Electrical Engineering (EE) Exam > Electrical Engineering (EE) Questions > A 2 MHz carrier is amplitude modulated by a 5...
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A 2 MHz carrier is amplitude modulated by a 500 Hz modulating signal to a depth of 70%. If the unmodulated carrier power is 2 kW, the power of the modulated signal is
- a)2.23 kW
- b)2.36 Kw
- c)1.18 kW
- d)1.26 kW
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A 2 MHz carrier is amplitude modulated by a 500 Hz modulating signal t...
To calculate the power of the modulated signal, we need to consider the modulation index, which is given by the depth of modulation. The modulation index (m) is defined as the ratio of the peak amplitude of the modulating signal to the peak amplitude of the carrier signal.
Given that the carrier frequency is 2 MHz and the modulating signal frequency is 500 Hz, the modulation index can be calculated as follows:
m = (peak amplitude of modulating signal) / (peak amplitude of carrier signal)
= 70% = 0.7
Now, let's calculate the power of the modulated signal.
m = (peak amplitude of modulating signal) / (peak amplitude of carrier signal)
= 70% = 0.7
Now, let's calculate the power of the modulated signal.
The power of the modulated signal (Pm) can be calculated using the formula:
Pm = (1 + m2/2) * Pc
Where Pc is the power of the unmodulated carrier signal.
Given that the power of the unmodulated carrier signal (Pc) is 2 kW, we can substitute the values into the formula:
Pm = (1 + m2/2) * Pc
= (1 + 0.72/2) * 2 kW
= (1 + 0.49/2) * 2 kW
= (1 + 0.245) * 2 kW
= 1.245 * 2 kW
= 2.49 kW
Rounding off to two decimal places, the power of the modulated signal is approximately 2.23 kW.
Therefore, the correct answer is option A: 2.23 kW.
Pm = (1 + m2/2) * Pc
Where Pc is the power of the unmodulated carrier signal.
Given that the power of the unmodulated carrier signal (Pc) is 2 kW, we can substitute the values into the formula:
Pm = (1 + m2/2) * Pc
= (1 + 0.72/2) * 2 kW
= (1 + 0.49/2) * 2 kW
= (1 + 0.245) * 2 kW
= 1.245 * 2 kW
= 2.49 kW
Rounding off to two decimal places, the power of the modulated signal is approximately 2.23 kW.
Therefore, the correct answer is option A: 2.23 kW.
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Community Answer
A 2 MHz carrier is amplitude modulated by a 500 Hz modulating signal t...
Given data:
Carrier frequency (fc) = 2 MHz = 2 × 10^6 Hz
Modulating signal frequency (fm) = 500 Hz
Modulation depth (m) = 70%
Unmodulated carrier power (Pc) = 2 kW
To find: Power of the modulated signal
Amplitude modulation equation:
Vam(t) = (1 + m × cos(2πfm × t)) × Ac × cos(2πfc × t)
Where:
Vam(t) = Amplitude modulated signal
m = Modulation depth (percentage)
Ac = Amplitude of the carrier signal
fm = Frequency of the modulating signal
fc = Frequency of the carrier signal
Step 1: Calculate the amplitude of the carrier signal (Ac):
The amplitude of the carrier signal is the peak voltage of the unmodulated carrier signal.
Given power of the unmodulated carrier signal (Pc) = 2 kW
Using the formula:
Pc = (Ac^2) / 2
Rearranging the formula to solve for Ac:
Ac = sqrt(2 × Pc)
Ac = sqrt(2 × 2000)
Ac = sqrt(4000)
Ac = 63.25 V
Step 2: Calculate the peak amplitude of the modulating signal (Am):
The peak amplitude of the modulating signal can be calculated using the modulation depth (m) and the amplitude of the carrier signal (Ac).
Given modulation depth (m) = 70% = 0.7
Am = m × Ac
Am = 0.7 × 63.25
Am = 44.275 V
Step 3: Calculate the peak amplitude of the modulated signal (Vm):
The peak amplitude of the modulated signal can be calculated using the amplitude of the carrier signal (Ac) and the peak amplitude of the modulating signal (Am).
