Wave Phenomena Physics 15c

Lecture 8 LC Transmission Line

Wave Reflection

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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

Example: parallel wire transmission line

  Wave velocity (in vacuum)

Z =

LC

cw =

ΔxL

ΔxC

=1ε0µ0

= c

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?

Energy Energy is in two forms   Electrostatic energy in C   Magnetic energy in L

We know both from 15b

We also know

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

Vn

qn In

Δx EC =

12

CV 2

EL =

12

LI2

V = ZI =

LC

I EC =

12

CLC

I⎛

⎝⎜

⎞

⎠⎟

2

=12

LI2 = EL

Energy Density Total energy density is given by

  Multiply by the wave velocity to get the energy transfer rate

For a normal mode

  Average energy density:

  Average energy transfer rate:

dEdx

=2 × EC

Δx=

CΔx

V 2 =CΔx

VLC

I =LCΔx

VI =VIcw

dEdt

= cw

dEdx

=VI That’s power, as we should expect

V(x,t) =V0 cos(kx −ω t)

dEdx

=12

V0I0

cw

dEdt

=12

V0I0

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

Lorentz Force Magnetic field between the wires is

Lorentz force is   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

I

d – r

r

B =

µ0I2πr

+µ0I

2π(d − r )

F = I×B B

F = IB dra

d −a

∫ =µ0I

2

2π1r+

1d − r

⎛⎝⎜

⎞⎠⎟

dra

d −a

∫

=µ0I

2

πln d − a

a=

LΔx

I2

LΔx

=µ0

πln d − a

a

From last lecture

Lorentz Force

  Using the impedance again

We can see this as the waves transmitting force   Which is also the rate of momentum transfer

We find (momentum density) = (energy density)/(velocity)

F

F =

LΔx

I2 =LΔx

VIZ

=LΔx

CL

VI =LCΔx

VI =VIcw

F

F =

dpdt

=VIcw

momentum density =dpdx

=VIcw

2

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

  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

12

LΔx

I2 =12µπ

ln da

I2

12

CΔx

V 2 =12

πεln(d / a)

V 2

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

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

Coaxial Cable L and C can be calculated using Physics 15b again   For inner and outer radii of a and b

  Derivation is left to the problem set

Impedance and wave velocities are

LΔx

=µ0

2πln b

a

CΔx

=2πε0

ln b / a( )

Z =

LC

=1

2πµ0

ε0

ln ba

cw =

ΔxLC

=1ε0µ0

= c

Real World Coaxial Cable Actual coaxial cables are filled with insulator   Such as polyethylene

  Non-magnetic

  Impedance is typically 50 Ω (or 75 Ω)

  For κ = 2.25, b/a = 3.49   This is how the coaxial cables are designed

ε =κε0 ≈ 2.25ε0 κ = dielectric constant

µ = µ0

Z =

12π

µε

ln ba=

12π

µ0

κε0

ln ba=

377(Ω)2π κ

ln ba= 50(Ω)

cw =

1εµ

=1

κε0µ0

≈c

2.25= 2 ×108m/s

vacuum impedance

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

  Direction is perpendicular to both E and B   For both parallel-wire and coaxial cables,

Poynting vector points the direction of wave transmission

S = E ×H = E ×

Bµ

Power Density In a coaxial cable, E and H at radius r is given by

  ρ is the charge density of the inner wire   I is the current on the inner wire

Poynting vector is   Integrate this

E =

ρ2πε0r

H =I

2πr

S =

ρI4π 2ε0r

2

S dAA∫∫ =

ρI4π 2ε0r

2 r dr dϕa

b

∫0

2π

∫ =ρI

2πε0

1r

dra

b

∫ =ρI

2πε0

ln ba

= ρIΔxC

=QIC

=VI

CΔx

=2πε0

ln b / a( )Power carried by

the EM 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

Imagine a finite-length cable without termination   Signal V = V(x – cwt) 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

V(x – cwt)

Reflection at Open End Let’s call the forward and backward going signals V+(x – cwt) and V- (x + cwt)

  At x = L, the total current must be zero

V+ and V- have the same amplitude and polarity

V+(x − cwt) = ZI+(x − cwt) V−(x + cwt) = −ZI−(x + cwt)

I+(L − cwt) + I−(L + cwt) = 0

V+(L − cwt) −V−(L + cwt) = 0

V+(x – cwt)

V-(x + cwt)

V−(L + cwt) =V+(L − cwt)

Watch sign!

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

Exactly what happens at x = L?

V+ = ZI+ V− = −ZI−

V+

V-

I+

I-

Mirror Waves Imagine the cable continued beyond x = L   Imagine a backward going pulse is coming from x = 2L   Two pulses meet at x = L

  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

I+

I-

I- V+

V-

V-

Reflection at Shorted End

What if the end is short-circuited   This time, the voltage is forced to be 0 at x = L

  The reflected wave has the same amplitude but opposite polarity as the original wave

  Current is given by

V(x – cwt)

V+(L − cwt) +V−(L + cwt) = 0

V+(x – cwt)

V-(x + cwt)

I−(L + cwt) = I+(L − cwt) V+ = ZI+ V− = −ZI−

Mirror Waves This time, the voltage is canceling out at x = L   Current doubles at x = L

V+

V-

V- I+

I-

I-

Partial Reflection

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

V(x – cwt)

R ≠ Z =

LC

Partial Reflection At x = L, V and I must satisfy Ohm’s law

  Use V+ = ZI+ and V− = −ZI−

  Solve for V−

  Setting R = ∞ or 0 gives open/shorted-end solutions

V+(L − cwt) +V−(L + cwt) = R(I+(L − cwt) + I−(L + cwt))

V+(L − cwt) +V−(L + cwt) = R

ZV+(L − cwt) − R

ZV−(L + cwt)

V−(L + cwt) = R − Z

R + ZV+(L − cwt)

I−(L + cwt) = −

R − ZR + Z

I+(L − cwt)

General solutions for reflection

Energy and Power Power in the reflected waves is

  “–” 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

Energy conservation:

V− =

R − ZR + Z

V+

I− = −

R − ZR + Z

I+ P− =V−I− = −

R − ZR + Z

⎛⎝⎜

⎞⎠⎟

2

V+I+ = −R − ZR + Z

⎛⎝⎜

⎞⎠⎟

2

P+

PR =

(V+ +V− )2

R=

1R

2RR + Z

⎛⎝⎜

⎞⎠⎟

2

V+2 =

4R(R + Z)2 V+ZI+ =

4RZ(R + Z)2 P+

P− + PR =

R − ZR + Z

⎛⎝⎜

⎞⎠⎟

2

P+ +4RZ

(R + Z)2 P+ = P+

Summary Energy and momentum in LC transmission lines

  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

Pabsorbed =

4RZ(R + Z)2 Pinput

Preflected =

(R − Z)2

(R + Z)2 Pinput

S = E ×H

dEdx

=VIcw

V− =

R − ZR + Z

V+ I− = −

R − ZR + Z

I+

P =

dEdt

=VI

dpdx

=VIcw

2 F =

dpdt

=VIcw