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 = ± F Zv � Mechanical impedance is defined by = K ρ � For transverse/longitudinal waves: Z [ or ] T l � 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

Goals For Today � Define Fourier integral � Fourier series is defined for repetitive functions � Discreet values of frequencies contribute ∞ ∑ ( ) = + ω + ω f t ( ) a a cos t b sin t 0 n n n n = n 1 � 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

Looking Back � In Lecture #5, we solved the wave equation ∂ ∂ 2 2 ξ = ξ 2 ( , ) x t c ( , ) x t w ∂ ∂ 2 2 t x ω = c ξ = ξ ± ω i kx ( t ) � Normal-mode solutions � ( , ) x t e w k 0 � Using Fourier series, we can make any arbitrary waveform with linear combination of the normal modes � Example: forward-going repetitive waves π 2 n ω = ∞ n ∑ T ( ) ξ = − = − ω + − ω ( , ) x t f x ( c t ) a cos( k x t ) b sin( k x t ) w n n n n n n = c ω k = n 1 n w n � Non-repetitive waves also OK if we make T � ∞ � This makes ω continuous A little math work needed

Fourier Series � For repetitive function f ( t ) π ∞ 2 n ∑ ( ) = + ω + ω ω = f t ( ) a a cos t b sin t 0 n n n n n T = n 1 1 = ∫ 2 = ∫ 2 T T T ∫ = ω ω a f ( t ) dt a f t ( )cos tdt b f t ( )sin tdt 0 n n n n T T T 0 0 0 � Express cos ω n t and sin ω n t with complex exponentials − + ∞ ∞   a ib a ib ∑ ∑ ( ) ω − ω ω + ω = + i t i t n n n n a cos t b sin t e e   n n n n n n 2 2   = = n 1 n 1 + + − ∞  a ib   a ib  1 ∑ ∑ − ω − ω = + i t i t m m e n n e     m n  2   2  =−∞ = m n 1 = − = = − ω = − ω m n a a b b m n m n m n

Fourier Series Sum includes n = 0 + ∞ a ib = ∑ = − ω = i t F a F n n f t ( ) F e � Define and n 0 0 n n 2 =−∞ n � How do we calculate F n ?   1 2 2 1 T T T ∫ ∫ ∫ ω = ω + ω = i t F f t ( )cos tdt i f t ( )sin tdt f t e ( ) d t n   n n n 2  T T  T 0 0 0 = ∫ 1 T F f t dt ( ) same 0 T 0 � It’s useful later if I shift the integration range here = ∫ 1 T 2 OK because f ( t ) ω i t F f t e ( ) dt n n is repetitive T − T 2 � Now we take it to the continuous limit…

Fourier Integral ∞ π = ∑ = ∫ 1 2 n T 2 − ω i t ω i t ω = f t ( ) F e F f t e ( ) dt n n n n n T T − T 2 =−∞ n � Make T � ∞ ∞ ∞ F π 2 ∑ ∑ − ω − ∆ ω = i t = ∆ ω in t f t ( ) lim F e lim n e ∆ ω ≡ n n ∆ ω →∞ →∞ T T T =−∞ =−∞ n n T F e ∞ ∫ = − ω ω i t lim d n π 2 −∞ →∞ T T 1 ∞ ∞ ∫ ∫ = ω − ω ω i t ω ≡ = ω F ( ) e d i t F ( ) lim F f t e ( ) dt n π π −∞ 2 2 →∞ −∞ T � F ( ω ) is the Fourier integral of f ( t ) 1 = ∫ ∞ ∞ ∫ ω = ω ω − ω ω i t i t F ( ) f t e ( ) dt f t ( ) F ( ) e d π 2 −∞ −∞

Fourier Integral 1 = ∫ ∞ ∞ ∫ ω = ω ω − ω ω i t i t F ( ) f t e ( ) dt f t ( ) F ( ) e d π 2 −∞ −∞ � 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 ( ω ) and f ( t ) are two equally-good representations of a same function � f ( t ) is in the time domain � F ( ω ) is in the frequency domain

