A second look at waves

• Travelling waves• A first look at Amplitude Modulation (AM)• Stationary and reflected waves• Lossy waves: dispersion & evanescence

I think this is the MOST IMPORTANT of my eight lectures, and I am determined not to rush it, even if it means that we don’t cover the whole of lecture 8

Recap on the wave equation

)()(),(

2

22

2

2

ctxgctxftxu

Tc

xuc

tu

−++=

=

∂∂

=∂∂

solution sAlembert'd' studied We

interpret could westring, 1D the for

:is equation waveThe

µ

)()(),( kxtjkxtj BeAetxu +− += ωω :solution this of instance particular astudy Now we

bothnot but or either fix can you, Given

ifonly and if equation wave theof solutiona is ),( that followsit ly,surprisingnot So, and

: thatfind will you,,, derive youIf22

ωω

ω

kckc

txuukuuu

uuuu

xxtt

tttxxx

=

−=−=

2

22

2

2

),|,(

xu

ktu

ktxu

∂∂

⎟⎠⎞

⎜⎝⎛=

∂∂ ω

ω equation wavethe satisfies it and function a such have youely,Alternativ

Space & time variables

wavethe ofvelocity phase the call we, Since

phase the determines term The wavethe of amplitude the is constant The

on econcentratinitially we,simplicity For

cck

kxtA

Aetxf kxtj

=

−

= −

ωω

ω

)(

),( )(

ntdisplaceme spatial 2 a for wavesof number the is and number wavethe called is

h wavelengtthe to equates This

spatially cycles wavethe Evidently,

term spatial the on econcentrat to order in fix First,

π

πλ

π

kk

k

kxtt

2

20

=

=

wavethe offrequency circular the called is

is wavethe of period the above, As aspect. temporal the on econcentrat to order in fix Now

ωωπ2

0

=

=

T

Txx

⎩⎨⎧⎩⎨⎧

h wavelengt,number wave,

space, with associated variables

period ,frequencycircular ,

, with timeassociated variables

λ

ω

kx

Tt

ψ

t

T

Low Circular Frequency, ω

High Circular Frequency, ω

Period, T

Circular frequency

ψ

z

λ

Low Wave Number, k

High Wave Number, k

Wavelength, λ

Wave number

Travelling wave

Travelling wave*)()(

)( )(

kxtfkxtfAekxtf kxtj

by right the to of shift a to scorrespond that Note on econcentrat to continue We

ωωω ω

−=− −

* Change coordinates or read Kreysig page 592

f(ωt)

ωt

f(ωt)

ck

kxtf

ckt

xxkt

xxkttkxtt

=−

==

−=+−+=−

ωω

ωδδ

δωδδδωω

δ

speed withright the to moves that wavea is is, That

same the stays shape the but by increases time that Suppose

)(

0))()(()(

kx

Forwards and backwards( )kxtjAtxf −= ωψ exp),(

Corresponds to a wave that propagates, or travels, in the forward x direction

( )kxtjBtxb += ωψ exp),(Corresponds to a wave that propagates, or travels, in the backward x direction

Similarly:

),(),(),( txtxtx bf ψψψ +=General solution Consists of a forward and backward moving wave

Wave Velocity: Phase Velocity

The phase of a wave was defined to be the term ( )kxt −= ωφ

(Recall that phase φ wraps around every 2π)

Differentiating phase: kdxdtd −= ωφ

ck

vp ==ω

This is why we called c the Phase Velocity

If we are moving with the wave, at the wave velocity, then ( ) 0=φd

ckdt

dxkdxdt ===−ωω or 0It follows that:

The string wave: a non-dispersive wave

µω Tck

vp ===

In this case, vp does not depend on wave properties (such as wave number, k, or angular frequency, ω)

A wave for which the phase velocity is independent of wave properties is called non-dispersive

Earlier, we saw that:

We will return to a discussion of dispersive waves after a brief discussion of standing waves

Dispersive vs non-dispersive waves

Non-dispersive wave (left) and dispersive wave (right)

Standing WavesStanding waves occur when two travelling waves of equal amplitude and speed, but opposite direction, superpose

)()(),( kxtjkxtj BeAetx +− += ωωψIf the amplitudes are equal, A = B

( ))()(),( kxtjkxtj eeAtx +− += ωωψ[ ]jkxjkxtj eeAe += −ω

[ ]kxjkxkxjkxAe tj sincossincos −++= ω

kxAe tj cos2 ω=

We found that, in general:

kxAetx tj cos2),( ωψ =

Nodes occur for2

)12(0cos π+=⇒= nkxkx

Antinodes occur for π)12(1cos +=⇒= nkxkx

All points for which vibrate in phase with each other, and in antiphase with those for which

