Transmission Lines 1 Introduction When the source radiates in a wide area, the energy spreads out. The radiated energy is not guided and the transmission of energy through radiation is inefficient. Directive antenna would have huge dimensions in the order of the wave length of the broadcasting electromagnetic waves. We study the transverse electromagnetic (TEM) waves guided by transmission lines. The characteristics of TEM waves guided by transmission lines are the same as those for a uniform plane wave propagating in an unbounded dielectric medium. The three most common types of guiding structures: 1. Parallel-plate transmission line. At microwave frequencies, parallel-plate transmission lines can be fabricated on a dielectric substrate using print-circuit technology. They are often called striplines. 2. Two-wire transmission line: power and telephone lines. 3. Coaxial transmission line: TV cables and the input cables to high-frequency precision measuring instruments. 2 Parallel-Plate Transmission Line (Geometric Model) Physical Model 1 ∂2E ω 2 ∇2E − = 0 since E ∝ eiωt Æ ∇2E + E = 0 (∇2E + k 2E = 0 ) c2 ∂t 2 c2 0 Assume it’s a plane wave propagate in the z with polarization in y direction. d 2 ~ ~ E + k 2E = 0 Æ E = yˆE e−ik0 z+iωt (+z), E = yˆE e+ik0 z+iωt (-z) dz2 y 0 y 0 0 y d ~ −ikz+iωt 1 1 ~ −ikz+iωt E = yˆE0e , H = ()zˆ × yˆ E0e µ v z 0 1 ~ ω H = −xˆ E e−ikz+iωt = H xˆ (k = = ω µε ) µ 0 x v ε When the TEM wave propagates in the parallel-plate transmission line, charge and current may be induced in the plates. Crossing the boundary from dielectric medium to the perfect conduction plates: r r ∇ × E = 0 Æ E// D = E// C = 0 , ∇ ⋅ H = 0 H ⊥D = H ⊥C = 0 y E d σ++++ z 0 lower plate r at y = 0 (lower plate) nˆ = yˆ , ∇ ⋅ D = ρ r Æ E⊥C = 0 & εE⊥D ⋅ nˆ = σ L ~ −ikz+iωt r ~ −ikz+iωt E = yˆE0e Æ σ L = E ⋅ yˆ = εE0e K r r r 1 ~ ∇ × H = J Æ H w − H w = K w Æ K = zˆ E e−ikz+iωt D C L L µ 0 ε Use an Ampere’s loop to obtain the surface current. at y = d (upper plate) nˆ = −yˆ ~ r 1 ~ r σ = −εE e−ikz+iωt = −σ & K = −zˆ E e−ikz+iωt = zˆH = −K U 0 L U µ 0 x L ε In the dielectric media, the electric and magnetic fields satisfy the Maxwell’s eqs: r r y r ∂B r ∂E ∇ × E = − & ∇ × B = µε ∂t ∂t d r r r r z Æ ∇ × E = −iωµH & ∇ × H = iωεE 0 ~ ~ r r E = yˆEy ()z,t , H = xˆH x (z,t) (E →V ,H → σ → I ) basic differential equations iˆ ˆj k ∂ ∂ ∂ dE dH = −iωµH Æ y = iωµH & x = iωεE ∂x ∂y ∂z x dz x dz y 0 Ey 0 Use Maxwell’s equation to get the two basic differential equations. w d dE d r y dy = iωµ H dy , since K = zˆH Æ H = K y ∫ ∫ x U D x U 0 dz 0 d z dV ()z d dV (z) Æ − = iωµK ()z d = iωµ ()K ()z w = iωLI(z) Æ − = iωLI(z) dz U w u dz d dV V L' dI I()z = K ()z w & L = µ is the inductance per unit length ( − ~ ~ ) u w dz l l dt Use integration to obtain the voltage and current relation, and to calculate the inductance per unit length from our physical model. KU w w w dH x y dx = iωε Eydx ∫0 dz ∫ 0 d H z d d d ()H w = ()K w = I()z = iωεE w dz x dz U dz y Æ l w w = −iωε ()− E d = −iωε V ()z = −iωCV ()z d y d dI()z w Æ − = iωCV (z) where C = ε is the capacitance per unit length dz d I 1 dQ 1 d C' dV ~ ~ ()C'V ~ l l dt l dt l dt Use integration to obtain the capacitance per unit length. Time harmonic transmission line equations: Derived differential dV ()z dI()z equation − = iωLI(z) & − = iωCV (z) dz dz d 2V ()z d 2I(z) Æ = −ω 2LCV (z) & = −ω 2LCI(z) dz 2 dz2 ω V ()z = V e−ikz , k = ω LC = = ω µε , V (z,t) = V e−ikz+iωt 0 v 0 dV (z) − dV / dz ikV I()z = I e−ikz , I()z,t = I e−ikz+iωt − = iωLI(z) Æ I()z = = 0 0 dz iωL iωL Use the derived differential equations to obtain the current and voltage variation