Wave Phenomena Physics 15c Lecture 11 Fourier Analysis (H&L - - PowerPoint PPT Presentation

Wave Phenomena

Physics 15c

Lecture 11 Fourier Analysis

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

What We Did Last Time

Studied reflection of mechanical waves

Similar to reflection of electromagnetic waves Mechanical impedance is defined by

For transverse/longitudinal waves: Useful in analyzing reflection

Studied standing waves

Created by reflecting sinusoidal waves Oscillation pattern has nodes and antinodes Musical instruments use standing waves to produce their

distinct sound Z T F Zv ± = [ or ]

l

K ρ =

Goals For Today

Define Fourier integral

Fourier series is defined for repetitive functions

Discreet values of frequencies contribute

Extend the definition to include non-repetitive functions

Sum becomes an integral

Discuss pulses and wave packets

Sending information using waves Signal speed and bandwidth Connection with Quantum Mechanics

( )

1

( ) cos sin

n n n n n

f t a a t b t ω ω

∞ =

= + +

∑

Looking Back

In Lecture #5, we solved the wave equation

Normal-mode solutions Using Fourier series, we can make any arbitrary waveform

with 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 2 2 2 2

( , ) ( , )

w

x t c x t t x ξ ξ ∂ ∂ = ∂ ∂

( )

( , )

i kx t

x t e

ω

ξ ξ

±

=

w

c k ω =

( )

1

( , ) ( ) cos( ) sin( )

w n n n n n n n

x t f x c t a k x t b k x t ξ ω ω

∞ =

= − = − + −

∑

2

n

n T π ω =

n w n

k c ω =

A little math work needed

Fourier Series

For repetitive function f(t)

Express cosωnt and sinωnt with complex exponentials

( )

1

( ) cos sin

n n n n n

f t a a t b t ω ω

∞ =

= + +

∑ ∫

=

T

dt t f T a ) ( 1 2 ( )cos

T n n

a f t tdt T ω = ∫ 2 ( )sin

T n n

b f t tdt T ω = ∫ 2

n

n T π ω =

( )

1 1 1 1

cos sin 2 2 2 2

n n m n

i t i t n n n n n n n n n n i t i t m m n n m n

a ib a ib a t b t e e a ib a ib e e

ω ω ω ω

ω ω

∞ ∞ − = = − ∞ − − =−∞ =

− +   + = +     + +     = +        

∑ ∑ ∑ ∑

m n

ω ω = −

m n

b b = −

m n

a 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…

( )

n

i t n n

f t F e

ω ∞ − =−∞

= ∑ 1 ( )

T

F f t dt T = ∫ 1 2 2 1 ( )cos ( )sin ( ) 2

n

T T T i t n n n

F f t tdt i f t tdt f t e d T T T

ω

ω ω   = + =    

∫ ∫ ∫

2

n n n

a ib F + = F a =

same

2 2

1 ( )

n

T i t n T

F f t e dt T

ω −

= ∫

OK because f(t) is repetitive Sum includes n = 0

t

Fourier Integral

Make T ∞

F(ω) is the Fourier integral of f(t)

2

n

n T π ω = ( )

n

i t n n

f t F e

ω ∞ − =−∞

= ∑

2 2

1 ( )

n

T i t n T

F f t e dt T

ω −

= ∫ ( ) lim lim lim 2 ( )

n

i t in t n n T T n n i t n T i t

F f t F e e T F e d F e d

ω ω ω ω

ω ω ω π ω ω

∞ ∞ − − ∆ →∞ →∞ =−∞ =−∞ ∞ − −∞ →∞ ∞ − −∞

= = ∆ ∆ = =

∑ ∑ ∫ ∫

2 T π ω ∆ ≡ 1 ( ) lim ( ) 2 2

i t n T

T F F f t e dt

ω

ω π π

∞ −∞ →∞

≡ =

∫

( ) ( )

i t

f t F e d

ω

ω ω

∞ − −∞

= ∫ 1 ( ) ( ) 2

i t

F f t e dt

ω

ω π

∞ −∞

=

∫

Fourier Integral

Fourier integral F(ω) is

A decomposition of f(t) into different frequencies An alternative, complete representation of f(t)

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

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

( ) ( )

i t

f t F e d

ω

ω ω

∞ − −∞

= ∫ 1 ( ) ( ) 2

i t

F f t e dt

ω

ω π

∞ −∞

=

∫

F(ω) and f(t) are two equally-good representations

• f a same function

Warning

Different conventions exist in Fourier integrals

• Watch out when you read other textbooks

( ) ( )

i t

f t F e d

ω

ω ω

∞ − −∞

= ∫ 1 ( ) ( ) 2

i t

f t F e d

ω

ω ω π

∞ − −∞

=

∫

1 ( ) ( ) 2

i t

F f t e dt

ω

ω π

∞ −∞

=

∫

( ) ( )

i t

F f t e dt

ω

ω

∞ −∞

= ∫

and and

1 ( ) ( ) 2

i t

f t F e d

ω

ω ω π

∞ − −∞

=

∫

and

1 ( ) ( ) 2

i t

F f t e dt

ω

ω π

∞ −∞

=

∫

Square Pulse

Consider a short pulse with unit area

F(ω) is a bunch of little ripples

around ω = 0

Height is 1/2π Area is 1/T

T 1 T

1 2 2

( )

