Transmission Lines in Frequency Domain for Communication Systems

ECE 3317 Applied Electromagnetic Waves Prof. David R. Jackson Fall 2024 Notes 9 Transmission Lines (Frequency Domain) 1

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.) Examples: 100BASET, 10GBASET 2

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 also write j t ( ) = ( ) v t e v dt 1 1 = j t ( ) ( ) v t Re v e d j t = ( ) ( ) v t v e d 2 0 ( ) ( ) ( ) = * since v v Jean-Baptiste-Joseph Fourier 3

Frequency Domain (cont.) 1 j t = ( ) Re ( ) v t v d e A collection of phasor-domain signals! 0 Phasor ( ) v t 1.0 A pulse is resolved into a collection (spectrum) of infinite sinusoidal waves with different frequencies, amplitudes, and phases. t W V t Time 4

Frequency Domain (cont.) Example: Rectangular pulse = j t ( ) ( ) v t e v dt ( ) v t 1.0 t W ( ) ( ) = sinc /2 v W W ( ) x x sin ( ) x sinc 5

Frequency Domain (cont.) System ( ) H ( ) ( ) Phasor domain: V V in out In the frequency domain, the system has a transfer function H( ): ( ) out V ( ) ( ) = H V in System ( ) t ( ) ( ) t H v Time domain: v in out The time-domain response of the system to an input signal is: 1 ( ) = j t ( ) ( ) t Re v H v e d out in 0 6

Frequency Domain (cont.) Summary Phasordomain: ( ) ( ) V V ( ) out H System in ( ) t ( ) ( ) t H v v in out 1 ( ) = j t ( ) ( ) t Re v H v e d out in 0 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 is one reason why the phasor domain is so important! This applies for transmission lines also! 7

Telegraphers Equations 2 2 v v t v ( ) + = ( ) 0 RG v RC LG LC 2 2 z t L z R z ( ) + , i z t ( ) , v z t G z C z - z z 8

Frequency Domain j To convert to the phasor domain, we use: t 2 2 v v t v ( ) + = ( ) 0 RG v RC LG LC 2 2 z t 2 V z ( ) ( ) 2 + = ( ) 0 RG V j RC LG V j LCV 2 or 2 d V dz ( ) ( ) = + + 2 ( ) RG V j RC LG V LC V 2 9

Frequency Domain (cont.) 2 d V dz ( ) ( ) = + + 2 ( ) RG j RC LG LC V 2 Note that + + = + j L G + j C 2 ( ) ( )( ) RG j RC LG LC R = = + + = = Z Y R G j L j C series impedance / length parallel admittance / length 2 d V dz = ( ) ZY V We can therefore write: 2 10

Telegraphers Equations 2 d V dz = ( ) ZY V 2 = + j L Z R = + j C Y G L z R z + ( ) V z G z C z - z z 11

Frequency Domain (cont.) 2 d V dz = ( ) ZY V 2 Define = + j L G + j C 2 ( )( ) ZY R Then 2 d V dz = 2 ( ) V 2 Solution: + = + z z ( ) V z Ae Be Note: We have an exact solution, even for a lossy line, in the phasor domain! 12

Propagation Constant Convention: We choose the (complex) square root to be the principal branch: + j L G + j C ( )( ) R (lossy case) is called the propagation constant, with units of [1/m] Review of principal branch of square root: Note: = /2 /2 /2 j c c e ( ) /2 j = c c e c Re 0 13

Propagation Constant (cont.) = + j L G + j C ( )( ) R = + j Denote: = propagation constant [1/m] = attenuation constant [nepers/m] = phase constant [radians/m] Choosing the principal branch means that 0 Re 0 14

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: ( ) = + j L G + j C = ( )( ) 1 R LC = j LC (lossless case) Hence Hence, we have that = = 0 LC Note: = 0 for a lossless line. 15

Propagation Constant (cont.) Physical interpretation of waves: + = = z ( ) z ( ) z V V Ae Be (forward traveling wave) Note: The waves must decay in the direction of propagation. + z (backward traveling wave) = + j + j z = z ( ) z V Ae e Forward traveling wave: (decaying as it travels) + + j z = z ( ) z V Be e (decaying as it travels) Backward traveling wave: 16

Propagation Wavenumber Alternative notations: = + j (propagation constant) = zk j (propagation wavenumber) = Note: jk z + j z = = = jk z z z ( ) z V Ae Ae Ae e z 17

Forward Wave + j z = z ( ) z V Ae e Forward traveling wave: = j A A e Denote + j z = j z ( ) z V A e e e Then In the time domain we have: ( ) z e + + j t = ( , ) Re v z t V + = + z ( , ) z t cos( ) v A e t z Hence, we have 18

