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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
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
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
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
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
−=
+=+=⇒=