ψ

z

Antinodes

Nodest = T1

t = T2A standing wave

0cos >kx0cos <kx

Nodes and Antinodes of a plucked string

Example 1: plucked string*

0)(2

)(22

02

)(

≡

⎪⎩

⎪⎨

⎧

<<−

<<=

xg

LxLxLLk

LxxLk

xf if

if

nBxg n all for , Since 0*,0)( =≡

*Kreysig, 8th Edn, page 593

HLT consult or relationsity orthogonalapply either for solve To nB

k0 L

Fourier coefficients

2212 8)1(

πnkB n

n−−=

6-p545Kreysig,ority,orthogonal HLT, Using

We need to extend the string to a periodic function

We know that the solution is a sum of sin terms, so choose an odd function periodic extension:

⎥⎦⎤

⎢⎣⎡ −+−= ...5cos5sin

513cos3sin

31cossin

118),( 2222 t

Lcx

Lt

Lcx

Lt

Lcx

Lktxu ππππππ

π

This is a standing wave

Lctn

Lxn

nk

nth

πππ 2

)12(cos2

)12(sin)12(

1822

+++

⋅

is expression the in term The

x in cosines ofsum a indeed is this that see wemidpoint the to scoordinate Changing

Impedance boundary

optics inlaw sSnell' from thisknow already you wavethe slower the material, the denser the that us tells this

wave,string 1D a of case the in

velocity phase withright the to travels it

wave,travelling a to scorrespond this that seen have We wavethe again Consider

11

11

)(11

1),(

µ

ω

Tc

ck

v

eAtxy

p

xkwtj

=

==

= −

Now we study what happens when there is a sudden change in mass per unit length, m, at x=0: an impedance boundary

This situation arises in many important practical cases:- sound waves moving from one medium to another- waves in a fluid with suddenly changing properties- ultrasound waves crossing a tissue boundary

Impedance boundaryleft the from atboundary impedance an approaches wavethe that Suppose 0),( )(

111 == − xeAtxy xktj ω

ψ

x

µ1 µ2y2

y1

change tdon' that y)(reasonabl assume We

and that Recall

right the to and left the to is string the ofdensity the that Suppose

ω

ωµ

ωµ

µµ

,2

221

11

21

Tk

cTk

cT====

What happens at the impedance boundary?

Some of the energy is absorbed (attenuated)Some of the energy is transmitted in to the different material x > 0 and some of the energy is reflected in the direction -x

22

11

11

ck

ckTc ωω

µ===

We assume initially that all the waves have the same frequency, that is, that the waves propagate in a non-dispersive medium

)()(1

11 ),( xktjxktj eRIetxy +− += ωω

)(2

2),( xktjTetxy −= ω

Put y1 = yincident + yreflected

y2 = ytransmitted

I - amplitude of incident waveT - amplitude of transmitted waveR - amplitude of reflected wave

where I

RT

Boundary Conditions

At x = 0 ttyty ∀= ),0(),0( 21

xty

xty

∂∂

=∂

∂ ),0(),0( 21

(1)

(2)

This is a statement of conservation of energy

I

RT

Evidently, the wave is continuous, and moves continuously at x=0

So (1) gives tjtjtj TeeRIe ωωω =+

TRI =+

TjkRjkIjk 211 −=+−From which we find:

TkRIk 21 )( =−so

tj

tjtj

Tejkx

ty

eRjkIejkx

ty

ω

ωω

22

111

),0(

.),0(

−=∂

∂

+−=∂

∂I

RT

21

12kk

kIT

+=

21

21

kkkk

IR

+−

=

Remembering that TRI =+ we get

and

Applying boundary condition 2

21

12kk

kIT

+=

21

21

kkkk

IR

+−

=The reflected wave is only in phase with the incident wave if k1 > k2

That is, the reflected wave is only in phase if the incident wave is in a denser medium

The transmitted wave is always in phase with the incident wave

ck ω

= 2121

2121 µµµµ

>⇒<⇒<⇒>TTcckkRecall that: so that

Interpreting these two equations

Coefficient of TransmissionThis is defined to be the ratio of the magnitudes of the transmitted and the incident waves

21

12kk

kIT

+==τ

soTTck

µω

µ

ωω===

21

12µµ

µτ

+=We know that

Given the Characteristic Impedance for a string TZ µ=

21

12ZZ

Z+

=τ the coefficient of transmission for a waveWe have

Coefficient of ReflectionThis is defined to be the ratio of the magnitudes of the reflected and the incident waves

