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Unformatted text preview: EECS 216 SOLUTIONS TO PROBLEM SET #8 Winter 2008 1. Sampling rate > 2(maximum frequency in Hertz); interval < 1 RATE = 2 π ω MAX . Total energy= R ∞∞  x ( t )  2 dt = 1 2 π R ∞∞  X ( ω )  2 dω (Parseval’s theorem). 1a. ω MAX = 3 π =1.5 Hertz → rate > 3 Hertz and interval < 1 3 second. 1a. Energy → ∞ since the cosine has infinite energy and it has average power = 1 2 . 1b. ω MAX = π + 12 π = 13 π =6.5 Hertz → rate > 13 Hertz and interval < 1 13 second. 1b. Energy=( 1 2 π 2 ) 2 2(2 π ) = π 3 4 =7.75 since sin( πt ) 2 t = π 2 sin( πt ) πt . Note modulation doubles frequency support but halves amplitude → halves energy. 1c. ω MAX = 4 π =2 Hertz → rate > 4 Hertz and interval < 1 4 second. 1c. Energy= 1 2 π R 4 π 4 π  1 jω +4  2 dω = 1 2 π R 4 π 4 π 1 ω 2 +16 dω = 1 8 π tan 1 ω 4  4 π 4 π = 1 4 π tan 1 π =0.100. 1c. Energy of e 4 t u ( t ) is R ∞ e 8 t dt = 1 8 =0.125. 80% of energy is at frequencies < 2 Hertz....
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This homework help was uploaded on 04/04/2008 for the course EECS 216 taught by Professor Yagle during the Winter '08 term at University of Michigan.
 Winter '08
 Yagle
 Frequency, Hertz

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