1. 16.360 Lecture 5 Last lecture: Transmission line parameters Types of transmission lines Lumped-element model Transmission line equations Telegrapher’s equations

2. 16.360 Lecture 5 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)

3. 16.360 Lecture 5 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)

4. 16.360 Lecture 5 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)

5. 16.360 Lecture 5 - 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)

6. 16.360 Lecture 5 - 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)

7. 16.360 Lecture 5 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’),

8. 16.360 Lecture 5 Today: Wave propagation on a Transmission line Characteristic impedance Standing wave and traveling wave Lossless transmission line Reflection coefficient

9. 16.360 Lecture 5 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

10. 16.360 Lecture 5 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.

11. 16.360 Lecture 5 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

12. 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:

13. 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)

14. 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

15. 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

16. 16.360 Lecture 5 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

17. 16.360 Lecture 5 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  =  =  

18. 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 +

19. 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.

20. 16.360 Lecture 5 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 + 

21. 16.360 Lecture 5 Next lecture Standing wave Input impedance