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16.360 Lecture 5 Last lecture: Transmission line parameters Types of transmission lines Lumped-element model Transmission line equations Telegrapher’s equations
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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)
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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 jtjt )/dt ),= Re(i jtjt e jj - 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)
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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 jtjt )/dt ),= Re(V jtjt e jj - 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)
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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)
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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)
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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’),
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16.360 Lecture 5 Today: Wave propagation on a Transmission line Characteristic impedance Standing wave and traveling wave Lossless transmission line Reflection coefficient
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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 zz i(z) = I 0 (15) + e -z-z + - I0I0 e zz
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16.360 Lecture 5 Wave equations V(z) = V 0 (14) + e -z-z + - V0V0 e zz i(z) = I 0 (15) + e -z-z + - I0I0 e zz where: + V0V0 - V0V0 andare determined by boundary conditions. + I0I0 - I0I0 andare related to - V0V0 + V0V0 andby characteristic impedance Z 0.
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16.360 Lecture 5 recall: - d V(z)/ dz = R’ i (z) + j L’ i (z), (4) V(z) = V 0 (14) + + - V0V0 e zz i(z) = I 0 (15) + e -z-z + - I0I0 e zz e -z-z e -z-z V0V0 + e zz V0V0 - - = ( R’ + j L’) i (z), (16) i (z) = ( R’ + j L’) e -z-z (V0(V0 + e zz V0V0 - - ) I0I0 + = V0V0 + I0I0 - = -- V0V0 - (17) (18) Characteristic impedance Z 0
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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:
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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 zz i(z) = I 0 (15) + e -z-z + - I0I0 e zz e -z-z (19) (20)
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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’) = jj L’C’ = + j , = L’C’ = 20 rad/m
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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’) = jj L’C’ = 0 = L’C’ lossless line
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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
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16.360 Lecture 5 For TEM transmission line : Vp = L’C’ 1 L’C’ = = 1 = L’C’ = = rrrr c Z0Z0 = L’ C’ summary : V(z) = V 0 + + - V0V0 e jzjz i(z) = I 0 + e -j z + - I0I0 e jzjz e = rrrr c Vp = L’C’ 1 = =
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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 jzjz - e -j z e + V0V0 Z0Z0 e jzjz - 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 +
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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.
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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 +
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16.360 Lecture 5 Next lecture Standing wave Input impedance
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