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Final-2006-solutions - EE 261 The Fourier Transform and its...

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EE 261 The Fourier Transform and its Applications Fall 2006 Final Exam Solutions Notes: There are 7 questions for a total of 120 points Write all your answers in your exam booklets When there are several parts to a problem, in many cases the parts can be done independently, or the result of one part can be used in another part. Please be neat and indicate clearly the main parts of your solutions
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1. (15 points) Let f ( t ) be a periodic signal of period 1. One says that f ( t ) has half-wave symmetry if f ( t - 1 2 )= - f ( t ) . (a) Sketch an example of a signal that has half-wave symmetry. (b) If f ( t ) has half-wave symmetry and its Fourier series is f ( t ± n = -∞ c n e 2 πint show that c n =0if n is even. Hint: - c n = - ² 1 0 e - 2 πint f ( t ) dt = ² 1 0 e - 2 πint f ( t - 1 2 ) dt . Solution: (a) A simple example is f ( t ) = sin(2 πt ). The graphs of sin(2 ) and sin(2 π ( t - 1 2 )) are shown. - 0.5 - 1 1 1 1 - 0.5 - 1 1 1 1 Algebraically, sin(2 π ( t - 1 2 )) = sin(2 - π - sin2 πt. 1
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(b) The hint says - c n = - ± 1 0 e - 2 πint f ( t ) dt = ± 1 0 e - 2 πint f ( t - 1 2 ) dt . We make a change of variable u = t - 1 2 in the second integral: ± 1 0 e - 2 πint f ( t - 1 2 ) dt = ± 1 / 2 - 1 / 2 e - 2 πin ( u + 1 2 ) f ( u ) du = ± 1 / 2 - 1 / 2 e - 2 πinu e - 2 πin 1 2 f ( u ) du = e - πin ± 1 / 2 - 1 / 2 e - 2 πinu f ( u ) du = e - πin c n , (because we can integrate over any cycle to compute c n ) . Thus - c n = e - πin c n . If n is even then e - πin = 1 and we have - c n = c n , hence c n =0 . A slightly different route to the same end is as follows. Again it uses the substitution u = t - 1 2 in an integral. c n = ± 1 0 e - 2 πint f ( t ) dt = ± 1 / 2 0 e - 2 πint f ( t ) dt + ± 1 1 / 2 e - 2 πint f ( t ) dt = ± 1 / 2 0 e - 2 πint f ( t ) dt - ± 1 1 / 2 e - 2 πint f ( t - 1 2 ) dt = ± 1 / 2 0 e - 2 πint f ( t ) dt - ± 1 / 2 0 e - 2 πin ( u + 1 2 ) f ( u ) du = ± 1 / 2 0 e - 2 πint f ( t ) dt - e - πin ± 1 / 2 0 e - 2 πinu f ( u ) du , and if n is even the integrals cancel, giving c n . 2
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2. (20 points) Sampling using the derivative Suppose that f ( t ) is a bandlimited signal with F f ( s )=0for | s |≥ 1 (bandwidth 2). According to the sampling theorem, knowing the values f ( n ) for all integers n (sampling rate of 1) is not sufficient to interpolate the values f ( t ) for all t . However, if in addition one knows the values of the derivative f ± ( n ) at the integers then there is an interpolation formula with a sampling rate of 1. In this problem you will derive that result. Let F ( s )= F f ( s ) and let G ( s 1 2 πi ( F f ± )( s sF ( s ). (a) For 0 s 1 show that (III * F )( s F ( s )+ F ( s - 1) * G )( s sF ( s )+( s - 1) F ( s - 1) and then show that F ( s )=(1 - s )(III * F )( s )+(III * G )( s ) . (b) For - 1 s 0 show that * F )( s F ( s F ( s +1) * G )( s sF ( s s F ( s and then show that F ( s )=(1+ s )(III * F )( s ) - * G )( s ) . (c) Using parts (a) and (b) show that for all s , -∞ <s< , F ( s )=Λ( s )(III * F )( s ) - Λ ± ( s )(III * G )( s ) , where Λ( s ) is the triangle function Λ( s ± 1 -| s | , | s |≤ 1 , 0 , | s 1 . (d) From part (c) derive the interpolation formula f ( t ² n = -∞ f ( n )sinc 2 ( t - n ² n = -∞ f ± ( n )( t - n )sinc 2 ( t - n ) . Solutions: (a) Since F ( s ) is zero outside of | s 1, if we restrict s to lie between 0 and 1 we only get a few shifts F ( s - n ) that are nonzero, namely, * F )( s ² n = -∞ F ( s - k ) = ··· + F ( s +2)+ F ( s ³ ´µ =0 + F ( s F ( s - 1) + F ( s - 2) + ³ ´µ =0 = F ( s F ( s - 1) 3
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Since G ( s )= sF ( s ) it’s the same thing; only two nonzero terms in III * G : (III * G )( s ±
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Final-2006-solutions - EE 261 The Fourier Transform and its...

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