PS-1-2007-Solutions

# PS-1-2007-Solutions - EE 261 The Fourier Transform and its...

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Unformatted text preview: EE 261 The Fourier Transform and its Applications Fall 2007 Solutions to Problem Set One 1. Some practice with geometric sums and complex exponentials Well 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 = 1, and p and q are any integers, show that q 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 n =0 e 2 in/N and explain your answer geometrically. (c) Derive the formula N k =- N e 2 ikt = sin(2 t ( N + 1 / 2)) sin( t ) Solution (a) Were supposed to show q n = p w n = w p- w q +1 1- w . We can derive this from the usual formula, one that youve probably seen and should certainly know: N n =0 w n = 1- w N +1 1- w . 1 We have q 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 lets 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 | &lt; 1. If so then w q +1 0 as q and n = p w n = w p 1- w . If p =- then, but contrast, well get w p 0 if | w | &gt; 1. Assuming this, q 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 Id prefer q n =- w n = w q +1 w- 1 . Note that if w is real and &gt; 1 then this is positive, as it should be. If p =- and q = wed have to have both | w | &lt; 1 and | w | &gt; 1 to get convergence, an absurd condition. Thus we conclude n =- w n never converges. Heres a summary: q n = p w n = q- p + 1 , w = 1 w p- w q +1 1- w , w = 1 ,- &lt; p , q &lt; w q +1 w- 1 , | w | &gt; 1 , p =- , q &lt; w p 1- w , | w | &lt; 1 ,- &lt; p , q = 2 (b) From the formula in Part (a) we obtain N- 1 n =0 e 2 ik/N = 1- e 2 iN/N 1- e 2 i/N = 1- e 2 i 1- e 2 i/N = 0 This answer also makes complete sense geometrically. The general complex exponential re i can be thought of as a vector in the complex plane, of length r and at an angle counterclockwise from the real axis. Thus w = e 2 i/n is a vector of length 1 at an angle of 2 /n . Similarly, w k = e 2 ik/n has length 1 and is at an angle of 2 k/n . The points are distinct and equidistantly spaced 2 /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 ik/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....
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PS-1-2007-Solutions - EE 261 The Fourier Transform and its...

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