T T T

t f t t  <  =  >  

2 2

1 1 1 ( ) ( ) sin 2 2 2

T i t i t T

T F f t e dt e dt T T

ω ω

ω ω π π πω

∞ −∞ −

= = =

∫ ∫

Fourier

ω 2 T π 1 (0) 2 F π =

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 transformation

Example: Gaussian function

1 ( ) sin 2 T F T ω ω πω =

“width” 2

T π

2 2

2

1 ( ) 2

t T

f t e T π

−

=

2 2

2

1 ( ) 2

T

F e

ω

ω π

−

= T 1 T

Sending Information

Consider sending information using waves

Voice in the air Voice converted into EM signals on a phone cable Video signals through a TV cable

You can’t do it with pure sine waves cos(kx – ωt)

It just goes on Completely predictable No information You need waves that change patterns with time

What you really need are pulses

Pulse width T determines the speed

Pulses must be separated by at least T

Amplitude Modulation

Audio signals range from 20 to 20 kHz

Too low for efficient radio transmission Use a better frequency and modulate amplitude

Modulated waves are no longer pure sine waves

What is the frequency composition?

Carrier wave Audio signal Amplitude-modulated waves

Wave Packet

Consider carrier waves modulated by a pulse

This makes a short train of waves

A wave packet

T = 1/(20 kHz) for audio signals

Fourier integral is

T ( ) f t

2 2

( )

i t T T

e t f t t

ω −

 <  =  >  

2 2

( ) 1 1 ( ) sin 2 ( ) 2

T i t i t T

T F e e dt

ω ω

ω ω ω π π ω ω

− −

− = = −

∫

Wave Packet

Similar to the square pulse

Width is 2π/T Centered at ω = ω0 This is called the bandwidth of your radio station This limits how close the frequencies of radio stations can be

You need 20 kHz for HiFi audio

It’s more like 5 kHz in commercial AM stations

( ) 1 ( ) sin ( ) 2 T F ω ω ω π ω ω − = − ω ω 2 T π 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 / second

Determined by the pulse width in the time domain Translated into bandwidth in the frequency domain We say “bandwidth” to mean “speed of communication”

“Broadband” means “fast communication”

Each medium has its maximum bandwidth

You can split it into smaller bandwidth “channels”

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

You want to minimize the bandwidth of each channel

Telephones carry only between 400 and 3400 Hz

Delta Function

Take the square pulse again

Make it narrower by T → 0 The height grows 1/T → ∞

We get an infinitely narrow pulse with unit area

Dirac’s delta function δ(t) For any function f(t)

T 1 T ( ) t t t δ ∞ =  =  ≠  ( ) 1 t dt δ

∞ −∞

=

∫

( ) ( ) (0) f t t dt f δ

∞ −∞

=

∫

( ) ( ) ( ) f t t t dt f t δ

∞ −∞

− =

∫

and

Delta Function

What is the Fourier integral of δ(t)?

δ(t) contains all frequencies equally

1 1 ( ) ( ) 2 2

i t

F t e dt

ω

ω δ π π

∞ −∞

= =

∫

You can get this also by making T 0 in

1 ( ) sin 2 T F T ω ω πω =

1 ( ) 2

i t

t e d

ω

δ ω π

∞ − −∞

=

∫

Another way of defining δ(t)

Pure Sine Waves

Consider pure sine waves with angular frequency ω0

( )

i t

f t e

ω −

=

( )

1 1 ( ) ( ) 2 2

i t i t i t

F e e dt e dt

ω ω ω ω

ω δ ω ω π π

∞ ∞ − − −∞ −∞

= = = −

∫ ∫

t ω ω ( ) f t ( ) F ω

How Things Fit Together

T infinite t width 1/T infinite ω width F(ω) f(t) Finite pulse and everything else uniform δ(t) δ pulse δ(ω0 – ω) uniform Sinusoidal ω domain t domain Waveform

Pure sine waves and δ pulses are the two extreme

cases of all waves

Everything falls in between Widths 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 ω

Fourier

2 2

( ) ( ) t f t dt t f t dt

∞ −∞ ∞ −∞

= ∫

∫

2 2

( ) ( ) F d F d ω ω ω ω ω ω

∞ −∞ ∞ −∞

= ∫

∫

( )

( )

2 2

t t t ∆ = −

( )

( )

2 2

ω ω ω ∆ = −

r.m.s. = root mean square

Arbitrary Signal Width

( )

( ) ( )

2 2 2 2 2

( ) ( ) t t f t dt t t t f t dt

∞ −∞ ∞ −∞

− ∆ = − = ∫

∫

( )

( ) ( )

2 2 2 2 2

( ) ( ) F d F d ω ω ω ω ω ω ω ω ω

∞ −∞ ∞ −∞

− ∆ = − = ∫

∫

What can we do with this mess??

