ECE 6340 Intermediate EM Waves. Fall 2016 Prof. David R. Jackson Dept. of ECE. Notes 15

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ECE 634 Intermediate EM Waves Fall 6 Prof. David R. Jackson Dept. of ECE Notes 5

Attenuation Formula Waveguiding system (WG or TL): S z Waveguiding system Exyz (,, ) = E( xye, ) = E( xye, ) e γz jβz αz Hxyz (,, ) = H( xye, ) = H( xye, ) e γz jβz αz At z = : At z = z : * Pf () = ( E H ) zˆ ds S * α z P ( ) ( ) ˆ f z = E H e z ds S

Attenuation Formula (cont.) Hence P ( z) = P () e f α z Re P ( z) = Re P () e so f f f α z P f ( z) = P () f z e α If α z P f ( z) P () ( α z) f = P () α z P () f f 3

Attenuation Formula (cont.) P ( z) P () α z P () f f f so α P f () P ( z) z P () f f S z = z = z From conservation of energy: P () P ( z) z P l ( z/ ) f f d where l Pd ( z) = power dissipated per length at point z 4

Attenuation Formula (cont.) Hence l z d ( z/ ) α P z P () f As z : d () α = P P f () l Note: Where the point z = is located is arbitrary. 5

Attenuation Formula (cont.) General formula: α = P l d ( z) P ( z ) f This is a perturbational formula for the conductor attenuation. The power flow and power dissipation are usually calculated assuming the fields are those of the mode with PEC conductors. z P f ( z) 6

Attenuation on Transmission Line The current of the TEM mode flows in the z direction. Attenuation due to Conductor Loss α = α c J sz z l d α = P P f C A C = C + C A B C B L z 7

Attenuation on Line (cont.) Power dissipation due to conductor loss: Power flowing on line: P f = Z I (Z is assumed to be approximately real.) S P = R J z l d s sz S = ( z) Rs J sz ( ) dl z = C z R J s C sz ( ) ds dl C A A I C= C A + C B C B B 8

Attenuation on Line (cont.) Hence α c R = J dl ( ) s sz Z I CA+ CB 9

R on Transmission Line R z L z I C z G z z Ignore G for the R calculation (α = α c ): l d α c = P P f P l d P f = = RI Z I

R on Transmission Line (cont.) We then have α = c R Z Hence R = α c ( Z ) Substituting for α c, R = Rs J () sz l dl I C

Total Attenuation on Line Method # α = α + α c d α d = α TEM When we ignore conductor loss to calculate α d, we have a TEM mode. k = β jα = k = k jk TEM z d so α = k d Hence, α = α + k c

Total Attenuation on Line (cont.) Method # α = Reγ ( R jωl G jωc ) = Re ( + )( + ) where R G = α ( c Z ) ( ωc) ε c = ε c The two methods give approximately the same results. 3

Example: Coax Coaxial Cable z I a A b I ε r B α R s c = J sz ( ) dl + J sz ( ) dl Z I C C A B A) B) J J sz sz = = I π a I π b 4

Example (cont.) Hence Also, π π R s I I αc = adφ bdφ + Z I πa πb Z Rs = + Z πa πb η b = a ln π εr Hence α c = R b s b ε r + a b η ln a (nepers/m) 5

Example (cont.) Calculate R: R = α ( Z ) c Rs = + ( Z) Z πa πb Rs = + π a b = + πσδ a b 6

Example (cont.) R = + π aσδ π bσδ This agrees with the formula obtained from the DC equivalent model. (The DC equivalent model assumes that the current is uniform around the boundary, so it is a less general method.) a δ δ b DC equivalent model of coax 7

Internal Inductance An extra inductance per unit length L is added to the TL model in order to account for the internal inductance of the conductors. This extra (internal) inductance consumes imaginary (reactive) power. The external inductance L accounts for magnetic energy only in the external region (between the conductors). This is what we get by assuming PEC conductors. L= L + L Internal inductance L z L z R z C z G z 8

