114_1_sol_midterm_2010_Part2

# 114_1_sol_midterm_2010_Part2 - EE114 Introduction to Speech...

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Unformatted text preview: EE114 Introduction to Speech and Image Processing Winter Quarter, 2010 ( 29 ( 29 2 2 2 2 2 2 2 2 2 1 2 1 1 j j H a e e a j j a π π π β β β β β β β- = + + + = +- + = + ( 29 ( 29 ( 29 ( 29 2 2 2 2 2 2 2 1 1 2 1 j j H a e e a a β β β β β β- = + + + = + + = + We now equate the ratio of the square of the magnitudes of the magnitude responses to 5 9 , and solve for the possible values of β . ( 29 ( 29 ( 29 ( 29 ( 29 ( 29( 29 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 5 9 1 1 5 9 1 1 5 1 2 9 9+9 5 10 5 4 10 5 =0 1 , 2 H a a H a a π β β β β β β β β β β β β β β β β β β 2 + = = + + = + + = + + = + +- + 4 = 0 2- + 2 = 0 2-1- 2 = = 2 2 Therefore, the two-tap filter can take the form of ( 29 ( 29 [ ] 1 1, , or 1, 2 2 h n a h n a = = where a is any real nonzero constant. The frequency response of 1 1, 2 is shown below. EE114 Introduction to Speech and Image Processing...
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114_1_sol_midterm_2010_Part2 - EE114 Introduction to Speech...

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