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Prof. David R. Jackson

ECE 3317. Prof. David R. Jackson. Spring 2013. Notes 9 Transmission Lines ( Frequency Domain). Frequency Domain. Why is the frequency domain important?. Most communication systems use a sinusoidal signal (which may be modulated ).

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Prof. David R. Jackson

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  1. ECE 3317 Prof. David R. Jackson Spring 2013 Notes 9 Transmission Lines(Frequency Domain)

  2. Frequency Domain Why is the frequency domain important? • Most communication systems use a sinusoidal signal (which may be modulated). (Some systems, like Ethernet, communicate in “baseband”, meaning that there is no carrier.)

  3. Frequency Domain (cont.) Why is the frequency domain important? • A solution in the frequency-domain allows to solve for an arbitrary time-varying signal on a lossy line (by using the Fourier transform method). Fourier-transform pair For a physically-realizable (real-valued) signal, we can write Jean-Baptiste-Joseph Fourier

  4. Frequency Domain (cont.) A collection of phasor-domain signals! t W A pulse is resolved into a collection (spectrum) of infinite sine waves.

  5. Frequency Domain (cont.) Example: rectangular pulse t W

  6. Frequency Domain (cont.) System Phasor domain: In the frequency domain, the system has a transfer function H(): System Time domain: The time-domain response of the system to an input signal is:

  7. Frequency Domain (cont.) System If we can solve the system in the phasor domain (i.e., get the transfer function H()), we can get the output for any time-varying input signal. This applies for transmission lines

  8. Telegrapher’s Equations RDz LDz CDz GDz Dz z Transmission-line model for a small length of transmission line: C = capacitance/length [F/m] L = inductance/length [H/m] R = resistance/length [/m] G = conductance/length [S/m]

  9. Telegrapher’s Equations (cont.) • Take the derivative of the first TE with respect to z. • Substitute in from the second TE. Recall: There is no exact solution in the time domain, for the lossy case.

  10. Frequency Domain To convert to the phasor domain, we use: or

  11. Frequency Domain (cont.) Note that = series impedance/length = parallel admittance/length Then we can write:

  12. Frequency Domain (cont.) Define Then Solution: Note: We have an exact solution, even for a lossy line!

  13. Propagation Constant Convention: we choose the (complex) square root to be the principle branch: (lossy case)  is called the propagation constant, with units of [1/m] Principle branch of square root: Note:

  14. Propagation Constant (cont.) Denote:  = propagation constant [1/m] = attenuation constant [nepers/m] = phase constant [radians/m] Choosing the principle branch means that

  15. Propagation Constant (cont.) For a lossless line, we consider this as the limit of a lossy line, in the limit as the loss tends to zero: (lossless case) Hence Hence we have that Note:  = 0 for a lossless line.

  16. Propagation Constant (cont.) Physical interpretation of waves: (forward traveling wave) (backward traveling wave) (This interpretation will be shown shortly.) Forward traveling wave:

  17. Propagation Wavenumber Alternative notations: (propagation constant) (propagation wavenumber)

  18. Forward Wave Forward traveling wave: Denote Then In the time domain we have: Hence we have

  19. Forward Wave (cont.) Snapshot of Waveform: The distance  is the distance it takes for the waveform to “repeat” itself in meters. = wavelength

  20. Wavelength The wave “repeats” (except for the amplitude decay) when: Hence:

  21. Attenuation Constant The attenuation constant controls how fast the wave decays.

  22. Phase Velocity The forward-traveling wave is moving in the positive z direction. Consider a lossless transmission line for simplicity (= 0): t = t1 t = t2 v= velocity z [m] Crest of wave:

  23. Phase Velocity (cont.) The phase velocityvp is the velocity of a point on the wave, such as the crest. Set Take the derivative with respect to time: Hence Note: This result holds also for a lossy line.) We thus have

  24. Phase Velocity (cont.) Let’s calculate the phase velocity for a lossless line: Also, we know that Hence (lossless line)

  25. Backward Traveling Wave Let’s now consider the backward-traveling wave (lossless, for simplicity): t = t1 t = t2 v= velocity z [m]

  26. Attenuation in dB/m Attenuation in dB: Use the following logarithm identity: Therefore Hence we have:

  27. Attenuation in dB/m (cont.) Final attenuation formulas:

  28. Example: Coaxial Cable a b z Copper conductors (nonmagnetic: =0) a = 0.5 [mm] b = 3.2 [mm] r = 2.2 tand = 0.001 ma= mb = 5.8107 [S/m] f = 500 [MHz] (UHF) (skin depth) Dielectric conductivity is specified in terms of the loss tangent: d = effective conductivity of the dielectric material

  29. Example: Coaxial Cable (cont.) a b z Relation between G and C Hence Note: The loss tangent of practical insulating materials (e.g., Teflon) is approximately constant over a wide range of frequencies.

  30. Example: Coaxial Cable (cont.) a b z Ignore loss for calculating Z0: a = 0.5 [mm] b = 3.2 [mm] r = 2.2 tand = 0.001 ma= mb = 5.8107 [S/m] f = 500 [MHz] (UHF)

  31. Example: Coaxial Cable (cont.) a b z a = 0.5 [mm] b = 3.2 [mm] r = 2.2 tand = 0.001 ma= mb = 5.8107 [S/m] f = 500 [MHz] (UHF) R = 2.147 [/m] L = 3.713 E-07 [H/m] G = 2.071 E-04 [S/m] C = 6.593 E-11 [F/m]  = 0.022 +j (15.543) [1/m]  = 0.022 [nepers/m]  = 15.543 [rad/m]  = 0.404 [m] attenuation = 0.191 [dB/m]

  32. Current Use the first Telegrapher equation: Next, use so

  33. Current (cont.) Hence we have Solving for the phasor current I, we have

  34. Characteristic Impedance Define the (complex) characteristic impedance Z0: Lossless: Then we have

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