Presentation on theme: "Sheng-Fang Huang. 11.7 Fourier Integral Consider the periodic rectangular wave ƒ L (x) of period 2L > 2 given by The left part of Fig. 277 shows this."— Presentation transcript:
11.7 Fourier Integral Consider the periodic rectangular wave ƒ L (x) of period 2L > 2 given by The left part of Fig. 277 shows this function for 2L = 4, 8, … as well as the nonperiodic function ƒ(x), which we obtain from ƒ L if we let L → ∞,
Amplitude Spectrum Consider the Fourier coefficients of ƒ L as L increases. Since ƒ L is even, b n = 0 for all n. For a n, This sequence of Fourier coefficients is called the amplitude spectrum of ƒ L because ｜ a n ｜ is the maximum amplitude of the a n cos (nπx/L).
Fig Waveforms and amplitude spectra
(See Fig. 277) For increasing L these amplitudes become more and more dense on the positive w n - axis, where w n = nπ/L. For 2L = 4, 8, 16 we have 1, 3, 7 amplitudes per “half-wave” of the function (2 sin w n )/(Lw n ). Hence, for 2L = 2 k we have 2 k-1 – 1 amplitudes per half- wave. These amplitudes will eventually be everywhere dense on the positive w n -axis (and will decrease to zero).
From Fourier Series to Fourier Integral Consider any periodic function ƒ L (x) of period 2L that is represented by a Fourier series what happens if we let L → ∞? We should expect an integral (instead of a series) involving cos wx and sin wx with w no longer restricted to integer multiples w = w n = nπ/L of π/L but taking all values.
If we insert a n and b n, and denote the variable of integration by v, the Fourier series of ƒ L (x) becomes We now set
Then 1/L = Δw/π, and we may write the Fourier series in the form (1) Let L → ∞ and assume that the resulting nonperiodic function is absolutely integrable on the x-axis; that is, the following limits exist:
1/L → 0, and the value of the first term on the right side of (1) → zero. Also Δw = π/L → dw. The infinite series in (1) becomes an integral from 0 to ∞, which represents ƒ(x), namely, (3) If we introduce the notations (4)
Fourier integral we can write this in the form (5) This is called a representation of ƒ(x) by a Fourier integral.
Fourier Integral THEOREM 1 If ƒ(x) is piecewise continuous in every finite interval and has a right-hand derivative and a left-hand derivative at every point and if the integral exists, then ƒ(x) can be represented by a Fourier integral with A and B given by (4). At a point where ƒ(x) is discontinuous the value of the Fourier integral equals the average of the left- and right-hand limits of ƒ(x) at that point.
Applications of Fourier Integrals Example 2: Single Pulse, Sine Integral Find the Fourier integral representation of the function
Sine Integral The case x = 0 is of particular interest. If x = 0, then (7) gives (8*) We see that this integral is the limit of the so-called sine integral (8) as u → ∞. The graphs of Si(u) and of the integrand are shown in Fig. 279.
Fig Sine integral Si(u) and integrand
In the case of the Fourier integral, approximations are obtained by replacing ∞ by numbers a. Hence the integral (9) which approximates ƒ(x).
Gibbs Phenomenon We might expect that these oscillations disappear as a → ∞. However, with increasing a, they are shifted closer to the points x = ±1. This unexpected behavior is known as the Gibbs phenomenon.
Fourier Cosine Integral and Fourier Sine Integral If ƒ(x) is an even function, then B(w) = 0 and (10) The Fourier integral (5) then reduces to the Fourier cosine integral
Fourier Cosine Integral and Fourier Sine Integral If ƒ(x) is an odd function, then A(w) = 0 and (12) The Fourier integral (5) then reduces to the Fourier cosine integral
11.8 Fourier Cosine and Sine Transforms Fourier Cosine Transform For an even function ƒ(x), the Fourier integral is the Fourier cosine integral (1) We now set A(w)=, where c suggests “cosine.” Then from (1b), writing v = x, we have (2) and from (1a), (3)
Fourier Sine Transform Similarly, for an odd function ƒ(x), the Fourier integral is the Fourier sine integral (4) We now set B(w)=, where s suggests “sine.” From (4b), writing v = x, we have (5) This is called the Fourier sine transform of ƒ(x). From (4a) (6)
Fourier Sine Transform Equation (6) is called the inverse Fourier sine transform of. The process of obtaining from ƒ(x) is also called the Fourier sine transform or the Fourier sine transform method. Other notations are and and for the inverses of and, respectively.
