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# mt2007sol - EE 261 The Fourier Transform and its...

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EE 261 The Fourier Transform and its Applications Fall 2007 Solutions to Midterm Exam There are 5 questions for a total of 110 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) Let f ( x ) be a real, periodic function of period 1. The autocorrelation of f with itself is the function ( f ± f )( x ) = ± 1 0 f ( y ) f ( y + x ) dy . (a) Show that f ± f is also periodic of period 1. (b) If f ( x ) = ² n = -∞ ˆ f ( n ) e 2 πinx show that the Fourier series of ( f ± f )( x ) is ( f ± f )( x ) = ² n = -∞ | ˆ f ( n ) | 2 e 2 πinx . From the deﬁnition of autocorrelation ( f ± f )( x + 1) = ± 1 0 f ( y ) f ( y + x + 1) dy = ± 1 0 f ( y ) f ( y + x ) dy since f ( y + x + 1) = f ( y + x ) by periodicity of f . This shows rhat f ± f is periodic of period 1. For part (b) we plug the Fourier series of f into the deﬁnition of autocorrelation: ( f ± f )( x ) = ± 1 0 f ( y ) f ( y + x ) dy = ± 1 0 ³ ² n = -∞ ˆ f ( n ) e 2 πiny ´ ³ ² m = -∞ ˆ f ( m ) e 2 πim ( y + x ) ´ dy = ± 1 0 ³ ² n = -∞ ˆ f ( n ) e 2 πiny ´ ³ ² m = -∞ ˆ f ( m ) e 2 πimy e 2 πimx ´ dy = ± 1 0 ² n,m = -∞ ˆ f ( n ) ˆ f ( m ) e 2 πiny e 2 πimy e 2 πimx dy Now swap the summation and integration ± 1 0 ² n,m = -∞ ˆ f ( n ) ˆ f ( m ) e 2 πiny e 2 πimy e 2 πimx dy = ² n,m = -∞ ˆ f ( n ) ˆ f ( m ) e 2 πimx ± 1 0 e 2 πiny e 2 πimy dy = ² n,m = -∞ ˆ f ( n ) ˆ f ( m ) e 2 πimx ± 1 0 e 2 πi ( n + m ) y dy We’ve seen that integral of exponentials. It will only be nonzero if n + m = 0, i.e., if m = - n , in which case it integrates to 1. Thus what remains is ² n = -∞ ˆ f ( n ) ˆ f ( - n ) e 2 πinx . 2
ˆ f ( - n ) = ˆ f ( n ) . With this the sum becomes

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## This note was uploaded on 12/04/2011 for the course EE 263 at Stanford.

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mt2007sol - EE 261 The Fourier Transform and its...

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