Wave PhenomenaPhysics 15c

Lecture 11Fourier Analysis

(H&L Sections 13.1–4)(Georgi Chapter 10)

What We Did Last Time

Studied reflection of mechanical wavesSimilar to reflection of electromagnetic wavesMechanical impedance is defined by

For transverse/longitudinal waves:Useful in analyzing reflection

Studied standing wavesCreated by reflecting sinusoidal wavesOscillation pattern has nodes and antinodesMusical instruments use standing waves to produce their distinct sound

Z TF Zv±=

[ or ] lK ρ=

Goals For Today

Define Fourier integralFourier series is defined for repetitive functions

Discreet values of frequencies contribute

Extend the definition to include non-repetitive functionsSum becomes an integral

Discuss pulses and wave packetsSending information using wavesSignal speed and bandwidthConnection with Quantum Mechanics

( )01

( ) cos sinn n n nn

f t a a t b tω ω∞

=

= + +∑

Looking Back

In Lecture #5, we solved the wave equation

Normal-mode solutions Using Fourier series, we can make any arbitrary waveformwith linear combination of the normal modes

Example: forward-going repetitive waves

Non-repetitive waves also OK if we make T ∞This makes ω continuous

2 22

2 2( , ) ( , )wx t c x tt x

ξ ξ∂ ∂=

∂ ∂( )

0( , ) i kx tx t e ωξ ξ ±= wckω

=

( )1

( , ) ( ) cos( ) sin( )w n n n n n nn

x t f x c t a k x t b k x tξ ω ω∞

=

= − = − + −∑2

nn

Tπω =

n w nk c ω=

A little math work needed

Fourier Series

For repetitive function f(t)

Express cosωnt and sinωnt with complex exponentials

( )01

( ) cos sinn n n nn

f t a a t b tω ω∞

=

= + +∑

∫=T

dttfT

a00 )(1

0

2 ( )cosT

n na f t tdtT

ω= ∫ 0

2 ( )sinT

n nb f t tdtT

ω= ∫

2n

nTπω =

( )1 1

1

1

cos sin2 2

2 2

n n

m n

i t i tn n n nn n n n

n n

i t i tm m n n

m n

a ib a iba t b t e e

a ib a ibe e

ω ω

ω ω

ω ω∞ ∞

−

= =

− ∞− −

=−∞ =

− + + = +

+ + = +

∑ ∑

∑ ∑

m nω ω= −m nb b= −m na a=m n= −

Fourier Series

Define and

How do we calculate Fn?

It’s useful later if I shift the integration range here

Now we take it to the continuous limit…

( ) ni tn

nf t F e ω

∞−

=−∞

= ∑

0 0

1 ( )T

F f t dtT

= ∫

0 0 0

1 2 2 1( )cos ( )sin ( )2

nT T T i t

n n nF f t tdt i f t tdt f t e dT T T

ωω ω = + = ∫ ∫ ∫

2n n

na ibF +

= 0 0F a=

same

2

2

1 ( ) nT i t

n TF f t e dt

Tω

−= ∫ OK because f(t)

is repetitive

Sum includes n = 0

t

Fourier Integral

Make T ∞

F(ω) is the Fourier integral of f(t)

2n

nTπω =( ) ni t

nn

f t F e ω∞

−

=−∞

= ∑2

2

1 ( ) nT i t

n TF f t e dt

Tω

−= ∫

( ) lim lim

lim2

( )

ni t in tnnT Tn n

i tnT

i t

Ff t F e e

T F e d

F e d

ω ω

ω

ω

ωω

ωπ

ω ω

∞ ∞− − ∆

→∞ →∞=−∞ =−∞

∞ −

−∞→∞

∞ −

−∞

= = ∆∆

=

=

∑ ∑

∫

∫

2Tπω∆ ≡

1( ) lim ( )2 2

i tnT

TF F f t e dtωωπ π

∞

−∞→∞≡ = ∫

( ) ( ) i tf t F e dωω ω∞ −

−∞= ∫

1( ) ( )2

i tF f t e dtωωπ

∞

−∞= ∫

Fourier Integral

Fourier integral F(ω) isA decomposition of f(t) into different frequenciesAn alternative, complete representation of f(t)

