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By Myung H. Kwack

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EXAMPLE To =4 h(1) (2-17) 2-4 Consider the function (2-20) (2-22) 18 Chap. 2 THE FOURIER TRANSFORM Sec. 2-3 THE FOUJUER TRANSFORM 19 By means of Condition 2, the Fourier transform pair HIli hIlI· 2Afo sin 1:°11 2Af, sin (2n/ot) o 2n/ot 01 B ~f H(/) = A 1/1

From Condition 2 the Fourier transform of h(t) exists and is given by f~ 2Af, sin (2ft/ot) e- 12• /1 dt H(f) = _~ ~ f~ sin (2n/ot)[cos (2n/t) n _~ t = 1 2n/ot 0 j sin (2n/t)] dt = ~ f~ sin (2n/ot) cos (2n/t) dt n _~ t (2-23) The imaginary term integrates to zero since the integrand term is an odd function. 1. 1. S(t)e- 12 • ,t dt = Ke O = K r~ [~]eI2,'t d/ = r~ K cos ~2n/t) d/ + j f~ K sin (2nft) d/ (2-30) 2nt (2-31) Because the integrand of the second integral is an odd function, the integral is zero; the first integral is meaningless unless it is interpreted in the sense of distribution theory.

Fl·H(f) (3-20) The· time-shifted Fourier transform pair is 2A h(1 - 10 ) EXAMPLE ·To H(f)e-/hfl' (3-21) 3-6 A pictorial description of this pair is illustrated in Fig. 3-4. As shown, timeshifting results in a change in the phase angle (J(f) = tan-I[/(f)/R(f)]. Note that time-shifting does not alter the magnitude of the Fourier transform. ;Hl(f) (3-22) 4h(4t) H(~ 4 4A where H(f) has been assumed (0 be real for simplicity. These results are easily extended to the case of H(f), a complex function.

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Families of normal maps in several variables and classical theorems in complex analysis by Myung H. Kwack


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