f(x) cos dx L L f(x) sin L + b n sin a n cos
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1 Chapter Fourier Series and Transforms. Fourier Series et f(x be an integrable functin on [, ]. Then the fourier co-ecients are dened as a n b n f(x cos f(x sin The claim is that the function f then can be written as f(x a 0 + n [ a n cos dx n 0, dx n > 0. (. + b n sin ]. The rhs is called the fourier series of f. Using fourier trick, one can obtain the expressions for a n and b n. Multiplying both sides of the expression by cos ( mπx for some m>0 and integrating one gets ( mπx f(x cos dx a ( 0 mπx ( mπx ( mπx ] cos dx + [a n cos cos dx + b n cos sin dx n 0+ [a n δ m,n +0]a m n This proves the formula given by Equation.. Example. et f(x x on [ π, π]. Find fourier series. 8
2 The function is odd, hence for all n and Thus the fourier series is b n π a n π π π π π x cos(nxdx 0 x sin(nxdx n ( n+ f(x [sin (x sin (x+ ] Figure.: Fourier series with term, 3 terms and 00 terms Two points can be noted from the example. The convergence is not uniform, some points converge faster than other points.. At x ±π the fourier series does not converge to f. Example 3. et f(x x / where x [, ]. This is a trianglular function. The function is even, thus fourier series will contain onlycos terms. b n f(x sin dx 0. and for n>0, a n f(x cos dx 0 ( + x cos dx + ( x cos dx 0 n π ( ( n 4 n π odd n 0 even n 9
3 Simillarly, a 0. The fourier series is f(x + 4 π [cos ( πx + ( ] 3πx 9 cos +. Dirichlet Condition Denition 4. A function f on [, ] is said to be satisfy Dirichlet conditions if f has nite number of extrema, f has nite number of discontinuities, f is abosultely integrable, f is bounded. Theorem 5. Suppose a function f satises Dirichlet conditions. Then the fourier series of f converges to f at points where f is continuous. The fourier series converges to the midpoint of discontinuity at points where f is discontinuous. Example 6. A square function f is dened as { 0 x [, 0] f(x x [0,] The fourier coecients are a 0, a n 0 { b n nπ odd n 0 even n The fourier series is + ( πx π sin + ( 3πx 3π sin + Clearly, at x 0series has a value 0.5 which is equal to [f(0+ + f(0 ] /. Theorem 7. If a function f on [, ] is square integrable, that is f(x dx exists, then the fourier series of f converges to f almost everywhere. 0
4 Exponential Fourier Series Substituting cos x [e x + e x ] / and sin x [e x e x ] /i, the fourier series of a function f can be rewritten as [ a n cos f(x a 0 + n n ( c n exp i nπx + b n sin ] where (a n ib n n>0 c n (a n + ib n n<0 a 0 n 0 Or, c n ( f(x exp i nπx Typically, if the argument of f is a time variable, then ω n nπ/ are called frequencies. The fourier coecients are usually labelled by frequencies, that is c n is written as c ωn. Example 8. Exponential fourier coecients for triangular function are given by c 0 c n 0 c n (n π The following graph shows the fourier coecients as a function of frequencies. 0.5 CΩ 5 Π 3 Π Π Π 3 Π 5 Π Ω. Fourier Transform For an interval [, ], the fourier frequencies are given by ω n nπ/. When, the fourier frequencies become a continuous variable. To get this idea, redene the fourier coecients c ωn f(x exp ( iω n x
5 Then the fourier series becomes Now, let ω π/, then f(x n c ωn exp (iω n x f(x n c ωn exp (iω n x ω c(ω exp (iωx dω as The last step is just the denition of Riemann integral. The function c(ω is called as fourier transform of f(x. It is denoted as c F (f. Dierent text books dene fourier transform dierently by placing the constat factor dierently. In this note, fourier transform is dened as c(ω f(x f(x exp ( iωx dx c (ω exp (iωx dω. Example 9. et f(x for x a, and is 0 otherwise. If g is the FT of f, then g(ω a a a sin (ωa ωa f(x exp ( iωx dx exp ( iωx dx f x g Ω a a x 4 Π Π Π 4 Π Ω Here is another expample that illustrates the uncertainty principle.
6 Example 30. et f(x a/π exp[ ax ]. Ifg is the FT of f, then g(ω a π f(x exp ( iωx dx a π e ω /4a exp ( ω 4a exp ( ax iωx dx ( exp a ( x + iω dx a The width of the gaussian function is dened by parameter a. Thusf(x is broad if a is small. And g(ω is narrow and sharp for small a. Here is a very useful theorem called Parseval theorem. Theorem 3. If g is the fourier transform of a square integrable function f then f (x dx g(ω dω. Uncertainty Relation The uncertainty relation is built into the fourier transform. It may be interpreted dierently depending on the situation in which it is applied. Here we state the theorem without proof. Theorem 3. Suppose g is a fourier transform of a square integrable function f that is normalized to unity, that is. et µ f σf µ g σg then σ f σ g. f(x dx x f(x dx (x µ f f(x dx ω f(ω dω (ω µ ω f(ω dω The quantities σ are called uncertainties. 3
7 Example 33. et f(x (a/π /4 exp[ ax /]. Then f(x dx Clearly, µ f 0and If g is the FT of f, then Note that Now µ g 0. Then Then, σ f g(ω σ g x f(x dx a ( /4 exp ( ω. πa a g(ω dω ω g(ω dω a σ f σ g 4
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