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PS-2-2009-solutions

PS-2-2009-solutions - 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 Two 1. (10 points) A famous sum You cannot go through life knowing about Fourier series and not know the application to evaluating a very famous sum. Let S ( t ) be the sawtooth function, that is S ( t ) = t for 0 t 1 and then periodized to have period 1. Show that the Fourier series for S ( t ) is 1 2 summationdisplay n = −∞ ,n negationslash =0 1 2 πin e 2 πint and use it to show that summationdisplay n =1 1 n 2 = π 2 6 . Solution: The zeroth Fourier coefficient is the average value: c 0 = integraldisplay 1 0 te 2 πi 0 t dt = integraldisplay 1 0 t dt = 1 2 . You can also see this from the graph of S ( t ). When n negationslash = 0 we calculate c n = integraldisplay 1 0 S ( t ) e 2 πint dt = integraldisplay 1 0 te 2 πint dt = 1 2 πin parenleftbigg te 2 πint + 1 2 πin e 2 πint parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle 1 0 = 1 2 πin parenleftbigg e 2 πin + 1 2 πin e 2 πin 1 2 πin parenrightbigg = 1 2 πin parenleftbigg 1 + 1 2 πin 1 2 πin parenrightbigg = i 2 πn 1
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The key is to apply Rayleigh’s theorem, summationdisplay n = −∞ | c n | 2 = integraldisplay 1 0 | S ( t ) | 2 dt On the left-hand side, we have summationdisplay n = −∞ | c n | 2 = vextendsingle vextendsingle vextendsingle vextendsingle 1 2 vextendsingle vextendsingle vextendsingle vextendsingle 2 + summationdisplay n = −∞ n negationslash =0 vextendsingle vextendsingle vextendsingle vextendsingle i 2 πn vextendsingle vextendsingle vextendsingle vextendsingle 2 = 1 4 + summationdisplay n = −∞ n negationslash =0 1 4 π 2 n 2 = 1 4 + 2 summationdisplay n =1 1 4 π 2 n 2 For the integral of S ( t ) 2 we have integraldisplay 1 0 | S ( t ) | 2 dt = integraldisplay 1 0 t 2 dt = 1 3 Combining the two sides, we obtain 1 4 + 2 summationdisplay n =1 1 4 π 2 n 2 = 1 3 summationdisplay n =1 1 4 π 2 n 2 = 1 24 summationdisplay n =1 1 n 2 = π 2 6 2. (30 points) Some practice with symmetry Considerations of symmetry arise frequently in Fourier analysis, e.g., evenness or oddness of a signal. It is helpful to introduce the reversed signal , defined by f ( t ) = f ( t ) . Thus f is even if and only if f = f f is odd if and only if f = f For any signal, f + f is even and f f is odd, and thus any signal can be decomposed into its even and odd parts: f = 1 2 ( f + f ) + 1 2 ( f f ) = f even + f odd . We do not assume that f is real-valued. Suppose that a square integrable function f , defined on −∞ < x < , is written as a sum of its even and odd parts as f ( x ) = f even ( x ) + f odd ( x ). 2
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(a) (10) Show that integraldisplay −∞ | f ( x ) | 2 dx = integraldisplay −∞ | f even ( x ) | 2 dx + integraldisplay −∞ | f odd ( x ) | 2 dx (b) (10) Find an expression of a similar kind for integraldisplay −∞ f ( x ) f ( x ) dx.
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