Lecture 34. Fourier Transforms

Transcription

1 Lecture 34 Fourier Transforms In this section, we introduce the Fourier transform, a method of analyzing the frequency content of functions that are no longer τ-periodic, but which are defined over the entire real line, i.e, f : R R. The Fourier transform involves integrals over such functions. In order for such integrals to be finite, it will be necessary that the functions f(x 0 as x ±. At first, this might seem to be a rather strict condition on f, but when you consider that in practice, all signals have a beginning and an end in other words, they have finite support it is not too strict at all. The relevant section of the AMATH 3 Course Notes is Section 5.4, The Fourier transform and Fourier integral. In this particular course, we shall have covered the following subsections from the Course Notes: Section 5.4.: The definition Section 5.4.: Calculating Fourier transforms using the definition Section 5.4.3: Properties of the Fourier transform Section 5.4.4: Parseval s formula for a non-periodic function Recall from earlier in this course, the idea of extending Fourier series associated with -periodic functions to functions which are τ-periodic, i.e., a function f(t such that f(t + τ = f(t, ( for some τ > 0. This was done rather efficiently by means of the following complex Fourier series, It where f(x = n= ω 0 = τ c n e inω 0x, ( (3

2 is the fundamental frequency. The complex Fourier coefficients, c n, n Z, are given by the formula, and satisfy c n = τ τ/ τ/ f(t e inω 0t dt, (4 c n = c n. (5 The complex-valued basis functions, e n (t = e inω 0t = e int/τ, (6 satisfy the orthogonality condition (complex inner product e n, e m = τ/ τ/ e n (t e m (t dt = τ δ mn. (7 Let us first make the following tiny modification of the above results. We shall let τ = L, (8 so that our functions are now L-periodic and are defined over the primary interval [ L, L]. This means that the following set of complex-valued functions, u n (t = e inπt/l, n {,,, 0,,, } (9 forms an orthogonal basis for the complex-valued space of L-periodic functions L [ L, L]. In the special case L = π, we have the standard Fourier series. The Fourier series expansion of a function f L [ L, L] is given by f(x = a n e inπx/l, (0 n= where the equality is understood in the L sense, i.e., convergence in the norm. The coefficients a n are given by a n = L f(te inπt/l dt. ( L L

3 We mention here that Eq. (0 essentially represents an expansion of the function f(x in terms of its frequency components, with the frequencies given by ω n = nπ L. ( Let us now substitute Eq. ( for the a n into Eq. (0: f(x = ( L f(te nπt/l dt L n= L ( L = L n= f(te n(x tπt/l dt L e inπx/l. (3 The idea is now to take the limit L so that the support, [ L, L], of our functions f, becomes the real line R. From Eq. ( one might think that this implies that all frequencies ω n 0. Letting n at the same time, however, will still produce frequencies of arbitrarily large magnitude. More important, however, is that the spacing between consecutive frequencies will go to zero as L. This spacing is given by ω = ω n+ ω n = π L. (4 From this relation, it follows that the factor /(L in Eq. (3 becomes L = ω Substitution of these expressions into (3 yields f(x = ( L f(te iωn(x t dt ω n= L = F(x, ω n, L ω. (6 n= The sum on the RHS may be interpreted as a Riemann sum involving the function F(x, ω, L = L L (5 f(t e iω(x t dt. (7 The sample points ω n are equally spaced over the entire real line. As such, the Riemann sum approximates the integration of the variable ω on R. In the limit L, ω 0, and we claim that the Riemann sum converges to the integral F(x, ω dω, (8 3

4 where F(x, ω = (Note that x is fixed. Substitution into Eq. (6 yields the following, f(x = = = A few comments regarding this result: F(x, ω dω f(t e iω(x t dt. (9 f(t e iω(x t dt dω ( f(t e iωt dt e iωx dω. (0. The term e iωx looks like a basis element - the summation over the index n, therefore over the discrete frequencies ω n, has been replaced by an integration over ω.. The term in brackets, f(te iωt dt, ( looks like a complex inner product, i.e., f, e iωt, ( which means that it could be interpreted as a a Fourier series coefficient. 3. From these two observations, the RHS has the form of a continuous expansion. The term in brackets, viewed as a Fourier coefficient, is called the Fourier transform of f and denoted as F(ω: We say that F(ω = f(te iωt dt. (3 F(ω is the Fourier transform (FT of f(t, or F = F(f. Eq. (0 may then be written as follows: f(t = F(ωe iωt dω. (4 The above equation defines the inverse Fourier transform (IFT of F(ω, i.e., f = F (F. We mention here that these definitions of the Fourier transform and its inverse are also adopted by the MAPLE and MATLAB programming languages. 4

