41. A binary symmetric channel (BSC) has a transition probability of $$\frac{1}{8}.$$ If the binary transmit symbol X is such that $$P\left( {X = 0} \right) = \frac{9}{{10}},$$ then the probability of error for an optimum receiver will be
42. Choose the correct one from among the alternatives a, b, c, dafter matching an item from Group 1 with the most appropriate item in Group 2.
Group 1
Group 2
1. FM
P. Slope overload
2. DM
Q. H-law
3. PSK
R. Envelope detector
4. PCM
S. Capture effect
T. Hilbert transform
U. Matched filter
| Group 1 | Group 2 |
| 1. FM | P. Slope overload |
| 2. DM | Q. H-law |
| 3. PSK | R. Envelope detector |
| 4. PCM | S. Capture effect | T. Hilbert transform | U. Matched filter |
43. A source emits bit 0 with probability $$\frac{1}{3}$$ and bit 1 with probability $$\frac{2}{3}.$$ The emitted bits are communicated to the receiver. The receiver decides for either 0 or 1 based on the received value R. It is given that the conditional density functions of R as
\[\begin{gathered}
{f_{R|0}}\left( r \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{1}{4},}&{ - 3 \leqslant r \leqslant 1} \\
{0,}&{{\text{otherwise;}}}
\end{array}} \right. \hfill \\
{f_{R|1}}\left( r \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{1}{6},}&{ - 1 \leqslant r \leqslant 5} \\
{0,}&{{\text{otherwise;}}}
\end{array}} \right. \hfill \\
\end{gathered} \]
The minimum decision error probability is
\[\begin{gathered} {f_{R|0}}\left( r \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{4},}&{ - 3 \leqslant r \leqslant 1} \\ {0,}&{{\text{otherwise;}}} \end{array}} \right. \hfill \\ {f_{R|1}}\left( r \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{6},}&{ - 1 \leqslant r \leqslant 5} \\ {0,}&{{\text{otherwise;}}} \end{array}} \right. \hfill \\ \end{gathered} \]
The minimum decision error probability is
44. The minimum sampling frequency (in samples/sec) required to reconstruct the following signal from its samples without distortion
$$x\left( t \right) = 5{\left( {\frac{{\sin 2\pi 1000t}}{{\pi t}}} \right)^3} + 7{\left( {\frac{{\sin 2\pi 1000t}}{{\pi t}}} \right)^2}$$
would be
$$x\left( t \right) = 5{\left( {\frac{{\sin 2\pi 1000t}}{{\pi t}}} \right)^3} + 7{\left( {\frac{{\sin 2\pi 1000t}}{{\pi t}}} \right)^2}$$
would be