Wave Phenomena Physics 15c. Lecture 10 Fourier Transform

Transcription

1 Wave Phenomena Physics 15c Lecture 10 Fourier ransform

2 What We Did Last ime Reflection of mechanical waves Similar to reflection of electromagnetic waves Mechanical impedance is defined by For transverse/longitudinal waves: Useful in analyzing reflection Standing waves Created by reflecting sinusoidal waves Oscillation pattern has nodes and antinodes F = ±Zv Z = [ or K]ρ l Musical instruments use standing waves to produce their distinct sound

3 Goals For oday Define Fourier transform Fourier series is defined for repetitive functions Discrete values of frequencies contribute f (t) = a 0 + 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 ( a n cosω n t + b n sinω n t) n=1 Connection with Quantum Mechanics

4 Looking Back In Lecture #4, we solved the wave equation 2 t ξ(x,t) = c 2 w x ξ(x,t) 2 c = ω w k Normal-mode solutions ξ(x,t) = ξ 0 e i(kx±ωt ) Using Fourier series, we can make arbitrary waveform with linear combination of normal modes 2 2 Example: forward-going repetitive waves ( ) n=1 f (x c w t) = a n cos(k n x ω n t) + b n sin(k n x ω n t) Non-repetitive waves also OK if we make his makes ω continuous Sum turns into integral We need a little math work to make this happen ω n = n k n = ω n c w

5 Fourier Series Start with a repetitive function f(t) a 0 = 1 0 f (t) = a 0 + f (t)dt a n = 2 ( a n cosω n t + b n sinω n t) n=1 0 f (t)cosω n t dt b n = 2 0 ω n = n f (t)sinω n t dt Express cos ω n t and sin ω n t with complex exponentials ( a n cosω n t + b n sinω n t) = n=1 n=1 a n ib n 2 e iω n t + a n + ib n 2 e iω n t Re-express the first term with m = n, a m = a n, b m = b n, ω m = ω n = 1 m= a m + ib m 2 e iω m t + n=1 a n + ib n 2 e iω n t

6 Fourier Series Sum includes n = 0 Define F n = a + ib n n and 2 F = a 0 0 How do we calculate F n? F n = f (t)cosω n t dt + i 2 0 f (t) = n= f (t)sinω n t dt = 1 F n e iω n t 0 f (t)e iω n t dt F 0 = 1 0 f (t)dt same It s useful later if I shift the integration range here F n = f (t)e iω n t dt OK because f(t) is repetitive Now we take it to the continuous limit

7 Fourier ransform We have f (t) = Bring f (t) = lim = lim n= n= F n e iω n t F n e iω n t and = lim F n e iωt dω F n = F n Δω e inδωt n= Let s define the Fourier transform of f(t) by F(ω) lim F = 1 n f (t)e iω n t dt Δω f (t)e iωt dt ω n = n Δω hen the original function is f (t) = F(ω)e iωt dω

8 Confusion Different conventions exist for Fourier transform We ve got: f (t) = F(ω)e iωt dω, F(ω) = 1 f (t)e iωt dt Most physics textbooks use this Symmetric: f (t) = 1 F(ω)e iωt dω, F(ω) = 1 f (t)e iωt dt Advanced physics books and Mathematica (default) Frequency in Hz: Signal processing and Wikipedia and a few others f (t) = F(ν)e iνt dν, F(ν) = f (t)e iνt dt We ll use the first one, but be careful when you read other textbooks

9 Fourier ransform f (t) = F(ω)e iωt dω F(ω) = 1 Fourier transform 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)e iωt dt F(ω) and f(t) are two equally-good representations of a same function We often say that f(t) is in the time domain F(ω) is in the frequency domain

10 Square Pulse Consider a short pulse with unit area Fourier transform is F(ω) = 1 f (t)e iωt dt = 1 is a bunch of ripples around x = 0 f (t) = 1 for t < 2 0 for t > 2 sinc(x) sin(x) x sin(x) sinc(0) = lim = 1 x 0 x sinc(nπ) = sin(nπ) = 0 for n 0 nπ 2 e iωt dt = 1 2 πω sinω 2 = 1 sinc ω 2 1

11 Fourier of Square Pulse Pulse of duration he zero-crossing points are sinc ω 2 = 0 he width of F(ω) is inversely proportional to the width of f(t) (width in t) (width in ω) = const his is a general feature of Fourier transformation More about this later F(ω) = 1 sinc ω 2 ω 2 = nπ ω = 2nπ 0 ω Peak height is given by F(0) = 1 f (t)dt = total area below f (t)

12 Sending Information We send information using waves Voice in the air Voice converted into EM signals on a phone cable You can t do it with pure sine waves It just goes on Completely predictable No information Waves must change patterns with time What you really need are pulses Pulse width determines the speed Narrower pulses carry more information per unit time