5 Other definitions of the Fourier transform It turns out that there are a number of other definitions of the Fourier transform and associated inverse. For example, some books (including the book, Applied Partial Differential Equations by R. Haberman, which has been used in the AMATH 353 Course, Partial Differential Equations II, employ e iωt instead of e iωt in Eq. (78 and therefore e iωt instead of e iωt in Eq. (4. There is another definition of the FT in which the factor in Eq. (0 is divided symmetrically between the two integrals, leading to the following definition of the Fourier transform, F(ω = and its associated inverse Fourier transform, f(t = f(te iωt dt, (5 F(ωe iωt dω. (6 This symmetric formulation is adopted in many mathematical treatises. One reason is that in this formulation, the Fourier transform is norm preserving. And to complicate matters even further, it is convenient in the engineering literature not to use the angular frequency ω (radians/unit time, but rather the wavenumber k, the number of cycles per unit time. (For example, we normally think of the range of human hearing to be something like 0-0,000 cycles per second and not its equivalent in radians per second. These two frequencies are related as follows, ω = k, (7 since there are radians/cycle. Using the definition in Eq. (78, the resulting Fourier transform is F(k = f(te ikt dt. (8 From the change of variable ω = k, we have dω = dk, so that the inverse Fourier transform becomes f(t = F(ωe ikt dk. (9 Note that neither the FT nor the inverse FT have any factors in front of the integrals, which is most convenient. 5

6 Here is a summary of the formulas from Fourier series for τ-periodic functions and Fourier transforms of functions on R as employed in this course: τ-periodic functions (ω 0 = /τ square-integrable functions on R c n = τ Fourier coefficients τ/ f(te inω 0t dt F(ω = Fourier transform τ/ f(te iωt dt Fourier series f(t = c n e inω 0t f(t = Fourier integral F(ωe iωt dω. Two noteworthy comments:. In the Fourier series representation for a τ-periodic function f(t on [ τ/, τ/], the coefficient c n characterizes the component of f that oscillates at the frequency ω n = nω 0 = nπ τ. The Fourier series is a summation of these components over discrete frequencies ω n.. In the Fourier transform representation for a function f(t on (,, the coefficient F(ω measures the component of f that oscillates at the frequency ω. The Fourier transform is an integration of components over continuous frequencies ω. We now present a few simple examples to illustrate some basic points. Examples:. The rectangular window function (also known as the boxcar function in signal processing, defined as follows, A plot is given at the left in the figure below., t < W(t =, 0, t >. (30 6

7 We compute the Fourier transform as follows, F(ω = = = / / iω W(te iωt dt e iωt dt [ e iωt ] / / = iω [e iω/ e iω/ ] (3 = [(cos(ω/ i sin(ω/ (cos(ω/ + i sin(ω/] (3 iω = ( i sin(ω/ iω = ω sin(ω/ = sin(ω/ ω/ ( ω = sinc. (33 Here, we have used the mathematical definition of the sinc function, sin x sinc(x =, x 0, x, x = 0. (34 A plot is shown at the right in the figure below x x Left: Plot of the rectangular window function, W(t, Example. Right : Its Fourier transform F(ω. The boxcar function defined above is piecewise constant, with a nonzero constant value over ( /, /. (The constant value of zero outside ( /, / will not contribute to 7

