hw3 - and decade, There are seven logical channels, all...

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Zhou, HuanXian (ID: 0225418) HW3 EE136 Instructor: Kang Xi 8.9 Twenty-four voice signals are to be multiplexed and transmitted over twisted pair. What is the bandwidth required for FDM? Assuming a bandwidth efficiency of 1 bps/Hz, what is the bandwidth required for TDM using PCM? A single voice signal has a maximum frequency of 4 kHz, therefore, the required bandwidth for FDM is 24 x 4 = 96 kHz. With PCM, each voice signal requires a data rate of 64 kbps, for a total data rate of 24 x 64 = 1.536 Mbps. At 1 bps/Hz, this requires a bandwidth of 1.536 MHz. 8.13 Ten 9600-bps lines are to be multiplexed using TDM. Ignoring overhead bits, what is the total capacity required for synchronous TDM? Assuming that we wish to limit average line utilization of 0.8, and assuming that each line is busy 50% of the time, what is the capacity required for statistical TDM? Synchronous TDM: 9600 bps x 10 = 96 kbps Statistical TDM: 9600 bps x 10 x 0.5/0.8 = 60 kbps 9.7 Figure 9.12, base on one in [BELL00], depicts a simplified scheme for CDMA encoding
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Unformatted text preview: and decade, There are seven logical channels, all using Pass with a spreading code of 7 bit Sequence, the signals from all source combine at the receiver so that two positive or two negative values reinforce a positive and negative value cancel. To decode a given channel, the receiver multiplies the incoming composite signal by the spreading code for that channel, sums the result, and assigns binary 1 for a positive value and binary 0 for a negative value. a. What are the spreading codes for the seven channels? b. Determine the receiver output measurement for channel 1 and the bit value assigned. C. Repeat part (b) for channel 2. a. C0 = 1110010; C1 = 0111001; C2 = 1011100; C3 = 0101110; C4 = 0010111; C5 = 1001011; C6 = 1100101 b. C1 output = (-5)+1+1+(-3)+(-1)+3+(-3) = -7; bit value = 0 c. C2 output = 5+(-1)+1+(-3)+1+3+3 = +9; bit value = 1...
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