325_Sp2011_15_Transmission_lines

# 325_Sp2011_15_Transmission_lines - 15 Transmission lines...

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© Copyright Dean P. Neikirk 2004-2009 Mikhail Belkin, EE 325, ECE Dept., UT Austin 1 15. Transmission lines, Smith chart EE325 Mikhail Belkin

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© Copyright Dean P. Neikirk 2004-2009 Mikhail Belkin, EE 325, ECE Dept., UT Austin 2 Telegraphist’s equations Consider a long piece of a transmission line (e.g., a coax) we can call it a “wire-pair” one wire carrying a time-varying current out and the other carrying the return current one wire is at some time varying voltage relative to the other We will use circuit-based approach for current and voltage. Alternatively, Maxwell’s equations may be solved for E and H with boundary conditions defined by conductors. what might the equivalent circuit look like for a short segment of a long wire-pair? we expect something related to current and magnetic fields: inductance, resistance (if wires have finite conductivity) we would expect something related to voltage and charge: capacitance, and conductance between the wires if the dielectric is leaky z direction V(z, t) I(z, t)
© Copyright Dean P. Neikirk 2004-2009 Mikhail Belkin, EE 325, ECE Dept., UT Austin 3 z δ Z( ϖ ) Y( ϖ ) z Z( ϖ ) Y( ϖ ) z Z( ϖ ) Y( ϖ ) z ( 29 V z ( 29 V z z + ( 29 I z ( 29 I z z + Z( ϖ ) Y( ϖ ) Z in Z 1 z ( 29 V z ( 29 I z Z( ϖ ) Y( ϖ ) Z in Z 1 z direction V(z, t) I(z, t) Equivalent circuit using ‘lumped elements’ (i.e. elements with δ z<< λ ) Equivalent circuit Individual element:

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© Copyright Dean P. Neikirk 2004-2009 Mikhail Belkin, EE 325, ECE Dept., UT Austin 4 Equivalent circuit so a reasonable guess for the equivalent circuit of a short segment of our wire pair would be here R, L, C, and G are per unit length values 1 2 R z δ × ÷ z 1 2 R z × ÷ 1 2 L z × ÷ 1 2 L z × ÷ ( 29 C z × ( 29 G z × ( 29 V z ( 29 ( 29 V z z V z V + = + ( 29 I z ( 29 ( 29 I z z I z I + = + I
© Copyright Dean P. Neikirk 2004-2009 Mikhail Belkin, EE 325, ECE Dept., UT Austin 5 Circuit response in time harmonic form, using phasors and using Kirchhoff’s laws, we have: 1 2 R z δ × ÷ z 1 2 L z × ÷ ( 29 C z × ( 29 G z × ( 29 V z ( 29 ( 29 V z z V z V + = + ( 29 I z ( 29 ( 29 I z z I z I + = + I ( 29 ( 29 ( 29 ( 29 1 1 2 2 1 1 2 2 V z R z j L z I z R z j L z I z I V z V ϖ = × + × ÷ ÷ + × + × + ÷ ÷   + + [ ] ( 29 { } [ ] 1 0 2 2 z R j L I z I V = + + + ÷ [ ] ( 29 { } [ ] ( 29 2 V R j L I z I R j L I z z - = + + + [ ] ( 29 dV R j L I z dz = - + [ ] ( 29 [ ] ( 29 [ ] ( 29 2 2 V I G z j C z V z V z G j C V z z G j C V z z δ δ = × + × - + = ÷ = - + + ≈ - + [ ] ( 29 dI G j C V z dz = - +

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• Fall '08
• Brown
• Transmission line, Impedance matching, Austin Motor Company, Dean P. Neikirk, Mikhail Belkin, Copyright Dean P.

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325_Sp2011_15_Transmission_lines - 15 Transmission lines...

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