One can convert f(t) into F(ω) and vice versa

f(t) is in the time domainF(ω) is in the frequency domain

( ) ( ) i tf t F e dωω ω∞ −

−∞= ∫

1( ) ( )2

i tF f t e dtωωπ

∞

−∞= ∫

F(ω) and f(t) are two equally-good representationsof a same function

Warning

Different conventions exist in Fourier integrals

Watch out when you read other textbooks

( ) ( ) i tf t F e dωω ω∞ −

−∞= ∫

1( ) ( )2

i tf t F e dωω ωπ

∞ −

−∞= ∫

1( ) ( )2

i tF f t e dtωωπ

∞

−∞= ∫

( ) ( ) i tF f t e dtωω∞

−∞= ∫

and

and

1( ) ( )2

i tf t F e dωω ωπ

∞ −

−∞= ∫ and

1( ) ( )2

i tF f t e dtωωπ

∞

−∞= ∫

Square Pulse

Consider a short pulse with unit area

F(ω) is a bunch of little ripplesaround ω = 0Height is 1/2πArea is 1/T

T

1T

12

2

( )0

TT

T

tf t

t <= >

2

2

1 1 1( ) ( ) sin2 2 2

Ti t i t

T

TF f t e dt e dtT T

ω ω ωωπ π πω

∞

−∞ −= = =∫ ∫Fourier

ω0

2Tπ

1(0)2

Fπ

=

Pulse Width

Pulse of duration T

The shorter the pulse, the wider the F(ω)(width in t) × (width in ω) = 2π = const

This is a general feature of Fourier transformationExample: Gaussian function

1( ) sin2TF

Tωω

πω= “width”

2Tπ

2

221( )2

tTf t e

Tπ

−=

2 2

21( )2

T

F eω

ωπ

−=

T1T

Sending Information

Consider sending information using wavesVoice in the airVoice converted into EM signals on a phone cableVideo signals through a TV cable

You can’t do it with pure sine waves cos(kx – ωt)It just goes on Completely predictable No informationYou need waves that change patterns with time

What you really need are pulsesPulse width T determines the speed

Pulses must be separated by at least T

Amplitude Modulation

Audio signals range from 20 to 20 kHzToo low for efficient radio transmissionUse a better frequency and modulate amplitude

Modulated waves are no longer pure sine wavesWhat is the frequency composition?

Carrier wave

Audio signal

Amplitude-modulatedwaves

Wave Packet

Consider carrier waves modulated by a pulseThis makes a short train of waves

A wave packet

T = 1/(20 kHz) for audio signals

Fourier integral is

T

( )f t

02

2

( )0

i t T

T

e tf t

t

ω− <= >

02 02

0

( )1 1( ) sin2 ( ) 2

T i t i t

T

TF e e dtω ω ω ωωπ π ω ω

−

−

−= =

−∫

Wave Packet

Similar to the square pulseWidth is 2π/TCentered at ω = ω0

This is called the bandwidth of your radio stationThis limits how close the frequencies of radio stations can be

You need 20 kHz for HiFi audioIt’s more like 5 kHz in commercial AM stations

0

0

( )1( ) sin( ) 2

TF ω ωωπ ω ω

−=

−

ω0ω

2Tπ

To send pulses every T second, your signal must have a minimum spread of 2π/T in ω, which corresponds to 1/T in frequency

Bandwidth

Speed of information transfer = # of pulses / secondDetermined by the pulse width in the time domainTranslated into bandwidth in the frequency domainWe say “bandwidth” to mean “speed of communication”

“Broadband” means “fast communication”

Each medium has its maximum bandwidthYou can split it into smaller bandwidth “channels”

