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Fourier Series & the Fourier Transform

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Fourier Series & the Fourier Transform Fourier Series & The Fourier Transform What is the Fourier Transform? Fourier Cosine Series for even functions and Sine Series for odd functions The continuous limit: the Fourier transform (and its inverse) The spectrum Some examples and theorems ∞ ∞ 1 ft()= F (ω )exp( iωω td ) F(ωω )=−ft ( ) exp( itdt ) 2π ∫ ∫ −∞ −∞ What do we want from the Fourier Transform? We desire a measure of the frequencies present in a wave. This will lead to a definition of the term, the “spectrum.” Plane waves have only one frequency, ω. Light electric field Time This light wave has many frequencies. And the frequency increases in time (from red to blue). It will be nice if our measure also tells us when each frequency occurs. Lord Kelvin on Fourier’s theorem Fourier’s theorem is not only one of the most beautiful results of modern analysis, but it may be said to furnish an indispensable instrument in the treatment of nearly every recondite question in modern physics. Lord Kelvin Joseph Fourier, our hero Fourier was obsessed with the physics of heat and developed the Fourier series and transform to model heat-flow problems. Anharmonic waves are sums of sinusoids. Consider the sum of two sine waves (i.e., harmonic waves) of different frequencies: The resulting wave is periodic, but not harmonic. Essentially all waves are anharmonic. Fourier decomposing functions Here, we write a square wave as a sum of sine waves. Any function can be written as the sum of an even and an odd function Let f(x) be any function. E(-x) = E(x) E(x) Ex()≡ [ f () x+− f ( x )]/2 O(x) O(-x) = -O(x) Ox()≡−− [ f () x f ( x )]/2 ⇓ f(x) fx()=+ Ex () Ox () Fourier Cosine Series Because cos(mt) is an even function (for all m), we can write an even function, f(t), as: ∞ 1 f(t) = F cos(mt) π ∑ m m =0 where the set {Fm; m = 0, 1, … } is a set of coefficients that define the series. And where we’ll only worry about the function f(t) over the interval (–π,π). The Kronecker delta function ⎧1 if mn= δ mn, ≡ ⎨ ⎩0 if mn≠ Finding the coefficients, Fm, in a Fourier Cosine Series ∞ 1 f ()tFmt= cos( ) Fourier Cosine Series: π ∑ m m=0 To find Fm, multiply each side by cos(m’t), where m’ is another integer, and integrate: π ∞ π 1 f (t) cos(m' t) dt = F cos(mt) cos(m' t) dt ∫ π ∑ ∫ m − π m= 0 − π π π if m= m ' But: ⎧ cos(mt ) cos( m ' t ) dt =≡⎨ π δ mm,' ∫ ⎩0'if m≠ m −π π ∞ 1 So: ft()cos( mtdt ') = Fπ δ Å only the m’ = m term contributes ∫ π ∑ mmm,' −π m=0 π Dropping the ‘ from the m: Å yields the Fm = ft()cos( mtdt ) ∫ coefficients for −π any f(t)! Fourier Sine Series Because sin(mt) is an odd function (for all m), we can write any odd function, f(t), as: ∞ 1 f (t) = F ′ sin(mt) π ∑ m m= 0 where the set {F’m; m = 0, 1, … } is a set of coefficients that define the series. where we’ll only worry about the function f(t) over the interval (–π,π). Finding the coefficients, F’m, in a Fourier Sine Series ∞ 1 Fourier Sine Series: f (t) = F ′ sin(mt) π ∑ m m= 0 To find Fm, multiply each side by sin(m’t), where m’ is another integer, and integrate: ππ 1 ∞ f (tmtdt ) sin( ' )= Fmtmtdt′ sin( ) sin( ' ) ∫∫∑ m π m=0 But: −− ππ π ⎧π if m= m' sin(mt ) sin( m ' t ) dt =≡⎨ π δmm,' ∫ 0'if m≠ m −π ⎩ π So: 1 ∞ ft()sin( mtdt ') = F′ πδ Å only the m’ = m term contributes ∫ π ∑ mmm,' −π m=0 π Dropping the ‘ from the m: Fftmtdt′ = ()sin( ) Å yields the coefficients m ∫ −π for any f(t)! Fourier Series So if f(t) is a general function, neither even nor odd, it can be written: 11∞∞ ′ f ()tFmtFmt=+∑∑mm cos( ) sin( ) ππmm==00 even component odd component where F = f (t) cos(mt) dt and F ′ = f (t) sin(mt) dt m ∫ m ∫ We can plot the coefficients of a Fourier Series 1 Fm vs. m .5 0 52510 15 20 30 m We really need two such plots, one for the cosine series and another for the sine series. Discrete Fourier Series vs. Continuous Fourier Transform Fm vs. m Let the integer m become a F(m) real number and let the coefficients, Fm, become a function F(m). m Again, we really need two such plots, one for the cosine series and another for the sine series. The Fourier Transform Consider the Fourier coefficients. Let’s define a function F(m) that incorporates both cosine and sine series coefficients, with the sine series distinguished by making it the imaginary component: F(m) ≡ F – iF’ = f ()cos(tmtdt ) − ift()sin( mtdt ) m m ∫ ∫ Let’s now allow f(t) to range from –∞ to ∞, so we’ll have to integrate from –∞ to ∞, and let’s redefine m to be the “frequency,” which we’ll now call ω: ∞ The Fourier Fftitdt()ωω=− ()exp( ) ∫ Transform −∞ F(ω) is called the Fourier Transform of f(t). It contains equivalent information to that in f(t). We say that f(t) lives in the time domain, and F(ω) lives in the frequency domain. F(ω) is just another way of looking at a function or wave. The Inverse Fourier Transform The Fourier Transform takes us from f(t) to F(ω). How about going back? Recall our formula for the Fourier Series of f(t) : 11∞∞ f ()tFmtFmt=+ cos( )' sin( ) ∑∑mm ππmm==00 Now transform the sums to integrals from –∞ to ∞, and again replace Fm with F(ω). Remembering the fact that we introduced a factor of i (and including a factor of 2 that just crops up), we have: ∞ 1 Inverse f (tFitd )= (ω ) exp(ωω ) Fourier 2π ∫ Transform −∞ Fourier Transform Notation There are several ways to denote the Fourier transform of a function. If the function is labeled by a lower-case letter, such as f, we can write: f(t) → F(ω) If the function is labeled by an upper-case letter, such as E, we can write: E()tEt→ Y { ()} or: Et()→ E% (ω ) Sometimes, this symbol is ∩ used instead of the arrow: The Spectrum We define the spectrum, S(ω), of a wave E(t) to be: 2 SEt()ω ≡ Y {()} This is the measure of the frequencies present in a light wave. Example: the Fourier Transform of a rectangle function: rect(t) 1/2 1 Fitdtit(ωω )=−= exp( ) [exp( − ω )]1/2 ∫ −1/2 −1/2 −iω 1 =−−/2)][exp(iiωω / 2) exp( −iω 1−−/2)exp(iiωω / 2) exp( = (/2)ω 2i sin(ω/2) F(ω) =≡/2)sinc(ω (/2)ω Imaginary F {rect(t) } Component = 0 =/2)sinc(ω ω Example: the