8 F(ω. As such, one would expect that its largest frequency component is at ω = 0. That being said, one would also expect that the discontinuities at x = ±/ will have to be accomodated, accounting for the high-frequency components.. The function cos 3t, π t π, f(t = 0, otherwise. This may be viewed as a clipped audio signal, obtained by multiplying the function cos(3t, t R by a suitably scaled version of the rectangular window function of Example. We begin to compute its Fourier transform as follows, F(ω = = π π f(te iωt dt (35 cos 3t e iωt dt. (36 There are at least two possible paths to take to compute the above integral. We could either (i express the cosine term as a linear combination of complex exponentials or (ii express the complex exponential as a linear combination of cos ωt and sin ωt using Euler s formula. Here, we ll use Method (i: F(ω = π π π [e i3t + e i3t ] e iωt dt = e i(3 ωt + e i(3+ωt dt π = [ i 3 ω ei(3 ωt ] π 3 + ω e i(3+ωt π = [ i 3 ω [ei(3 ωπ e i(3 ωπ ] ] 3 + ω [e i(3+ωπ e i(3+ωπ ] = [ ] (i sin((3 ωπ + (i sin((3 + ωπ i 3 ω 3 + ω = sin((3 ωπ + 3 ω 3 + ω Now simplify the sin functions, sin((3 + ωπ. (37 sin((3 ωπ = sin(3π cos(ωπ cos(3π sin(ωπ = sin(ωπ sin((3 + ωπ = sin(3π cos(ωπ + cos(3π sin(ωπ = sin(ωπ, (38 8

9 so that [ F(ω = 3 ω ] sin(ωπ 3 + ω = ω sin(πω 9 ω. (39 Plots of the clipped audio signal, f(t, and its Fourier transform, F(ω, are shown in the figure below. The largest frequency components of F(ω are at ±3, as expected since the original function has a cos(3t component, at least over a finite time interval. (The fact that F(ω is finite at ω = ±3 is left as an exercise t t Left: Plot of f(t, Example. (The vertical lines at the discontinuities x = ±π are artifacts of the plotting routine. Right : Fourier transform F(ω. 3. The triangular peak function, f(t = π + t, π t π, π t, 0 < t π, 0, otherwise. (40 Its Fourier transform is given by (Exercise F(ω = cos(πω ω. (4 There are some other noteworthy observations to be made regarding the above examples:. The Fourier transform in Examples and decay as F(ω = O(/ω as ω. 9

10 t t Left: Plot of f(t, Example 3. Right : Fourier transform F(ω.. The Fourier transform in Example 3 decays as F(ω = O(/ω as ω. As in the case of Fourier series coefficients which represent a discrete Fourier transform the faster decay rate of Example 3 is due to the higher degree of regularity of the function f(t: it is continuous at all x R, whereas the function in Example is piecewise continuous. Here are some Fourier transforms F(ω of standard functions f(t (taken from Table 5. of AMATH 3 Course Notes. f(t W(t sinc(t F(ω ( ω sinc ( ω πw e t + ω + t πe ω e t / e ω / 0

11 Lecture 35 Fourier transforms (cont d Some properties of Fourier transforms Here we list some of the more important properties of Fourier transforms. In what follows, we assume that the functions f(t and g(t are differentiable as often as necessary, and that all necessary integrals exist. (This implies that f(t 0 as t. Regarding notation: f (n (t denotes the nth derivative of f w.r.t. t; F (n (ω denotes the nth derivative of F w.r.t. ω. The first few properties are those listed in the AMATH 3 Course Notes, pp Linearity of the FT operator and the inverse FT operator: F(f + g = F(f + F(g F(cf = c F(f, c C (or R, (4 F (f + g = F (f + F (g F (cf = c F (f, c C (or R, (43 These properties follow naturally from the integral definition of the FT.. Fourier transform of a scaled function ( Scaling Theorem : For a b 0, Proof: F(f(bt = = b = b F(f(bt = b F ( ω b f(bte iωt dt = b F ( ω b f(se iωs/b ds f(se i( ω bs ds. (44 (s = bt, dt = ds, etc. b. (45

12 The appearance of the absolute value in the denominator arises from the fact that the integration limits are switched in the case that b is negative. Switching them again in order to integrate from to produces a b in this case, which is equivalent to b. We ll examine this result in greater detail below. 3. Fourier transform of a modulation ( Frequency Shift Theorem : F(e iω0t f(t = F(ω ω 0. (46 Proof: F(e iω0t f(t = = e iω 0t f(te iωt dt f(te i(ω ω 0t dt = F(ω ω 0. (47 4. Fourier transform of a translation ( Spatial Shift Theorem : F(f(t a = e iωa F(ω. (48 Proof: F(f(t a = = f(t ae iωt dt f(se iω(s+a ds = e iωa f(se iωs ds (s = t a, dt = ds, etc. = e iωa F(ω. (49 The following properties are not listed in the AMATH 3 Course Notes, but are presented here for completeness. 5. Fourier transform of a product of f with t n : F(t n f(t = i n F (n (ω. (50

