A signal bandlimited to 5 kHz is sampled at a rate higher than Nyquist rate such that the guard band is 3 kHz. The graph shows the probability of resulting samples. The data rate will be _ (in kbps).
24.74
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Solution
The correct option is A 24.74 Guard Band=fs−2fm=3Hz ⇒fz=(2×5+3)kHz fz=13kHz ∴Symbol rate,r=fs=13ksymbols/sec Entropy,H=−∑4m=1P(xi)log2P(x1) =−[0.4log20.4+0.25log20.25+0.2log20.2+0.15log20.15] =1.903bits/symbol Date rate=R.H=13×1.903kbps=24.74kbps