Wave Phenomena Physics 15c

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1 Wave Phenomena Physics 5c Lecture Fourier Analysis (H&L Sections 3. 4) (Georgi Chapter )

2 What We Did Last ime Studied reflection of mechanical waves Similar to reflection of electromagnetic waves Mechanical impedance is defined by F = ± Zv 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 = [ or K] ρ l

3 Goals For oday Define Fourier integral Fourier series is defined for repetitive functions Discreet values of frequencies contribute = + n n + n n n= 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 ( ω ω ) f() t a a cos t b sin t Connection with Quantum Mechanics

4 Looking Back In Lecture #5, we solved the wave equation ξ( xt, ) = c w ξ( xt, ) t x ω i( kx± ωt) c Normal-mode solutions ξ( xt, ) = ξe w = k Using Fourier series, we can make any arbitrary waveform with linear combination of the normal modes Example: forward-going repetitive waves ( ) ξ( xt, ) = f( x ct) = a cos( kx ω t) + bsin( kx ω t) w n n n n n n n= Non-repetitive waves also OK if we make his makes ω continuous ω = k A little math work needed n = π n c ω n w n

5 Fourier Series For repetitive function f(t) a = f ( t) dt = + n n + n n n= ( ω ω ) f() t a a cos t b sin t an f t tdt = ()cosω n n = ω = π n b f()sin t ωntdt Express cosω n t and sinω n t with complex exponentials an ibn iωnt an + ibn iωnt ancosωnt+ bnsin ωnt = e + e n am + ibm iωmt an + ibn iωnt = e + e m= n= ( ) n= = m = n a m = a n b m = b n ω m = ω n n

6 Fourier Series Sum includes n = an + ibn Define Fn and How do we calculate F n? iω nt Fn = ()cos ()sin () f t ω ntdt + i f t ω ntdt = f t e d t F = f() t dt = It s useful later if I shift the integration range here iω nt Fn = f() t e dt Now we take it to the continuous limit F = a same i n f() t = Fne ω n= OK because f(t) is repetitive t

7 Fourier Integral i nt f() t Fne ω = iω nt Fn = f() t e dt n= Make iω nt Fn in ωt f() t = lim Fn e = lim e ω ω n= = lim Fe n π = iωt iωt F( ω) e dω d ω n= F(ω) is the Fourier integral of f(t) iωt f() t = F( ω) e dω ω = n ω π n π iωt F( ω) lim Fn = f( t) e dt π π iωt F( ω) = f( t) e dt π

8 Fourier Integral iωt f() t = F( ω) e dω 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 iωt F( ω) = f( t) e dt π F(ω) and f(t) are two equally-good representations of a same function

9 Warning Different conventions exist in Fourier integrals iωt f() t F( ω) e d = () = ( ) π iωt f t F ω e d iωt f() t F( ω) e d Watch out when you read other textbooks ω and ω iωt F( ω) = f( t) e dt π and = ω and π F( ω) = f( t) e iωt dt iωt F( ω) = f( t) e dt π

10 Square Pulse t < f() t = t > iωt iωt ω F( ω) = f( t) e dt e dt sin π = π = πω Consider a short pulse with unit area Fourier F(ω) is a bunch of little ripples around ω = Height is /π Area is / F() = π π ω

11 Pulse Width ω F( ω) = sin π width πω he shorter the pulse, the wider the F(ω) Pulse of duration (width in t) (width in ω) = π = const his is a general feature of Fourier transformation Example: Gaussian function f() t = π e t F( ω) = e π ω

12 Sending Information Consider sending information using waves Voice in the air Voice converted into EM signals on a phone cable Video signals through a V 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 determines the speed Pulses must be separated by at least

13 Amplitude Modulation Audio signals range from to khz oo low for efficient radio transmission Use a better frequency and modulate amplitude Carrier wave Audio signal Modulated waves are no longer pure sine waves What is the frequency composition? Amplitude-modulated waves

14 Wave Packet Consider carrier waves modulated by a pulse his makes a short train of waves A wave packet f() t = ω e t < = /( khz) for audio signals Fourier integral is i t t > f() t iω ( t iωt ω ω) F( ω) = e e dt = sin π π( ω ω)

15 Wave Packet ( ω ω) F( ω) = sin πω ( ω ) Similar to the square pulse Width is π/ Centered at ω = ω o send pulses every second, your signal must have a minimum spread of π/ in ω, which corresponds to / in frequency his is called the bandwidth of your radio station his limits how close the frequencies of radio stations can be You need khz for HiFi audio ω It s more like 5 khz in commercial AM stations π ω

