γ = τ. (8)

8 Let s Just Use a Different Notation Electric fields on both sides are given as, Ei = V + 1 e γ z,1 z ˆx, Et = V 2 e γ z,2 z ˆx, and Er = V 1 eγ z,1 z ˆx. Similarly, magnetic fields on both sides are given as, Hi = I + 1 e γ z,1 z ˆx = V+ 1 Z 1 e γ z,1 z ŷ, 0 x Ht = I 2 e γ z,2 z ˆx = V 2 Z 2 e γ z,2 z ŷ, and 0 Hr = I 1 eγ z,1 z ˆx = V 1 Z 1 e γ z,1 z ŷ. 0 From the boundary conditions (at z = 0), V V 1 = V 2, and V + 1 Z 1 0 V 1 Z 1 0 = V 2 Z 2. (9) 0 From (9), Γ = V 1 V + 1 τ = V 2 V 1 = Z2 0 Z1 0 Z 2 0 +, (10) Z1 0 = 2Z2 0 Z 2 0 +, and (11) Z Γ = τ. (12)

9 Analogy LHS: V tot (z < 0) = V + 1 e γ z,1 z + V 1 eγ z,1 z I tot (z < 0) = I + 1 e γ z,1 z + I 1 eγ z,1 z x = V+ 1 Z 0,1 e γ z,1 z V 1 Z 0,1 e γ z,1 z RHS: V tot (z > 0) = V 2 e γ z,2 z + V + 2 eγ z,2 z + + I tot (z > 0) = I 2 e γ z,2 + I + 2 eγ z,2 z No Power Miss-match - - = V 2 e γ z,2 z V + 2 e γ z,2 z Z 0,2 Z 0,2 }{{} this term is zero here here

10 Outline 1 Free Space as a TX Line 2 TX Line Connected to a Load 3 Some Special Cases 4 Smith Chart 5 Impedance Matching 6 Z 0 &β 7 Problems

11 TX Line Connected to a Load + - V (z) = I (z) = V + e }{{ γz } + V e }{{ γz } incident wave reflected wave I + e }{{ γz } + I e }{{ γz } incident wave reflected wave = V+ Z 0 e γz V Z 0 e γz (13)

12 Reflection Coefficient (Γ/Γ 0 ) - Definition + - Γ (z) = voltage of the reflected wave voltage of the incident wave = V e γz V + e ( γz V ) = e 2γz V + = Γ 0 e 2γz (14) Γ 0 = Γ (z = 0) = V V + = I I +. (15)

13 (Γ/Γ 0 ) - Derivation + - From (13) V (z = 0) I (z = 0) = Z L = V+ + V ( V + Z 0 = Z 0 By re-arranging the above equation, one gets ) V Z V V + 1 V V + Γ 0 = Z L Z 0 Z L + Z 0 Γ (z) = = Z 0 ( 1 + Γ0 1 Γ 0 ( ZL Z 0 Z L + Z 0 ). ) e 2γz (16)

14 Input Impedance (Z in ) - Definition + - { V (z) = V + e γz + V e γz I (z) = V+ Z 0 e γz V Z 0 e γz Z in = V (z) I (z) ( V = + e γz + V e γz ) Z 0 V + e γz V e γz ( 1 + Γ0 e = 2γz ) Z 0 1 Γ 0 e 2γz

15 Z in - Lossless Case - Derivation ( 1 + Γ0 e Z in = 2γz ) Z 0 1 Γ 0 e 2γz ( ) = Z ZL Z 0 Z L +Z e j2βz ( 0 ) ZL, since γ = jβ, for lossless case Z 1 0 Z L +Z e j2βz 0 [ ZL + Z 0 + (Z L Z 0) e = j2βz ] Z 0 Z L + Z 0 (Z L Z 0) e j2βz [ ZL + Z 0 + Z L e = j2βz Z 0 e j2βz ] Z 0 Z L + Z 0 Z L e j2βz + Z 0 e j2βz [ ( ) ( ZL 1 + e j2βz + Z ) ] 0 1 e j2βz = Z 0 Z L (1 e j2βl ) + Z 0 (1 + e j2βl ) ( ) = Z 0 Z 1 e L + Z j2βz 0 1+e ( j2βz ) 1 e Z 0 + Z j2βz L 1+e [ j2βz ] ZL jz 0 tan βz = Z 0 Z 0 jz L tan βz

16 Voltage Standing Wave Ratio (VSWR) - Definition VSWR is defined as V max V min. Since voltage along a TX line is V (z) = V + e jβz + V e jβz, it can be derived from the Schwartz inequality principle that V (z) = V + e jβz + V e jβz V + e jβz V + e jβz V + + V = V max. Similarly, it can also be derived that V (z) = V + e jβz + V e jβz V + e jβz V e jβz V + V = V min.

17 Voltage Standing Wave Ratio (VSWR) - Derivation So, VSWR is given as VSWR = V max V min = V+ + V V + V = 1 + V V + 1 V V + = 1 + Γ 0 1 Γ 0.

