University of Connecticut OpenCommons@UConn Chemistry Education Materials Department of Chemistry 10-24-2006 Fourier Series, Examples and the Fourier Integral Carl W. David University of Connecticut, [email protected] Follow this and additional works at: https://opencommons.uconn.edu/chem_educ Part of the Chemistry Commons Recommended Citation David, Carl W., "Fourier Series, Examples and the Fourier Integral" (2006). Chemistry Education Materials. 28. https://opencommons.uconn.edu/chem_educ/28 Fourier Series, Examples and the Fourier Integral C. W. David Department of Chemistry University of Connecticut Storrs, Connecticut 06269-3060 (Dated: October 24, 2006) I. SYNOPSIS Series? Yes. Consider the domain −L/2 ≤ τ ≤ +L/2, and write The Fourier Integral is introduced by converting a n=∞ a0 X 2nπx 2nπx Fourier series, in complex form, into the integral. Some f(x) = + a sin + b cos 2 n L n L examples are then given. n=1 (2.5) where we use lower case letters for the constants in this II. INTRODUCTION case. The same argument that got us the coefficients before, works here, and we find We chose to introduce Fourier Series using the Par- 2 Z L/2 2nπx ticle in a Box solution from standard elementary quan- an = f(x) sin dx (2.6) tum mechanics, but, of course, the Fourier Series ante- L −L/2 L dates Quantum Mechanics by quite a few years (Joseph Fourier, 1768-1830, France). 2 Z L/2 2nπx Normal discussion of Fourier Series starts with a do- bn = f(x) cos dx (2.7) main for the independent variable (here x) from −π ≤ L −L/2 L x ≤ π and considers replicating functions (such as sine with and cosine) which map partly on this domain, and yet re- ally extend over the domain −∞ ≤ x ≤ +∞, replicating 2 Z L/2 0πx 1 Z L/2 themselves every 2π. a0 = f(x) sin dx = f(x)dx L −L/2 L L −L/2 So, assume we have a function f(x) in the domain −π ≤ (2.8) x ≤ π which may be replicating itself as noted above. The Fourier Series for f(x) is then given by Converting to Complex Form n=∞ A0 X We now use DeMoivre’s Theorem, to re-write these equa- f(x) = + (A sin nx + B cos nx) (2.1) 2 n n tions in a form more suited to the purposes at hand, i.e., n=1 passing to the Fourier Integral. To repeat the derivation of the minimum error (above) here would require us to come to grips with the idea that 1 Z L/2 eı2nπx/L − e−ı2nπx/L an = f(x) dx (2.9) sin x and cos x are orthogonal to each other. These in- 2L −L/2 ı tegrals are trivial, over the domain in question, whether using ‘x’ or ‘nx’. All one really needs is DeMoivre’s The- orem and some familiarity in using it. 1 Z L/2 eı2nπx/L + e−ı2nπx/L bn = f(x) dx (2.10) The coefficients are determinable (using this orthogo- L −L/2 2 nality) via with R π f(x) sin nxdx −π L/2 An = R π 2 (2.2) 1 Z −π sin nxdx a0 = f(x)dx (2.11) L −L/2 R π f(x) cos nxdx −π Adding, we have Bn = (2.3) R π cos2 nxdx −π Z L/2 1 ı2nπx/L and cn = bn + ıan = f(x)e dx (2.12) R π 2L −L/2 −π f(x)dx A0 = R π (2.4) and, subtracting, we have −π dx (where the last term is often thought of as the “direct Z L/2 ∗ 1 −ı2nπx/L current” equivalent value, when translating the original cn = bn − ıan = f(x)e dx (2.13) f(x) into a language of voltage (f) versus time (x)). 