13 This is easily seen for the case n = by taking the derivative of F(ω in Eq. (78: F (ω = = = d dω = i f(te iωt dt f(t d dω [ e iωt ] dt (Leibniz Rule f(t( ite iωt dt tf(te iωt dt = if(tf(t. (5 Repeated applications of the differentiation operator produces the result, from which the property follows. 6. Inverse Fourier transform of a product of F with ω n : F (n (ω = F(t n f(t, (5 F (ω n F(ω = ( i n f (n (t. (53 Here, we start with the definition of the inverse FT in Eq. (4 and differentiate both sides repeatedly with respect to t. The reader will note a kind of reciprocity between this result and the previous one. 7. Fourier transform of an nth derivative: This is a consequence of 3. above. 8. Inverse Fourier transform of an nth derivative: This is a consequence of. above. F(f (n (t = (iω n F(ω. (54 F (F (n (ω = ( it n f(t. (55 3

14 Many, if not all, of the above properties are useful for the computation of more complicated Fourier transforms. Indeed, there are often several ways to compute a Fourier transform one method may be much easier than the others. Let us now return to examine a couple of these properties in more detail. Fourier transform of a modulation ( (Frequency Shift Theorem : F(e iω0t f(t = F(ω ω 0. (56 As mentioned in the Course Notes, multiplication of a given signal f(t by the oscillatory complex exponential e iω0t in the time domain produces a shift in the frequency domain. The following may help to understand this result. Suppose that the signal f(t contains the oscillatory frequency ω, which means that it will include the term Ae iωt (57 in its Fourier expansion. This means that the ω-content of f is A. Equivalently, the amplitude A will appear at frequency ω in the Fourier transform F(ω of f(t. This means that the modified signal g(t = e iω 0t (58 will have the modified term, e iω 0t Ae iωt = Ae i(ω+ω 0t, (59 in its Fourier expansion. The amplitude A will now appear at frequency ω + ω 0, which is equivalent to shifting the Fourier transform F(ω to the right by ω 0. There are two interesting consequences of this Shift Theorem. We may be interested in computing the FT of the product of either cosω 0 t or sin ω 0 t with a function f(t. In this case, we express these trigonometric functions in terms of appropriate complex exponentials and then employ the above Shift Theorem. First of all, we start with the relations cosω 0 t = [ e iω 0 t + e ] iω 0t sin ω 0 t = [ e iω 0 t e ] iω 0t. (60 i 4

15 From these results and the Frequency Shift Theorem, we have where F(ω denotes the FT of f(t. F(cosω 0 tf(t = [F(ω ω 0 + F(ω + ω 0 ] F(sin ω 0 tf(t = i [F(ω ω 0 + F(ω + ω 0 ], (6 To show how these results may be helpful in the computation of FTs, let us revisit Example presented in the previous lecture, namely the computation of the FT of the function f(t defined as the function cos 3t, but restricted to the interval [ π, π]. We may view f(t as a product of two functions, i.e., where f(t = cos 3t b(t, (6, t < π, b(t = 0, t > π. The boxcar function b(t is a scaled version of the rectangular window function W(t discussed in the previous Lecture. In fact, b(t = W (63 ( t, (64 As such, we could use the Scaling Theorem (next result to be examined to compute the FT of b(t. But here, we ll simply compute it from the definition: B(ω = From the Frequency Shift Theorem, =. π π W(te iωt dt e iωt dt = sinc(πω. (65 F(ω = F(f(t = [B(ω 3 + B(ω + 3], (66 where B(ω = sinc(πω = sin(πω πω. (67 5

16 is the FT of the boxcar function b(t. Substitution into the preceeding equation yields [ ] sin(π(ω 3 sin(π(ω + 3 F(ω = π ω 3 ω + 3 = ω sin(πω ω 9 = ω sin(πω 9 ω, (68 in agreement with the result obtained for Example earlier. That being said, it is probably easier now to understand the plot of the Fourier transform that was presented with Example. Instead of trying to decipher the structure of the plot of the function sin(πω divided by the polynomial 9 ω, one can more easily visualize a sum of two shifted sinc functions. Fourier transform of a scaled function ( Scaling Theorem : For a b 0, F(f(bt = ( ω b F. (69 b Proof: F(f(bt = = b = b f(bte iωt dt = b F ( ω b f(se iωs/b ds f(se i( ω bs ds (s = bt, dt = ds, etc. b. (70 The appearance of the absolute value in the denominator arises from the fact that the integration limits are switched in the case that b is negative. Switching them again in order to integrate from to produces a b in this case, which is equivalent to b. Let s try to get a picture of this result. Suppose that b >. Then the graph of the function g(t = f(bt is obtained from the graph of f(t by contracting the latter horizontally toward the y-axis by a factor of /b, as sketched in the plots below. On the other hand, the graph G(ω = F ( ω b is obtained from the graph of F(ω by stretching it outward away by a factor of b from the y-axis, as sketched in the next set of plots below. 6

