1. 16.360 Lecture 2 Transmission lines 1.Transmission line parameters, equations 2.Wave propagations 3.Lossless line, standing wave and reflection coefficient 4.Input impedence 5.Special cases of lossless line 6.Power flow 7.Smith chart 8.Impedence matching 9.Transients on transmission lines

2. 16.360 Lecture 2 1.Transmission line parameters, equations Vg(t) V BB’ (t) V AA’ (t) A A’ B’ B L V AA’ (t) = Vg(t) = V0cos(  t), V BB’ (t) = V AA’ (t-t d ) = V AA’ (t-L/c) = V0cos(  (t-L/c)), V BB’ (t) = V AA’ (t) Low frequency circuits: Approximate result V BB’ (t) = V AA’ (t)

3. 16.360 Lecture 2 1.Transmission line parameters, equations Vg(t) V BB’ (t) V AA’ (t) A A’ B’ B L V BB’ (t) = V AA’ (t-t d ) = V AA’ (t-L/c) = V0cos(  (t-L/c)) = V0cos(  t- 2  L/ ), Recall:  =c, and  = 2  If >>L, V BB’ (t)  V0cos(  t) = V AA’ (t), If <= L, V BB’ (t)  V AA’ (t), the circuit theory has to be replaced.

4. 16.360 Lecture 2 1.Transmission line parameters, equations Vg(t) V BB’ (t) V AA’ (t) A A’ B’ B L  = 2  f  t = 0.06  e. g:  = 1GHz, L = 1cm Time delay  t = L/c = 1cm /3x10 10 cm/s = 30 ps Phase shift V BB’ (t) = V AA’ (t)  = 2  f  t = 0.6   = 10GHz, L = 1cm Time delay  t = L/c = 1cm /3x10 10 cm/s = 30 ps Phase shift V BB’ (t)  V AA’ (t)

5. 16.360 Lecture 2 Transmission line parameters Vg(t) V BB’ (t) V AA’ (t) A A’ B’ B L time delay V BB’ (t) = V AA’ (t-t d ) = V AA’ (t-L/v p ), Reflection: the voltage has to be treat as wave, some bounce back power loss: due to reflection and some other loss mechanism, Dispersion: in material, V p could be different for different wavelength

6. 16.360 Lecture 2 Types of transmission lines Transverse electromagnetic (TEM) transmission lines B E E B a) Coaxial lineb) Two-wire linec) Parallel-plate line d) Strip linee) Microstrip line

7. 16.360 Lecture 2 Types of transmission lines Higher-order transmission lines a) Optical fiber b) Rectangular waveguidec) Coplanar waveguide

8. 16.360 Lecture 2 Lumped-element Model Represent transmission lines as parallel-wire configuration Vg(t) V BB’ (t) V AA’ (t) A A’ B’ B zz zz zz Vg(t) R’  z L’  z G’  z C’  z R’  z L’  z G’  z R’  z C’  z L’  z G’  z C’  z

9. 16.360 Lecture 2 Transmission line equations Represent transmission lines as parallel-wire configuration V(z,t) R’  z L’  z G’  z C’  z V(z+  z,t) V(z,t) = R’  z i (z,t) + L’  z  i (z,t)/  t + V(z+  z,t), (1) i (z,t) i (z+  z,t) i (z,t) = G’  z V(z+  z,t) + C’  z  V(z+  z,t)/  t + i (z+  z,t), (2)

10. 16.360 Lecture 2 Transmission line equations V(z,t) = R’  z i (z,t) + L’  z  i (z,t)/  t + V(z+  z,t), (1) V(z,t) R’  z L’  z G’  z C’  z V(z+  z,t) i (z,t) i (z+  z,t) -V(z+  z,t) + V(z,t) = R’  z i (z,t) + L’  z  i (z,t)/  t -  V(z,t)/  z = R’ i (z,t) + L’  i (z,t)/  t, (3) Rewrite V(z,t) and i (z,t) as phasors, for sinusoidal V(z,t) and i (z,t) : V(z,t) = Re( V(z) e jtjt ), i (z,t) = Re( i (z) e jtjt ),

