10Fall_hw09soln - ECE210 - Fall 2010 - Homework 09...

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Unformatted text preview: ECE210 - Fall 2010 - Homework 09 Solutions 1. Problem 6.13 from text Solution: (a) ∂P e ∂ ˆ a = ∂ ∂ ˆ a 1 T Z T e ( t ) 2 dt = 2 T Z T e ( t ) ∂ ∂ ˆ a ( e ( t )) dt = 2 T Z T e ( t ) ∂ ∂ ˆ a ( f N ( t )- f ( t )) dt = 2 T Z T e ( t )( 1 2 ) dt = 1 T Z T e ( t ) dt 1 T Z T e ( t ) dt = 1 T Z T ( f N ( t )- f ( t )) dt = 1 T Z T ˆ a 2 + N X n =1 ˆ a n cos( nω t ) + ˆ b n sin( nω t ) ! dt- 1 T Z T f ( t ) dt (1) = 1 T Z T ˆ a 2 dt- 1 T Z T f ( t ) dt = ˆ a 2- 1 T Z T f ( t ) dt (Where (1) is true because cos( nω t ) and sin( nω t ) integrate to zero over integer multiples of the period T corresponding to ω .) For 1 ≤ m ≤ N , for ˆ a m , the derivation is similar ∂P e ∂ ˆ a m = ∂ ∂ ˆ a m 1 T Z T e ( t ) 2 dt = 2 T Z T e ( t ) ∂ ∂ ˆ a m ( e ( t )) dt = 2 T Z T e ( t ) ∂ ∂ ˆ a m ( f N ( t )- f ( t )) dt = 2 T Z T e ( t ) cos( mω t ) dt 2 T Z T e ( t ) cos( mω t ) dt = 2 T Z T ( f N ( t )- f ( t )) cos( mω t ) dt = 2 T Z T ˆ a 2 + N X n =1 ˆ a n cos( nω t ) + ˆ b n sin( nω t ) ! cos( mω t ) dt- 2 T Z T f ( t ) cos( mω t ) dt (2) = 2 T Z T ˆ a m cos 2 ( mω t ) dt- 2 T Z T f ( t ) cos( mω t ) dt = 2 T Z T ˆ a m 1 2 (1 + cos(2 mω t ))...
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This note was uploaded on 03/05/2011 for the course ECE 210 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.

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10Fall_hw09soln - ECE210 - Fall 2010 - Homework 09...

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