quiz3_08

quiz3_08 - ECEN 487 QUIZ 3 NAME_KEY 1 Use Cauchey’s...

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Unformatted text preview: ECEN 487 QUIZ 3 NAME:___________KEY_________________________ 1. Use Cauchey’s residue theorem to compute the inverse z transform of X ( z) = ( z 2)( z + j 0.15)( z (z + 2 )( z 3 j ) for n > 0 only. The desired region of e )( z + 7)( z 0.1 j + 2) convergence includes a circular contour about the origin with radius r = 0.5. For n > 0 there are no poles of X(z) zn-1 enclosed by C: | z | = 0.5. Therefore x[n] = 0 for n > 0. This region of convergence yields a left sided inverse. 2. Let xc(t) = cos(2 fxt), where fx = 10,000 Hz. This signal is sampled by an analog to digital converter (ADC) with a sample rate of 8 kHz with no anti-alias filtering, and then played back out by a digital to analog converter (DAC) at the same sample rate, but with antialiasing filters which block all frequencies above 4 kHz. What frequecy will be heard at playback? 1 Xc ( j [ T + 2T k ]) Recall cos(2 f x t ) [( 2 f x ) + ( + 2 f x )] . X (e j ) = T k= X (e j ) = ([ T + 2T k ] 2 f x ) + ([ T + 2T k ] + 2 f x ) k= T = 8000 k= ([8000 ([ + (2 )k ] + 8000(2 )k ] 10000(2 )) + 5 4 ([8000 k= + 8000(2 )k ] + 10000(2 )) = k= (2 )) + ([ + (2 )k ] + 5 (2 )) = 4 ( ( 2 k +1) 2 )+ ( + (2 k +1) 2 ) This has impulses at 3 5 7 = ± ,± ,± ,± ,L . Only the = ± terms are in 2 2 2 2 2 < < . The output signal is: the primary spectrum range of Xo ( j ) = T [ (T = [( 2T 2 ) + (T + 2 )] 2000(2 )) + ( + 2000(2 ))] ) + ( + 2T )] = [ ( xo [ n ] = cos(2 (2000)t ). Your hear a 2 kHz tone! ...
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quiz3_08 - ECEN 487 QUIZ 3 NAME_KEY 1 Use Cauchey’s...

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