11. 16.360 Lecture 2 Transmission line equations V(z,t) R’  z L’  z G’  z C’  z V(z+  z,t) i (z,t) i (z+  z,t) Recall: di(t)/dt = Re(d i e jtjt )/dt ),= Re(i jtjt e jj -  V(z,t)/  z = R’ i (z,t) + L’  i (z,t)/  t, (3) - d V(z)/ dz = R’ i (z) + j  L’ i (z), (4)

12. 16.360 Lecture 2 Transmission line equations Represent transmission lines as parallel-wire configuration V(z,t) R’  z L’  z G’  z C’  z V(z+  z,t) V(z,t) = R’  z i (z,t) + L’  z  i (z,t)/  t + V(z+  z,t), (1) i (z,t) i (z+  z,t) i (z,t) = G’  z V(z+  z,t) + C’  z  V(z+  z,t)/  t + i (z+  z,t), (2)

13. 16.360 Lecture 4 Transmission line equations V(z,t) R’  z L’  z G’  z C’  z V(z+  z,t) i (z,t) i (z+  z,t) - i (z+  z,t) + i (z,t) = G’  z V (z +  z,t) + C’  z  V(z +  z,t)/  t -  i(z,t)/  z = G’ V (z,t) + C’  V (z,t)/  t, (5) Rewrite V(z,t) and i (z,t) as phasors, for sinusoidal V(z,t) and i (z,t) : V(z,t) = Re( V(z) e jtjt ), i (z,t) = Re( i (z) e jtjt ), i (z,t) = G’  z V(z+  z,t) + C’  z  V(z+  z,t)/  t + i (z+  z,t), (2)

14. 16.360 Lecture 2 Transmission line equations V(z,t) R’  z L’  z G’  z C’  z V(z+  z,t) i (z,t) i (z+  z,t) Recall: dV(t)/dt = Re(d V e jtjt )/dt ),= Re(V jtjt e jj -  i(z,t)/  z = G’ V (z,t) + C’  V (z,t)/  t, (6) - d i(z)/ dz = G’ V (z) + j  C’ V (z), (7)

15. 16.360 Lecture 2 V(z,t) R’  z L’  z G’  z C’  z V(z+  z,t) i (z,t) i (z+  z,t) - d i(z)/ dz = G’ V(z) + j  C’ V (z), (7) - d V(z)/ dz = R’ i (z) + j  L’ i (z), (4) Telegrapher’s equation in phasor domain Take d /dz on both sides of eq. (4) - d² V(z)/ dz² = R’ d i (z)/dz + j  L’ d i (z)/dz, (8)

16. 16.360 Lecture 2 - d i(z)/ dz = G’ V(z) + j  C’ V (z), (7) - d V(z)/ dz = R’ i (z) + j  L’ i (z), (4) Telegrapher’s equation in phasor domain substitute (7) to (8) d² V(z)/ dz² = ( R’ + j  L’) (G’+ j  C’)V(z), - d² V(z)/ dz² = R’ d i (z)/dz + j  L’ d i (z)/dz, (8) d² V(z)/ dz² - ( R’ + j  L’) (G’+ j  C’)V(z) = 0, (9) or d² V(z)/ dz² -  ² V(z) = 0, (10)  ² = ( R’ + j  L’) (G’+ j  C’), (11)

17. 16.360 Lecture 2 V(z,t) R’  z L’  z G’  z C’  z V(z+  z,t) i (z,t) i (z+  z,t) - d i(z)/ dz = G’ V(z) + j  C’ V (z), (7) - d V(z)/ dz = R’ i (z) + j  L’ i (z), (4) Telegrapher’s equation in phasor domain Take d /dz on both sides of eq. (7) - d² i(z)/ dz² = G’ d V (z)/dz + j  C’ d V (z)/dz, (12)

