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Unformatted text preview: 18.03 Problem Set 7 Fall 2009 Solutions 1. (6 points) Suppose f and g are 2 periodic functions which are piecewise continuous (so that their Fourier coefficients a n ( f ) , a n ( g ) , and so on are all defined). Define a new function ( f * g )( t ) = 1 2 Z  f ( s ) g ( t s ) ds. a) Show that f * g is 2 periodic. ( f * g )( t + 2 ) = 1 2 Z  f ( s ) g ( t + 2  s ) ds = 1 2 Z  f ( s ) g ( t s ) ds = ( f * g )( t ) , where the middle step is true because g is 2 periodic. b) Find a formula for the complex Fourier coefficients of c n ( f * g ) in terms of c n ( f ) and c n ( g ) . We have c n ( f * g ) = 1 2 Z  ( 1 2 Z  f ( s ) g ( t s ) ds ) e int dt = 1 2 Z  f ( s ) ( 1 2 Z  g ( t s ) e int dt ) ds. Let u := t s . Then c n ( f * g ) = 1 2 Z  f ( s ) ( 1 2 Z  s  s g ( u ) e in ( s + u ) du ) ds = 1 2 Z  f ( s ) ( 1 2 Z  s  s g ( u ) e inu du ) e ins ds = 1 2 Z  f ( s ) c n ( g ) e ins ds = c n ( g ) 1 2 Z  f ( s ) e ins ds = c n ( f ) c n ( g ) . c) Suppose that f and f 1 are the square wave and sawtooth functions defined in Problem Set 6. Calculate the convolution f * f 1 . Recall that c 2 n ( f ) = c 2 n ( f 1 ) = 0. We also have a 2 n +1 ( f ) = ( 1) n 4 (2 n + 1) and b 2 n +1 ( f 1 ) = ( 1) n 4 (2 n + 1) 2 (and b 2 n +1 ( f ) = a 2 n +1 ( f 1 ) = 0). Thus f * f 1 is the function whose Fourier coefficients are given by c 2 n ( f * f 1 ) = 0 and c 2 n +1 ( f * f 1 ) = a 2 n +1 ( f ) 2 ib 2 n +1 ( f 1 ) 2 = i 4 ( 1) n 4 (2 n + 1) ( 1) n 4 (2 n + 1) 2 = 4 i (2 n + 1) 3 2 and c (2 n +1) ( f * f 1 ) = a 2 n +1 ( f ) 2 ib 2 n +1 ( f 1 ) 2 = 4 i (2 n + 1) 3 2 . (This expression looks very close to the expression for c n ( f 2 ). Can you give a nice description of the graph of f * f 1 in terms of the graph of f 2 ?) 2. (8 points) The string of a certain musical instrument has length L . Think of it as lying on the x axis, with one end fixed at the origin (0 , 0) and the other end fixed at ( L, 0) ....
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 Spring '09
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