21

21

kkkk

IR

+−

==ρ

soTTck

µω

µ

ωω===

21

21

µµµµ

ρ+

−=We know that

We define the Characteristic Impedance for a string TZ µ=

21

21

ZZZZ

+−

=ρ - the coefficient of reflection for a waveSo that

Reflection of wavestransmission from less dense to more dense

5.0

22

2

4

2

1

112

1

1

=

===

=

=

ZZ

ZTTZ

:is That

so and

that so

left the to material the than denser times 4

isboundary impedance the of right the to material the that Suppose

2

2

2

µµ

µµ

µµ

21

12ZZ

Zincident

dtransmitte+

=21

21

ZZZZ

incidentreflected

+−

=

Previous Example Z1/Z2 = 1/2

32

121

1=

+=

incidentdtransmitte

31

121

121

=+

−=

incidentreflected

ψ

x

y2

y1TZ µ=

Recall: and

1Z

2Z

Reflection of wavestransmission from denser to far less dense

2

22

2

4

2

1

112

1

1

=

===

=

=

ZZ

ZTTZ

:is That

so and

that so

left the to material the as dense as times 1/4only

isboundary impedance the of right the to material the that Suppose

2

2

2

µµ

µµ

µµ

21

12ZZ

Zincident

dtransmitte+

=21

21

ZZZZ

incidentreflected

+−

=

ψ

x

y2

y1TZ µ=

Recall: and

The second example Z1/Z2 = 2

34

124

=+

=incident

dtransmitte31

1212

=+−

=incidentreflected

1Z

2Z

Impedance boundaries

Left ventricle and myocardium Cyst in the breast

Dispersive or Lossy (damped) wave

What’s a reasonable model for the decay in the amplitude A of the wave (as a function of x)?

Exponential decay xAexA λ−=)(

Forward moving wave (again)

Recall that a forward moving wave can be described by

( )kxtjeAV −= ω

where

λπω 2

==c

k

In the cases we have considered previously, A was a constant. Now we let A(x) decay exponentially.

Lossy wave

x

V

Now consider a damped, or lossy, wave, whose amplitude A decays as the wave propagates, say according to:

factor" loss" the is where lAexA lx−=)(

Complex wave number

( )

( )

])([

)(

xjlktj

kxtjlx

kxtj

AeeAe

exAV

−−

−−

−

=

=

=

ω

ω

ω

)( jlkl

− :complex is number wavethe , factor loss with,lossy wave a for is, That

1D string in a viscous fluidWe introduced the wave equation by studying the 1D string. We now re-visit the string but this time place it in a viscous medium so that it loses energy to the medium. This causes damped vibrations, familiar to you from the first year.

x δx

β

α

2T

),( txu

Re-doing the analysis that led to the wave equation, the equilibrium equation becomes

damping viscous to scorrespond term the where,t

xtx

xTt

x∂

∂∆

∂∂

−∂∂

∆=∂∂

∆ψβψβψψµ 2

2

2

2

Damped wave equation

2

2

2

2

xT

tt ∂∂

=∂

∂+

∂∂ ψ

µψµβψ

2

22

2

2

xc

tt ∂∂

=∂

∂Γ+

∂∂ ψψψ

Straightforwardly, the damped wave equation for the 1D string becomes:

Slightly more generally, for lossy wave problems

Substituting into this equation the general solution for a forward travelling wave (recalling that the wave number κ=(k-jl) is from now on generally complex)

( )xtjeAV κω −=

VcVjVj 222)( κωω −=Γ+We find that:

222 κωω cj =Γ− is called the Dispersion Relation

( )lkjlkcj 22222 −−=Γ− ωωEquating real and imaginary parts

( )2222 lkc −=ω

klc22=Γω

from which 2

2 21

2⎟⎠⎞

⎜⎝⎛⋅

Γ⋅=

Γ=

ckkcl ω

ωω

( ) ⎥⎦⎤

⎢⎣⎡ Γ+±⎟

⎠⎞

⎜⎝⎛= 2

22 11

21 ωω

ck

in to the dispersion relation:

(the loss factor)

Analysing the dispersion relation

)( jlk −=κ number wave(complex) the substitute We

Extreme case 1: Light Damping 0≈Γ

22 ⎟

⎠⎞

⎜⎝⎛≈

ck ω

ck

cl

ω±≈

Γ±≈

2

so

that is, the loss factor is small

( )

( )kxtjlx

xtj

eeAeAtx

−−

−

=

=ω

κωψ ),(

Not surprisingly, the situation is almost that of an undamped wave.