18. 16.360 Lecture 2 - d i(z)/ dz = G’ V(z) + j  C’ V (z), (7) - d V(z)/ dz = R’ i (z) + j  L’ i (z), (4) Telegrapher’s equation in phasor domain substitute (4) to (12) d² i(z)/ dz² = ( R’ + j  L’) (G’+ j  C’)i(z), d² i(z)/ dz² - ( R’ + j  L’) (G’+ j  C’) i(z) = 0, (9) or d² i(z)/ dz² -  ² i(z) = 0, (13)  ² = ( R’ + j  L’) (G’+ j  C’), (11) - d² i(z)/ dz² = G’ d V (z)/dz + j  C’ d V (z)/dz, (12)

19. 16.360 Lecture 2 Wave equations d² i(z)/ dz² -  ² i(z) = 0, (13) d² V(z)/ dz² -  ² V(z) = 0, (10)  =  + j ,  = Re ( R’ + j  L’) (G’+ j  C’),  = Im ( R’ + j  L’) (G’+ j  C’),

20. 16.360 Lecture 2 Wave equations d² i(z)/ dz² -  ² i(z) = 0, (13) d² V(z)/ dz² -  ² V(z) = 0, (10)  =  + j ,  = Re ( R’ + j  L’) (G’+ j  C’),  = Im ( R’ + j  L’) (G’+ j  C’), V(z) = V 0 (14) Solving the second order differential equation + e -z-z + - V0V0 e zz i(z) = I 0 (15) + e -z-z + - I0I0 e zz

21. 16.360 Lecture 2 Wave equations V(z) = V 0 (14) + e -z-z + - V0V0 e zz i(z) = I 0 (15) + e -z-z + - I0I0 e zz where: + V0V0 - V0V0 andare determined by boundary conditions. + I0I0 - I0I0 andare related to - V0V0 + V0V0 andby characteristic impedance Z 0.

22. 16.360 Lecture 2 recall: - d V(z)/ dz = R’ i (z) + j  L’ i (z), (4) V(z) = V 0 (14) + + - V0V0 e zz i(z) = I 0 (15) + e -z-z + - I0I0 e zz e -z-z e -z-z V0V0 + e zz V0V0 - - = ( R’ + j  L’) i (z), (16) i (z) = ( R’ + j  L’)  e -z-z (V0(V0 + e zz V0V0 - - ) I0I0 + =  V0V0 + I0I0 - = -- V0V0 - (17) (18) Characteristic impedance Z 0

23. 16.360 Lecture 5 Characteristic impedance Z 0 I0I0 + = ( R’ + j  L’)  V0V0 + I0I0 - = -- V0V0 - (17) (18) = ( R’ + j  L’)  + Z 0  Define characteristic impedance Z 0 I0I0 + V0V0 =  = ( R’ + j  L’) (G’+ j  C’) ( R’ + j  L’) (G’+j  C’) recall:

24. 16.360 Lecture 5 Summary:  = ( R’ + j  L’) (G’+ j  C’) + Z 0  I0I0 + V0V0 = ( R’ + j  L’) (G’+j  C’) V(z) = V 0 (14) + + - V0V0 e zz i(z) = I 0 (15) + e -z-z + - I0I0 e zz e -z-z (19) (20)

25. 16.360 Lecture 5 Example, an air line : solution: R’ = 0 , G’ = 0 / , Z 0 = 50 ,  = 20 rad/m, f = 700 MHz L’ = ? and C’ = ? Z0Z0 = ( R’ + j  L’) (G’+j  C’) = j  L’ j  C’ = 50   = ( R’ + j  L’) (G’+ j  C’) = jj L’C’  =  + j ,  =  L’C’ = 20 rad/m

26. 16.360 Lecture 5 lossless transmission line : = ( R’ + j  L’) (G’+ j  C’)  =  + j , If R’<< j  L’ and G’ << j  C’,  = ( R’ + j  L’ ) (G’+ j  C’) = jj L’C’  = 0  =  L’C’ lossless line

27. 16.360 Lecture 2 lossless transmission line :  = 0  =  L’C’ lossless line Z0Z0 = ( R’ + j  L’) (G’+j  C’) = j  L’ j  C’ Z0Z0 = L’ C’  = 2  / = 2  /  =  L’C’ 1 Vp =  /  = L’C’ 1