Forward Wave (cont.) + = + z ( , ) z t cos( ) v A e t z Snapshot of Waveform: t = 0 g z A e z The distance gis the distance it takes for the waveform to repeat itself in meters. g= guided wavelength 19

Wavelength The wave repeats (except for the amplitude decay) when: = 2 g 2 = Hence: g Note: This equation can be used to find g if we already know : 2 2 2 = = = ( ) ( ) g Im + j L G + j C Im ( )( ) R 20

Wavelength (cont.) Lossless case: 2 2 2 f 1 1 1 f c c = = = = = = = = = 0 d f g d 2 LC LC f LC f r r r r Summary for lossless case: = g d = 0 d = wavelength in dielectric d r r c f = 0 = wavelength in free-space (air) 0 c 8 2.99792458 10 m/s 21

Attenuation Constant The attenuation constant controls how fast the wave decays. + = + z ( , ) z t cos( ) v A e t z = z envelope A e t = 0 g z A e z ( ) ( ) = = + j L G + j C Re Re ( )( ) R 22

Phase Velocity The forward-traveling wave is moving in the positive z direction. Consider a sinusoidal wave moving on a transmission line (shown in the figure below for a lossless line ( = 0) for simplicity): + = + z ( , ) cos( ) v z t A e t z = = t t t t ( ) p v = velocity + 1 , v z t 2 m z + = 0 t z Crest of wave: 23

Phase Velocity (cont.) The phase velocity vphase is the velocity of a point on the wave, such as the crest. = = constant t z Set dz dt = 0 Take the derivative with respect to time: dz dt Hence = We thus have = phase v Note: This result holds for a general lossy line. ( ) = + j L G + j C Recall : Im ( )( ) R 24

Phase Velocity (cont.) Let s calculate the phase velocity for a lossless line: 1 LC Lossless line: = = = phase v = = 0 LC LC 1 c = = LC Also, we know that 2 d Recall: = phase v c Hence (lossless line) For a lossless line, any signal travels at the speed cd. (So this must be true for a sinusoidal signal!) d 1 1 1 c = = = Recall: c d 0 0 r r r r 25

Backward Traveling Wave Let s now consider the backward-traveling wave (shown in the figure below for a lossless line ( = 0) for simplicity): + + + j z j + + j z = = = B e z z z = ( ) z j V Be Be e B e e e B Denote + = + + z ( , ) cos( ) v z t B e t z ( ) = = t t t t p v = velocity , v z t 1 2 m z This wave has the same phase velocity, but it travels backwards. 26

Group Velocity The group velocity vgroup is the velocity of a pulse. gv z Note: We have (derivation omitted): For a lossy line, the pulse will be distorting as it propagates down the line. d d = group v Note: for a lossless line we have: = = = = 1/ 1/ group v phase v LC c d ( ) = = Lossless line: / 1/ LC d d LC 27

Attenuation in dB/m + j z = = = jk z z z ( ) z V Ae Ae Ae e z + ( ) (0) V V z = = z dB 20log 20log ( ) e Gain in dB: 10 10 + ln ln10 x x = log Use the following logarithm identity: 10 ( ) z ln e ( ) 20 ln10 z = = = dB 20 20 z Therefore, the gain is: ln10 ln10 20 ln10 = Attenuation [dB/m] Hence, we have: 28

Attenuation in dB/m (cont.) Final attenuation formulas: 20 ln10 = Attenuation [dB/m] ( ) Attenuation 8.686 [dB/m] 29

Example: Coaxial Cable 2 = F/m C 0 b a r Copper conductors (nonmagnetic: m= 0 ) ln b a = ln H/m L 0 = = 0.5 mm r a a 2 3.2 mm 2.2 = = b 2 b = S/m G d z b a = ln r TV coax = tan 0.001 = d 5.8 10 S/m 7 ma mb 1 1 ( ) = + /m R UHF 500 MHz f 2 2 a b ma ma mb mb Note: 2 2 The loss tangent of the dielectric is called tan d. = = ma mb ma ma mb mb (skin depth of the two conductors) 30

Example: Coaxial Cable (cont.) Dielectric conductivity is often specified in terms of the loss tangent: = tan d d d 0 r d = effective conductivity of the dielectric material* Note: The loss tangent of practical insulating materials (e.g., Teflon) is approximately constant over a wide range of frequencies. (For example, tan d 0.001for Teflon.) *The effective conductivity accounts for the actual conductivity as well as molecular friction and other effects. 31

