(Solved): c) The Shannon capacity formula for the maximum bitrate that can be supported ...

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c) The Shannon capacity formula for the maximum bitrate that can be supported by a noisy channel is: $$C=B?{\log }_2(1+S/N)=B?{\log }_2\left(\frac{S+N}{N}\right)$$ i) Find this Shannon capacity if the analogue bandwidth is $$10\mathrm{kHz}$$ and the noise is such that $$95\%$$ of the Gaussian distributed noise values lie within $$+/?1$$ millivolt. Assume the signal power is equivalent to that in a sinusoid with an amplitude of $$1\mathrm{mW}$$. ii) Originally the signal data was such that $$\mathrm{p}(1)=0.5$$ and $$\mathrm{p}(0)=0.5$$. The signal is NRZ ASK. Now the data has changed such that $$\mathrm{p}(1)=0.9$$ and $$\mathrm{p}(0)=0.1$$ Recalculate the Shannon capacity. iii) If we use bipolar ASK instead of unipolar ASK, what would be the effect of changing the ratio of 1's and 0's on the signal power and the Shannon Capacity? Explain your answer. iv) Why are we, in practice, not likely to achieve the Shannon Capacity as given by the formula above?