ps07sol - 18.03 Problem Set 7 Fall 2009 Solutions 1(6...

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18.03 Problem Set 7 Fall 2009 Solutions 1. (6 points) Suppose f and g are 2 π -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 π Z π - π f ( s ) g ( t - s ) ds. a) Show that f * g is 2 π -periodic. ( f * g )( t + 2 π ) = 1 2 π Z π - π f ( s ) g ( t + 2 π - s ) ds = 1 2 π Z π - π f ( s ) g ( t - s ) ds = ( f * g )( t ) , where the middle step is true because g is 2 π -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 π Z π - π ( 1 2 π Z π - π f ( s ) g ( t - s ) ds ) e - int dt = 1 2 π Z π - π f ( s ) ( 1 2 π Z π - π g ( t - s ) e - int dt ) ds. Let u := t - s . Then c n ( f * g ) = 1 2 π Z π - π f ( s ) ( 1 2 π Z π - s - π - s g ( u ) e - in ( s + u ) du ) ds = 1 2 π Z π - π f ( s ) ( 1 2 π Z π - s - π - s g ( u ) e - inu du ) e - ins ds = 1 2 π Z π - π f ( s ) c n ( g ) e - ins ds = c n ( g ) 1 2 π Z π - π f ( s ) e - ins ds = c n ( f ) c n ( g ) . c) Suppose that f 0 and f 1 are the square wave and sawtooth functions defined in Problem Set 6. Calculate the convolution f 0 * f 1 . Recall that c 2 n ( f 0 ) = c 2 n ( f 1 ) = 0. We also have a 2 n +1 ( f 0 ) = ( - 1) n 4 (2 n + 1) π and b 2 n +1 ( f 1 ) = ( - 1) n 4 (2 n + 1) 2 π (and b 2 n +1 ( f 0 ) = a 2 n +1 ( f 1 ) = 0).
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Thus f 0 * f 1 is the function whose Fourier coefficients are given by c 2 n ( f 0 * f 1 ) = 0 and c 2 n +1 ( f 0 * f 1 ) = a 2 n +1 ( f 0 ) 2 - ib 2 n +1 ( f 1 ) 2 = - i 4 ( - 1) n 4 (2 n + 1) π ( - 1) n 4 (2 n + 1) 2 π = - 4 i (2 n + 1) 3 π 2 and c - (2 n +1) ( f 0 * f 1 ) = a 2 n +1 ( f 0 ) 2 ib 2 n +1 ( f 1 ) 2 = 4 i (2 n + 1) 3 π 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 0 * 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) . When the string is vibrating, the bit at position x is displaced (say in the y direction) by a small amount y ( x, t ) ; the amount of displacement depends both on the position x and on the time t . Suppose that the vibration is started by applying to the string some initial displacement function
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