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In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6?
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In a communication system operating at wavelength 800nm, only one perc...
Given:
- Wavelength of the communication system = 800 nm
- Signal bandwidth available = 1% of source frequency
- Bandwidth of TV signals = 6 MHz
To Find:
The number of channels accommodated for transmitting TV signals of bandwidth 6 MHz.
Solution:
Step 1: Calculate the source frequency
The source frequency can be calculated using the formula:
Speed of light (c) = Wavelength (λ) x Frequency (f)
Given wavelength (λ) = 800 nm = 800 x 10^(-9) m
Speed of light (c) = 3 x 10^8 m/s (approx.)
Plugging in the values, we get:
3 x 10^8 = (800 x 10^(-9)) x f
Simplifying the equation, we find:
f = (3 x 10^8) / (800 x 10^(-9))
f = 3.75 x 10^14 Hz
So, the source frequency is 3.75 x 10^14 Hz.
Step 2: Calculate the signal bandwidth
The signal bandwidth available is given as 1% of the source frequency.
Signal bandwidth = 1% of (3.75 x 10^14 Hz)
Signal bandwidth = (1/100) x (3.75 x 10^14 Hz)
Signal bandwidth = 3.75 x 10^12 Hz
Step 3: Calculate the number of channels
The number of channels that can be accommodated can be calculated using the formula:
Number of channels = Signal bandwidth / Bandwidth per channel
Given bandwidth per channel = 6 MHz = 6 x 10^6 Hz
Plugging in the values, we get:
Number of channels = (3.75 x 10^12 Hz) / (6 x 10^6 Hz)
Simplifying the equation, we find:
Number of channels = 6.25 x 10^5
Therefore, the number of channels accommodated for transmitting TV signals of bandwidth 6 MHz is 6.25 x 10^5. Hence, the correct option is C).
- Wavelength of the communication system = 800 nm
- Signal bandwidth available = 1% of source frequency
- Bandwidth of TV signals = 6 MHz
To Find:
The number of channels accommodated for transmitting TV signals of bandwidth 6 MHz.
Solution:
Step 1: Calculate the source frequency
The source frequency can be calculated using the formula:
Speed of light (c) = Wavelength (λ) x Frequency (f)
Given wavelength (λ) = 800 nm = 800 x 10^(-9) m
Speed of light (c) = 3 x 10^8 m/s (approx.)
Plugging in the values, we get:
3 x 10^8 = (800 x 10^(-9)) x f
Simplifying the equation, we find:
f = (3 x 10^8) / (800 x 10^(-9))
f = 3.75 x 10^14 Hz
So, the source frequency is 3.75 x 10^14 Hz.
Step 2: Calculate the signal bandwidth
The signal bandwidth available is given as 1% of the source frequency.
Signal bandwidth = 1% of (3.75 x 10^14 Hz)
Signal bandwidth = (1/100) x (3.75 x 10^14 Hz)
Signal bandwidth = 3.75 x 10^12 Hz
Step 3: Calculate the number of channels
The number of channels that can be accommodated can be calculated using the formula:
Number of channels = Signal bandwidth / Bandwidth per channel
Given bandwidth per channel = 6 MHz = 6 x 10^6 Hz
Plugging in the values, we get:
Number of channels = (3.75 x 10^12 Hz) / (6 x 10^6 Hz)
Simplifying the equation, we find:
Number of channels = 6.25 x 10^5
Therefore, the number of channels accommodated for transmitting TV signals of bandwidth 6 MHz is 6.25 x 10^5. Hence, the correct option is C).
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In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6?
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In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6?.
In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth.The number of channels accommodated for transmitting TV signals of bandwidth 6MHz are A) 3.75*10^6 B) 4.87 * 10^5 C) 6.25 * 10^5 D) 3.86* 10^6?.
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