Wave Phenomena Physics 15c Lecture 8 LC transmission line (H&L - PowerPoint PPT Presentation

Wave Phenomena Physics 15c Lecture 8 LC transmission line (H&L Section 9.2)

What We Did Last Time � Studied sound in solid, liquid and gas � Young’s modulus Y , bulk modulus M B and volume density ρ v determine everything � Ideal gas is particularly simple: only need molecular mass � Analyzed transverse waves on string � Equation of motion looks the same as longitudinal waves � Just replace elastic modulus K with tension T � Solution has the same characteristics as well � Energy and momentum densities are also identical

Goals for Today � Explore another, totally different type of waves: Electromagnetic waves on an LC transmission line � Start from coupled LC oscillators � Find and solve the wave equation � Calculate the wave velocity � Introduce impedance � Discuss real examples – parallel wires, coaxial cables � What about the energy and momentum?

Coupled LC Oscillators � Imagine an infinite array of inductors and capacitors L L L L L L V n I n q n C C C C C C 0V Electrical analogue of coupled pendulums

Charge, Voltage and Current � Voltage across C L L L V n V n +1 q n = n V I n-1 I n q n C C C � Voltage across L dI − = − V V L n + 1 n n dt � Charge conservation Similar to the dq coupled pendulums − − = I I n n 1 n dt 2 d V 1 ( ) � Equation of “motion” =  − − −  C n V V ( V V )   + − n 1 n n n 1 2 dt L

Going Continuous � Assume we have L ’s and C ’s L L L V n V n +1 at every ∆ x I n-1 I n q n C C � Replace indices with position V n → I n → q n → V ( x ) I ( x ) q ( x ) ∆ x � The equations become: ∂ ∂ dI I x ( ) V x ( ) − = − − = + ∆ − ∆ V V L n L V x ( x ) V x ( ) x + 1 n n ∂ ∂ dt t x ∂ ∂ dV V x ( ) I x ( ) − = = − ∆ − − ∆ n I I C C I x ( x ) I x ( ) x − 1 n n ∂ ∂ dt t x

Wave Equation ∂ ∂ ∂ ∂ L I x t ( , ) V x t ( , ) C V x t ( , ) I x t ( , ) − = − = ∆ ∂ ∂ ∆ ∂ ∂ x t x x t x � L / ∆ x is the linear inductance density (Henry/m) � C / ∆ x is the linear capacitance density (Farad/m) � Combine them to get the wave equation ∂ ∆ ∂ ∂ ∂ ∂ 2 2 2 x V ( x , t ) I ( x , t ) C V ( x , t ) − = = − ∂ ∂ ∂ ∂ ∆ ∂ ∂ 2 2 t x L x t x x t ∂ ∂ 2 2 V ( x , t ) L C V ( x , t ) For the = ∂ ∆ ∆ ∂ voltage 2 2 x x x t

How About the Current? � For the current I ( x , t ) ∂ ∂ ∂ ∂ L I x t ( , ) V x t ( , ) C V x t ( , ) I x t ( , ) − = − = ∆ ∂ ∂ ∆ ∂ ∂ x t x x t x ∂ ∂ ∂ ∆ ∂ ∂ 2 2 2 L I ( x , t ) V ( x , t ) x I ( x , t ) − = = − ∂ ∆ ∂ ∂ ∂ ∂ ∂ 2 2 t x t x t C x x ∂ ∂ 2 2 L C I ( x , t ) I ( x , t ) = ∆ ∆ ∂ ∂ 2 2 x x t x V ( x , t ) and I ( x , t ) satisfy the same wave equation

EM vs. Mechanical � Correspondences between the constants in the electromagnetic and mechanical wave equations: L ↔ � Inductance ↔ mass m � Capacitance ↔ spring constant ↔ C 1 / k S � Compare the wave equations ∂ ∂ ∂ ξ ∂ ξ 2 2 2 2 L C V ( x , t ) V ( x , t ) ( x , t ) ( x , t ) = ρ = K l ∆ ∆ ∂ ∂ ∂ ∂ 2 2 2 2 x x t x t x m k S ∆ x ∆ x

Normal Mode Solutions ∂ ∂ ∂ ∂ 2 2 2 2 L C V ( x , t ) V ( x , t ) L C I ( x , t ) I ( x , t ) = = ∆ ∆ ∂ ∂ ∆ ∆ ∂ ∂ 2 2 2 2 x x t x x x t x � We know the normal mode solution = ± ω = ± ω V ( x , t ) V exp( i ( kx t )) I ( x , t ) I exp( i ( kx t )) 0 0 � Throw them into the wave equations = ω ∆ ∆ x x L C = ω = c w 2 2 k ∆ ∆ k L C x x K for mechanical waves ρ l

Voltage vs. Current = ω ∆ ∆ x x = c w k L C � The solutions for V ( x , t ) and I ( x , t ) must be related ∂ ∂ ∂ ∂ L I x t ( , ) V x t ( , ) C V x t ( , ) I x t ( , ) − = − = ∆ ∂ ∂ ∆ ∂ ∂ x t x x t x � Throw in = ± ω = ± ω V ( x , t ) V exp( i ( kx t )) I ( x , t ) I exp( i ( kx t )) 0 0 to the first equation: ω L V L L ω = = = i I V ik m 0 m m 0 0 ∆ ∆ x I xk C 0 � Voltage and current are proportional to each other � Sounds like Ohm’s law