16 Bandwidth Speed of information transfer = # of pulses / second Determined by the pulse width in the time domain ranslated 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 V 75 MHz / 6 MHz = 5 channels You want to minimize the bandwidth of each channel elephones carry only between 4 and 34 Hz

17 Delta Function ake the square pulse again Make it narrower by he height grows / We get an infinitely narrow pulse with unit area Dirac s delta function δ(t) δ () t t = = t and For any function f(t) δ () tdt= f() t δ () t dt = f() f t δ t t dt = f t () ( ) ( )

18 Delta Function What is the Fourier integral of δ(t)? iωt F( ω) = δ( t) e dt = π π δ(t) contains all frequencies equally iωt δ () t = e dω Another way of defining δ(t) π You can get this also by making in ω F( ω) = sin πω

19 Pure Sine Waves Consider pure sine waves with angular frequency ω f() t = i t e ω F ω = e e dt e dt δω ω π = π = iωt i( ) t ( ) i ω t ω ω ( ) f() t F( ω) t ω ω

20 How hings Fit ogether Waveform t domain t width ω domain ω width Sinusoidal uniform infinite δ(ω ω) δ pulse δ(t) uniform infinite Finite pulse and everything else f(t) F(ω) / 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?

21 Arbitrary Signal Width Now we consider a signal with an arbitrary shape f () t F( ω) Fourier Let s define the average time and the average frequency t = t f() t f() t dt dt ω = ω F( ω) dω F( ω) dω Because (energy density) (amplitude) Now we define the r.m.s. widths in t and ω ( t) = ( t t ) ( ω) = ( ω ω ) r.m.s. = root mean square

22 Arbitrary Signal Width ( ) ( ) ( ) () t = t t = f() t dt t t f t dt ( ) ( ) ( ) F( ) ω = ω ω = F( ω) dω ω ω ω dω What can we do with this mess?? We can express F(ω) with f(t) as F ω dω f t f se dωdtds 4π * = f () t f ( s) δ ( t s) dtds π = f() t dt π * iω ( t s) ( ) = ( ) ( ) = π ( ) ( ) i ω F ω f t e t dt

23 Arbitrary Signal Width Next we take iωt f() t = F( ω) e dω d [ f () t ] i F ( ) e iωt d with t dt = ω ω ω i t d ω ωf( ω) e dω = i f ( t) dt We can use this to construct Differentiate * ( ) ( ) = ( ) ( )( ) ( ) ( ) d = i ω f() t dt π dt [ ] ω ω F ω dω ω ω F ω ω ω F ω δ ω ω dωdω * i( ω ω ) t = ( ω ω ) F( ω) ( ω ω ) F ( ω ) e dωdω d π t

24 Arbitrary Signal Width Now we have ( t) = ( ) t t f() t dt f() t Here comes the trick: we calculate the integral I ( κ ) It s a positive number divided by a positive number κ is a real number dt ( ω ) = d i ω f() t dt dt f() t d t t iκ i ω f() t dt dt = > f() t dt dt

25 Arbitrary Signal Width I ( κ) ( t) κ ( ω) κ = + + he integral in the denominator becomes κ d d dt dt ( ) + ( ) * * t t f () t f () t f () t t t f () t dt Integrate the first term in parts d d dt dt * * ( t t ) f() t + κ ( t t ) f() t f () t + f() t ( t t ) f () t dt κ = because the pulse has a finite extent d * d * ( t t ) f() t iκ i ω f () t + iκ i ω f() t ( t t ) f () t dt dt dt f() t d * d * ( tf () t ) f () t + f () t tf () t dt = κ f () t dt dt dt dt

26 Arbitrary Signal Width We ve come a long way Now we got I ( ) ( t) ( ) κ = + κ ω κ > If a quadratic function of κ is always positive, ( ) ( ) D= 4 t ω < finally! t ω > For any signal, the product of the r.m.s. widths t and ω in the time and frequency domain is greater than /

27 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 ikx f( x) = ikx F( k) e dk Fk ( ) = f ( xe ) dx π In particular, x k > for any signal traveling in space Why is it important?

28 Uncertainty Principle In Quantum Mechanics, particles are wave packets Unlike a classical particle, wave packet has a length he position cannot be determined more accurately than x Momentum is related to the wavenumber by p = hk his means h = x p= h x k > h h π Planck s constant = J s Heisenberg s Uncertainty Principle

29 Summary Defined Fourier integral iωt f() t = iωt F( ω) e dω F( ω) = f( t) e dt π f(t) and F(ω) represent a function in time/frequency domains Analyzed pulses and wave packets ime 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 t ω > Connection with Heisenberg s Uncertainty Principle in QM

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