17 y y y = f(t y = f(bt A A t t t t /b t /b t Left: Graph of f(t, with two points t and t, where f(t = A. Right: Graph of f(bt for b >. The two points at which f(t = A have now been contracted toward the origin and are at t /b and t /b, respectively. y y A y = F(ω A y = F(ω/b ω ω ω bω bω ω Left: Graph of Fourier transform F(ω, with two points ω and ω, where F(ω = A. Right: Graph of F(ω/b for b > 0. The two points at which F(ω = A have now been expanded away from the origin and are at bω and bω, respectively. The contraction of the graph of f(t along with an expansion of the graph of F(ω is an example of the complementarity of time (or space and frequency domains. We shall return to this point shortly. As in the case of discrete Fourier series, the magnitude F(ω of the Fourier transform of a function must go to zero as ω. Assume once again that b >. Suppose most of the energy of a Fourier transform F(ω of a function f(t is situated in the interval [ ω 0, ω 0 ]: For ω values outside this interval, F(ω is negligible. But the Fourier transform of the function f(bt is now F(ω/b, which means that it lies in the interval [ bω 0, bω 0 ], which represents an expansion of the interval [ ω 0, ω 0 ]. This implies that the FT of f(bt contains higher frequencies than the FT of f(t. Does this make sense? 7

18 The answer is, Yes, because the compression of the graph of f(t to produce f(bt will produce gradients of higher magnitude the function will have greater rates of decrease or increase. As a result, it must have higher frequencies in its FT representation. (We saw this in the case of Fourier series. Of course, in the case that 0 < b <, the situation is reversed. The graph of f(bt will be a horizontally-stretched version of the graph of f(t, and the corresponding FT, F(ω/b will be a horizontally-contracted version of the graph of F(ω. Let s now go back and compute the FT of the boxcar function b(t from the previous example using the Scaling Formula and our result for the FT of the window function W(t from the previous lecture. Recall that b(t = W Using the Scaling Theorem, with b = /(, we have ( t. (7 ( B(ω = sinc ω = sinc(πω, (7 which is in agreement with our previous calculation. 8

19 Parseval s Formula for a non-periodic function Here we recall the complex-valued Fourier series for a τ-periodic function f(t: where f(t = n= Also recall Parseval s formula associated with this expansion, τ τ/ τ/ c n e inω 0t, (73 ω 0 = τ. (74 f(t dt = n= c n. (75 There exists a Parseval formula between non-periodic functions f(t on R and their Fourier transforms F(ω. It is as follows, Derivation: The Fourier transform of f(t is defined as f(t dt = F(ω dω. (76 F(ω = and the inverse Fourier transform (reconstructing f from F is f(t = f(t e iωt dt (77 F(ω e iωt dω. (78 Multiply both sides of Eq. (78 by f(t and integrate from to : f(t dt = t= ( ω= f(tf(ω e iωt dω dt t= ω= = ω= ( t= F(ω f(t e iωt dt dω (see note below ω= t= = ω= F(ωF(ωdω ω= = F(ω dω. (79 Note: The interchange of order of integration is valid subject to suitable restrictions on f and F (Fubini s Theorem. 9

20 Note that we may express Parseval s formula as follows, f, f = F, F, (80 where, denotes the following complex-valued inner product on R, f, g = f(t g(tdt. (8 (Since f(t is normally assumed to be real-valued, the complex conjugate may be viewed as irrelevant, but the inner product must also apply to the complex-valued Fourier transform F(ω as well. Recalling that Parseval s formula may also be written as Appendix: Plancherel s formula f, f = f = f(t dt, (8 f = F, or f = F. (83 In fact, Parseval s formula may be generalized to the following result, known as Plancherel s formula: Let f and g be defined on R and let their Fourier transforms be denoted as F(ω and G(ω, respectively. Then where, denotes the complex inner product on R, i.e., f, g = F, G, (84 f(tg(t dt = F(ωG(ω dω. (85 In the special case that f = g, implying that F = G, Plancherel s formula reduces to Parseval s formula. Proof of Plancherel s formula Plancherel s formula is proved in much the same way as was Parseval s formula. We first express the function f(t in terms of the inverse Fourier transform, i.e., f(t = F(ωe iωt dω. (86 0