28. 16.360 Lecture 2 For TEM transmission line : Vp = L’C’ 1 L’C’ =  =  1  =  L’C’ =   = rrrr c Z0Z0 = L’ C’ summary : V(z) = V 0 + + - V0V0 e jzjz i(z) = I 0 + e -j  z + - I0I0 e jzjz e = rrrr c Vp = L’C’ 1  =  =  

29. 16.360 Lecture 5 Voltage reflection coefficient : Vg(t) VLVL A z = 0 B l ZLZL z = - l Z0Z0 ViVi i(z) = V(z) = V 0 + + - V0V0 e jzjz - e -j  z e + V0V0 Z0Z0 e jzjz - V0V0 Z0Z0 V L = + = - V0V0 + V0V0 V(z) z = 0 - + V0V0 Z0Z0 - V0V0 Z0Z0 i(z) z = 0 i L = = ZLZL = VLVL iLiL = + - V0V0 + V0V0 - + V0V0 Z0Z0 - V0V0 Z0Z0 + V0V0 - V0V0 = ZLZL Z0Z0 - ZLZL Z0Z0 +

30. 16.360 Lecture 5 Voltage reflection coefficient : - V0V0 + V0V0 = ZLZL Z0Z0 - ZLZL Z0Z0 +    Current reflection coefficient : - i0i0 + i0i0 = -  i  - V0V0 + V0V0 = -  Notes : 1.|  |  1, how to prove it? 2.If Z L = Z 0,  = 0. Impedance match, no reflection from the load Z L.

31. 16.360 Lecture 2 An example : A’ z = 0 A Z 0 = 100  R L = 50  C L = 10pF f = 100MHz Z L = R L + j/  C L = 50 –j159 = ZLZL Z0Z0 - ZLZL Z0Z0 + 

32. 16.360 Lecture 2 Standing wave Input impedance i(z) = V(z) = V 0 + + - V0V0 e jzjz - e -j  z e + V0V0 Z0Z0 e jzjz - V0V0 Z0Z0 - V0V0 + V0V0 with  = i(z) = V(z) = V 0 ( ) + +  e jzjz - e -j  z (e(e + V0V0 Z0Z0 e jzjz  ) |V(z)| = | V 0 | | | + e -j  z |||| e jzjz + e jrjr = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2

33. 16.360 Lecture 2 Standing wave i(z) = V(z) = V 0 + + - V0V0 e jzjz - e -j  z e + V0V0 Z0Z0 e jzjz - V0V0 Z0Z0 - V0V0 + V0V0 with  = i(z) = V(z) = V 0 ( ) + +  e jzjz - e -j  z (e(e + V0V0 Z0Z0 e jzjz  ) |V(z)| = | V 0 | | | + e -j  z |||| e jzjz + e jrjr = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2

34. 16.360 Lecture 2 Standing wave i(z) = V(z) = V 0 ( ) - + +  e jzjz e -j  z (e(e + V0V0 Z0Z0 e jzjz  ) |i(z)| = | V 0 | /|Z0|| | + e -j  z |||| e jzjz - e jrjr = | V 0 |/|Z 0 | [ 1+ |  |² - 2|  |cos(2  z +  r )] + 1/2 = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2 |V(z)|

35. 16.360 Lecture 2 Special cases = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2 |V(z)| 1.Z L = Z 0,  = 0 = | V 0 | + |V(z)| 2. Z L = 0, short circuit,  = -1 |V(z)| |V0| + - -3 /4- /2- /4 = | V 0 | [ 2 + 2cos(2  z +  )] + 1/2 |V(z)| 2|V0| + - -3 /4- /2- /4

36. 16.360 Lecture 2 Special cases = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2 |V(z)| 3. Z L = , open circuit,  = 1 = | V 0 | [ 2 + 2cos(2  z )] + 1/2 |V(z)| 2|V0| + - -3 /4- /2- /4

