Wave Phenomena Physics 15c Lecture 9 LC transmission line (H&L - PowerPoint PPT Presentation

Wave Phenomena Physics 15c Lecture 9 LC transmission line (H&L Section 9.3, 9.4, 9.6) Wave Reflection (H&L Chapter 6)

What We Did Last Time � Studied waves on an LC transmission line � Mechanism is totally different � Same wave equation � Voltage and current are proportional L Z = � Impedance is a convenient concept C � Example: parallel wire transmission line � Wave velocity (in vacuum) ∆ ∆ x x 1 = = = c w c � Energy and momentum ε µ L C 0 0 � Transfer rates for normal mode are Energy = 1 V I 1 0 0 Momentum V I and 0 0 2 c Velocity 2 w

Goals for Today � Find where the energy is in electromagnetic waves � It’s not in carried through the wires � Introduce Poynting vector � Coaxial cable � Another example of an LC transmission line � Wave reflection � What happens at the end of a transmission line?

Where is the Energy? � Energy of waves on an LC L L L transmission line is in L and C � Where is “in” L and C ? C C � Energy in L is stored in the form of magnetic field � It’s around, but not in, the wire � Magnetic energy density for a parallel-wire transmission line is It depends on the µ of material µ 1 L 1 d = 2 2 surrounding the wires I ln I ∆ π 2 x 2 a

Where is the Energy? L L L � Energy in C is stored in the form of electric field C C � It’s between, but not in, the wire � Electric energy density for a parallel-wire transmission line is πε It depends on the ε of material 1 C 1 = 2 2 V V surrounding the wires ∆ 2 x 2 ln( d / a ) � The energy is not carried by the physical wires, but in the space around them as electromagnetic field � Wires are guiding the waves, but not transmitting them

Parallel Wire Transmission Line � Parallel wire transmission line is surrounded by E and B fields � The energy is carried in these fields. � They extend to infinity � It’s a leaky way of carrying energy � Radiation loss becomes a problem

Coaxial Cable � A wire surrounded by a cylindrical conductor makes a coaxial cable � Electromagnetic field is confined between the inner and outer conductors � Magnetic field circles � Electric field runs radially � It’s leak-tight

Inductance and Capacitance � L and C can be calculated using Physics 15b again � Derivation is left to the problem set πε µ C 2 L b a and b are the radii of the = = 0 0 ln ( ) ∆ ∆ π inner and outer conductors x ln b / a x 2 a � Looks very similar to the parallel-wire transmission line µ πε L d C = = 0 ln 0 ( ) ∆ π ∆ x a x ln d / a × 2 b ↔ d

Parallel Wire � Coaxial � One can replace one of the two wires with a metal sheet in the middle � The sheet reflects the electric field from the wire so that it looks as if there is a mirror image of the wire � E and B field exist only above the sheet � Wrap the sheet around the wire � Coaxial cable � Factor 2 here and there due to the wrapping process

Impedance and Velocity � For coaxial cables πε µ 2 C L b = = 0 0 ln ( ) ∆ ∆ π x ln b / a x 2 a � Impedance and wave propagation velocities are µ L 1 b Depends weakly = = Z 0 ln π ε C 2 a on the radii 0 1 1 = ∆ = = c w x c speed of light ε µ LC 0 0

Real World Coaxial Cable � Actual coaxial cables are filled with insulator ε = κε ≈ 2 ε κ = � Such as Teflon™ dielectric constant 0 0 µ = µ � Non-magnetic � 0 1 1 c = = ≈ ≈ × 8 c 2 10 m/s w εµ κε µ 2 0 0 � Impedance is usually 50 Ω (or 75 Ω ) vacuum impedance µ µ Ω 1 b 1 b 377 ( ) b = = = = Ω 0 Z ln ln ln 50 ( ) π ε π ε π 2 a 2 2 a a 2 2 0 = b / a 3 . 25 This is how they are designed

Energy Density � Energy is carried in the space between the conductors � It’s like an air-filled pipe in which sound is traveling � The cable/pipe looks like the medium, but the true medium is inside them! � Electromagnetic field holds the energy � Can we quantify?

