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 MB and volume density

ρv 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

C L C L C L C L C L C L

0V Vn qn In Electrical analogue of coupled pendulums

Charge, Voltage and Current

Voltage across C Voltage across L Charge conservation

Equation of “motion”

L C L C L

Vn qn In In-1 Vn+1 C q V

n n =

dt dI L V V

n n n

− = −

+1 1 n n n

dq I I dt

− −

=

( )

2 1 1 2

1 ( )

n n n n n

d V C V V V V dt L

+ −

= − − −    

Similar to the coupled pendulums

Going Continuous

Assume we have L’s and C’s

at every ∆x

Replace indices with position

The equations become:

L C L C L

Vn qn In In-1 Vn+1 ) (x V Vn → dt dI L V V

n n n

− = −

+1

( ) ( ) ( ) V x C I x x I x t ∂ = − ∆ − ∂

∆x

) (x I In → ) (x q qn → ( ) ( ) ( ) I x L V x x V x t ∂ − = + ∆ − ∂ dt dV C I I

n n n

= −

−1

( ) V x x x ∂ ∆ ∂ ( ) I x x x ∂ − ∆ ∂

Wave Equation

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

( , ) ( , ) L I x t V x t x t x ∂ ∂ − = ∆ ∂ ∂ ( , ) ( , ) C V x t I x t x t x ∂ ∂ − = ∆ ∂ ∂

2 2 2 2 2

) , ( ) , ( ) , ( t t x V x C x t t x I x t x V L x ∂ ∂ ∆ − = ∂ ∂ ∂ = ∂ ∂ ∆ − x ∂ ∂ t ∂ ∂

2 2 2 2

) , ( ) , ( t t x V x C x L x t x V ∂ ∂ ∆ ∆ = ∂ ∂ For the voltage

How About the Current?

For the current I(x, t)

2 2 2 2 2

) , ( ) , ( ) , ( x t x I C x t x t x V t t x I x L ∂ ∂ ∆ − = ∂ ∂ ∂ = ∂ ∂ ∆ − ( , ) ( , ) L I x t V x t x t x ∂ ∂ − = ∆ ∂ ∂ ( , ) ( , ) C V x t I x t x t x ∂ ∂ − = ∆ ∂ ∂ x ∂ ∂ t ∂ ∂

2 2 2 2

) , ( ) , ( x t x I t t x I x C x L ∂ ∂ = ∂ ∂ ∆ ∆

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:

Inductance ↔ mass Capacitance ↔ spring constant

Compare the wave equations

m L ↔

S

k C / 1 ↔

2 2 2 2

) , ( ) , ( x t x V t t x V x C x L ∂ ∂ = ∂ ∂ ∆ ∆

2 2 2 2

) , ( ) , ( x t x K t t x

l

∂ ∂ = ∂ ∂ ξ ξ ρ x m ∆ x kS∆

Normal Mode Solutions

We know the normal mode solution

Throw them into the wave equations

2 2 2 2

) , ( ) , ( x t x V t t x V x C x L ∂ ∂ = ∂ ∂ ∆ ∆

2 2 2 2

) , ( ) , ( x t x I t t x I x C x L ∂ ∂ = ∂ ∂ ∆ ∆ )) ( exp( ) , ( t kx i V t x V ω ± = )) ( exp( ) , ( t kx i I t x I ω ± =

2 2

k x C x L = ∆ ∆ ω C x L x k cw ∆ ∆ = = ω

l

K ρ for mechanical waves

C x L x k cw ∆ ∆ = = ω

Voltage vs. Current

The solutions for V(x, t) and I(x, t) must be related

Throw in

to the first equation:

Voltage and current are proportional to each other

Sounds like Ohm’s law

L i I V ik x ω = ∆ m )) ( exp( ) , ( t kx i V t x V ω ± = )) ( exp( ) , ( t kx i I t x I ω ± = C L xk L I V m m = ∆ = ω ( , ) ( , ) L I x t V x t x t x ∂ ∂ − = ∆ ∂ ∂ ( , ) ( , ) C V x t I x t x t x ∂ ∂ − = ∆ ∂ ∂

Impedance

C L I V e I e V t x I t x V

t kx i t kx i

m = = =

± ± ) ( ) (

) , ( ) , (

ω ω

Ohm’s law: V = RI We could call this resistance

We call it impedance Z The name implies that, unlike a real

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

C L Z =

) (

) , ( ) , (

t kx i

e V t x ZI t x V

ω ±

= = m + if forward-going; − if backward-going

Impedance

C L C L C L C L C L C L

I(t) V(t)

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 indistinguishable from a simple resistor to the circuit driving it V(t) C L R =

Impedance Matching

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

C L C L C L C L C L

I(t) V(t) C L R =

Propagation Velocity

LC x C x L x k cw 1 ∆ = ∆ ∆ = = ω

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

cw 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

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

radius a distance d

Inductance

Inductance is defined by

the magnetic field

Suppose we ran current I

and –I on these wires

Calculate the magnetic flux in the d × ∆x rectangle Magnetic field at distance r and d – r from the two wires is Integrate between the wires to calculate the flux