18 Instantaneous & Time Average Power Instantaneous power corresponding to the above set of voltage & current is defined as P inst (t) = v 0 i 0 cos (ωt + φ 1) cos (ωt + φ 2) Time average power is defined as = v 0i 0 2 [cos (ωt + φ 1 + ωt + φ 2) + cos (ωt + φ 1 ωt φ 2)] = v 0i 0 2 [cos (2ωt + φ 1 + φ 2) + cos (φ 1 φ 2)] (17) P avg = 1 T 0 ˆ T0 0 P inst dt = v 0i 0 2 cos (φ 1 φ 2) (18)

19 Time Average Power - Complex Notation v real = v 0 cos (ωt + φ 1) v complex = V = v 0 e j(ωt+φ 1) i real = i 0 cos (ωt + φ 2) i complex = I = i 0 e j(ωt+φ 2) P avg = v 0 i 0 2 cos (φ 1 φ 2) P avg = 1 2 Re (VI ) = v 0 i 0 2 cos (φ 1 φ 2) In the above, does the equation 1 2 Re (VI ) remind you of some thing?... Isn t it very similar to the ( ) complex Poynting vector 1 2 Re E H that you study in EMT course?!

20 Power Transfer along a TX Line - Definitions + - V (z) = I (z) = V + e }{{ jβz } + V e }{{ jβz } incident wave reflected wave I + e }{{ jβz + I } e jβz = V+ e }{{} jβz V e jβz Z 0 Z 0 incident wave reflected wave From the above set of equations, P total = 1 2 Re [V (z)] [I (z)] = 1 2 Re P incident = 1 2 Re ( V + e jβz) ( I + e jβz) = 1 P reflected = 1 2 Re ( V e jβz) ( I e jβz) = 1 V + 2 Z 0 V 2 Z 0 2 Re ( V +) ( I +) 1 = 2 2 Re ( V ) ( I ) 1 = 2 V+ V e j2βz + V+ V e j2βz Z 0 Z 0 }{{} this term is imaginary V + 2 Z 0 V 2 Z 0. (19)

21 Return Loss - Definition + - V (z) = I (z) = V + e }{{ jβz } + V e }{{ jβz } incident wave reflected wave I + e }{{ jβz + I } e jβz = V+ e }{{} jβz V e jβz Z 0 Z 0 incident wave reflected wave So, from (19) [ ] P total = 1 2 Re [V (z)] [I (z)] = 1 2 Re V + 2 V 2 Z 0 Z 0 = P incident P reflected (20) From the above equation, return loss is defined as RL = P incident P reflected = V+ 2 V 2 = 1 Γ 0 2 = 1 Γ (z) 2 = 20 log (Γ 0) in db (21)

22 Important Formulas - Summary Γ (z = l) = Γ 0 e j2βl, where Γ 0 = Z L Z 0 Z L +Z 0 ( ) 1+Γ Z in (z = l) = Z 0 e j2βl [ ] ZL +jz 0 = Z 0 tan βl 0 Z 0 +jz L tan βl VSWR = 1+ Γ 0 1 Γ 0 1 Γ 0 e j2βl P total = 1 2 Re [V (z)] [I (z)] = 1 2 Re [ V + 2 Z 0 V 2 Z 0 ] = P incident P reflected RL = P incident P = V+ 2 reflected V 2 = 1 Γ 0 2 = 1 Γ(z) 2 = 20 log (Γ 0) in db

23 Outline 1 Free Space as a TX Line 2 TX Line Connected to a Load 3 Some Special Cases 4 Smith Chart 5 Impedance Matching 6 Z 0 &β 7 Problems

24 Short Circuit & Open Circuit [ ] ZL + jz 0 tan βl Z in (z = l) = Z 0 Z 0 + jz L tan βl From the above formula and L Hospital s rule, one gets Z SC in = jz 0 tan βl and Also, one can notice from the above equations that Z OC in = jz 0 cot βl. (22) Z SC in Z OC in = Z 2 0.

25 Quarter-wave (λ/4) Transformer + - [ ] ZL + jz 0 tan βl Z in (z = l) = Z 0 Z 0 + jz L tan βl From the above formula and L Hospital s rule, when l = λ/4, Z λ/4 in = Z2 0 Z L. (23) In other words, Z λ/4 in Z L = Z 2 0.

26 Outline 1 Free Space as a TX Line 2 TX Line Connected to a Load 3 Some Special Cases 4 Smith Chart 5 Impedance Matching 6 Z 0 &β 7 Problems

27 This is how a Smith Chart Looks Like...

28 Would you Like to See a More Complicated Smith Chart?...

29 Impedance Circles Z in Z 0 = r + jx = 1 + Γ 1 Γ r = 1 Γ2 r Γ 2 i (1 Γ r) 2 + Γ 2 i 2Γ x = i (1 Γ r) 2 + Γ 2 i [ Γ r [Γ r 1] 2 + r 1 + r ] 2 [ 1 + Γ 2 i = 1 + r [ Γ i 1 x ] 2 = [ 1 x ] 2 ] 2