as a function of z and t. V ()z V0 ωLI0 L µd / w d µ The impedance: Z0 = = = = = = (not divided by I()z I0 kI0 C εw/ d w ε length) The velocity of propagation along the line is: ω 1 1 1 v = = = = k LC ()µd / w (εw/ d )µε Obtain impedance and speed of propagation waves. Microstrip Lines The development of solid-state microwave devices and systems has led to the wide spread use of a form of parallel-plate transmission lines called microstrip lines or simply striplines. w d The microstrip lines are closer to the parallel-plate transmission lines if w >> d . Lossy Parallel-Plate Transmission Lines r r r r Q ∫ D ⋅ ds ∫εE ⋅ ds C = = r r = r r V − ∫ E ⋅ dl − ∫ E ⋅ dl r r r r V − ∫ E ⋅ dl − ∫ E ⋅ dl ε C ε R = = r = r Æ RC = Æ = I ∫ J ⋅ dsr ∫σE ⋅ dsr σ G σ Get surface impedance and resistance from the concept of power loss. When the capacitance between the two conductors, the permittivity and the conductivity are know, we have E G σ w w H = (C = ε ) Æ G = σ S C ε d d K E 1 Averaged Power Dissipation: p = −yˆp = Re{zˆE × xˆH } (Poynting’s avg surface 2 z x theorem) Upper plate: KU = H x E Ez Define a surface impedance: ZSurface = = KSurface H x ZSurface = RSurface + iX Surface Obtain the surface impedance from the concept of attenuated waves in a perfect conductor. A good conductor is a medium for which σ >> ωε . r r ~ ~ ~ ∂2 r ∂E ∂2E E()z,t = E ei(−kz+ωt ), E = µ σ + µ ε Æ 0 ∂z2 c c ∂t c c ∂t 2 2 2 − k = iµcσ cω − µcε cω ~ iµcσ cω E E ωµ ωµ µ ω i µ ω ⎛ 1 1 ⎞ µ ω = = c ~ c = c = c ⎜ + i ⎟ = c ()1+ i H 1 k k σ 1 σ 2σ c E c − iµcσ cω c c ⎝ 2 2 ⎠ c µc ω Ez Ez µcω Æ ZSurface = RSurface + iX Surface = = = ()1+ i KU H x 2σ c Obtain resistance of a parallel-plate transmission line. 1 2 1 2 1 2 1 2 ⎛ RSurface ⎞ pSurface = Re{}KU ZSurface = KU RSurfecae , P = wpSurface = wKU RSurface = I ⎜ ⎟ 2 2 2 2 ⎝ w ⎠ The effective series resistance per unit length for both plates of a parallel-plate transmission line of width w is ⎛ RSurface ⎞ 2 µcω R = 2⎜ ⎟ = (two times from the viewpoint of power dissipation) (per ⎝ w ⎠ w 2σ c unit length) E Elw Vw Rw Z ≡ = = = (not divided by length) surface K Kwl Il l 3 General Transmission-Line Equations -Q (Electronic Circuit Model) +Q r r r r Q ∫ D ⋅ ds ∫εE ⋅ ds C = = r r = r r V − ∫ E ⋅ dl − ∫ E ⋅ dl r r r r V − ∫ E ⋅ dl − ∫ E ⋅ dl ε C ε R = = r = r Æ RC = Æ = I ∫ J ⋅ dsr ∫σE ⋅ dsr σ G σ The capacitor is the capability of storing charges per volt. The conductor is the current flow per volt. σ w w G = C (C = ε in parallel plate transmission lines) Æ G = σ ε d d i(z,t) R∆z L∆z C∆z i(z,t) G∆z i(z+∆z,t) v(z,t) v(z+∆z,t) ∂i()z,t v()z,t − R∆zi()z,t − L∆z = v()z + ∆z,t ∂t ∂v()z,t ∂i()z,t − = Ri()z,t + L ∂z ∂t ∂v()z,t i()z,t − G∆zv()z,t − C∆z = i()z + ∆z,t ∂t ∂i()z,t ∂v()z,t − = Gv()z,t + C ∂z ∂t let v()z,t = Re{V (z)eiωt } and i()z,t = Re{I(z)eiωt } dV ()z dI(z) Æ − = ()R + iωL I(z) and − = ()G + iωC V (z) dz dz d 2V d ⎛ dI ⎞ = ()− ()R + iωL I = (R + iωL)⎜− ⎟ = ()R + iωL (G + iωC)V (z) dz 2 dz ⎝ dz ⎠ d 2V let V ()z = e−kz & = k 2V (z) Æ k 2 = (R + iωL)(G + iωC) Æ dz 2 k = ()R + iωL (G + iωC) let k = α + iβ = (R + iωL)(G + iωC) (α : attenuation constant, β : phase constant) + −kz − +kz + −kz − +kz + −kz+iωt − +kz+iωt V ()z = V0 e +V0 e and I()z = I0 e + I0 e (v(z,t) = V0 e +V0 e and + −kz+iωt − +kz+iωt i()z,)t = I0 e + I0 e dV ()z − = (R + iωL)I(z) Æ kV +e−kz − kV −ekz = (R + iωL)(I +e−kz + I −ekz ) dz 0 0 0 0 + − V0 V0 R + iωL Æ − = − − = I0 I0 k R + iωL R + iωL V /lz characteristic impetance: Z0 = = (∝ ) (not divided by length) k G + iωC I /lz Example: Demonstrate the analog between the wave characteristics on a transmission line and uniform plane waves in a lossy medium.