Radio wave frequencies Regulated by the governmentCable TV 750 MHz / 6 MHz = 125 channels

You want to minimize the bandwidth of each channelTelephones carry only between 400 and 3400 Hz

Delta Function

Take the square pulse againMake it narrower by T → 0The height grows 1/T → ∞

We get an infinitely narrow pulse with unit areaDirac’s delta function δ(t)

For any function f(t)

T

1T

0( )

0 0t

tt

δ∞ =

= ≠( ) 1t dtδ

∞

−∞=∫

( ) ( ) (0)f t t dt fδ∞

−∞=∫ 0 0( ) ( ) ( )f t t t dt f tδ

∞

−∞− =∫

and

Delta Function

What is the Fourier integral of δ(t)?

δ(t) contains all frequencies equally

1 1( ) ( )2 2

i tF t e dtωω δπ π

∞

−∞= =∫

You can get this also bymaking T 0 in

1( ) sin2TF

Tωω

πω=

1( )2

i tt e dωδ ωπ

∞ −

−∞= ∫ Another way of defining δ(t)

Pure Sine Waves

Consider pure sine waves with angular frequency ω00( ) i tf t e ω−=

0 0( )0

1 1( ) ( )2 2

i t i ti tF e e dt e dtω ω ωωω δ ω ωπ π

∞ ∞− −

−∞ −∞= = = −∫ ∫

t

ω0ω

( )f t ( )F ω

How Things Fit Together

T0infinitet width

1/Tinfinite0ω width

F(ω)f(t)Finite pulse and everything else

uniformδ(t)δ pulseδ(ω0 – ω)uniformSinusoidalω domaint domainWaveform

Pure sine waves and δ pulses are the two extreme cases of all waves

Everything falls in betweenWidths in t and ω are inversely proportional to each other

Wait… Did I prove it?

Arbitrary Signal Width

Now we consider a signal with an arbitrary shape

Let’s define the average time and the average frequency

Because (energy density) ∝ (amplitude)2

Now we define the r.m.s. widths in t and ω

( )f t ( )F ωFourier

2

2

( )

( )

t f t dtt

f t dt

∞

−∞∞

−∞

= ∫∫

2

2

( )

( )

F d

F d

ω ω ωω

ω ω

∞

−∞∞

−∞

= ∫∫

( ) ( )22t t t∆ = − ( ) ( )22ω ω ω∆ = −

r.m.s. = root mean square

Arbitrary Signal Width

( ) ( )( )2 2

22

2

( )

( )

t t f t dtt t t

f t dt

∞

−∞∞

−∞

−∆ = − = ∫

∫

( ) ( )( )2 2

22

2

( )

( )

F d

F d

ω ω ω ωω ω ω

ω ω

∞

−∞∞

−∞

−∆ = − = ∫

∫

What can we do with this mess??

We can express F(ω) with f(t) as2 * ( )

2

*

2

1( ) ( ) ( )41 ( ) ( ) ( )

21 ( )

2

i t sF d f t f s e d dtds

f t f s t s dtds

f t dt

ωω ω ωπ

δπ

π

∞ ∞ ∞ ∞ −

−∞ −∞ −∞ −∞

∞ ∞

−∞ −∞

∞

−∞

=

= −

=

∫ ∫ ∫ ∫

∫ ∫

∫

12( ) ( ) i tF f t e dtωπω

∞

−∞= ∫

Arbitrary Signal Width

Next we take

We can use this to construct

( ) ( ) i tf t F e dωω ω∞ −

−∞= ∫

Differentiatewith t [ ]( ) ( ) i td f t i F e d

dtωω ω ω

∞ −

−∞= − ∫

[ ]( ) ( )i t dF e d i f tdt

ωω ω ω∞ −

−∞=∫

( ) ( ) ( ) ( )

( ) ( ) 0

2 2 *

( )*

2

( ) ( ) ( )

1 ( ) ( )21 ( )

