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The Fourier Transform

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The Fourier Transform Chapter 6 The Fourier Transform In this chapter we will turn to the study of Fourier transforms. These will provide an integral representation of functions de¯ned on the real line. Such functions can represent analog signals. Analog signals are continuous signals which may be sums over a continuous set of frequencies, as opposed to the sum over discrete frequencies, which Fourier series were used to represent. We will see how to rewrite our trigonometric Fourier series as complex exponential series. Then we will extend our series to signals with in¯nite periods. In later chapters we will see the connection between analog and digital signals. 6.1 Complex Exponential Fourier Series We ¯rst recall the trigonometric Fourier series representation of a function de¯ned on [¡¼; ¼] with period 2¼. The Fourier series is given by 1 a0 X f(x) » + (an cos nx + bn sin nx) ; (6.1) 2 n=1 where the Fourier coe±cients were found as Z 1 ¼ an = f(x) cos nx dx; n = 0; 1;:::; ¼ ¡¼ 133 134 CHAPTER 6. THE FOURIER TRANSFORM Z 1 ¼ bn = f(x) sin nx dx; n = 1; 2;::: (6.2) ¼ ¡¼ In order to derive the exponential Fourier series, we replace the trigonometric functions with exponential functions and collect like terms. This gives à à ! à !! 1 inx ¡inx inx ¡inx a0 X e + e e ¡ e f(x) » + an + bn 2 n=1 2 2i µ ¶ µ ¶ a X1 a ¡ ib X1 a + ib = 0 + n n einx + n n e¡inx: (6.3) 2 n=1 2 n=1 2 The coe±cients can be rewritten by de¯ning 1 c = (a + ib ); n = 1; 2;:::: (6.4) n 2 n n Then we also have that 1 c¹ = (a ¡ ib ); n = 1; 2;:::: (6.5) n 2 n n This gives our representation as X1 X1 a0 inx ¡inx f(x) » + c¹ne + cne : 2 n=1 n=1 Reindexing the ¯rst sum, by letting k = ¡n, we can write a ¡1X X1 f(x) » 0 + c¹ e¡ikx + c e¡inx: 2 ¡k n k=¡1 n=1 Now, we can de¯ne cn =c ¹¡n; n = ¡1; ¡2;:::: a0 Finally, we note that we can take c0 = 2 : So, we can write the complex exponential Fourier series representation as X1 ¡inx f(x) » cne ; (6.6) ¡1 6.1. COMPLEX EXPONENTIAL FOURIER SERIES 135 where 1 c = (a + ib ); n = 1; 2;::: n 2 n n 1 c = (a ¡ ib ); n = ¡1; ¡2;::: n 2 ¡n ¡n a c = 0 : (6.7) 0 2 Given such a representation, we would like to write out the integral forms of the coe±cients, cn. So, we replace the an's and bn's with their integral representations and replace the trigonometric functions with complex exponential functions. Doing this, we have for n = 1; 2;:::: 1 cn = (an + ibn) 2 · Z Z ¸ 1 1 ¼ i ¼ = f(x) cos nx dx + f(x) sin nx dx 2 ¼ ¡¼ ¼ ¡¼ Z à ! Z à ! 1 ¼ einx + e¡inx i ¼ einx ¡ e¡inx = f(x) dx + f(x) dx 2¼ ¡¼ 2 2¼ ¡¼ 2i Z 1 ¼ = f(x)einx dx (6.8) 2¼ ¡¼ It is a simple matter to determine the cn's for other values of n. For n = 0, we have that Z ¼ a0 1 c0 = = f(x) dx: 2 2¼ ¡¼ For n = ¡1; ¡2;:::, we ¯nd that Z ¼ Z ¼ 1 ¡inx 1 inx cn =c ¹n = f(x)e dx = f(x)e dx: 2¼ ¡¼ 2¼ ¡¼ Therefore, for all n we have show that Z ¼ 1 inx cn = f(x)e dx: (6.9) 2¼ ¡¼ We have converted our trigonometric series for functions de¯ned on [¡¼; ¼] to a complex exponential series in Equation (6.6) with Fourier coe±cients given by (6.9). We can easily extend the above analysis to other intervals. For example, for x 2 [¡L; L] the Fourier trigonometric series is 1 µ ¶ a0 X n¼x n¼x f(x) » + an cos + bn sin 2 n=1 L L 136 CHAPTER 6. THE FOURIER TRANSFORM with Fourier coe±cients Z 1 L n¼x an = f(x) cos dx; n = 0; 1;:::; L ¡L L Z 1 L n¼x bn = f(x) sin dx; n = 1; 2;:::: L ¡L L This can be rewritten in an exponential Fourier series of the form X1 ¡in¼x=L f(x) » cne ¡1 with Z L 1 in¼x=L cn = e dx: 2¼ ¡L 6.2 Exponential Fourier Transform Both the trigonometric and complex exponential Fourier series provide us with representations of a class of functions in term of sums over a discrete set of frequencies for functions of ¯nite period. On intervals [¡L; L] the period is 2L. Writing the arguments in terms of frequencies, we have n¼ n 2¼f = L , or the sums are over frequencies f = 2L : This is a discrete, or countable, set of frequencies. We would now like to extend our interval to x 2 (¡1; 1) and to extend the discrete set of frequencies to a continuous set of frequencies. One can do this rigorously, but it amounts to letting L and n get large and keeping n L ¯xed. We de¯ne ! = 2¼f and the sum over a continuous set of frequencies becomes an integral. Formally, we arrive at the Fourier transform Z 1 F [f] = f^(!) = f(x)ei!x dx: (6.10) ¡1 This is a generalization of the Fourier coe±cients (6.9). Once we know the Fourier transform, then we can reconstruct our function using the inverse Fourier transform, which is given by Z 1 F ¡1[f] = f(x) = f^(!)e¡i!x d!: (6.11) ¡1 6.2. EXPONENTIAL FOURIER TRANSFORM 137 We note that it can be proven that the Fourier transform exists when f(x) is absolutely integrable, i.e., Z 1 jf(x)j dx < 1: ¡1 Such functions are said to be L1. The Fourier transfrom and inverse Fourier transform are inverse operations. This means that F ¡1[F [f]] = f(x) and F [F ¡1[f^]] = f^(!): We will now prove the ¯rst of these equations. The second follows in a similar way. This is done by inserting the de¯nition of the Fourier transform into the inverse transform de¯nition and then interchanging the orders of integration. Thus, we have Z 1 1 F ¡1[F [f]] = F [f]e¡i!x d! 2¼ ¡1 Z ·Z ¸ 1 1 1 = f(»)ei!» d» e¡i!x d! 2¼ Z¡1 Z ¡1 1 1 1 = f(»)ei!(»¡x) d»d! 2¼ ¡1 ¡1 Z ·Z ¸ 1 1 1 = ei!(»¡x) d! f(») d»: (6.12) 2¼ ¡1 ¡1 In order to complete the proof, we need to evaluate the inside integral, which does not depend upon f(x). This is an improper integral, so we will de¯ne Z L i!x DL(x) = e d! ¡L and compute the inner integral as Z 1 i!(»¡x) e d! = lim DL(» ¡ x): ¡1 L!1 We can compute DL(x): A simple evaluation yields Z L i!x DL(x) = e d! ¡L 138 CHAPTER 6. THE FOURIER TRANSFORM 8 6 4 2 ±4 ±3 ±2 ±10 1 2 3 4 x Figure 6.1: A plot of the function DL(x) for L = 4. 80 60 40 20 ±4 ±20 2 4 x Figure 6.2: A plot of the function DL(x) for L = 40. ei!x = jL ix ¡L eixL ¡ e¡ixL = 2ix 2 sin xL = : (6.13) x We can graph this function. As x ! 0, DL(x) ! 2L. For large x, The function tends to zero. A plot of this function is in Figure 6.1. For large L the peak grows and the values of DL(x) for x 6= 0 tend to zero as show in Figure 6.2. We note that in the limit L !, DL(x) = 0 for x 6= 0 and it is in¯nite at 6.2. EXPONENTIAL FOURIER TRANSFORM 139 4 3 2 1 ±1 ±0.8 ±0.6 ±0.4 ±0.20 0.2 0.4 0.6 0.8 1 x Figure 6.3: A plot of the functions fn(x) for n = 2; 4; 8. x = 0. However, the area is constant for each L. In fact, Z 1 DL(x) dx = 2¼: ¡1 This behavior can be represented by the limit of other sequences of functions. De¯ne the sequence of functions ( 1 0; jxj > n fn(x) = n 1 2 ; jxj < n This is a sequence of functions as shown in Figure 6.4. As n ! 1, we ¯nd the limit is zero for x 6= 0 and is in¯nite for x = 0. However, the area under each member of the sequences is one. Thus, the limiting function is zero at most points but has area one. The limit is not really a function. It is a generalized function. It is called the Dirac delta function, which is de¯ned by 1. ±(x) = 0 for x 6= 0. R 1 2. ¡1 ±(x) dx = 1: Before returning to the proof, we state one more property of the Dirac delta function, which we will prove in the next section. We have that Z 1 ±(x ¡ a)f(x) dx = f(a): ¡1 140 CHAPTER 6. THE FOURIER TRANSFORM Returning to the proof, we now have that Z 1 i!(»¡x) e d! = lim DL(» ¡ x) = 2¼±(» ¡ x): ¡1 L!1 Inserting this into (6.12), we have Z ·Z ¸ 1 1 1 F ¡1[F [f]] = ei!(»¡x) d! f(») d»: 2¼ Z¡1 ¡1 1 1 = 2¼±(» ¡ x)f(») d»: 2¼ ¡1 = f(x): (6.14) Thus, we have proven that the inverse transform of the Fourier transform of f is f.
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