Skin Inductance (cont.) Imaginary (reactive) power per meter consumed by the extra inductance: Circuit model: Skin-effect formula: PI = ( ω L) I P ( ) I = X s J sz dl C A + C B Equate L z L z R z I C z G z 9

Skin Inductance (cont.) Hence: ( ) ω L = X s J sz dl I = = ( ) R s J sz dl I R C C A A + C + C B B

Skin Inductance (cont.) ω L= R Hence X = R or L = R ω

Summary of High-Frequency Formulas Assumption: δ << a for Coax HF R a HF R b = = π aσδ π bσδ R = R + R HF HF HF a b HF a ( HF L ) a X = ω = HF b ( HF L ) b X = ω = π aσδ π bσδ L = L + L HF HF HF a b

Low Frequency (DC) Coax Model At low frequency (DC) we have: R = R + R DC DC DC a b L = L + L DC DC DC a b DC R a = ( ) σ πa DC R b = σ πbt ( ) t = c - b a c b Derivation omitted L = DC L a DC b = + µ 8π 4 c c ln µ b b π 4 3c ( ) ( ) c b c b 3

Tesche Model This empirical model combines the low-frequency (DC) and the high-frequency (HF) skin-effect results together into one result by using an approximate circuit model to get R(ω) and L(ω). F. M. Tesche, A Simple model for the line parameters of a lossy coaxial cable filled with a nondispersive dielectric, IEEE Trans. EMC, vol. 49, no., pp. -7, Feb. 7. Note: The method was applied in the above reference for a coaxial cable, but it should work for any type of transmission line. (Please see the Appendix for a discussion of the Tesche model.) 4

Twin Lead a y Twin Lead x Assume uniform current density on each conductor (h >> a). a y h DC equivalent model δ x R h + π aσδ π aσδ 5

Twin Lead a y Twin Lead x R π aσδ + π aσδ = π aσδ h or R R s π a (A more accurate formula will come later.) 6

Wheeler Incremental Inductance Rule y ˆn x R = Rs J () sz l dl I C A B Wheeler showed that R could be expressed in a way that is easy to calculate (provided we have a formula for L ): R = Rs µ L n L is the external inductance (calculated assuming PEC conductors) and n is an increase in the dimension of the conductors (expanded into the active field region). H. Wheeler, "Formulas for the skin-effect," Proc. IRE, vol. 3, pp. 4-44, 94. 7

Wheeler Incremental Inductance Rule (cont.) The boundaries are expanded a small amount n into the field region. ˆn y Field region x n A B PEC conductors L = external inductance (assuming perfect conductors). R = Rs µ L n 8

Wheeler Incremental Inductance Rule (cont.) Derivation of Wheeler Incremental Inductance rule R = Rs J () sz l dl I C y ˆn Field region (S ext ) x n W H = 4 4 L I WH = µ H ds S ext We then have µ L = H ds I S ext Hence µ L n H dl ( ) = I C L µ µ = H dl = J l dl n I I () sz C C A B PEC conductors 9

Wheeler Incremental Inductance Rule (cont.) ˆn R = Rs J () sz l dl I C n y Field region (S ext ) x From the last slide, A B PEC conductors L n µ L = J () () sz l dl J sz l dl = I C I µ C n Hence R = Rs µ L n 3

Wheeler Incremental Inductance Rule (cont.) Example : Coax L µ b = π a ln a b = + L L L µ b µ b = + ( ) = ( b) n a b π a a π a a µ π a b R = Rs µ L n R= R s + πa πb 3

Wheeler Incremental Inductance Rule (cont.) Example : Twin Lead a y ε, µ x From image theory (or conformal mapping): C L = πε LC = µε cosh h a µ h = π a cosh Z h η h = π a cosh η h Z ln, a << h π a 3

