HW_6(483)S09_Solns

HW_6(483)S09_Solns - EE 483 Introduction to Digital Signal Processing Spring 2009 Homework#6 Solutions 6.1 The frequency response of the LTI

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1 EE 483 – Introduction to Digital Signal Processing Spring 2009 Homework #6 Solutions 6.1 The frequency response of the LTI discrete-time system is given by . e a e a e a e a ) e ( H m m m jm j ϖ + - - ϖ + - ϖ + - ϖ - ϖ + + + = ) 3 j( 1 ) 2 j( 2 ) 1 j( 2 1 There is no value of m that will make ) e ( H j ϖ a real function of ϖ . 6.2 ( 29 ] 1 [ ) ]( 0 [ ] 1 [ ) 1 ]( 0 [ ) ( 2 h e e h e e h e h e H j j j j j j + + + + + + = + + + + + + = - - - - ω ω ω ω ω ω ]). 1 [ cos ] 0 [ 2 ( h h e j + + + = - ω ω Thus, we require 1 1 7 0 0 2 7 0 = + = ] [ h ) . cos( ] [ h ) e ( H . j and . 0 ] 1 [ h ) 4 . 0 cos( ] 0 [ h 2 ) e ( H 4 . 0 j = + = Solving these two equations we get 3.2006 - = ] 0 [ h and . ] 1 [ h 5.896 = 6.3 α + α - = α + α + α + α - = α + α + α + α - = ϖ + ϖ - = ϖ + ϖ = ϖ 1 1 ) 1 )( 1 ( ) 1 )( 1 ( 1 2 1 ) 1 ( 4 1 ) cos( 1 ) ( cos 1 ) cos( 1 ) sin( 2 tan 2 2 2 2 2 2 2 2 2 c c c c c Because ϖ 2 tan 0 c for π ϖ c 0 , and , 2 tan 1 2 tan 1 ϖ + ϖ - = α c c 1 0 α . Therefore the pole is inside the unit circle which, since we assume the filter is causal, implies stability. 6.4 . ) 1 ( 1 2 1 2 1 ) ( 2 1 2 1 α + α + β - + β - α + = - - - - z z z z z H BS Thus, ) cos( ) 1 ( 2 ) 2 cos( 2 ) 1 ( 1 ) 2 cos( 2 ) cos( 8 4 2 2 1 ) ( 2 2 2 2 2 2 2 ϖ α + β - ϖ α + α + α + β + ϖ + ϖ β - β + α
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This note was uploaded on 02/02/2010 for the course EE 483 taught by Professor Mitra during the Fall '08 term at USC.

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HW_6(483)S09_Solns - EE 483 Introduction to Digital Signal Processing Spring 2009 Homework#6 Solutions 6.1 The frequency response of the LTI

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