Example 1: Fourier Cosine and Fourier Sine Transforms Find the Fourier cosine and Fourier sine transforms of the function Solution:
Example 2: Fourier Cosine Transform of the Exponential Function Find Solution.
Linearity, Transforms of Derivatives The Fourier cosine and sine transforms are linear operations, (7)
Cosine and Sine Transforms of Derivatives THEOREM 1 Let ƒ(x) be continuous and absolutely integrable on the x-axis, let ƒ'(x) be piecewise continuous on every finite interval, and let let ƒ(x) → 0 as x → ∞. Then (8)
Formula (8a) with ƒ' instead of ƒ gives (when ƒ', ƒ'' satisfy the respective assumptions for ƒ, ƒ' in Theorem 1) hence by (8b) (9a) Similarly, (9b)
Example 3: An Application of the Operational Formula (9) Find the Fourier cosine transform (e -ax ) of ƒ(x) = e -ax, where a > 0. Solution.
11.9 Fourier Transform. Discrete and Fast Fourier Transforms The complex Fourier integral is (4)
Fourier Transform and Its Inverse Writing the exponential function in (4) as a product of exponential functions, we have (5) The expression in brackets is a function of w, is denoted by, and is called the Fourier transform of ƒ; writing v = x, we have (6)
With this, (5) becomes (7) and is called the inverse Fourier transform of. Another notation for the Fourier transform is so that The process of obtaining the Fourier transform (ƒ) = from a given ƒ is also called the Fourier transform or the Fourier transform method.
Example 1: Fourier Transform Find the Fourier transform of ƒ(x) = 1 if ︱ x ︱ < 1 and ƒ(x) = 0 otherwise. Solution. Using (6) and integrating, we obtain As in (3) we have e iw = cos w + i sin w, e -iw = cos w – i sin w, and by subtraction e iw – e -iw = 2i sin w. Substituting this in the previous formula, we see that i drops out and we obtain the answer
Example 2: Fourier Transform Find the Fourier transform (e -ax ) of ƒ(x) = e -ax if x > 0 and ƒ(x) = 0 if x 0. Solution.
Linearity. Fourier Transform of Derivatives Linearity of the Fourier Transform THEOREM 2 The Fourier transform is a linear operation; that is, for any functions ƒ(x) and g(x) whose Fourier transforms exist and any constants a and b, the Fourier transform of aƒ + bg exists, and (8)
Fourier Transform of the Derivative of ƒ(x) THEOREM 3 Let ƒ(x) be continuous on the x-axis and ƒ(x) → 0 as ︱ x ︱ → ∞. Furthermore, let ƒ'(x) be absolutely integrable on the x-axis. Then (9)
Convolution The convolution ƒ * g of functions ƒ and g is defined by (11) Taking the convolution of two functions and then taking the transform of the convolution is the same as multiplying the transforms of these functions (and multiplying them by ):
Convolution Theorem THEOREM 4 Suppose that ƒ(x) and g(x) are piecewise continuous, bounded, and absolutely integrable on the x-axis. Then (12)
Convolution Theorem By taking the inverse Fourier transform on both sides of (12), writing and as before, and noting that and 1/ in (12) and (7) cancel each other, we obtain (13)
Discrete Fourier Transform (DFT) The function f(x) is given only in terms of values at finitely many points. Dealing with sampled values, we can replace Fourier transform by the so-called discrete Fourier transform. Let f(x) be periodic with the period 2π. Assume N measurements are taken over the interval 0 ≦ x ≦ 2π at regular spaced points
Discrete Fourier Transform (DFT) We now to determine a complex trigonometric polynomial that interpolates f(x) at the nodes. That is, Hence, we must determine the coefficients c 0, …, c N-1 Multiply by and sum over k from 0 to N-1 Denote […] by r. For n = m, r = e 0 = 1. The sum of these terms over k equals N.
Discrete Fourier Transform (DFT) For n≠m we have r ≠1 and by the formula for a geometric sum: Because This shows that the right side of (17) equals c m N. Thus, we obtain the desired coefficient formula:
Discrete Fourier Transform (DFT) It is practical to drop the factor 1/N from c n and define the discrete Fourier transform of the given signal to be the vector with components This is the frequency spectrum of the signal.
Fourier Matrix In vector notation,, where the N×N Fourier matrix F N =[e nk ] has the entries [given in (18)]:
Example 4 Let N = 4 measurements (sample values) be given. Then w = e -2πi/N = e -πi/2 = –i and thus w nk = (–i) nk. Let the sample values be, say f = [ ] T. Then by (18) and (19), (20)