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PS-1-Solutions-2009

PS-1-Solutions-2009 - EE 261 The Fourier Transform and its...

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EE 261 The Fourier Transform and its Applications Fall 2009 Solutions to Problem Set One 1. Some practice with geometric sums and complex exponentials (5 points each) We’ll make much use of formulas for the sum of a geometric series, especially in combination with complex exponentials. (a) If w is a real or complex number, w 6 = 1, and p and q are any integers, show that q X n = p w n = w p - w q +1 1 - w . (Of course if w = 1 then the sum is q n = p 1 = q + 1 - p .) Discuss the cases when p = -∞ or q = . What about p = -∞ and q = + ? (b) Find the sum N - 1 X n =0 e 2 πin/N and explain your answer geometrically. (c) Derive the formula N X k = - N e 2 πikt = sin(2 πt ( N + 1 / 2)) sin( πt ) Solution (a) We’re supposed to show q X n = p w n = w p - w q +1 1 - w . We can derive this from the usual formula, one that you’ve probably seen – and should certainly know: N X n =0 w n = 1 - w N +1 1 - w . 1
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We have q X n = p w n = w p + w p +1 + w p +2 + · · · + w q - 1 + w q = w p (1 + w + w 2 + · · · + w q - p ) = w p 1 - w q - p +1 1 - w = w p - w q +1 1 - w Now let’s consider the cases when one or the other (or both) limits are infinite. When will the series converge? If q = then to get convergence we have to assume that | w | < 1. If so then w q +1 0 as q → ∞ and X n = p w n = w p 1 - w . If p = -∞ then, but contrast, we’ll get w p 0 if | w | > 1. Assuming this, q X n = -∞ w n = - w q +1 1 - w . It looks a little funny to me to write it this way, with the minus sign, so I’d prefer q X n = -∞ w n = w q +1 w - 1 . Note that if w is real and > 1 then this is positive, as it should be. If p = -∞ and q = we’d have to have both | w | < 1 and | w | > 1 to get convergence, an absurd condition. Thus we conclude X n = -∞ w n never converges. Here’s a summary: q X n = p w n = q - p + 1 , w = 1 w p - w q +1 1 - w , w 6 = 1 , -∞ < p , q < w q +1 w - 1 , | w | > 1 , p = -∞ , q < w p 1 - w , | w | < 1 , -∞ < p , q = 2
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(b) From the formula in Part (a) we obtain N - 1 X n =0 e 2 πik/N = 1 - e 2 πiN/N 1 - e 2 πi/N = 1 - e 2 πi 1 - e 2 πi/N = 0 This answer also makes complete sense geometrically. The general complex exponential re can be thought of as a vector in the complex plane, of length r and at an angle θ counterclockwise from the real axis. Thus w = e 2 πi/n is a vector of length 1 at an angle of 2 π/n . Similarly, w k = e 2 πik/n has length 1 and is at an angle of 2 πk/n . The points are distinct and equidistantly spaced 2 π/n radians apart around the unit circle. Consider the n points equally spaced around the unit circle as vertices of a regular n -gon, and the e 2 πik/n as vectors from 0 to the vertices. The (vector) sum of the points is then the perimeter of the polygon. Viewed as a closed loop, the vector sum is the zero vector. (c) Again appealing to the formula from Part (a), N X k = - N e 2 πikt = e - 2 πiNt - e 2 πi ( N +1) t 1 - e 2 πit Now factor out e πit from both the numerator and denominator: e - 2 πiNt - e 2 πi ( N +1) t 1 - e 2 πit = e πit e πit e - 2 πi ( N + 1 2 ) t - e 2 πi ( N + 1 2 ) t e - πit - e πit
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