JEE PYQ: Communication System Question 27
Question 27 - 2019 (09 Jan 2019 Shift 2)
In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwidth. The number of channels accommodated for transmitting TV signals of band width 6 MHz are:
(1) $3.75 \times 10^6$
(2) $3.86 \times 10^6$
(3) $6.25 \times 10^5$
(4) $4.87 \times 10^5$
Show Answer
Answer: (3)
Solution
$f = \frac{3 \times 10^8}{8 \times 10^{-7}} = 3.75 \times 10^{14}$ Hz. 1% $= 3.75 \times 10^{12}$ Hz $= 3.75 \times 10^6$ MHz. Channels $= \frac{3.75 \times 10^6}{6} = 6.25 \times 10^5$.
BETA