Warning � Different conventions exist in Fourier integrals = ∫ 1 ∞ ∞ ∫ ω − ω ω ω = ω i t i t f t ( ) F ( ) e d F ( ) f t e ( ) dt and � π 2 −∞ −∞ 1 = ∫ ∞ ∞ ∫ = ω − ω ω ω ω i t i t f t ( ) F ( ) e d F ( ) f t e ( ) dt and � π 2 −∞ −∞ 1 1 ∞ ∞ ∫ ∫ = ω − ω ω ω = ω i t i t f t ( ) F ( ) e d F ( ) f t e ( ) dt and � π π 2 −∞ 2 −∞ � Watch out when you read other textbooks

Square Pulse 1 � Consider a short pulse with unit area T  < t  1 T =  T 2 f t ( ) > 0 t T   T 2 ω 1 1 1 T ∞ T 2 ∫ ∫ ω = ω = ω = i t i t F ( ) f t e ( ) dt e dt sin Fourier π π πω 2 2 T T 2 −∞ − T 2 1 � F ( ω ) is a bunch of little ripples = F (0) π around ω = 0 2 � Height is 1/2 π π 2 � Area is 1/ T T ω 0

Pulse Width ω π 1 T “width” 2 � Pulse of duration T � ω = F ( ) sin 2 πω T T � The shorter the pulse, the wider the F ( ω ) � (width in t ) × (width in ω ) = 2 π = const � This is a general feature of Fourier transformation � Example: Gaussian function 2 t 2 2 1 ω T − 1 − = 2 f t ( ) e ω = 2 T F ( ) e 2 π π 2 T 2 1 T 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 – ω t ) � 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 Carrier wave Audio signal Amplitude-modulated waves � Modulated waves are no longer pure sine waves � What is the frequency composition?

Wave Packet � Consider carrier waves modulated by a pulse � This makes a short train of waves � A wave packet  − ω i t < e t  T 0 =  2 f t ( ) T > 0 t T   2 � T = 1/(20 kHz) for audio signals f t ( ) � Fourier integral is ω − ω 1 1 ( ) T T 2 ∫ − ω ω = ω = i t i t F ( ) e e dt sin 0 0 π π ω − ω 2 ( ) 2 − T 2 0

Wave Packet ω − ω 1 ( ) T ω = F ( ) sin 0 π ω − ω ( ) 2 π 2 0 � Similar to the square pulse T � Width is 2 π / T ω ω � Centered at ω = ω 0 0 To send pulses every T second, your signal must have a minimum spread of 2 π / T in ω, which corresponds to 1/ T in frequency � 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

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 1 � Take the square pulse again T � Make it narrower by T → 0 � The height grows 1/ T → ∞ T � We get an infinitely narrow pulse with unit area � Dirac’s delta function δ ( t ) ∞ =  t 0 ∞ ∫ δ =  δ = ( ) t ( ) t dt 1 and ≠ 0 t 0  −∞ � For any function f ( t ) ∞ ∞ ∫ ∫ δ = δ − = f t ( ) ( ) t dt f (0) f t ( ) ( t t dt ) f t ( ) 0 0 −∞ −∞

Delta Function � What is the Fourier integral of δ ( t )? You can get this also by 1 1 ∞ making T � 0 in ∫ ω = δ ω = i t F ( ) ( ) t e dt ω 1 T π π 2 2 −∞ ω = F ( ) sin 2 πω T � δ ( t ) contains all frequencies equally 1 ∞ ∫ δ = − ω ω i t ( ) t e d Another way of defining δ ( t ) π 2 −∞

Pure Sine Waves � Consider pure sine waves with angular frequency ω 0 = − ω i t f t ( ) e 0 1 1 ∞ ∞ ∫ ∫ ω = − ω ω = ω ω − = δ ω − ω i t i t i ( ) t F ( ) e e dt e dt ( ) 0 0 0 π π 2 2 −∞ −∞ F ω f t ( ) ( ) t ω ω 0