13 Amplitude Modulation Audio signals range from 20 Hz to 20 khz oo 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 We ve seen this in homework

14 Wave Packet Consider carrier waves modulated by a pulse his makes a short train of waves A wave packet f (t) = = 1/(20 khz) would work for audio signals Fourier integral is e iω 0 t for t < 2 0 for t > 2 f (t) F(ω) = 1 2 e iω 0 t 1 e iωt dt = 2 π(ω ω 0 ) sin (ω ω ) 0 2 = sinc (ω ω ) 0 2

15 Wave Packet F(ω) = sinc (ω ω ) 0 2 Similar to the square pulse Width is / Centered at ω = ω 0 o send pulses every second, your signal must have a minimum spread of / in ω, which corresponds to 1/ in frequency his is called the bandwidth of your radio station his 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 ω 0 ω

16 Bandwidth Speed of information transfer = # of pulses / second Determined by the pulse width in the time domain ranslated into bandwidth ~1/ 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 (see next page) Analog cable V 750 MHz / 6 MHz = 125 channels You want to minimize the bandwidth of each channel elephones carry only between 400 Hz and 3400 Hz Cell phones and VOIP use special speech-compression techniques

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18 Delta Function ake the square pulse again Make it narrower by 0 he height grows 1/ We get an infinitely narrow pulse with unit area Dirac s delta function δ(t) δ(t) = t = 0 0 t 0 For any function f(t) and δ(t) dt = 1 1 f (t)δ(t)dt = f (0) f (t)δ(t t 0 )dt = f (t 0 )

19 Delta Function What is the Fourier transform of δ(t)? Calculating directly F(ω) = 1 δ(t)e iωt dt = We can also use the square-pulse result and take 0 limit δ(t) contains all frequencies equally: Consider the inverse: Fourier transform of sine/cosine is a delta function f (t) = e iω 0 t F(ω) = 1 e i(ω ω 0 )t dt = δ(ω 0 ω) 1 1 F(ω) = lim 0 sinc ω 2 = 1 δ(t) = 1 e iωt dω

20 How hings Fit ogether Waveform t domain t width domain width Sinusoidal uniform infinite δ(ω 0 ω) 0 δ pulse δ(t) 0 uniform infinite Finite pulse and everything else Pure sine waves and δ pulses are the two extreme cases of all waves Everything falls in between f(t) F(ω) 1/ Widths in time t and angular frequency ω are inversely proportional to each other he last point deserves a more detailed look

21 Arbitrary Signal Now we consider a signal with an arbitrary shape f (t) Fourier F(ω) Let s define the average time and the average frequency t = t f (t) 2 dt f (t) 2 dt ω = ω F(ω) 2 dω We weight the averages by f(t) 2 and F(ω) 2 because (energy density) (amplitude) 2 F(ω) 2 dω

22 Width of Arbitrary Signal Next, we define the r.m.s. (root mean square) widths ( Δt) 2 = ( t t ) 2 = ( t t ) 2 f (t) 2 dt f (t) 2 dt ( Δω ) 2 = ( ω ω ) 2 = ( ω ω ) 2 F(ω) 2 dω F(ω) 2 dω Δt and Δω measure the spread of f(t) and F(ω)

23 Uncertainty heorem It can be proven (see supplement on the course web site) that ΔtΔω 1 2 for arbitrary waveform and its Fourier transform Provided that f(t) and F(ω) don t extend infinitely, i.e. lim t ± ( f (t) 2 ) = 0 lim ω ± Always true for real physical waves ( F(ω) 2 ) = 0 For any signal, the product of the r.m.s. widths Δt and Δω in the time and frequency domain is at least 1/2

24 Space and Wavenumber We have studied Fourier transformation in time t and frequency ω We can also do the same in position x and wavenumber k Everything works the same way f (x) = F(k)e ikx dk F(k) = 1 f (x)e ikx dx In particular, ΔxΔk 1 2 Why is it important? for any waves in space

25 Uncertainty Principle In Quantum Mechanics, particles are wave packets Unlike a classical particle, a wave packet has a length he position cannot be determined more accurately than Δx Momentum is related to the wavenumber by p = k = h Planck s constant = J s his means ΔxΔp = ΔxΔk 2 Position and momentum of a particle cannot be determined simultaneously better than Δx Δp = h/4π his is Heisenberg s Uncertainty Principle

26 Summary Defined Fourier transform f (t) = F(ω)e iωt dω F(ω) = 1 f(t) and F(w) represent a function in time and frequency domains Analyzed pulses and wave packets ime resolution Δt and bandwidth Δω related by Can be proven for arbitrary waveform Connection with Heisenberg s Uncertainty Principle in QM Rate of information transmission bandwidth Dirac s δ(t) a limiting case of infinitely fast pulse δ(t) = 1 e iωt dω f (t)e iωt dt ΔtΔω 1 2