PS-2-2009-solutions

PS-2-2009-solutions - EE 261 The Fourier Transform and its Applications Fall 2009 Solutions to Problem Set Two 1(10 points A famous sum You cannot

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 = integraldisplay 1 te − 2 πi t dt = integraldisplay 1 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 S ( t ) e − 2 πint dt = integraldisplay 1 te − 2 πint dt = − 1 2 πin parenleftbigg te − 2 πint + 1 2 πin e − 2 πint parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle 1 = − 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 The key is to apply Rayleigh’s theorem, ∞ summationdisplay n = −∞ | c n | 2 = integraldisplay 1 | 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 | S ( t ) | 2 dt = integraldisplay 1 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 )....
View Full Document

This note was uploaded on 11/28/2009 for the course EE 261 at Stanford.

Page1 / 17

PS-2-2009-solutions - EE 261 The Fourier Transform and its Applications Fall 2009 Solutions to Problem Set Two 1(10 points A famous sum You cannot

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online