SOLUTIONS TEST 02 : 2015
Q1.
The information transfer rate for a data channel is defined as the speed at which
binary information (bits) can be transferred from source to destination.
Units of Information Transfer Rate --> bits/second
The correct definition of symbol rate (sometimes called baud rate) is the rate at which
the signal state changes when observed in the communications channel and is not
necessarily equal to the information transfer rate.
Units of Symbol rate --> symbols/second (baud)
The Shannon–Hartley capacity limit for error-free communication is given by:
Channel capacity C = B.log2(S / N + 1) bits/second
For example, if a system uses four frequencies to convey pairs of bits through a
channel, and the frequency (symbol) is changed every 0.5 ms, then:
Symbol rate = 1 / 0.5 = 2000 symbols/second (2000 baud)
Q2. Power and bandwidth efficiency
For a system transmitting at maximum capacity, C, the average signal power, S,
measured at the receiver input, can be written as S = Eb.C, where Eb is the
average received energy per bit.
The average noise power, N, can also be redefined as, N = N0.B, where N0 is
the noise power density (Watts/Hz)
Using these definitions, the Shannon–Hartley theorem can be written in the
form:
C/B = log2[1 + Eb.C/N0.B]
The ratio C/B represents the bandwidth efficiency of the system in bits/second/Hz.
The larger the ratio, the greater the bandwidth efficiency. The ratio Eb/N0 is a measure
of the power efficiency of the system. The smaller the ratio, the less energy used by
each bit (and consequently for each symbol), to be detected successfully in the
presence of a given amount of noise. Choosing a power-efficient modem type is
particularly important in cellular handsets, for example, where the designer is trying to
maximi
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Q3. The formula for channel capacity in a baseband channel (i.e. the useable bandwidth
extends from 0 Hz to B Hz is given by:
Channel Capacity C = 2Blog2M
Thus for 28.8 kbps in 3.1 kHz, the number of symbol states required M is:
M = antilog228800 / (2 x 3100) = 22.8 or 23 symbol states
Sixty-four signalling states can support log264 = 6 bits per symbol. In a baseband
channel the maximum signalling rate is equal to twice the available bandwidth. For a
2048 Mbps link, the symbol rate is thus 2048000 / 6 = 341.3 kbaud and the bandwidth
required is 341000 / 2 = 170.7 KHz.
Q4. The Shannon–Hartley theorem can be written as: C/B = log2[1 + Eb.C/N0.B]
Now for Eb/N0 = –0.6 dB, the ratio Eb/N0 = 100.6/10 = 0.871.
The bandwidth efficiency C/B that can be supported is thus:
C/B = log2[1 + 0.871C/B] therefore: C/B = 0.6 bits/second/Hz (approximately).
In a bandwidth of 3400 Hz, the system can thus operate at an information transfer rate
of 3400 x 0.6 = 2040 bps.
Q5.
Q6.