37. 16.360 Lecture 2 Voltage maximum = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2 |V(z)| | V 0 | [ 1+ |  |], + |V(z)| max = when 2  z +  r = 2n . –z =  r /4  + n /2 n = 1, 2, 3, …, if  r <0 n = 0, 1, 2, 3, …, if  r >= 0

38. 16.360 Lecture 2 Voltage minimum = | V 0 | [ 1+ |  |² + 2|  |cos(2  z +  r )] + 1/2 |V(z)| | V 0 | [ 1 - |  |], + |V(z)| min = when 2  z +  r = (2n+1) . –z =  r /4  + n /2 + /4 Note: voltage minimums occur /4 away from voltage maximum, because of the 2  z, the special frequency doubled.

39. 16.360 Lecture 2 Voltage standing-wave ratio VSWR or SWR 1 - |  | |V(z)| min S  |V(z)| max = 1 + |  | S = 1, when  = 0, S = , when |  | = 1,

40. 16.360 Lecture 2 An example Vg(t) VLVL A z = 0 B l ZLZL z = - l Z0Z0 ViVi Voltage probe S = 3, Z 0 = 50 ,  l min = 30cm, l min = 12cm, Z L =?  l min = 30cm,  = 0.6m, S = 3,  |  | = 0.5, Solution: -2  l min +  r = - ,   r = -36º,  , and Z L.

41. 16.360 Lecture 2 Input impudence Vg(t) VLVL A z = 0 B l ZLZL z = - l Z0Z0 ViVi Zg IiIi Z in (z) = V(z) I(z) = + +  e jzjz e -j  z V0V0 ( ) + -  e jzjz e V0V0 ( ) Z0Z0 = + (1  e j2  z ) - (1  e j2  z ) Z0Z0 + (1  e -j2  l ) - (1  e -j2  l ) Z0Z0 Z in (-l) =

42. 16.360 Lecture 2 An example A 1.05-GHz generator circuit with series impedance Zg = 10-  and voltage source given by Vg(t) = 10 sin(  t +30º) is connected to a load ZL = 100 +j5-  through a 50- , 67-cm long lossless transmission line. The phase velocity is 0.7c. Find V(z,t) and i(z,t) on the line. Solution: Since, Vp = ƒ, = Vp/f = 0.7c/1.05GHz = 0.2m.  = 2  /,  = 10 .  = (Z L -Z 0 )/(Z L +Z 0 ),  = 0.45exp(j26.6º) + (1  e -j2  l ) - (1  e -j2  l ) Z0Z0 Z in (-l) = = 21.9 + j17.4  V 0 [ exp(-j  l)+  exp(j  l) ] + = Z in (-l) + Zg Z in (-l) Vg

43. 16.360 Lecture 2 short circuit line Z L = 0,  = -1, S =  Vg(t) VLVL A z = 0 B l Z L = 0 z = - l Z0Z0 Zg IiIi Z in sc i(z) = V(z) = V 0 ) - e jzjz + (e(e -j  z (e(e + V0V0 Z0Z0 e jzjz ) = -2jV 0 sin(  z) + = 2V 0 cos(  z)/Z 0 + Z in = V(-l) i(-l) = jZ 0 tan(  l)

44. 16.360 Lecture 2 short circuit line Z in = V(-l) i(-l) = jZ 0 tan(  l) If tan(  l) >= 0, the line appears inductive,j  L eq = jZ 0 tan(  l), If tan(  l) <= 0, the line appears capacitive,1/j  C eq = jZ 0 tan(  l), l = 1/  [  - tan (1/  C eq Z 0 )], The minimum length results in transmission line as a capacitor:

45. 16.360 Lecture 2 An example: tan (  l) = - 1/  C eq Z 0 = -0.354, Choose the length of a shorted 50-  lossless line such that its input impedance at 2.25 GHz is equivalent to the reactance of a capacitor with capacitance Ceq = 4pF. The wave phase velocity on the line is 0.75c. Solution: Vp = ƒ,   = 2  / = 2  ƒ/Vp = 62.8 (rad/m)  l = tan (-0.354) + n , = -0.34 + n ,