2

i t

F d F F d d

F F e d d d

di f t dtdt

ω ω

ω ω ω ω ω ω ω ω ω ω δ ω ω ω ω

ω ω ω ω ω ω ω ωπ

ωπ

∞ ∞ ∞

−∞ −∞ −∞

∞ ∞ ∞ − −

−∞ −∞ −∞

∞

−∞

′ ′ ′ ′ ′− = − − −

′ ′ ′ ′= − −

= −

∫ ∫ ∫

∫ ∫ ∫

∫

t

Arbitrary Signal Width

Now we have

Here comes the trick: we calculate the integral

It’s a positive number divided by a positive numberκ is a real number

( )( ) 2

2

2

( )

( )

t t f t dtt

f t dt

∞

−∞∞

−∞

−∆ = ∫

∫( )

2

2

2

( )

( )

di f t dtdt

f t dt

ωω

∞

−∞

∞

−∞

− ∆ =

∫

∫

( )

2

2

( )0

( )

dt t i i f t dtdt

If t dt

κ ωκ

∞

−∞

∞

−∞

− − − = >

∫

∫

Arbitrary Signal Width

The integral in the denominator becomes

Integrate the first term in parts

( ) ( )* *( ) ( ) ( ) ( )d dt t f t f t f t t t f t dtdt dt

κ∞

−∞

− + − ∫

( ) ( )2 * *( ) ( ) ( ) ( ) ( )d dt t f t t t f t f t f t t t f t ddt dt

∞ ∞

−∞−∞

− + − − + − ∫

= 0 because the pulse has a finite extent

( ) 2* *( ) ( ) ( ) ( ) ( )d dtf t f t f t tf t dt f t dtdt dt

κ κ∞ ∞

−∞ −∞

− + = − ∫ ∫

( ) ( ) ( )( ) ( )* *

2 22

2

( ) ( ) ( ) ( )

( )

d dt t f t i i f t i i f t t t f t dtdt dt

I tf t dt

κ ω κ ωκ κ ω

∞

−∞

∞

−∞

− − − + − − − = ∆ + ∆ +

∫

∫

( ) tκ κ

Arbitrary Signal Width

We’ve come a long wayNow we gotIf a quadratic function of κ is always positive,

( ) ( ) ( )2 22 0I tκ κ ω κ= ∆ + ∆ − >

( ) ( )2 21 4 0D t ω= − ∆ ∆ <12

t ω∆ ∆ >finally!

For any signal, the product of the r.m.s. widths ∆t and ∆ωin the time and frequency domain is greater than 1/2

Space and Wavenumber

We have studied Fourier transformation in time t and frequency ω

We can also do it in space x and wavenumber kEverything works the same way

In particular, for any signal traveling in space

Why is it important?

( ) ( ) ikxf x F k e dk∞ −

−∞= ∫

1( ) ( )2

ikxF k f x e dxπ

∞

−∞= ∫

12

x k∆ ∆ >

Uncertainty Principle

In Quantum Mechanics, particles are wave packetsUnlike a classical particle, wave packet has a lengthThe position cannot be determined more accurately than ∆x

Momentum is related to the wavenumber by

This means

p k= h2hπ

=h Planck’s constant = 6.63 × 10−34 J s

2x p x k∆ ∆ = ∆ ∆ >

hh Heisenberg’s Uncertainty Principle

Summary

Defined Fourier integral

f(t) and F(ω) represent a function in time/frequency domainsAnalyzed pulses and wave packets

Time resolution ∆t and bandwidth ∆ω related byProved for arbitrary waveform

Rate of information transmission ∝ bandwidthDirac’s δ(t) a limiting case of infinitely fast pulseConnection with Heisenberg’s Uncertainty Principle in QM

( ) ( ) i tf t F e dωω ω∞ −

−∞= ∫

1( ) ( )2

i tF f t e dtωωπ

∞

−∞= ∫

12

t ω∆ ∆ >