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An antenna pointing in a certain direction has a noise temperature of 50K. The ambient
temperature is 290K. The antenna is connected to pre-amplifier that has a noise figure of 2dB and
an available gain of 40 dB over an effective bandwidth of 12 MHz. The effective input noise
temperature Te for the amplifier and the noise power Pao at the output of the preamplifier,
respectively, are
temperature is 290K. The antenna is connected to pre-amplifier that has a noise figure of 2dB and
an available gain of 40 dB over an effective bandwidth of 12 MHz. The effective input noise
temperature Te for the amplifier and the noise power Pao at the output of the preamplifier,
respectively, are
- a)Te = 169.36K and Pao = 3.73x10–10 W
- b)Te = 170.8K and Pao = 4.56x10–10 W
- c)Te = 182.5K and Pao = 3.85x10–10 W
- d)Te = 160.62K and Pao = 4.6x10–10 W
Correct answer is option 'A'. Can you explain this answer?
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-21 W
b)Te = 169.36K and Pao = 3.73x10^-21 W
To find the effective input noise temperature Te for the amplifier, we can use the formula:
Te = (T0 + Teamp - 1) / Gamp
where T0 is the ambient temperature (290K), Teamp is the noise temperature of the amplifier (given as 50K), and Gamp is the available gain of the amplifier (40 dB).
Substituting the values into the formula:
Te = (290 + 50 - 1) / (10^(40/10)) = 169.36K
To find the noise power Pao at the output of the preamplifier, we can use the formula:
Pao = k * Te * B
where k is the Boltzmann constant (1.38 x 10^-23 J/K), Te is the effective input noise temperature, and B is the effective bandwidth (12 MHz).
Substituting the values into the formula:
Pao = (1.38 x 10^-23) * (169.36) * (12 x 10^6) = 3.73 x 10^-21 W
Therefore, the correct answer is Te = 169.36K and Pao = 3.73 x 10^-21 W.
b)Te = 169.36K and Pao = 3.73x10^-21 W
To find the effective input noise temperature Te for the amplifier, we can use the formula:
Te = (T0 + Teamp - 1) / Gamp
where T0 is the ambient temperature (290K), Teamp is the noise temperature of the amplifier (given as 50K), and Gamp is the available gain of the amplifier (40 dB).
Substituting the values into the formula:
Te = (290 + 50 - 1) / (10^(40/10)) = 169.36K
To find the noise power Pao at the output of the preamplifier, we can use the formula:
Pao = k * Te * B
where k is the Boltzmann constant (1.38 x 10^-23 J/K), Te is the effective input noise temperature, and B is the effective bandwidth (12 MHz).
Substituting the values into the formula:
Pao = (1.38 x 10^-23) * (169.36) * (12 x 10^6) = 3.73 x 10^-21 W
Therefore, the correct answer is Te = 169.36K and Pao = 3.73 x 10^-21 W.
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An antenna pointing in a certain direction has a noise temperature of 50K. The ambienttemperature is 290K. The antenna is connected to pre-amplifier that has a noise figure of 2dB andan available gain of 40 dB over an effective bandwidth of 12 MHz. The effective input noisetemperature Te for the amplifier and the noise power Pao at the output of the preamplifier,respectively, area)Te = 169.36K and Pao = 3.73x10–10 Wb)Te = 170.8K and Pao = 4.56x10–10 Wc)Te = 182.5K and Pao = 3.85x10–10 Wd)Te = 160.62K and Pao = 4.6x10–10 WCorrect answer is option 'A'. Can you explain this answer?
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An antenna pointing in a certain direction has a noise temperature of 50K. The ambienttemperature is 290K. The antenna is connected to pre-amplifier that has a noise figure of 2dB andan available gain of 40 dB over an effective bandwidth of 12 MHz. The effective input noisetemperature Te for the amplifier and the noise power Pao at the output of the preamplifier,respectively, area)Te = 169.36K and Pao = 3.73x10–10 Wb)Te = 170.8K and Pao = 4.56x10–10 Wc)Te = 182.5K and Pao = 3.85x10–10 Wd)Te = 160.62K and Pao = 4.6x10–10 WCorrect answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An antenna pointing in a certain direction has a noise temperature of 50K. The ambienttemperature is 290K. The antenna is connected to pre-amplifier that has a noise figure of 2dB andan available gain of 40 dB over an effective bandwidth of 12 MHz. The effective input noisetemperature Te for the amplifier and the noise power Pao at the output of the preamplifier,respectively, area)Te = 169.36K and Pao = 3.73x10–10 Wb)Te = 170.8K and Pao = 4.56x10–10 Wc)Te = 182.5K and Pao = 3.85x10–10 Wd)Te = 160.62K and Pao = 4.6x10–10 WCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An antenna pointing in a certain direction has a noise temperature of 50K. The ambienttemperature is 290K. The antenna is connected to pre-amplifier that has a noise figure of 2dB andan available gain of 40 dB over an effective bandwidth of 12 MHz. The effective input noisetemperature Te for the amplifier and the noise power Pao at the output of the preamplifier,respectively, area)Te = 169.36K and Pao = 3.73x10–10 Wb)Te = 170.8K and Pao = 4.56x10–10 Wc)Te = 182.5K and Pao = 3.85x10–10 Wd)Te = 160.62K and Pao = 4.6x10–10 WCorrect answer is option 'A'. Can you explain this answer?.
