Wave Phenomena Physics 15c. Lecture 8 LC Transmission Line Wave Reflection

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1 Wave Phenomena Physics 15c Lecture 8 LC Transmission Line Wave Reflection

2 Midterm Exam #1 Midterm #1 has been graded Class average = 80.4 Standard deviation = 14.6 Your exam will be returned in the section Contact your TF is you can t make this week s section!"#$%$&'()*(+&,#$( ''" (!" ('" )!" )'" *!" *'" +!" +'" #!!" If you are concerned about your grade, please me and we will discuss plans for the rest of the semester

3 What We Did Last Time Studied waves on an LC transmission line Mechanism is totally different Same wave equation Voltage and current are proportional Impedance is a convenient concept Z = Example: parallel wire transmission line L C Wave velocity (in vacuum) c w = Δx L Δx C = 1 ε 0 µ 0 = c

4 Today s Goals Energy and momentum density in the LC transmission line Transfer rate = power and force Discuss where the energy is in electromagnetic waves It s not in carried through the wires Recall Poynting vector from 15b/153 Also: coaxial cable Another example of an LC transmission line Wave reflection What happens at the end of a transmission line?

5 Energy Energy is in two forms Electrostatic energy in C Magnetic energy in L We know both from 15b E C = 1 2 CV 2 E L = 1 2 LI 2 q n V n I n Δx We also know V = ZI = L C I E = 1 C 2 C L C I Electrostatic energy in C = Magnetic energy in L This holds locally, i.e., at each point x at any time t Recall (kinetic energy) = (potential energy) in mechanical waves 2 = 1 2 LI 2 = E L

6 Energy Density Total energy density is given by de dx = 2 E C Δx Multiply by the wave velocity to get the energy transfer rate de dt = c w de dx = VI For a normal mode Average energy density: = C Δx V 2 = C Δx V L C I = LC Δx That s power, as we should expect V(x,t) = V 0 cos(kx ωt) de dx = 1 2 V 0 I 0 c w VI = VI c w Average energy transfer rate: de dt = 1 2 V 0 I 0

7 Momentum Waves on an LC transmission line do carry momentum But we do not see any mass moving with it We can only see that forces are needed/produced at the ends of the transmission line Let s drive a transmission line that is terminated To create waves, the wires from the driving circuit must run across the gap between the wires Same for the terminating resistor Current flows across the wires There is B field between the wires Lorentz force

8 Lorentz Force Magnetic field between the wires is B = µ I 0 2πr + µ 0 I 2π(d r) Lorentz force is F = I B It s parallel to the direction of the wires Backward (into the screen) at the driver Forward (out of the screen) at the terminator Integrate I B to calculate the total force F = d a IB dr = µ I 2 0 2π a = µ 0 I 2 π d a a 1 r + 1 d r dr B I d r ln d a a = L Δx I 2 L Δx = µ 0 π ln d a a r From last lecture

9 Lorentz Force F F Using the impedance again F = L Δx I 2 = L VI Δx Z = L C Δx L VI = LC VI VI = Δx c w We can see this as the waves transmitting force Which is also the rate of momentum transfer F = dp dt = VI c w momentum density = We find (momentum density) = (energy density)/(velocity) dp dx = VI c w 2

10 Where is the Energy? Energy of waves on an LC transmission line is in L and C Where is in L and C? Energy in L is stored in the form of magnetic field Energy in C is stored in the form of electric field Recall for the parallel-wire transmission line 1 L 2 Δx I 2 = 1 µ 2 π ln d a I 1 C 2 2 Δx V 2 = 1 πε 2 ln(d / a) V 2 Energy densities depend on the magnetic/electric properties µ and ε of the medium around the wires 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

11 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 It may interfere with E or B fields generated by nearby circuits Radiation loss becomes a problem at higher frequencies

12 Coaxial Cable One can replace one of the two wires with an infinite metal sheet in the middle Recall: image current method E and B field exist only above the sheet Wrap the sheet around the wire Coaxial cable Electromagnetic field is confined between the inner and outer conductors Magnetic field circles. Electric field runs radially

13 Coaxial Cable L and C can be calculated using Physics 15b again For inner and outer radii of a and b L Δx = µ 0 2π ln b a C Δx = 2πε 0 ln b / a ( ) Derivation is left to the problem set Impedance and wave velocities are Z = L C = 1 2π c w = µ 0 ε 0 Δx LC = 1 ε 0 µ 0 ln b a = c