) ( 2 2 r d I r I B − + = π µ π µ H/m 10 4 vacuum

• f

ty Permeabili

7 −

× = = π µ

I −I

r r d −

( ) ( )

d a a

B r drdx x B r dr

−

Φ = = ∆

∫∫ ∫

Inductance

[ ]

a d I a a d I r d r I dr r d r I Bdr

a d a a d a a d a

ln ln ln ln 2 1 1 2 π µ π µ π µ π µ ≈ − = − − =       − + =

− − −

∫ ∫

I −I

r r d −

a d I x ln π µ ∆ = Φ

Total magnetic flux inside the rectangle is Inductance L is defined by

LI = Φ

0 ln d

L x I a µ π Φ = = ∆ a d x L ln π µ = ∆

Capacitance

Capacitance is defined by

the electric field

Suppose we have charge densities

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 Integrate between the wires to calculate the potential

) ( 2 2 r d r E − + = πε ρ πε ρ F/m 10 85 . 8 vacuum

• f

ty Permittivi

12 −

× = = ε

r r d −

ρ −ρ

E ( )

d a a

V E r dr

−

= ∫

Capacitance

[ ]

a d a a d r d r dr r d r Edr V

a d a a d a a d a

ln ln ln ln 2 1 1 2 πε ρ πε ρ πε ρ πε ρ ≈ − = − − =       − + = =

− − −

∫ ∫

r

ρ

Total charge in a ∆x long piece is Capacitance C is defined by

x q ∆ = ρ CV q = a d C x ln πε ρ ρ = ∆

( )

a d x C / ln πε = ∆

r d −

E

−ρ

Got all what we need!

Impedance and Wave Velocity

Impedance is Wave velocity is

( )

a d x C / ln πε = ∆ a d x L ln π µ = ∆ a d a d C L Z ln ) ( 377 ln 1 π ε µ π Ω = = = Vacuum impedance c C x L x cw = = ∆ ∆ = 1 µ ε speed of light!

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

ε0 and µ0 becomes ε and µ of the medium Wave velocity

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

Wave velocity of an LC transmission line never

exceeds the speed of light

Now we feel better…

εµ / 1 =

w

c always smaller than c smaller cw

Energy

Energy is in two forms

Electrostatic energy in C Magnetic energy in L

Both are easy to calculate We know

Energies in C and L are identical

L C L C L

Vn qn In In-1 Vn+1

∆x

2

2 1 CV EC =

2

2 1 LI EL = I C L ZI V = =

L C

E LI I C L C E = =         =

2 2

2 1 2 1 Sounds familiar?

Energy

Consider a forward-going wave

Total energy density is given by Time average gives Multiply by the wave velocity to get the transfer rate

2 2 2

2 cos ( )

C

E dE CV C V kx t dx x x x ω × = = = − ∆ ∆ ∆ ) cos( ) , ( t kx V t x V ω − =

2

1 2 dE C V dx x = ∆ Average energy density

2 2 2 2

1 1 1 1 1 1 2 2 2 2 2

w

V dE C C C c V x V V V I dt x LC x L Z = = ∆ = = = ∆ ∆

Power

The energy transfer rate is

This makes perfect sense if we consider the power needed to

create this wave

Consider an AC power supply driving this

transmission line

It must produce The power is given by (voltage) x (current)

2 1 I V t V t ZI t V ω cos ) ( ) ( = = t I V t I t V ω

2

cos ) ( ) ( = ×

average

2 1 I V

Energy conservation

Momentum

Can electrical wires carry momentum?

What if they were feeding power to an electromagnet? We can sort-of argue that electricity carries force

Traveling waves on an LC transmission line do carry

momentum

But we cannot see any mass moving with it We can only see that forces are needed/produced at the ends

• f the transmission line

Let’s take a look at a parallel wire example…

Parallel Wire 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 ) ( 2 2 r d I r I B − + = π µ π µ = × F I B B

d a r a

dr

− =

= ×

∫

F I B

Lorentz Force

2 2 2

1 1 2 ln

d a d a a a

I F IBdr dr r d r I d L I a x µ π µ π

− − 

 = = +   −   ≈ = ∆

∫ ∫

a d x L ln π µ = ∆

The driving circuit gets this force backward The terminating resistor gets this force forward

We can see this as the waves transmitting force

F F

Energy vs. Momentum

The waves transmit force

This is also the momentum transfer rate

For normal mode waves, the driver must produce

The force is then

2

I x L F ∆ = t I x L F ω

2 2 0 cos

∆ = t V t ZI t V ω cos ) ( ) ( = =

w

c I V I x L

2

2 1 2 1 = ∆

average

I C L ZI V = = LC x cw 1 ∆ =

This is energy transfer rate divided by the velocity

Summary

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 =