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ENC3250 Solutions pg 29

# ENC3250 Solutions pg 29 - 7.2 The total number of tones or...

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-29- 7.2 The total number of tones, or individual channels is: W s / f d = (400 MHz)/(100 Hz) = 4 × 10 6 . The minimum number of PN bits = log 2 (4 × 10 6 ) = 22 where x indicates the smallest integer value not less than x. Source: [SKLA01] 7.3 W s = 1000 f d ; W d = 4 f d ; Using Equation 7.3 , G p = W s / W d = 250 = 24 dB 7.4 a. Period of the PN sequence is 2 4 – 1 = 15 b. MFSK c. L = 2 d. M = 2 L = 4 e. k = 3 f. slow FHSS g. 2 k = 8 h. Time 0 1 2 3 4 5 6 7 8 9 10 11 Input data 0 1 1 1 1 1 1 0 0 0 1 0 Frequency f 1 f 3 f 3 f 2 f 0 f 2 Time 12 13 14 15 16 17 18 19 Input data 0 1 1 1 1 0 1 0 Frequency f 1 f 3 f 2 f 2 7.5 a. Period of the PN sequence is 2 4 – 1 = 15 b. MFSK c. L = 2 d. M = 2 L = 4 e. k = 3 f. fast FHSS g. 2 k = 8 h. Same as for Problem 7.4 7.6 a. This is from the example 6.1. f 1 = 75 kHz 000 f 2 =125 kHz 001 f 3 = 175 kHz 010
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