14 Real World Coaxial Cable Actual coaxial cables are filled with insulator Such as polyethylene ε = κε ε 0 κ = dielectric constant µ = µ 0 c w = 1 εµ = 1 Non-magnetic κε 0 µ 0 c 2.25 = m/s Impedance is typically 50 Ω (or 75 Ω) vacuum impedance Z = 1 2π µ ε ln b a = 1 2π For κ = 2.25, b/a = 3.49 µ 0 κε 0 ln b a = 377(Ω) 2π κ ln b a = 50(Ω) This is how the coaxial cables are designed

15 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! Poynting vector gives us the density of energy flow S = E H = E B µ Direction is perpendicular to both E and B For both parallel-wire and coaxial cables, Poynting vector points the direction of wave transmission

16 Power Density In a coaxial cable, E and H at radius r is given by ρ E = H = I 2πε 0 r 2πr ρ is the charge density of the inner wire I is the current on the inner wire Poynting vector is Integrate this S da = A 2π b a S = ρi 4π 2 ε 0 r 2 ρi r dr dϕ = ρi 1 4π 2 2 ε 0 r 2πε 0 r dr = ρi ln b 2πε 0 a 0 = ρi Δx C = QI C = VI b a Power carried by the EM waves C Δx = 2πε 0 ln b / a ( )

17 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?

18 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 We call it reflection

19 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 w t) = ZI + (x c w t) V (x + c w t) = ZI (x + c w t) At x = L, the total current must be zero I + (L c w t) + I (L + c w t) = 0 V + (L c w t) V (L + c w t) = 0 V + and V - have the same amplitude and polarity V + (x c w t) V (L + c w t) = V + (L c w t) V - (x + c w t) Watch sign!

20 Reflection at Open End V + and V look the same V + goes from x = 0 to L V - comes back from x = L to 0 I + and I look flipped because V + V + = ZI + V = ZI Exactly what happens at x = L? I + V - I -

21 Mirror Waves Imagine the cable continued beyond x = L V + Imagine a backward going pulse is coming from x = 2L Two pulses meet at x = L V - I + I - V - I - I + and I cancel each other Meet I = 0 constraint at x = L V + and V add up Double the pulse height at x = L

22 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 V + (L c w t) +V (L + c w t) = 0 The reflected wave has the same amplitude but opposite polarity as the original wave Current is given by V + (x c w t) V - (x + c w t) V + = ZI + V = ZI I (L + c w t) = I + (L c w t)

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

24 Partial Reflection V(x c w t) R Z = L 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

25 Partial Reflection At x = L, V and I must satisfy Ohm s law V + (L c w t) +V (L + c w t) = R(I + (L c w t) + I (L + c w t)) Use V + = ZI + and V = ZI V + (L c w t) +V (L + c w t) = R Z V + (L c w t) R Z V (L + c w t) Solve for V V (L + c w t) = R Z R + Z V + (L c w t) I (L + c w t) = R Z R + Z I + (L c w t) Setting R = or 0 gives open/shorted-end solutions General solutions for reflection

26 Energy and Power Power in the reflected waves is P = V I = R Z R + Z sign because the power is flowing back If R = Z, P = 0 No reflection If R = or R = 0, P = P + Total reflection Power consumed in the resistor is P R = (V + +V )2 R = 1 R Energy conservation: P + P R = R Z R + Z 2 2R R + Z V + I + = R Z R + Z 2 V + 2 = 2 2 P + V = R Z R + Z V + I = R Z R + Z I + 4R (R + Z) V ZI = 4RZ (R + Z) P 2 + P + + 4RZ (R + Z) P = P 2 + +

27 Summary Energy and momentum in LC transmission lines de dx = VI c w The energy is carried by the EM field around the wires The wires guide the waves Poynting vector gives the power density Studied the wave reflection Determined by the impedance matching Power is reflected or absorbed according to P reflected = P = de dt (R Z)2 = VI V = R Z R + Z V + (R + Z) 2 P input dp dx = VI 2 c w P absorbed = S = E H I = R Z R + Z I + F = dp dt = VI 4RZ c w (R + Z) 2 P input