Extreme case 2. Heavy Damping

kcl 22

ωΓ=

( ) ⎥⎦⎤

⎢⎣⎡ Γ+±⎟

⎠⎞

⎜⎝⎛= 2

22 11

21 ωω

ck

1>>ωΓ

The dispersion relationships:

Heavy damping implies that

( )

lc

ck

=

Γ⎟⎠⎞

⎜⎝⎛≈

⎥⎦⎤

⎢⎣⎡ Γ+±⎟

⎠⎞

⎜⎝⎛=

that so

ωω

ωω

2

22

2

21

1121

2

22

2

2

xc

tt ∂∂

=∂

∂Γ+

∂∂ ψψψ

Phase & group velocity for a vibrating string in a viscous fluid

and the complex wave number was

Recall that the dispersion relation was 222 κωω cj =Γ−

jlk−=κ

( )2222 lkc −=ω :parts real equating

The damped equation was:

2

1 ⎟⎠⎞

⎜⎝⎛−==

klc

kvp

ω ,definitionBy

Phase velocity

2

2

2

1

22:

⎟⎠⎞

⎜⎝⎛−

==

==

kl

ckcv

kcdkd

dkdv

g

g

ω

ωωω

so and

but ,definitionBy

Group velocity

2cvv gp =Note that

Check what happens when we put the damping to zero 0=Γ

ck ω

=

0=l

As expected, these do not depend on the wave properties, so the medium is now non-dispersive

First:

also

ck

vp ==ω

cvg =

Phase velocity:

Group velocity

Consider two (rightwards travelling) waves, (a) and (b), which have different frequencies and wavelengths. The sum of the two waves at the same instant is shown in (c)

( )xktja

aaeAtx −= ωψ ),(

( )xktjb

bbeBtx −= ωψ ),(

bac ψψψ +=

a

apa k

v ω=

b

bpb k

v ω=

Sum of two waves

If the two component waves (a) and (b) have the same velocity, v, their sum (c) maintains the same shape and simply travels to the right at velocity v.

The non-dispersive case

If the velocities of waves (a) and (b) are different, they would then ‘slide’passed one another as they travelled , and their sum, (c) would change shape as the waves travelled along.

The dispersive case

If the wave is non-dispersive, then the sinusoidal component waves of different frequencies that make up the wave, travel with the same velocity.

Amplitude modulated (AM) waves The important thing in any communications system is to be able to send information from one place to another.

This means we have to find a way to impress that information on the radio wave in such a way that it can be recovered at the other end.

This process is known as modulation.

In order to modulate a radio wave, we have to change either or both of the two basic characteristics of the wave: the amplitude or the frequency. Here we just consider the amplitude.

Amplitude Modulated (AM) waves)( kxtjAeV −= ω

( ) ( )[ ]

( ) ( )[ ]

21

2

1

),(),(

),(

ψψψψ

ψωω

ωω

−==

=∆−−∆−

∆+−∆+

txAetx

Aetxxkktj

xkktj

We start, as usual, from a forward moving wave, which defines the carrier of the signal

From this, we create two waves, one by decrementing both ω and k; the other by incrementing them by the same amount, then consider the difference between these two waves:

kk <<∆<<∆ and where ,ωω

Carrier & signal

( ) ( )kxtkxt

kxtkxtkxtkxtA

∆−∆−=

∆−∆−−−∆−∆+−=

ωω

ωωωωψ

sinsin2

)]()cos[()]()cos[()Re(

Carrier Signal

),( txΨ

t

When the signal and carrier are transmitted through a dispersive medium, they travel at different speeds

ω )dispersive-(non ttanconsk

vp ==ω

e)(dispersiv )(ωpp vv =

kFor the dispersive wave, we can write

( )dkdk........

dkd

!k

dkdk ω∆ω∆ω∆ω∆ ≈++= 2

22

2For the amplitude modified wave

( ) ( )

( ) ⎟⎠⎞

⎜⎝⎛ −∆−=

∆−∆−=

dkdxtkxtA

kxtkxtAωωω

ωωψ

sinsin

sinsin

AM in a dispersive medium

Evidently, the carrier has velocity ω/k, which is equal to the phase velocity

kvp

ω=

Similarly, the signal has velocity

dkdvg

ω=

Phase Velocity

This is called the Group Velocity

For a non-dispersive medium, for which ω/k is constant, pvkdk

d==

ωω

For a dispersive medium,

λλ

λλ

ωω

ddv

vv

dkd

ddv

kvdkdv

kvdkdkv

ppg

pp

ppp

−=

+=+=⇒=