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527solns10 - 642:527 SOLUTIONS ASSIGNMENT 10 FALL 2007...

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642:527 SOLUTIONS: ASSIGNMENT 10 FALL 2007 Section 17.3: 4 (g) f ( x ) = | sin x | has period π (since sin( x + π ) = - sin x ) so its Fourier series will have the form FS f = a 0 + summationdisplay n =1 [ a n cos 2 nx + b n sin 2 nx ] . Since f ( x ) is even ( | sin( - x ) | = | - sin x | = | sin x | ) the coefficients b n will all vanish. Moreover, using sin θ = ( e - e ) / 2 i and cos θ = ( e + e ) / 2, we have a 0 = 1 π integraldisplay π 0 f ( x ) dx = 1 π integraldisplay π 0 sin( x ) dx = 2 π , a n = 2 π integraldisplay π 0 f ( x ) cos(2 nx ) dx = 2 π integraldisplay π 0 sin( x ) cos(2 nx ) dx = 1 2 integraldisplay π 0 bracketleftBig e (2 n +1) ix - e (2 n 1) ix + e ( 2 n +1) ix - e ( 2 n 1) ix bracketrightBig dx = - 1 2 π bracketleftbigg e (2 n +1) ix 2 n + 1 - e (2 n 1) ix 2 n - 1 - e (2 n 1) ix 2 n - 1 + e (2 n +1) ix 2 n + 1 bracketrightbigg π 0 = - 4 π (4 n 2 - 1) . That is, FS f = 2 π - 4 π summationdisplay n =1 cos 2 nx 4 n 2 - 1 . 16 (b) The complex Fourier series of f ( x ) is summationdisplay n = −∞ c n e iπnx with c n = 1 2 integraldisplay 2 0 f ( x ) e iπnx dx = 1 2 integraldisplay 2 0 e (1 iπn ) x dx = e 2 - 1 2(1 - iπn ) . 18 (c) The given function F ( t ) is periodic with period 2; it is even, so there are no sine terms in its Fourier series. The cosine coefficients are a 0 = 1 2 integraldisplay 2 0 F ( t ) dt = 5 2 , a n = integraldisplay 2 0 f ( t ) cos nπt dt = 2 integraldisplay 1 0 5 t cos nπt dt = braceleftbigg - 20 / ( ) 2 , n odd, 0 , n even so that, since f ( t ) is continuous, f ( t ) = 5 2 - summationdisplay n =1 , 3 , 5 ,... 20 n 2 π 2 cos nπt To find the steady-state solution x ′′ + x = F ( t ) we observe that the steady-state solution of x ′′ + x = 1 is x ( t ) = 1 and that the steady-state solution of x ′′ + x = cos( nπt ) is x ( t ) = cos( nπt ) / (( ) 2 + 1), and then superimpose: x ( t ) = 5 2 - summationdisplay n =1 , 3 , 5 ,...
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