Fourier Transform of a decaying exponential: exp(-at) (t > 0) ∞ Fatitdt(ωω)=∫ exp( − )exp( − ) 0 ∞∞ =−−=−+]∫∫exp(at iωω t ) dt exp( [ a i t ) dt 00 −−11+∞ =−+]=−∞−exp( [aitω ) [exp( ) exp(0)] ai++ωω0 ai −1 =−[0 1] ai+ ω 1 = ai+ ω 1 Fi(ω )=− ω − ia A complex Lorentzian! Example: the Fourier Transform of a Gaussian, exp(-at2), is itself! ∞ F {exp(−=at22 )}∫ exp( − at )exp( − iω t ) dt −∞ 2 ∝−exp(ω / 4a ) The details are a HW problem! exp(− at 2 ) exp(−ω 2 / 4a ) ∩ 0 t 0 ω Fourier Transform Symmetry Properties Expanding the Fourier transform of a function, f(t): ∞ F()ωωω=+∫ [Re{f ()ti} Im{f ()ttitdt}][cos( ) − sin( )] −∞ ∞ Expanding more, noting that: ∫ Ot() dt= 0 if O(t) is an odd function −∞ = 0 if Re{f(t)} is odd = 0 if Im{f(t)} is even ∞∞↓↓ F()ωω=+∫∫ Re{()}cos()ft tdt Im{()}sin() ft ω tdt ←Re{F(ω)} −∞ −∞ = 0 if Im{f(t)} is odd = 0 if Re{f(t)} is even ∞∞ ↓↓ +−i ∫∫Im{ f ()ttdti} cos(ωω ) Re{ f ()ttdt}sin( ) ←Im{F(ω)} −∞↑↑ −∞ Even functions of ω Odd functions of ω The Dirac delta function Unlike the Kronecker delta-function, which is a function of two integers, the Dirac delta function is a function of a real variable, t. δ(t) ⎧∞=if t 0 δ ()t ≡ ⎨ ⎩0 if t ≠ 0 t ⎧∞ if t = 0 The Dirac delta function δ ()t ≡ ⎨ ⎩0 if t ≠ 0 It’s best to think of the delta function as the limit of a series of peaked continuous functions. 2 fm(t) = m exp[-(mt) ]/√π δ(t) f3(t) f2(t) f1(t) t Dirac δ−function Properties δ(t) ∞ ∫ δ ()tdt= 1 t −∞ ∞∞ ∫∫δδ(t−=−= aftdt ) () ( t afadt ) () fa () −∞ −∞ ∞ ∫ exp(±=itdtωπδω ) 2 ( ) −∞ ∞ ∫ exp[±−itdt (ω ωπδωω′ ) ] = 2 ( −)′ −∞ The Fourier Transform of δ(t) is 1. ∞ ∫ δω(titdti ) exp(− )=− exp( ω [0]) = 1 −∞ δ(t) 1 t ω ∞ And the Fourier Transform of 1 is 2πδ(ω): ∫ 1exp(−itdtω )=) 2πδ ( ω −∞ 1 2πδ(ω) t ω The Fourier transform of exp(iω0 t) ∞ F exp(itωωω )=− exp( it ) exp( itdt ) {}00∫ ∞ −∞ =−−exp(itdt [ωω ] ) = 2(π δω− ω ) ∫ 0 0 −∞ exp(iω0t) Y {exp(iω0t)} Im t 0 ω Re t 0 ω0 0 The function exp(iω0t) is the essential component of Fourier analysis. It is a pure frequency. The Fourier transform of cos(ω0 t) ∞ F cos(ωωωttitdt )=− cos( ) exp( ) {}00∫ −∞ ∞ 1 =+−−[]exp(itωωω00 ) exp( it ) exp( itdt ) 2 ∫ −∞ ∞∞ 11 =−−+exp(i [ωω ] t ) dt exp( −+ i [ ωω ] t ) dt 22∫∫00 −∞ −∞ = π δω()−+ ω00 πδω () + ω F {cos(t )} cos(ω0t) ω0 t 0 ω −ω0 0 +ω0 The Modulation Theorem: The Fourier Transform of E(t) cos(ω0 t) ∞ F E(tt )cos(ωωω )=− Ettitdt ( )cos( ) exp( ) {}00∫ ∞ −∞ 1 =+−−E(titititdt )⎡⎤ exp(ωωω ) exp( ) exp( ) 2 ∫ ⎣⎦00 ∞∞−∞ 11 =−−+−+E()exp(ti [ωω ]) tdtEtitdt ()exp( [ ωω ]) 22∫∫00 −∞ −∞ 11 F {}Et()cos(ω t )=−++ E%% (ωω ) E ( ωω ) 0022 0 Et()cos(ω t ) F Et()cos(ω t ) Example: 0 { 0 } E(t) = exp(-t2) t -ω0 0 ω0 ω Scale Theorem The Fourier transform of a scaled function, f(at): F {(f at )}= F (/)/ω a a ∞ Proof: F {f ( at )}=−∫ f ( at ) exp( iω t ) dt −∞ Assuming a > 0, change variables: u = at ∞ F {(f at )}=−∫ f ()exp( u iω [/]) u a du / a −∞ ∞ =−∫ f ()exp([u iω /] audu ) / a −∞ = Faa(/)/ω If a < 0, the limits flip when we change variables, introducing a minus sign, hence the absolute value.
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