Prof. Ji Chen Notes 9 Transmission Lines (Frequency Domain) ECE Spring 2014

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

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). 3 For a physically-realizable (real-valued) signal, we can write Fourier-transform pair Jean-Baptiste-Joseph Fourier Frequency Domain (cont.)

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

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

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

Frequency Domain (cont.) 7 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

Telegrapher’s Equations RzRz LzLz GzGz CzCz z z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 ] 8

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. 9

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

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

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

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 ] 13 Principle branch of square root: Note:

14  = propagation constant [ 1/m ]  = attenuation constant [ nepers / m ]  = phase constant [ radians / m ] Denote: Choosing the principle branch means that 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) 15 Hence Propagation Constant (cont.) Hence we have that Note:  = 0 for a lossless line.

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

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

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

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

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

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

Phase Velocity The forward-traveling wave is moving in the positive z direction. Consider a lossless transmission line for simplicity (  = 0 ): z [m] v = velocity t = t 1 t = t 2 Crest of wave: 22

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

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

Backward Traveling Wave Let’s now consider the backward-traveling wave (lossless, for simplicity): z [m] v = velocity t = t 1 t = t 2 25

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

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

Example: Coaxial Cable a b z (skin depth) a = 0.5 [mm] b = 3.2 [mm]  r = 2.2 tan  d =  ma =  mb = 5.8  10 7 [S/m] f = 500 [MHz] (UHF) Dielectric conductivity is specified in terms of the loss tangent: Copper conductors (nonmagnetic:  =  0 ) 28  d = effective conductivity of the dielectric material

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

30 a b z Ignore loss for calculating Z 0 : a = 0.5 [mm] b = 3.2 [mm]  r = 2.2 tan  d =  ma =  mb = 5.8  10 7 [S/m] f = 500 [MHz] (UHF)

R = [  /m] L = E-07 [H/m] G = E-04 [S/m] C = E-11 [F/m]  = j (15.543) [1/m]  = [nepers/m]  = [rad/m] attenuation = [dB/m] = [m] Example: Coaxial Cable (cont.) 31 a b z a = 0.5 [mm] b = 3.2 [mm]  r = 2.2 tan  d =  ma =  mb = 5.8  10 7 [S/m] f = 500 [MHz] (UHF)

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

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

Characteristic Impedance Define the (complex) characteristic impedance Z 0 : Then we have 34 Lossless: