This preview shows page 1. Sign up to view the full content.
Unformatted text preview: -24-6.5For ASK, BT= (1 + r)R = (1.5)2400 = 3600 HzFor FSK, BT= 2 F + (1 + r)R = 2(2.5 x 103) + (1.5)2400 = 8600 Hz6.6For multilevel signaling BT= [(1 + r)/log2L]RFor 2400 bps QPSK, log2L = log24 = 2BT= (2/2)2400 = 2400 Hz, which just fits the available bandwidthFor 8-level 4800 bps signaling, log2L = log28 = 3BT = (2/3)(4800) = 3200 Hz, which exceeds the available bandwidth6.7As was mentioned in the text, analog signals in the voice band that represent digitaldata have more high frequency components than analog voice signals. Thesehigher components cause the signal to change more rapidly over time. Hence, DMwill suffer from a high level of slope overload noise. PCM, on the other hand, doesnot estimate changes in signals, but rather the absolute value of the signal, and isless affected than DM.6.8No. The demodulator portion of a modem expects to receive a very specific type ofwaveform (e.g., ASK) and would not produce meaningful output with voice input....
View Full Document
- Fall '11