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The un-modulated carrier power in an AM transmitter is 5kW. This carrier is modulated by a sinusoidal modulating signal. The maximum percentage of modulation is 50%. If it is reduced to 40%, then the maximum un-modulated carrier power (in kW) that can be used without overloading the transmitter is ___________
Correct answer is '5.19 to 5.23'. Can you explain this answer?
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The un-modulated carrier power in an AM transmitter is 5kW. This carri...
Total power when μ = 50% is
When μ = 40%
Total power remains 5.625
Pc = 5.22
When μ = 40%
Total power remains 5.625
Pc = 5.22
Most Upvoted Answer
The un-modulated carrier power in an AM transmitter is 5kW. This carri...
Solution:
Given Data:
Un-modulated Carrier Power, P_O = 5 kW
Maximum percentage of modulation, m_max = 50%
Maximum un-modulated carrier power, P_OM = ?
The formula for AM Power is given by:
P_AM = P_O(1 + m^2/2)
Where,
P_AM = AM Power
P_O = Un-modulated Carrier Power
m = Modulation Index
Calculation:
Case 1: When maximum percentage of modulation is 50%
Given that the maximum percentage of modulation is 50%.
Therefore, the modulation index can be calculated as:
m_max = 50% = 0.5
The maximum AM power can be calculated as:
P_AM_max = P_O(1 + m_max^2/2)
P_AM_max = 5 kW(1 + 0.5^2/2)
P_AM_max = 7.5 kW
Case 2: When maximum percentage of modulation is 40%
Given that the maximum percentage of modulation is 40%.
Therefore, the modulation index can be calculated as:
m = 40% = 0.4
The maximum AM power can be calculated as:
P_AM = P_O(1 + m^2/2)
P_AM = P_OM(1 + m^2/2)
5 kW(1 + 0.4^2/2) = P_OM(1 + 0.4^2/2)
P_OM = 5 kW(1 + 0.4^2/2)/(1 + 0.5^2/2)
P_OM = 5.19 to 5.23 kW (Approx)
Therefore, the maximum un-modulated carrier power that can be used without overloading the transmitter when maximum percentage of modulation is reduced to 40% is 5.19 to 5.23 kW.
Conclusion:
The maximum un-modulated carrier power that can be used without overloading the transmitter when the maximum percentage of modulation is reduced to 40% is 5.19 to 5.23 kW.
Given Data:
Un-modulated Carrier Power, P_O = 5 kW
Maximum percentage of modulation, m_max = 50%
Maximum un-modulated carrier power, P_OM = ?
The formula for AM Power is given by:
P_AM = P_O(1 + m^2/2)
Where,
P_AM = AM Power
P_O = Un-modulated Carrier Power
m = Modulation Index
Calculation:
Case 1: When maximum percentage of modulation is 50%
Given that the maximum percentage of modulation is 50%.
Therefore, the modulation index can be calculated as:
m_max = 50% = 0.5
The maximum AM power can be calculated as:
P_AM_max = P_O(1 + m_max^2/2)
P_AM_max = 5 kW(1 + 0.5^2/2)
P_AM_max = 7.5 kW
Case 2: When maximum percentage of modulation is 40%
Given that the maximum percentage of modulation is 40%.
Therefore, the modulation index can be calculated as:
m = 40% = 0.4
The maximum AM power can be calculated as:
P_AM = P_O(1 + m^2/2)
P_AM = P_OM(1 + m^2/2)
5 kW(1 + 0.4^2/2) = P_OM(1 + 0.4^2/2)
P_OM = 5 kW(1 + 0.4^2/2)/(1 + 0.5^2/2)
P_OM = 5.19 to 5.23 kW (Approx)
Therefore, the maximum un-modulated carrier power that can be used without overloading the transmitter when maximum percentage of modulation is reduced to 40% is 5.19 to 5.23 kW.
Conclusion:
The maximum un-modulated carrier power that can be used without overloading the transmitter when the maximum percentage of modulation is reduced to 40% is 5.19 to 5.23 kW.
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The un-modulated carrier power in an AM transmitter is 5kW. This carrier is modulated by a sinusoidal modulating signal. The maximum percentage of modulation is 50%. If it is reduced to 40%, then the maximum un-modulated carrier power (in kW) that can be used without overloading the transmitter is ___________Correct answer is '5.19 to 5.23'. Can you explain this answer?
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The un-modulated carrier power in an AM transmitter is 5kW. This carrier is modulated by a sinusoidal modulating signal. The maximum percentage of modulation is 50%. If it is reduced to 40%, then the maximum un-modulated carrier power (in kW) that can be used without overloading the transmitter is ___________Correct answer is '5.19 to 5.23'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about The un-modulated carrier power in an AM transmitter is 5kW. This carrier is modulated by a sinusoidal modulating signal. The maximum percentage of modulation is 50%. If it is reduced to 40%, then the maximum un-modulated carrier power (in kW) that can be used without overloading the transmitter is ___________Correct answer is '5.19 to 5.23'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The un-modulated carrier power in an AM transmitter is 5kW. This carrier is modulated by a sinusoidal modulating signal. The maximum percentage of modulation is 50%. If it is reduced to 40%, then the maximum un-modulated carrier power (in kW) that can be used without overloading the transmitter is ___________Correct answer is '5.19 to 5.23'. Can you explain this answer?.
The un-modulated carrier power in an AM transmitter is 5kW. This carrier is modulated by a sinusoidal modulating signal. The maximum percentage of modulation is 50%. If it is reduced to 40%, then the maximum un-modulated carrier power (in kW) that can be used without overloading the transmitter is ___________Correct answer is '5.19 to 5.23'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about The un-modulated carrier power in an AM transmitter is 5kW. This carrier is modulated by a sinusoidal modulating signal. The maximum percentage of modulation is 50%. If it is reduced to 40%, then the maximum un-modulated carrier power (in kW) that can be used without overloading the transmitter is ___________Correct answer is '5.19 to 5.23'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The un-modulated carrier power in an AM transmitter is 5kW. This carrier is modulated by a sinusoidal modulating signal. The maximum percentage of modulation is 50%. If it is reduced to 40%, then the maximum un-modulated carrier power (in kW) that can be used without overloading the transmitter is ___________Correct answer is '5.19 to 5.23'. Can you explain this answer?.
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