4.1 Binary data is transmitted through an Additive White Gaussian Noise (AWGN) c
ID: 1811831 • Letter: 4
Question
4.1 Binary data is transmitted through an Additive White Gaussian Noise (AWGN) channel with SNR=3.5 dB and bandwidth B. Channel coding is used to ensure reliable communications. Then:
i. What is the maximum bit rate that can be transmitted?
ii. If the bit rate is increased to 3B, how much must the channel SNR be increased to ensure reliable transmission?
4.2 Binary data is transmitted at the rate of Rb bits/sec over a channel occupying a bandwidth B and the channel SNR=3 dB. If the data bit rate is increased to 2.65Rb and the bandwidth is increased to 1.75B:
i. What would be the channel SNR for the new system?
ii. What channel bandwidth is required to keep the same channel signal-to-noise ratio?
Explanation / Answer
4.1.
SNR=3.5 dB (=2.24 in ratio)
i. Channel capacity is given by Shannon equation (3.1):
C = B · log2(1 + 2.24) = B · log2 (3.24)
= B · log10(3.24)/log2(2)
= 1.7B
Note the maximum bit rate for binary transmission that can be achieved with no errors
in an ideal channel (no noise) is 2B. In this example the bit rate is about 1.7B.
ii. C=3B=Blog2(1+SNR) where SNR represents the channel’s new signal-to-noise
ratio.
Thus (1+SNR)=23 =8, therefore, SNR=7=8.45 dB
The increase in the channel SNR=8.45−3.5=4.95 dB.
Note in this case, the bit rate is greater than 2B and the transmission of the data over
the channel is multi-level but the symbol rate is still 2B.
4.2
i. Substitute the SNR in equation (3.1):
SNR = 3 dB = 2 in ratio
So that for the first case: Rb =Blog2(1+2) and for the second case:
2.65 · Rb = 1.75Blog2(1 + SNR) where SNR is the channel SNR for the new system.
1 / 2.65 = B · log2(3) / 1.75B · log2(1 + SNR)
= log10(3) / 1.75 log10(1 + SNR)
Therefore, 1.75 log10(1+SNR)=2.65 log10(3). This gives
SNR = 4.28(=6.3 dB).
ii. If the channel signal-to-noise ratio is kept at 3 dB, the expanded bandwidth (B) is
computed from 1 / 2.65 = B / B' .
Thus B =2.65B compared with 1.75B in the first case.
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