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A receiver has been operating at 290 K and receiving a signal of bandwidth 400 kHz. The amplifier used in the receiver has an average output resistance of 1.5 kΩ. This will lead to Johnson noise voltage of ________ μV.
  • a)
    3.05
  • b)
    3.15
Correct answer is between '3.05,3.15'. Can you explain this answer?
Verified Answer
A receiver has been operating at 290 K and receiving a signal of band...
We have a relation between voltage generated, temperature, bandwidth and resistance offered as follows,
We have, k = 1.38 × 10-23 J/K,
T = 290 K, B = 400 × 103 Hz,
R = 1.5 × 103Ω
Thus, vn = 3.099 μV.
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Most Upvoted Answer
A receiver has been operating at 290 K and receiving a signal of band...


Calculation of Johnson noise voltage:

1. Given data:
- Temperature (T) = 290 K
- Bandwidth (B) = 400 kHz = 400 * 10^3 Hz
- Output resistance (R) = 1.5 kΩ = 1.5 * 10^3 Ω
- Boltzmann constant (k) = 1.38 * 10^-23 J/K

2. Formula for Johnson noise voltage (Vn):
Vn = sqrt(4 * k * T * R * B)

3. Substitute the values:
Vn = sqrt(4 * 1.38 * 10^-23 * 290 * 1.5 * 10^3 * 400 * 10^3)

4. Calculate the Johnson noise voltage:
Vn = sqrt(4 * 1.38 * 10^-23 * 290 * 1.5 * 10^3 * 400 * 10^3)
= sqrt(3.312 * 10^-10)
≈ 5.75 * 10^-6 V
≈ 5.75 μV

Therefore, the Johnson noise voltage of the amplifier in the receiver operating at 290 K with a signal bandwidth of 400 kHz and an average output resistance of 1.5 kΩ is approximately 5.75 μV.

Since the correct answer should be between 3.05 and 3.15 μV, it seems that there might have been a mistake in the calculation or data provided.
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A receiver has been operating at 290 K and receiving a signal of bandwidth 400 kHz. The amplifier used in the receiver has an average output resistance of 1.5 kΩ. This will lead to Johnson noise voltage of ________ μV.a)3.05b)3.15Correct answer is between '3.05,3.15'. Can you explain this answer?
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