We can express F(ω) with f(t) as

2 * ( ) 2 * 2

1 ( ) ( ) ( ) 4 1 ( ) ( ) ( ) 2 1 ( ) 2

i t s

F d f t f s e d dtds f t f s t s dtds f t dt

ω

ω ω ω π δ π π

∞ ∞ ∞ ∞ − −∞ −∞ −∞ −∞ ∞ ∞ −∞ −∞ ∞ −∞

= = − =

∫ ∫ ∫ ∫ ∫ ∫ ∫

1 2

( ) ( )

i t

F f t e dt

ω π

ω

∞ −∞

=

∫

Arbitrary Signal Width

Next we take

We can use this to construct

( ) ( )

i t

f t F e d

ω

ω ω

∞ − −∞

= ∫

Differentiate with t

[ ]

( ) ( )

i t

d f t i F e d dt

ω

ω ω ω

∞ − −∞

= − ∫

[ ]

( ) ( )

i t

d F e d i f t dt

ω

ω ω ω

∞ − −∞

=

∫

( ) ( ) ( )

( )

( ) ( )

2 2 * ( ) * 2

( ) ( ) ( ) 1 ( ) ( ) 2 1 ( ) 2

i t

F d F F d d F F e d d d d i f t dt dt

ω ω

ω ω ω ω ω ω ω ω ω ω δ ω ω ω ω ω ω ω ω ω ω ω ω π ω π

∞ ∞ ∞ −∞ −∞ −∞ ∞ ∞ ∞ − − −∞ −∞ −∞ ∞ −∞

′ ′ ′ ′ ′ − = − − − ′ ′ ′ ′ = − −   = −    

∫ ∫ ∫ ∫ ∫ ∫ ∫

t

Arbitrary Signal Width

Now we have

Here comes the trick: we calculate the integral

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

( )

( )

2 2 2

( ) ( ) t t f t dt t f t dt

∞ −∞ ∞ −∞

− ∆ = ∫

∫

( )

2 2 2

( ) ( ) d i f t dt dt f t dt ω ω

∞ −∞ ∞ −∞

  −     ∆ = ∫

∫

( )

2 2

( ) ( ) d t t i i f t dt dt I f t dt κ ω κ

∞ −∞ ∞ −∞

    − − −         = >

∫ ∫

Arbitrary Signal Width

The integral in the denominator becomes Integrate the first term in parts

( ) ( )

* *

( ) ( ) ( ) ( ) d d t t f t f t f t t t f t dt dt dt κ

∞ −∞

      − + −            

∫

( ) ( )

2 * *

( ) ( ) ( ) ( ) ( ) d d t t f t t t f t f t f t t t f t d dt dt

∞ ∞ −∞ −∞

      − + − − + −            

∫

= 0 because the pulse has a finite extent

( )

2 * *

( ) ( ) ( ) ( ) ( ) d d tf t f t f t tf t dt f t dt dt dt κ κ

∞ ∞ −∞ −∞

      − + = −            

∫ ∫

( ) ( ) ( )

( ) ( )

* * 2 2 2 2

( ) ( ) ( ) ( ) ( ) d d t t f t i i f t i i f t t t f t dt dt dt I t f t dt κ ω κ ω κ κ ω

∞ −∞ ∞ −∞

          − − − + − − −                     = ∆ + ∆ + ∫

∫

( )

t κ κ    

Arbitrary Signal Width

We’ve come a long way

Now we got If a quadratic function of κ is always positive,

( ) ( ) ( )

2 2 2

I t κ κ ω κ = ∆ + ∆ − >

( ) ( )

2 2

1 4 D t ω = − ∆ ∆ < 1 2 t ω ∆ ∆ >

finally!

For any signal, the product of the r.m.s. widths ∆t and ∆ω in 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 k Everything works the same way In particular, for any signal traveling in space

Why is it important?

( ) ( )

ikx

f x F k e dk

∞ − −∞

= ∫ 1 ( ) ( ) 2

ikx

F k f x e dx π

∞ −∞

=

∫

1 2 x k ∆ ∆ >

Uncertainty Principle

In Quantum Mechanics, particles are wave packets

Unlike a classical particle, wave packet has a length The position cannot be determined more accurately than ∆x

Momentum is related to the wavenumber by

This means

p k = h 2 h π = h

Planck’s constant = 6.63 × 10−34 J s

2 x p x k ∆ ∆ = ∆ ∆ > h h Heisenberg’s Uncertainty Principle

Summary

Defined Fourier integral

f(t) and F(ω) represent a function in time/frequency domains

Analyzed pulses and wave packets

Time resolution ∆t and bandwidth ∆ω related by

Proved for arbitrary waveform

Rate of information transmission ∝ bandwidth Dirac’s δ(t) a limiting case of infinitely fast pulse Connection with Heisenberg’s Uncertainty Principle in QM

( ) ( )

i t

f t F e d

ω

ω ω

∞ − −∞

= ∫ 1 ( ) ( ) 2

i t

F f t e dt

ω

ω π

∞ −∞

=

∫

1 2 t ω ∆ ∆ >