soln0 - EECS 451 SOLUTIONS TO PROBLEM SET #10 1 0 for 0 | |...

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EECS 451 SOLUTIONS TO PROBLEM SET #10 1. x ( t ) A / D 3 H ( e j ω ) 2 D / A y ( t ) H ( e )= ± 1f o r 0 ≤| ω | < π 3 0f o r π 3 < | ω |≤ π . Upsampling and downsampling in this order avoids aliasing. 2. PART AFTER SAMPLING AFTER FIRST AFTER SECOND FINAL ( a ) ( b ) 1200 1800 200 800 ( c ) 1600 2400 400 600 3a. Passband: 0 < f < 500 Hz. Transition band: 500 < f < 1000=1500–500 Hz. 3b. Transition band: 1 octave wide. Need stopband gain < 0.001=3(–20)=–60 dB. Need 10 capacitors or inductors to get (10)(–6 dB/octave)=–60 dB in an octave. 3c. Passband: 0 < f < 500 Hz. Transition band: 500 < f < 16000=11(1500)–500 Hz. Transition band: Now 16000 500 =32=5 octaves wide. Much easier to analog ±lter this! Now need only 2 capacitors or inductors to get (2)(–6 dB/octave)(5 octave)=–60 dB. 4a. H a ( s ) H a ( s )po les : o e jπk/N ,N– N - 1 2 k N+ N - 1 2 } = { o e j 2 πk/ (2 N ) ,0 k 2N–1 } . So H a ( s )
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