21 Now substitute for F(ω using the definition of the Fourier transform: f(t = Now take the inner product of f(t with g(t: f, g = f(se iωs ds e iωt dω. (87 f(xe iωs ds e iωt dω g(t dt. (88 We now assume that f and g are sufficiently nice so that Fubini s Theorem will allow us to rearrange the order of integration. The result is f, g = = and the theorem is proved. ( f(se iωs ds F(ωG(ω dω, ( g(te iωt dt = F, G, (89 dω

22 The Fourier transform of a Gaussian (in t-space is a Gaussian (in ω-space This is a fundamental result in Fourier analysis as well as a number of applications, including theoretical physics (quantum mechanics. To show it, consider the following function, f σ (t = σ t e σ. (90 You might recognize this function: It is the normalized Gaussian distribution function, or simply the normal distribution with zero-mean and standard deviation σ or variance σ. The factor in front of the integral normalizes it, since f σ (t dt = σ (The above result implies that the L norm of f is, i.e., f =. e t σ dt =. (9 y f(0 = (σ σ σ Sketch of normalized Gaussian function f σ (t = 0 t σ t e σ. Just in case you are not fully comfortable with this result, we mention that all of these results come from the following fundamental integral: e x dx = π. (9 From this result, we find that e Ax dx = π, (93 A by means of the change of variable y = Ax. For simplicity, let us compute the FT of the function g σ (t = e t σ. (94

23 Our desired FT for f σ (t will then be given by Then G σ (ω = = = F σ (ω = σ G σ(ω. (95 e t σ e iωt dt e t σ [cos(ωt i sin(ωt] dt e t σ cos(ωt dt. (96 The sin(ωt term will not contribute to the integration because it is an odd function: Its product with the even Gaussian function is an odd function which, when integrated over (,, yields a zero result. The resulting integral in (96 is not straightforward to evaluate because the antiderivative of the Gaussian does not exist in closed form. But we may perform a trick here. Let us differentiate both sides of the equation with respect to the parameter ω. The differentiation of the integrand is permitted by Leibniz Rule, so that G σ (ω = te t σ sin(ωt dt. The appearance of the t in the integrand now permits an antidifferentiation using integration by parts: Let u = sin(ωt and dv = t exp( t /(σ, so that v = σ exp( t /(σ and du = ω cos(ωt. Then G σ(ω = [ ] sin(ωte t σ σ ω e t σ cos(ωt dt. The first term is zero, because the Gaussian function e t σ 0 as t ±. And the integral on the right, which once again involves the cos(ωt function, along with the / factor, is our original G σ (ω function. Therefore, we have derived the result, G σ(ω = σ ωg σ (ω. (97 This is a first order DE in the function G(ω. It is also a separable DE and may be easily solved to yield the general solution (details left for reader: G σ (ω = De σ ω, (98 3

24 where D is the arbitrary constant. From the fact that G σ (0 = we have that D = σ, implying that e t σ dt = σ, (99 G σ (ω = σe σ ω. (00 From (95, we arrive at our goal, F σ (ω, the Fourier transform of the Gaussian function f σ (t, F σ (ω = e σ ω. (0 In summary, the Fourier transform F σ (ω of the Gaussian function f σ (t is a Gaussian in the variable ω. There is one fundamental difference, however, between the two Gaussians, F σ (ω and f σ (t. The standard deviation of f σ (t is σ. But the standard deviation of F σ (ω, i.e., the value of ω at which F σ (ω assumes the value e / F(0, is σ. In other words, if σ is small, so that the Gaussian f σ (t is a thin peak, then F σ (ω is broad. This relationship, which is a consequence of the complementarity of the time (or space and frequency domains, is sketched below. y y f(0 = (σ σ 0 σ t /σ 0 F σ(0 = ( Generic sketch of normalized Gaussian function f σ (t = σ t e σ, with standard deviation σ (left, and its Fourier transform F σ (ω = e σ ω with standard deviation /σ (right. /σ ω 4