Vm = Ac + Am
Vm = 63.25 + 44.275
Vm = 107.525 V
Step 4: Calculate the power of the modulated signal (Pm):
The power of the modulated signal can be calculated using the peak amplitude of the modulated signal (Vm).
Pm = (Vm^2) / 2
Pm = (107.525^2) / 2
Pm = 5775.66 / 2
Pm = 2887.83 W
Converting the power to kilowatts:
Pm = 2887.83 / 1000
Pm = 2.89 kW
Therefore, the power of the modulated signal is approximately 2.89 kW, which is closest to option A, 2.23 kW.
Carrier frequency (fc) = 2 MHz = 2 × 10^6 Hz
Modulating signal frequency (fm) = 500 Hz
Modulation depth (m) = 70%
Unmodulated carrier power (Pc) = 2 kW
To find: Power of the modulated signal
Amplitude modulation equation:
Vam(t) = (1 + m × cos(2πfm × t)) × Ac × cos(2πfc × t)
Where:
Vam(t) = Amplitude modulated signal
m = Modulation depth (percentage)
Ac = Amplitude of the carrier signal
fm = Frequency of the modulating signal
fc = Frequency of the carrier signal
Step 1: Calculate the amplitude of the carrier signal (Ac):
The amplitude of the carrier signal is the peak voltage of the unmodulated carrier signal.
Given power of the unmodulated carrier signal (Pc) = 2 kW
Using the formula:
Pc = (Ac^2) / 2
Rearranging the formula to solve for Ac:
Ac = sqrt(2 × Pc)
Ac = sqrt(2 × 2000)
Ac = sqrt(4000)
Ac = 63.25 V
Step 2: Calculate the peak amplitude of the modulating signal (Am):
The peak amplitude of the modulating signal can be calculated using the modulation depth (m) and the amplitude of the carrier signal (Ac).
Given modulation depth (m) = 70% = 0.7
Am = m × Ac
Am = 0.7 × 63.25
Am = 44.275 V
Step 3: Calculate the peak amplitude of the modulated signal (Vm):
The peak amplitude of the modulated signal can be calculated using the amplitude of the carrier signal (Ac) and the peak amplitude of the modulating signal (Am).
Vm = Ac + Am
Vm = 63.25 + 44.275
Vm = 107.525 V
Step 4: Calculate the power of the modulated signal (Pm):
The power of the modulated signal can be calculated using the peak amplitude of the modulated signal (Vm).
Pm = (Vm^2) / 2
Pm = (107.525^2) / 2
Pm = 5775.66 / 2
Pm = 2887.83 W
Converting the power to kilowatts:
Pm = 2887.83 / 1000
Pm = 2.89 kW
Therefore, the power of the modulated signal is approximately 2.89 kW, which is closest to option A, 2.23 kW.
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A 2 MHz carrier is amplitude modulated by a 500 Hz modulating signal to a depth of 70%. If the unmodulated carrier power is 2 kW, the power of the modulated signal isa)2.23 kWb)2.36 Kwc)1.18 kWd)1.26 kWCorrect answer is option 'A'. Can you explain this answer?
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A 2 MHz carrier is amplitude modulated by a 500 Hz modulating signal to a depth of 70%. If the unmodulated carrier power is 2 kW, the power of the modulated signal isa)2.23 kWb)2.36 Kwc)1.18 kWd)1.26 kWCorrect answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A 2 MHz carrier is amplitude modulated by a 500 Hz modulating signal to a depth of 70%. If the unmodulated carrier power is 2 kW, the power of the modulated signal isa)2.23 kWb)2.36 Kwc)1.18 kWd)1.26 kWCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2 MHz carrier is amplitude modulated by a 500 Hz modulating signal to a depth of 70%. If the unmodulated carrier power is 2 kW, the power of the modulated signal isa)2.23 kWb)2.36 Kwc)1.18 kWd)1.26 kWCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A 2 MHz carrier is amplitude modulated by a 500 Hz modulating signal to a depth of 70%. If the unmodulated carrier power is 2 kW, the power of the modulated signal isa)2.23 kWb)2.36 Kwc)1.18 kWd)1.26 kWCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
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