PS2-solutions

PS2-solutions - EE261 Raj Bhatnagar Summer 2010-2011 EE 261...

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Unformatted text preview: EE261 Raj Bhatnagar Summer 2010-2011 EE 261 The Fourier Transform and its Applications Problem Set 2 Due Friday 8 July 1. (15 points) Fourier Series Symmetries (a) 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 ) . 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 ) = X n =- c n e 2 int , show that c n = 0 if n is even. (c) Let g ( t ) be a periodic signal of period 1 . One says that g ( t ) has quarter-wave symmetry if g ( t- 1 4 ) =- ig ( t ) . If g ( t ) has quarter-wave symmetry and Fourier series coe cients d n , show that d n must be zero for n 6 = 1 + 4 k where k is an integer. Solution: (a) A simple example is f ( t ) = sin(2 t ) . The graphs of sin(2 t ) and sin(2 ( t- 1 2 )) are shown. Figure 1: sin2 t Figure 2: sin2 ( t- 1 / 2) Algebraically, sin(2 ( t- 1 2 )) = sin(2 t- ) =- sin2 t. (b) Apply the half-wave symmetry equation to- c n :- c n =- Z 1 e- 2 int f ( t ) dt = Z 1 e- 2 int f ( t- 1 2 ) dt. 1 We make a change of variable u = t- 1 2 in the second integral: Z 1 e- 2 int f ( t- 1 2 ) dt = Z 1 / 2- 1 / 2 e- 2 in ( u + 1 2 ) f ( u ) du = Z 1 / 2- 1 / 2 e- 2 inu e- 2 in 1 2 f ( u ) du = e- in Z 1 / 2- 1 / 2 e- 2 inu f ( u ) du = e- in c n , (because we can integrate over any cycle to compute c n ) . Thus- c n = e- in c n . If n is even then e- in = 1 and we have- c n = c n , hence c n = 0 . (c) We expand both sides of the quarter wave equation as a Fourier series: g ( t- 1 / 4) =- ig ( t ) X n =- d n e 2 in ( t- 1 / 4) =- i X n =- d n e 2 int X n =- d n e 2 int- in/ 2 = e- i/ 2 X n =- d n e 2 int X n =- ( d n e- in/ 2 ) e 2 int = X n =- ( e- i/ 2 d n ) e 2 int d n e- in/ 2 = e- i/ 2 d n This last equation holds automatically if n = 1 , 5 , 7 ,... . For other values of n , this equation can only be true if d n must be zero. 2. (20 points) Some practice with inner products In class we de ned the inner product between two periodic functions. In this problem, we'll extend this operation to non-periodic functions: Let f ( t ) and g ( t ) be two non-periodic functions and de ne their inner product as ( f,g ) = Z - f ( t ) g ( t ) dt. Also, de ne the reversed signal to be f- ( t ) = f (- t ) , and the delay or shift operator by a f ( t ) = f ( t- a ) . 2 (a) Show that if both f ( t ) and g ( t ) are reversed, the inner product is unchanged: ( f- ,g- ) = ( f,g ) . (b) Show that ( f- ,g ) = ( f,g- ) . (c) Show that ( a f, a g ) = ( f,g ) . (d) Show that ( a f,g ) = ( f,- a g ) ....
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PS2-solutions - EE261 Raj Bhatnagar Summer 2010-2011 EE 261...

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