Poynting Vector � Poynting vector S is defined by × B = × = S E H E µ � Direction is perpendicular to both E and B � For both parallel-wire and coaxial cables, E and B are always perpendicular to each other and to the cable itself � Poynting vector points the direction of wave transmission

Poynting Vector × B = × = S E H E µ � What is the unit for S ? � E – Volts/meter V A Watts = 2 m m m � H – Ampares/meter � It looks like “how much power is flowing in a unit area” � One may call it the power density � Let’s calculate this for a coaxial cable

Power Density � In a coaxial cable, E and H at radius r ρ is given by I = = E H 2 πε π r 2 r 0 � ρ is the charge density of the inner wire � I is the current on the inner wire ρ I � Poynting vector is = S π ε 2 2 4 r � Integrate this 0 π 2 b ρ ρ b ρ I I 1 I b ∫∫ ∫∫ ∫ = φ = = SdA rdrd dr ln π ε πε πε 2 2 4 r 2 r 2 a 0 0 0 A 0 a a

Power Density ρ I b ∫∫ = SdA ln πε 2 a 0 A πε C 2 = � Remember 0 ( ) ∆ x ln b / a ρ ∆ xI qI ∫∫ = = = SdA VI Power C C A Poynting vector gives the density (and the direction) of power carried by electromagnetic waves

Quick Summary � Electromagnetic waves in LC transmission lines carry energy and momentum � Energy (and momentum) is not carried by the wires, but by the electromagnetic field in/around them � The wires are just guiding the EM field � Poynting vector S = E × H = power density � This is not only for LC transmission lines � It will become important when we study electromagnetic radiation in free space = light and radio waves

Impedance Matching � Characteristic impedance of an LC transmission line is a very useful concept � It connects the voltage and the current: V = ZI � An infinite LC transmission line ≈ a Z Ω resistor � If we send a signal through a 50 Ω coaxial cable terminated with a 50 Ω resistor, it will be absorbed � Signal will vanish as if it went down on an infinite cable � What if we forgot to put the resistor? What if we terminated with 0 Ω ? How about 100 Ω ?

Reflection at Open End V ( x – c w t ) � Imagine a finite-length cable without termination � Signal V = V ( x – c w t ) travels on it � The current I is given by I = V / Z until it reaches the end � When the signal reaches x = L , there is no more wire � The current does not have place to go � Must flow back � This generates a backward-going wave Reflection

Reflection at Open End � Let’s call the forward and backward going signals V + ( x – c w t ) and V − ( x + c w t ) − = − + = − + V ( x c t ) ZI ( x c t ) V ( x c t ) ZI ( x c t ) + + − − w w w w − + + = Watch sign! I ( L c t ) I ( L c t ) 0 � At x = L , + − w w − − + = V ( L c t ) V ( L c t ) 0 + − w w + = − V ( L c t ) V ( L c t ) − + w w � V + and V − have the same amplitude and polarity V + ( x – c w t ) V − ( x + c w t )

Reflection at Open End V + � V + goes x = 0 → L � V − comes back x = L → 0 � Current is given by + = ZI = − V V ZI + − − I + V − What exactly is happening at L ? I −

Mirror Waves � Imagine the cable continued beyond x = L � Imagine a backward going pulse is coming from x = 2 L � Two pulses meet at x = L I + I − They cancel out so that the constraint ( I = 0) at x = L is satisfied I −

Mirror Waves � Do the same with V + and V − � They add up instead of canceling out V + V − The amplitude at x = L gets twice as big V −

Reflection at Shorted End V ( x – c w t ) � What if the end is short-circuited � This time, the voltage is forced to be 0 at x = L � This also results in a reflection � Let’s calculate

Reflection at Shorted End − + + = V ( L c t ) V ( L c t ) 0 � At x = L , + − w w + = − − V ( L c t ) V ( L c t ) − + w w � The reflected wave has the same amplitude but opposite polarity as the original wave V + ( x – c w t ) V − ( x + c w t ) � Current is given by + = ZI = − V V ZI + − − + = − I ( L c t ) I ( L c t ) − + w w

Mirror Waves � This time, the voltage is canceling out at x = L � Current doubles at x = L I + V + V − I − V − I −

Partial Reflection V ( x – c w t ) L ≠ = R Z C � An LC transmission line is terminated with a resistor that does not match its impedance � It’s easy to guess the outcome: partial reflection � Part of the energy is absorbed by the resistor � Part of the energy will come back as a reflected wave