An antenna pointing in a certain direction has a noise temperature of 50K. The ambienttemperature is 290K. The antenna is connected to pre-amplifier that has a noise figure of 2dB andan available gain of 40 dB over an effective bandwidth of 12 MHz. The effective input noisetemperature Te for the amplifier and the noise power Pao at the output of the preamplifier,respectively, area)Te = 169.36K and Pao = 3.73x10–10 Wb)Te = 170.8K and Pao = 4.56x10–10 Wc)Te = 182.5K and Pao = 3.85x10–10 Wd)Te = 160.62K and Pao = 4.6x10–10 WCorrect answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An antenna pointing in a certain direction has a noise temperature of 50K. The ambienttemperature is 290K. The antenna is connected to pre-amplifier that has a noise figure of 2dB andan available gain of 40 dB over an effective bandwidth of 12 MHz. The effective input noisetemperature Te for the amplifier and the noise power Pao at the output of the preamplifier,respectively, area)Te = 169.36K and Pao = 3.73x10–10 Wb)Te = 170.8K and Pao = 4.56x10–10 Wc)Te = 182.5K and Pao = 3.85x10–10 Wd)Te = 160.62K and Pao = 4.6x10–10 WCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An antenna pointing in a certain direction has a noise temperature of 50K. The ambienttemperature is 290K. The antenna is connected to pre-amplifier that has a noise figure of 2dB andan available gain of 40 dB over an effective bandwidth of 12 MHz. The effective input noisetemperature Te for the amplifier and the noise power Pao at the output of the preamplifier,respectively, area)Te = 169.36K and Pao = 3.73x10–10 Wb)Te = 170.8K and Pao = 4.56x10–10 Wc)Te = 182.5K and Pao = 3.85x10–10 Wd)Te = 160.62K and Pao = 4.6x10–10 WCorrect answer is option 'A'. Can you explain this answer?.
Solutions for An antenna pointing in a certain direction has a noise temperature of 50K. The ambienttemperature is 290K. The antenna is connected to pre-amplifier that has a noise figure of 2dB andan available gain of 40 dB over an effective bandwidth of 12 MHz. The effective input noisetemperature Te for the amplifier and the noise power Pao at the output of the preamplifier,respectively, area)Te = 169.36K and Pao = 3.73x10–10 Wb)Te = 170.8K and Pao = 4.56x10–10 Wc)Te = 182.5K and Pao = 3.85x10–10 Wd)Te = 160.62K and Pao = 4.6x10–10 WCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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Here you can find the meaning of An antenna pointing in a certain direction has a noise temperature of 50K. The ambienttemperature is 290K. The antenna is connected to pre-amplifier that has a noise figure of 2dB andan available gain of 40 dB over an effective bandwidth of 12 MHz. The effective input noisetemperature Te for the amplifier and the noise power Pao at the output of the preamplifier,respectively, area)Te = 169.36K and Pao = 3.73x10–10 Wb)Te = 170.8K and Pao = 4.56x10–10 Wc)Te = 182.5K and Pao = 3.85x10–10 Wd)Te = 160.62K and Pao = 4.6x10–10 WCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
An antenna pointing in a certain direction has a noise temperature of 50K. The ambienttemperature is 290K. The antenna is connected to pre-amplifier that has a noise figure of 2dB andan available gain of 40 dB over an effective bandwidth of 12 MHz. The effective input noisetemperature Te for the amplifier and the noise power Pao at the output of the preamplifier,respectively, area)Te = 169.36K and Pao = 3.73x10–10 Wb)Te = 170.8K and Pao = 4.56x10–10 Wc)Te = 182.5K and Pao = 3.85x10–10 Wd)Te = 160.62K and Pao = 4.6x10–10 WCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for An antenna pointing in a certain direction has a noise temperature of 50K. The ambienttemperature is 290K. The antenna is connected to pre-amplifier that has a noise figure of 2dB andan available gain of 40 dB over an effective bandwidth of 12 MHz. The effective input noisetemperature Te for the amplifier and the noise power Pao at the output of the preamplifier,respectively, area)Te = 169.36K and Pao = 3.73x10–10 Wb)Te = 170.8K and Pao = 4.56x10–10 Wc)Te = 182.5K and Pao = 3.85x10–10 Wd)Te = 160.62K and Pao = 4.6x10–10 WCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of An antenna pointing in a certain direction has a noise temperature of 50K. The ambienttemperature is 290K. The antenna is connected to pre-amplifier that has a noise figure of 2dB andan available gain of 40 dB over an effective bandwidth of 12 MHz. The effective input noisetemperature Te for the amplifier and the noise power Pao at the output of the preamplifier,respectively, area)Te = 169.36K and Pao = 3.73x10–10 Wb)Te = 170.8K and Pao = 4.56x10–10 Wc)Te = 182.5K and Pao = 3.85x10–10 Wd)Te = 160.62K and Pao = 4.6x10–10 WCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice An antenna pointing in a certain direction has a noise temperature of 50K. The ambienttemperature is 290K. The antenna is connected to pre-amplifier that has a noise figure of 2dB andan available gain of 40 dB over an effective bandwidth of 12 MHz. The effective input noisetemperature Te for the amplifier and the noise power Pao at the output of the preamplifier,respectively, area)Te = 169.36K and Pao = 3.73x10–10 Wb)Te = 170.8K and Pao = 4.56x10–10 Wc)Te = 182.5K and Pao = 3.85x10–10 Wd)Te = 160.62K and Pao = 4.6x10–10 WCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice GATE tests.
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