A binary encoded signal has a bit-rate of Rb = 28 kbps. An Alternate .pdf
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The document discusses the transmission bandwidth required for a binary encoded signal with a bit-rate of 28 kbps using alternate mark inversion line-code. It explains that the minimum bandwidth is half the bit-rate (14 kbps), while the maximum bandwidth is typically considered as the bit-rate itself (28 kbps). Therefore, the required transmission bandwidth would be 14 kbps.
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A binary encoded signal has a bit-rate of Rb = 28 kbps. An Alternate .pdf
- 1. A binary encoded signal has a bit-rate of Rb = 28 kbps. An Alternate Mark Inversion line-code is used to encode these bits for baseband transmission over copper wire. What would be the transmission bandwidth required to transmit this signal? B = 112 kHz B = infinity B = 28 kHz B = 56 kHz B = 14 kHz B = 7 kHz Solution Speed or bit rate = nfs = Rb Bandwidth minimum is Rb/2 Bandwidth maximum is Rb Normally bandwidth consider to be half of speed Bandwidth = 28/2 Bandwidth = 14 kbps