Example: Coaxial Cable (cont.) Relation between G and C 2 = F/m C 0 b a r ln r a = G C d 2 b = S/m G d 0 r b a z ln G C = d 0 r Hence Recall G C = tan = tan d d d d 0 r This relationship holds for any type of transmission line. 32

Example: Coaxial Cable (cont.) Characteristic impedance (ignore R and G for this): L C b a = = ln 0 Z 0 2 r Z = 75 [ ] r a 0 b Skin depth of metal: z 2 = = ( ) 0.5 mm a = = m 0 m 3.2 mm 2.2 = = b = = 6 2.955 10 [m] r m = tan 0.001 = d 5.8 10 S/m 7 Effective conductivity of dielectric: ma mb ( ) ( ) UHF 500 MHz f = tan 0 d r d = 5 6.12 10 [S/m] d 33

Example: Coaxial Cable (cont.) = = 0.5 mm a 3.2 mm 2.2 = = b = r = tan 0.001 = d r a 5.8 10 S/m 7 ma mb b ( ) UHF 500 MHz f z = + j L G + j C ( )( ) R Results for (R,L,G,C): ) ( = + 0.022 15.543 1/m j H/m = = = = 2.147 /m R = = 0.022 nepers/m 15.544 rad/m 7 3.713 10 L 4 2.071 10 S/m G = Attenuation 0.191 dB/m 0.404 m 11 6.593 10 F/m C g = 34

Current Use the first Telegrapher equation: v z i t = Ri L j t V z = j LI RI Next, use + = + z z ( ) V z Ae Be V z ( ) z + = z z Ae Be so 35

Current (cont.) Hence, we have + = j LI z z Ae Be RI Solving for the phasor current I, we have + = z z I Ae Be + j L R ( )( j L ) + + j C R j L G R + + = z z Ae Be + + G R j C j L + = z z Ae Be 36

Characteristic Impedance Define the (complex) characteristic impedance Z0in the frequency domain for a lossy line: + + R G j L j C Z 0 Then we have: 1 Z + = z z ( ) I z Ae Be 0 Note: In the time domain, we only defined Z0 for a lossless line. In the frequency domain, we can define it for a lossy line. 37

Characteristic Impedance (cont.) ( ) I z Note the reference directions! + - ( ) ( ) z + V z V z + ( ) z V I = Z The characteristic impedance is the ratio of the voltage to the current, for a wave traveling in the positive z direction. ( ) z 0 + Practical note: Even though Z0 is always complex for a practical line (due to loss), we usually neglect this and take it to be real. + + R G j L j C L C = Z 0 38

Summary of Solution Characteristic Impedance Lossy Lossless: + + R G j L j C L C = Z = Z 0 0 Voltage and Current + = + z z ( ) V z Ae Be 1 Z + = z z ( ) I z Ae Be 0 39

Numerical Solution for Lossy Line System Phasordomain: ( ) t ( ) ( ) t H v ( ) ( ) v V V in ( ) out out H in 1 ( ) = j t ( ) ( ) t Re This must be evaluated numerically (e.g., using MATLAB). v H v e d out in 0 Lossy transmission line: + - ( ) ( ) z = + z z z = V z V 0 0 z ( )( ) ( ) R j L G j C + + z = = z H e e 0 0 40

Numerical Solution for Lossy Line (cont.) Example: Propagation on a lossy microstrip line (From ECE 5317) y x w t z + r ( ) z sv t Probe h Probe Coax SIDE VIEW TOP VIEW Input signal: = 2.33 = = = = s r = 0.5 10 10 t 0 tan h 0.001 1.0 ( ) 0.787 mm 31mils ( ) t gv 2.35 mm w ( ) 0.0175 mm t "half oz" copper cladding t = 3.0 10 S/m 7 0t m 41

Numerical Solution for Lossy Line (cont.) Example: Propagation on a microstrip line z = 20 cm z = 1 cm = 2.33 = = = = r tan h 0.001 z = 100 cm ( ) 0.787 mm 31mils 2.35 mm w ( ) 0.0175 mm t "half oz" copper cladding = 3.0 10 S/m 7 m 42

Appendix: Summary of Formulas General Lossy Case + + R G j L j C = = tan d d Z d 0 0 r G C = tan + = + z z ( ) V z Ae Be d 1 Z + = z z ( ) I z Ae Be 2 = 0 g + j L G + j C ( )( ) R = phase v = + j ( ) = Attenuation 8.686 [dB/m] 43

Appendix: Summary of Formulas Lossless Case = g d L C = Z 2 0 = d j z + j z = + ( ) V z Ae Be = 0 d r r 1 Z c f j z + j z = ( ) I z Ae Be = 0 0 = = j phase v c d c = c = d LC r r c = 8 2.99792458 10 m/s 44