Figure 7 sketch of problem 2 part d problem 3 for

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Unformatted text preview: 11) (12) The same comments as for part c) apply here. Figure 7: Sketch of problem 2 part d. Problem 3 For better explanation, we draw a system diagram. Figure 8: System diagram of problem 3. where x(t) = repP (rect( 2t )) P (13) Note ellipses. These need to be included to obtain full credit. ... ... Figure 9: Plot of x(t). 5 This not really drawn to scale the intervals where the signal has value 0 should be of the same width as those where the signal has value 1. P 1 P 1 comb P ( sinc( f )) P 2 2 P 1 1 = comb P (sinc( f )) 2 2 X (f ) = F{x(t)} = (14) Figure 10: Plot of X (f ). a) P = 1ms, 1 P = 1kHz, fc = 4kHz. Z (f ) = HLP (f )X (f ) Figure 11: 6 (15) n ) (16) fs ω Zd (ω ) = fs repfs [Z ( fs )] (17) 2π 1 ωfs 1 ωfs ωfs 1 ωfs ωfs = fs repfs [ δ ( ) + (δ ( − 1000) + δ ( + 1000)) − (δ ( − 3000) + δ ( + 3000))] 2 2π π 2π 2π 3π 2π 2π z [n] = z (nTs ) = z ( Note δ ( ωfs ) = 2π 2π fs δ (ω ).Therefore Zd (ω ) = rep2π [πδ (ω ) + 2(δ (ω − 2000π 2000π 2 6000π 6000π ) + δ (ω + )) − (δ (ω − ) + δ (ω + ))] fs f...
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This note was uploaded on 01/16/2014 for the course ECE 438 taught by Professor Staff during the Fall '08 term at Purdue University-West Lafayette.

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