30 Impedance Circles Constant r Circles Constant x Circles

31 Admittance Circles Impedance Circles Plot Y in Y 0 = g + jb = 1 Γ 1 + Γ g = 1 Γ2 r Γ 2 i (1 + Γ r) 2 + Γ 2 i 2Γ b = i (1 + Γ r) 2 + Γ 2 i [ Γ r + [Γ r + 1] 2 + g 1 + g ] 2 + Γ 2 i = [ 1 [ Γ i + 1 b ] 2 = 1 + g [ 1 b ] 2 ] 2

32 Admittance Circles Constant g Circles Constant b Circles

33 Locating a given Load on the Smith Chart Z L = j100ω, Z 0 = 50Ω

34 Moving towards the Generator using the Constant VSWR Circle Γ (z = l) = Γ 0 e j2βl

35 Adding a Series Inductor

36 Adding a Series Capacitor

37 Adding a Parallel Inductor

38 Adding a Parallel Capacitor

39 Outline 1 Free Space as a TX Line 2 TX Line Connected to a Load 3 Some Special Cases 4 Smith Chart 5 Impedance Matching 6 Z 0 &β 7 Problems

40 For this section, see the PDF file uploaded separately...

41 Outline 1 Free Space as a TX Line 2 TX Line Connected to a Load 3 Some Special Cases 4 Smith Chart 5 Impedance Matching 6 Z 0 &β 7 Problems

42 Telegraph Equations From the above figure and Kirchhoff s circuital laws: V (z + z) V (z) = Z s I (z) z I (z + z) I (z) = Y p V (z + z) z (24) As z 0, the above equations become V (z + z) V (z) V (z, t) lim = = Z s I (z) z 0 z z lim z 0 I (z + z) I (z) z = I (z, t) z = Y p V (z + z) (25)

43 Telegraph Equations If places of Z s and Y p and interchanged along the z-direction as shown in the above figure, then using Kirchoff s circuital laws gives V (z + z) V (z) V (z, t) lim = = Z s I (z + z) z 0 z z lim z 0 I (z + z) I (z) z = I (z, t) z = Y p V (z) (26) In either case, z 0; so I (z + z) I (z) and V (z + z) V (z)... So, the equations (25) and (26) reduce to well known telegraph equations shown below: V (z) = Z s I (z) z (27) I (z) = Y p V (z) z (28)

44 Wave Equation Substituting (27) and (28) into each other gives d 2 V (z) dz 2 = Z s Y p V (z) = γ 2 V (z) d 2 I (z) dz 2 = Z s Y p I (z) = γ 2 I (z) (29) where γ = Z s Y p. Solving the above second order homogeneous linear differential equation gives V (z) = V + e γz + V e +γz I (z) = I + e γz + I e +γz. (30) In general, Z S = R + jωl and Y P = G + jωc, where R and G indicate conductor and dielectric losses, respectively. So, γ = (R + jωl) (G + jωc) (31)

45 Characteristic Impedance Z 0 Substituting (30) in (28) gives I (z) z = (I+ e γz + I e +γz ) z = γi + e γz + γi e +γz = Y p V (z) = Y p ( V + e γz + V e +γz) γi + = Y p V + V + I + = γ Y P = Zs Yp = R+jωL G+jωC = Z 0 γi = Y p V V I = γ Y P = Zs Yp = R+jωL G+jωC = Z 0

46 Special Cases (Lossless & Almost lossless) lossless: For lossless case, R = 0 and G = 0. So, for lossless case, { L Z 0 = C and γ = 0 + jω LC. Almost-lossless: For almost lossless case, R jωl and G jωc. In such case, Z 0 = R + jωl G + jωc = ( jωl jωc and ) R jωl + 1 L G jωc + 1 C ( ) ( ) R G γ = (R + jωl) (G + jωc) = jωl jωc jωl + 1 jωc + 1 = jω ( ) LC 1 + R jωl + G jωc RG ω 2 jω ( LC 1 + R LC jωl + jω [ LC ( R 2 jωl + G )] = 1 ( ) R + GZ 0 jωc 2 Z 0 } {{ } α +j ω LC }{{} β ) G jωc

47 Special Cases (Distortion-less) Distortion-less: For distortion-less case, R L = G C. In such case, Z 0 = R + jωl G + jωc = ( jωl jωc and ) R jωl + 1 L G jωc + 1 = C ( ) ( ) R G γ = (R + jωl) (G + jωc) = jωl jωc jωl + 1 jωc + 1 = jω ( ) LC 1 + 2R jωl R2 = ω 2 L 2 jω ( LC 1 + R ) 2 jωl = jω ( LC 1 + R jωl ) = R Z 0 }{{} α +j ω LC }{{} β In the above equation, one can see that β ω, which is the condition for distortion-less transmission.

48 Wave Velocity along a TX Line For all the cases we considered till now (i.e., loss-less, lossy, distortion-less, etc.) β = ω LC. So, velocity of a wave along the TX line is given as v = ω β = 1 LC Don t you think this equation is very similar to c = 1 µε?

49 Outline 1 Free Space as a TX Line 2 TX Line Connected to a Load 3 Some Special Cases 4 Smith Chart 5 Impedance Matching 6 Z 0 &β 7 Problems

50 Transmission Lines - Basics

51 Smith Chart - Problems

52 Transmission Line - Lumped Element Model