Impedance ± ω i ( kx t ) V ( x , t ) V e V L = = = 0 0 m ± ω i ( kx t ) I ( x , t ) I e I C 0 0 � Ohm’s law: V = RI � We could call this resistance � We call it impedance Z L Z = � The name implies that, unlike a real C resistor, it does not consume power � The transmission line is made entirely of reactive (inductive and capacitive) components � We can now write the solution as = m = ± ω i ( kx t ) V ( x , t ) ZI ( x , t ) V e 0 + if forward-going; − if backward-going

Impedance L L L L L L I ( t ) V ( t ) C C C C C C � V ( t ) and I ( t ) are related by V ( t ) = ZI ( t ) � From the wave generator (= AC power supply), it looks identical to a simple resistor I ( t ) An infinite LC transmission line is V ( t ) L indistinguishable from a simple R = C resistor to the circuit driving it

Impedance Matching L L L L L I ( t ) V ( t ) L C C C C C R = C � Cut an LC transmission line and attach a resistor � If R equals to the impedance, the finite-length LC transmission line would look as if infinitely long � Which in turn looks like a simple resistor of value R � The line is terminated with a matching impedance � Properly terminated LC transmission line is easy to drive, as it behaves just like a resistor

Propagation Velocity = ω ∆ ∆ x x 1 = = ∆ c w x k L C LC � Propagation velocity is proportional to ∆ x � We can choose ∆ x freely � Velocity can be made as big as we like � It cannot be true if we consider special relativity � c w can never be made faster than c � Where is the catch? � We must look at an actual example

Parallel Wire Transmission Line � Imagine a pair of copper distance d radius a wires running in parallel � Like a phone line � There are inductance and capacitance � We can calculate them using Physics 15b � It’s an easy exercise � Then we can calculate the wave velocity � Suppose the wires are strung in vacuum

Inductance r � Inductance is defined by I the magnetic field � Suppose we ran current I − I d − and – I on these wires r � Calculate the magnetic flux in the d × ∆ x rectangle � Magnetic field at distance r and d – r from the two wires is µ µ µ = I I Permeabili ty of vacuum = + B 0 0 0 π π − 2 r 2 ( d r ) − = π × 7 4 10 H/m � Integrate between the wires to calculate the flux − d a ∫∫ ∫ Φ = = ∆ B r drdx ( ) x B r dr ( ) a

Inductance − − r d a d a µ   I 1 1 ∫ ∫ = + 0   Bdr dr I π − 2 r d r   a a µ [ ] I − I − d a = − − 0 ln r ln d r d − π r 2 a µ − µ I d a I d = ≈ 0 ln 0 ln π π a a µ I d � Total magnetic flux inside the rectangle is Φ = ∆ 0 x ln π a Φ = � Inductance L is defined by LI µ Φ µ L d 0 ln d = = = ∆ 0 ln L x ∆ π π x a I a

Capacitance r � Capacitance is defined by ρ the electric field E −ρ � Suppose we have charge densities d − r ρ (C/m) and – ρ on these wires � Calculate the electric potential between the two wires � Electric field at distance r and d – r from the two wires is ρ ρ ε = Permittivi ty of vacuum = + E 0 πε πε − 2 r 2 ( d r ) − = × 12 8 . 85 10 F/m 0 0 � Integrate between the wires to calculate the potential − d a = ∫ V E r dr ( ) a

Capacitance − − d a d a ρ   1 1 r ∫ ∫ = = +   V Edr dr ρ πε − 2  r d r  0 a a E ρ [ ] −ρ − d a = − − ln r ln d r πε 2 a d − r 0 ρ − ρ d a d = ≈ ln ln πε πε a a 0 0 � Total charge in a ∆ x long piece is = ρ ∆ q x q = � Capacitance C is defined by CV ρ πε C d C ρ ∆ = = x ln 0 ( ) πε ∆ a x ln d / a Got all what 0 we need!

Impedance and Wave Velocity µ πε L d C Vacuum = = 0 ln 0 ( ) ∆ π ∆ x a x ln d / a impedance µ Ω L 1 d 377 ( ) d � Impedance is = = = Z 0 ln ln π ε π C a a 0 ∆ ∆ x x 1 speed of = = = � Wave velocity is c w c ε µ light! L C 0 0 A pair of parallel wires in vacuum transmit electromagnetic waves at the speed of light

Speed of Light � One can run the wires in other medium than vacuum � ε 0 and µ 0 becomes ε and µ of the medium = εµ c 1 / � Wave velocity always smaller than c w � Any electrical circuit has at least two wires � A parallel wire transmission line is the simplest possible � Anything more complex has more inductance and capacitance per unit length smaller c w � Wave velocity of an LC transmission line never exceeds the speed of light � Now we feel better…