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mt2010sol - EE 261 The Fourier Transform and its...

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EE 261 The Fourier Transform and its Applications Fall 2010 Midterm Exam October 27, 2010 There are five questions for a total of 75 points. Please write your answers in the exam booklet provided, and make sure that your answers stand out. Don’t forget to write your name on your exam book! 1
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1. (15 points) Convolution and Fourier series The convolution of two functions f ( t ) and g ( t ) of period 1 is defined by ( f g )( t ) = integraldisplay 1 0 f ( τ ) g ( t τ ) dτ. Suppose f ( t ) and g ( t ) have Fourier series f ( t ) = summationdisplay n = −∞ a n e 2 πint g ( t ) = summationdisplay n = −∞ b n e 2 πint , respectively. Find the Fourier series of ( f g )( t ) and explain why this implies that f g = g f . Solution Next we find the Fourier series for f g . We substitute the Fourier series expansions into the formula for convolution. ( f g )( t ) = integraldisplay 1 0 f ( τ ) g ( t τ ) = integraldisplay 1 0 parenleftBigg summationdisplay n = −∞ a n e 2 πinτ parenrightBiggparenleftBigg summationdisplay m = −∞ b m e 2 πim ( t τ ) parenrightBigg (Notice that we use a different summation index for the series for g . This is necessar = integraldisplay 1 0 summationdisplay n,m = −∞ a n b m e 2 πinτ e 2 πim ( t τ ) = integraldisplay 1 0 summationdisplay n,m = −∞ a n b m e 2 πimt e 2 πi ( n m ) τ = summationdisplay n,m = −∞ a n b m e 2 πimt integraldisplay 1 0 e 2 πi ( n m ) τ (swapping summation and integration) We’ve seen that integral many times by now: integraldisplay 1 0 e 2 πi ( n m ) τ = braceleftBigg 1 , n = m 0 , n negationslash = m Thus the only terms that survive are when n = m and hence summationdisplay n,m = −∞ a n b m e 2 πimt integraldisplay 1 0 e 2 πi ( n m ) τ = summationdisplay n = −∞ a n b n e 2 πint We have shown ( f g )( t ) = summationdisplay n = −∞ a n b n e 2 πint .
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