46. 16.360 Lecture 2 open circuit line Z L = 0,  = 1, S =  Vg(t) VLVL A z = 0 B l Z L =  z = - l Z0Z0 Zg IiIi Z in oc i(z) = V(z) = V 0 ) + e jzjz - (e(e -j  z (e(e + V0V0 Z0Z0 e jzjz ) = 2V 0 cos(  z) + = 2jV 0 sin(  z)/Z 0 + Z in = V(-l) i(-l) = -jZ 0 cot(  l) oc

47. 16.360 Lecture 2 Application for short-circuit and open-circuit Network analyzer Measure Z in oc Z in sc and Calculate Z 0 Measure S paremeters Z in = -jZ 0 cot(  l) oc Z in = jZ 0tan (  l) sc =Z0Z0 Z in sc Z in oc Calculate  l = -jZ0Z0 Z in sc Z in oc

48. 16.360 Lecture 2 Line of length l = n /2 = Z L tan(  l) = tan((2  / )(n /2)) = 0, + (1  e -j2  l ) - (1  e -j2  l ) Z0Z0 Z in (-l) = Any multiple of half-wavelength line doesn’t modify the load impedance.

49. 16.360 Lecture 2 Quarter-wave transformer l = /4 + n /2 = Z 0 ²/Z L  l = (2  / )( /4 + n /2) =  /2, + (1  e -j2  l ) - (1  e -j2  l ) Z0Z0 Z in (-l) = + (1  e -j  ) - (1  e -j  ) Z0Z0 = (1 +  ) Z0Z0 (1 -  ) =

50. 16.360 Lecture 2 An example: A 50-  lossless tarnsmission is to be matched to a resistive load impedance with Z L = 100  via a quarter-wave section, thereby eliminating reflections along the feed line. Find the characteristic impedance of the quarter-wave tarnsformer. Z L = 100  /4 Z 01 = 50  Z in = Z 0 ²/Z L = 50  Z 0 = (Z in Z L ) = (50*100) ½ ½

51. 16.360 Lecture 2 Matched transmission line: 1.Z L = Z 0 2.  = 0 3.All incident power is delivered to the load.

52. 16.360 Lecture 8 Instantaneous power Time-average power i(z) = V(z) = V 0 ( ) + +  e jzjz - e -j  z (e(e + V0V0 Z0Z0 e jzjz  ) At load z = 0, the incident and reflected voltages and currents: V = V 0 + i = i + V0V0 Z0Z0 i V = V 0 - r i = - V0V0 Z0Z0 r

53. 16.360 Lecture 2 Instantaneous power + i P(t) = v(t) i(t) = Re[V exp(j  t)] Re[ i exp(j  t)] i i = Re[|V 0 |exp(j  )exp(j  t)] Re[|V 0 |/Z 0 exp(j  )exp(j  t)] ++ + = (|V 0 |²/Z 0 ) cos²(  t +  ) + + - r P(t) = v(t) i(t) = Re[V exp(j  t)] Re[ i exp(j  t)] r r = Re[|V 0 |exp(j  )exp(j  t)] Re[|V 0 |/Z 0 exp(j  )exp(j  t)] +- + = - |  |²(|V 0 |²/Z 0 ) cos²(  t +  +  r ) + +

54. 16.360 Lecture 2 Time-average (|V 0 |²/Z 0 ) cos²(  t +  )dt + + Time-domain approach: P av = i T 1 0 T P (t)dt i = 22  0 T = (|V 0 |²/2Z 0 ) + P av r = -|  |² (|V 0 |²/2Z 0 ) + = P av i P av + P av r = (1-|  |²) (|V 0 |²/2Z 0 ) + Net average power:

55. 16.360 Lecture 2 Time-average Phasor-domain approach P av r = -|  |² (|V 0 |²/2Z 0 ) + = (1-|  |²) (|V 0 |²/2Z 0 ) + = (½)Re[V i*] P av P av = (1/2) Re[V 0 V 0 * /Z 0 ] i + + = (|V 0 |²/2Z 0 ) P av