Wheeler Incremental Inductance Rule (cont.) Example : Twin Lead (cont.) L µ h = π a cosh a y ε, µ x Note: By incrementing a, we increment both conductors simultaneously. h L L µ h µ h µ a = = cosh = = n a π a a π a a h π h a a R = Rs µ L n R = R s h h a π a h a 33

Example : Twin Lead (cont.) s h a R R a h a π = a x y h Wheeler Incremental Inductance Rule (cont.) 34 cosh h Z a η π = cosh h L a µ π = cosh C h a πε = ( ) tan G C ω δ = Summary

Attenuation in Waveguide We consider here conductor loss for a waveguide mode. S S c z A waveguide mode is traveling in the positive z direction. C z l d α c = P P f P l d = Rs J z = C S c R J s s s ( ) ds dl 35

Attenuation in Waveguide (cont.) or P l d = R ˆ s n H dl C Power flow: P = E H zˆ ds * f Re ( t t ) S Next, use ( ) E = Z ( zˆ H ) Z = Z or Z t WG WG TE TM t Hence P = Re Z ( zˆ H ) H zˆds WG * f t t S 36

Attenuation in Waveguide (cont.) Vector identity: A ( B C) = B( AC ) C( AB ) ( ) ( ) ( ) ( ) * * * * zˆ H ˆ ˆ ˆ ˆ ˆ ˆ t Ht z = Ht z Ht z = z Ht Ht + Ht z Ht z = H t Hence P f WG = Re Z Ht ds S Assume Z WG = real ( f > f c and no dielectric loss) P f WG = S t Z H ds 37

Attenuation in Waveguide (cont.) Then we have α c nˆ H dl R s C = WG Z H t ds S y S ˆn x C 38

Attenuation in Waveguide (cont.) Total Attenuation: α = α + α c d Calculate α d (assume PEC wall): so k = β jα = k k z c α = Im d k k c where ( ) k = ω µε c = k µ rε r jtanδ 39

Attenuation in Waveguide (cont.) TE Mode y µ = r k = k ε rc b ε rc a x η = η ε rc α c R s b fc = + b( Reη ) ( f / f ) a f c α d π = Im k a 4

Attenuation in db S z z = Waveguiding system (WG or TL) z α V( z) = V() e e z jβz V( z) α z db = log = log ( e ) V () Use log x = ln x ln 4

Attenuation in db (cont.) so α z ln( e ) db = ln = ( α z) ln Hence Attenuation = ln α [db/m] 4

Attenuation in db (cont.) or Attenuation = 8.6859 α [db/m] ( ) 43

Appendix: Tesche Model The series elements Z a and Z b (defined on the next slide) account for the finite conductivity, and give us an accurate R and L for each conductor at any frequency. Z a Z b C G L Z = Z + Z + jωl a b Y = G+ jωc z C = πε ε rc b ln a G ωc ε c = tanδ = L ε c µ b = ln π a 44

Appendix: Tesche Model (cont.) Z a Inner conductor of coax DC L a DC R a HF R a HF L a The impedance of this circuit is denoted as ( ) Z = R + jω L a a a Z b Outer conductor of coax DC L b The impedance of this circuit is denoted as DC R b HF R b HF L b Z = R + jω L ( ) b b b 45

Appendix: Tesche Model (cont.) At low frequency the HF resistance gets small and the HF inductance gets large. DC L a DC R a HF R a HF L a Inner conductor of coax DC L a DC R a HF R a HF L a 46

Appendix: Tesche Model (cont.) At high frequency the DC inductance gets very large compared to the HF inductance, and the DC resistance is small compared with the HF resistance. DC L a DC R a HF R a HF L a Inner conductor of coax HF Ra HF L a 47

Appendix: Tesche Model (cont.) The formulas are summarized as follows: DC R a = ( ) σ πa DC R b = σ πbt ( ) = DC L a µ 8π L 4 c c ln µ b b π 4 DC b = + 3c ( ) ( ) c b c b R HF a π aσδ ( HF ) HF = ω La = ( HF R ) b ω Lb = = π bσδ 48