2L −L/2 Anyway, assuming that we accept the 2π domain so that our Fourier Series can be rewritten as Fourier Series, can we go on to any “even” domain Fourier Typeset by REVTEX 2 n=∞ n=∞ a0 X 2nπx 2nπx X f(x) = + a sin + b cos = c eı2nπx/L (2.14) 2 n L n L n n=1 n=−∞ where we have changed the summation from running is done! That means that ‘x’ is a “dummy” variable. −∞ → −1 and then a0/2 and then 1 → −∞ to Which in turn means that we can re-write this equation −∞ → +∞. Remember, as Z L/2 1 −ı2πnx/L Z L/2 cn = f(x)e dx (2.15) 1 −ı2πnτ/L 2L −L/2 cn = f(τ)e dτ 2L −L/2 but this is deceptive (for future work) because of the pres- ence of ‘x’ in the integral. One needs to remember that Combining the last two equation into a single one we ‘x’ in this integral has disappeared once the integration have n=∞ ! ! X 1 Z L/2 f(x) = f(τ)eı2πnτ/Ldτ e−ı2nπx/L (2.16) 2L n=−∞ −L/2 n=∞ ! ! X 1 Z L/2 = f(τ)eı2πnτ/Ldτ e−ı2nπx/L (n + 1 − n) = 2L n=−∞ −L/2 n=∞ ! ! X 1 Z L/2 f(τ)eı2πnτ/Ldτ e−ı2nπx/L ∆n (2.17) 2L n=−∞ −L/2 which is, passing through the standard calculus procedure ! ! Z n=∞ 1 Z L/2 f(x) = f(τ)eı2πnτ/Ldτ e−ı2nπx/L dn (2.18) n=−∞ 2L −L/2 ! 1 Z n=∞ Z L/2 f(x) = f(τ)eı2πnτ/L−ı2nπx/L dτdn (2.19) 2L n=−∞ −L/2 Defining ω = 2πn/2L we have which, in the limit, L→ ∞ becomes dω = π/Ldn (2.20) 1 Z ω=∞ Z τ=∞ f(x) = f(τ)eıω(τ−x) dτdω (2.23) so, solving, we have 4π ω=−∞ τ=−∞ 1 dn = dω (2.21) We can split this into symmetric “pairs”: π/L so 1 Z ω=∞ f(τ) = √ (g(ω)eıωτ ) dω (2.24) Z ω=∞ Z τ=L ! 2π ω=−∞ 1 1 ıω(τ−x) f(x) = f(τ)e dτdω 1 Z λ=∞ 2L π/L ω=−∞ τ=−L g(ω) = √ f(τ)e−ıωτ dτ (2.25) (2.22) 2π λ=−∞ 3 III. AN EXAMPLE which becomes, upon integration Consider the Fourier Transform of the function A cos ω0τ in a range −a ≤ τ ≤ a and zero elsewhere. A Z a eı(ω0−ω)τ + e−ı(ω0+ω)τ This is a truncated cosine! We have g(ω) = √ dτ 2 Z a 2 2π −a 1 −ıωτ g(ω) = √ A cos ω0τe dτ −a 2π where the limits are recognizing that f(τ) (the cosine) is which becomes, upon integration zero outside the domain |τ| ≥ a, which is easily evaluated using DeMoivre’s Theorem (smile): ıω τ −ıω τ e 0 + e 0 ı(ω −ω)τ a −ı(ω +ω)τ a ! cos ω τ = A e 0 e 0 0 2 g(ω) = √ + 2 2π ı(ω0 − ω) −a −ı(ω0 + ω) −a so we have 1 Z a eıω0τ + e−ıω0τ g(ω) = √ A e−ıωτ dτ 2π −a 2 A eı(ω0−ω)a e−ı(ω0−ω)a e−ı(ω0+ω)a e+ı(ω0+ω)a g(ω) = √ − + − 2 2π ı(ω0 − ω) ı(ω0 − ω) −ı(ω0 + ω) −ı(ω0 + ω) A sin (ω − ω)a sin (ω + ω)a g(ω) = √ 0 − 0 must be true, identifying α in terms of problem variables. 2π (ω0 − ω) (ω0 + ω) ıτ α = 2B IV. THE CLASSIC EXAMPLE and, of course The classic Fourier Transform illustration is the Gaus- ıτ 2 α2 = sian, since the transform of a Gaussian turns out to be 2B a Gaussian itself. Then as one narrows one Gaussian, the other widens, illustrating the Heisenberg uncertainty This means that the integrand in question can be written principle relating ’x’ to ’p’ expectation values. We have as 2 −Bω2 −B(ω+ ıτ )2− τ g(ω) = Ae e 2B 4B2 defining a function g which is Gaussian in its dependence. Substituting, we have We then have ω=∞ Z ω=∞ 1 Z ıτ 2 τ2 1 −B (ω+ ) − 2 f(τ) = √ (g(ω)eıωτ ) dω f(τ) = √ Ae 2B 4B dω 2π ω=−∞ 2π ω=−∞ which would read, substituting g(ω), which allows us to move the constant factor out of the integrand, obtaining Z ω=∞ 1 −Bω2 ıωτ f(τ) = √ Ae e dω 2 Z ω=∞ 2π −B τ 1 B((ω+ ıτ )2) ω=−∞ f(τ) = e 4B2 √ Ae 2B dω 2π How do we do this integral? Recalling from high school ω=−∞ the “completing the square” idea, we write i.e., 2 ωτ 2 2 2 B ω − ı = B (ω + α) = B ω + 2ωα + α γ = (ω + ıτ/(2B)) B so, and ωτ ı = 2ωα B dγ = dω 4 So or, cleaning up, 2 Z ω=∞ −B τ 1 −Bγ2 f(τ) = Ae 4B2 √ e dγ 2π ω=−∞ Now, it is known that the Gauss integral: 2 A − τ ∞ r 4B Z 2 π f(τ) = √ e e−ax dx = ı 2B −∞ a where here x=γ and a=B, so 2 r A −B τ 1 π f(τ) = e 4B2 √ ı 2π B.