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Explain FDM and show how CCITT standards help in building the base band?
Frequency Division Multiplexing: This is the process of combining some information channels through shifting their signals to various frequency groups in the spectrum therefore they can be transmitted simultaneously over common transmission service.
The bandwidth of a telephone speech signal is quite less than 4 kHz. Where, the available bandwidth on unloaded cables pair is fine above 100 kHz. Therefore this is possible, by using modulation techniques, for divide up the cable bandwidth then a number of telephone speech paths can be carried at the same time along a single cable pair. The normal arrangement contains 24 telephone channels per cable pair, the modulation in 24 channels being carried out within the two stages. In the primary stage, 12 channels are multiplexed together to form that is commonly termed as a basic group. This basic group arrangement is demonstrated in figure. Each of the 12 telephone signals are single side-band amplitude modulated on to carriers spaced at 4 kHz intervals 64 kHz to 108 kHz. The lower side-band (LSB) is utilized in every case. The 12 base-band signals are thus translated in the frequency band from 60 kHz 108 kHz as demonstrated. The block diagram for the channel translating equipment is specified in figure. To form a 24-channel system, to basic groups are occupied together. One basic group (B) is transmitted directly as this stands. Another basic group (A) is amplitude modulated on to a carrier are 120 kHz and the lower side-band is occupied such as to take the frequency range from 12 kHz to 60 kHz as demonstrated in figure. by using two stages of modulation for basic group A, this is possible to decrease the physical size of the component needed for the LSB filters of the figure, because the lowest cut- off frequency needed is at 64 kHz quite than 12 kHz when the whole block of 24 channels were assembled together in one stage of modulation.
FIG - Basic Group Arrangement
FIG - Channel Translating Equipment
What is race around condition? Ans: Race Around Condition:- Jn Kn Q(n+1) output 0 1 0 1 0 0
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The excess 3 code of decimal number 26 is ? Ans. (26) 10 in BCD is (00100110) BCD Add 011 to all BCD 01011001 for excess - 3
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