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

Impedance is a convenient concept

Example: parallel wire transmission line

Wave velocity (in vacuum)

Energy and momentum

Transfer rates for normal mode are

C L Z = c C x L x cw = = ∆ ∆ = 1 µ ε 2 1 I V

w

c I V 2 1 and Momentum Velocity Energy =

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

transmission line is in L and C

Where is “in” L and 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

L C L C L

2 2

ln 2 1 2 1 I a d I x L π µ = ∆ It depends on the µ of material surrounding the wires

Where is the Energy?

Energy in C is stored in the

form of electric field

It’s between, but not in, the wire Electric energy density for a

parallel-wire transmission line is

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

L C L C L

2 2

) / ln( 2 1 2 1 V a d V x C πε = ∆ It depends on the ε of material surrounding the wires

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 Looks very similar to the parallel-wire transmission line

a b x L ln 2 π µ = ∆

( )

a b x C / ln 2 πε = ∆ a and b are the radii of the inner and outer conductors

( )

a d x C / ln πε = ∆ a d x L ln π µ = ∆ ×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

Impedance and wave propagation velocities are

( )

a b x C / ln 2 πε = ∆ a b x L ln 2 π µ = ∆ a b C L Z ln 2 1 ε µ π = = Depends weakly

• n the radii

c LC x cw = = ∆ = 1 1 µ ε speed of light

Real World Coaxial Cable

Actual coaxial cables are filled with insulator

Such as Teflon™ Non-magnetic Impedance is usually 50 Ω (or 75 Ω)

2ε κε ε ≈ = constant dielectric = κ µ µ = ) ( 50 ln 2 2 ) ( 377 ln 2 2 1 ln 2 1 Ω = Ω = = = a b a b a b Z π ε µ π ε µ π

8

1 1 2 10 m/s 2

w

c c εµ κε µ = = ≈ ≈ × 25 . 3 / = a b This is how they are designed

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!

Electromagnetic field holds

the energy

Can we quantify?

Poynting Vector

Poynting vector S is defined by

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 µ = × = × B S E H E

Poynting Vector

µ = × = × B S E H E

What is the unit for S?

E – Volts/meter 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

2

m Watts m A m V =

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

r E 2πε ρ = r I H π 2 =

2 2

4 r I S ε π ρ = a b I dr r I rdrd r I SdA

b a b a A

ln 2 1 2 4

2 2 2

πε ρ πε ρ φ ε π ρ

π

= = =

∫ ∫∫ ∫∫

Power Density

Remember

a b I SdA

A

ln 2 πε ρ =

∫∫

( )

a b x C / ln 2 πε = ∆ VI C qI C xI SdA

A

= = ∆ =

∫∫

ρ

Power Poynting vector gives the density (and the direction)

• f 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 – cwt)

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

Reflection

Reflection at Open End

Let’s call the forward and backward going signals

V+(x – cwt) and V− (x + cwt)

At x = L,

V+ and V− have the same amplitude and polarity

) ( ) ( t c x ZI t c x V

w w

+ − = +

− −

) ( ) ( t c x ZI t c x V

w w

− = −

+ +

) ( ) ( = + + −

− +

t c L I t c L I

w w

) ( ) ( = + − −

− +

t c L V t c L V

w w

) ( ) ( t c L V t c L V

w w

− = +

+ −

Watch sign!

V+(x – cwt) V−(x + cwt)

Reflection at Open End

V+ V−

V+ goes x = 0 → L V− comes back x = L → 0

Current is given by

+ + = ZI

V

− −

− = ZI V I+ I− What exactly is happening at L?

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

They cancel out so that the constraint (I = 0) at x = L is satisfied I+ I− 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 – cwt)

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

At x = L, The reflected wave has the same amplitude but opposite

polarity as the original wave

Current is given by

) ( ) ( = + + −

− +

t c L V t c L V

w w

) ( ) ( t c L V t c L V

w w

− − = +

+ −

V+(x – cwt) V−(x + cwt)

+ + = ZI

V

− −

− = ZI V ) ( ) ( t c L I t c L I

w w

− = +

+ −

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

V(x – cwt) C L Z R = ≠

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

Partial Reflection

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

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

)) ( ) ( ( ) ( ) (

1 1 1 1

t c L I t c L I R t c L V t c L V

w w w w

+ + − = + + −

− + − + + + = ZI

V

− −

− = ZI V ) ( ) ( ) ( ) (

1 1 1 1

t c L V Z R t c L V Z R t c L V t c L V

w w w w

+ − − = + + −

− + − +

) ( ) (

1 1

t c L V Z R Z R t c L V

w w

− + − = +

+ −

) ( ) (

1 1

t c L I Z R Z R t c L I

w w

− + − − = +

+ −

General solutions for reflection

Reflected Energy

+ −

+ − = V Z R Z R V

+ −

+ − − = I Z R Z R I

The energy transfer rate of the reflected wave is

“–” sign because the power is flowing back If R = ∞ or R = 0, P– = –P+ Total reflection Any other value of R reflects a part of the power

+ + + − − −

      + − − =       + − − = = P Z R Z R I V Z R Z R I V P

2 2

Absorbed Energy

V at x = L is given by adding the original and reflected

waves

The power consumed by the resistor is

Energy conservation holds

+ + − +

+ =       + − + = + = = V Z R R V Z R Z R V V L x V 2 1 ) (

+ + + +

+ = + =       + = = P Z R RZ ZI V Z R R V Z R R R R V P

R 2 2 2 2 2

) ( 4 ) ( 4 2 1

+ + + −

= + +       + − = + P P Z R RZ P Z R Z R P P

R 2 2

) ( 4

Summary

Studied deeper into the energy transfer on 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

= × S E H

input 2 2 reflected

) ( ) ( P Z R Z R P + − =

input 2 absorbed

) ( 4 P Z R RZ P + =