ENC3250 Solutions pg 29 - 7.2 The total number of tones or individual channels is Ws/fd =(400 MHz(100 Hz = 4 × 106 The minimum number of PN bits = log

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Unformatted text preview: 7.2 The total number of tones, or individual channels is: Ws /fd = (400 MHz)/(100 Hz) = 4 × 106 . 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 Ws = 1000 fd ; Wd = 4 fd ; Using Equation 7.3 , G p = Ws /Wd = 250 = 24 dB 7.4 a. Period of the PN sequence is 24 – 1 = 15 b. MFSK c. L = 2 d. M = 2L = 4 e. k = 3 f. slow FHSS g. 2k = 8 h. Time 0 1 2 3 4 5 Input data 0 1 1 1 1 1 Frequency f1 f3 f3 Time Input data Frequency 12 0 13 1 14 1 f1 15 1 f3 6 1 7 0 8 0 f2 16 1 10 1 f0 17 0 f2 9 0 18 1 11 0 f2 19 0 f2 7.5 a. Period of the PN sequence is 24 – 1 = 15 b. MFSK c. L = 2 d. M = 2L = 4 e. k = 3 f. fast FHSS g. 2k = 8 h. Same as for Problem 7.4 7.6 a. This is from the example 6.1. f1 = 75 kHz 000 f2 =125 kHz 001 f3 = 175 kHz 010 f4 = 225 kHz 011 f5 = 275 kHz 100 f6 = 325 kHz 101 f7 = 375 kHz 110 f8 = 425 kHz 111 b. We need three more sets of 8 frequencies. The second set can start at 475 kHz, with 8 frequencies separated by 50 kHz each. The third set can start at 875 kHz, and the fourth set at 1275 kHz. 7.7 a. C0 = 1110010; C1 = 0111001; C2 = 1011100; C3 = 0101110; C4 = 0010111; C5 = 1001011; C6 = 1100101 b. C1 output = –7; bit value = 0 c. C2 output = +9; bit value = 1 7.8 It is –1 for each of the other 6 channels. -29- ...
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