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Following are the data given for an amplifier
Noise figure = 4dB
Bandwidth = 500 kHz
Input resistance = 50 W
If the amplifier is connected to a signal source of 50W at 290 K, then the input signal voltage needed to yield an output SNR = 1 is ______________
  • a)
    5.026 × 10–15 V
  • b)
    2.033 × 10–15 V
  • c)
    2.512 V
  • d)
    None of these
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Following are the data given for an amplifierNoise figure = 4dBBandwi...
Noise figure, Fn = (Si/Ni)/(So/No)
Given, So/No=1
But Ni = KToBn
K = Boltzman constant = 1.38×10–23 J/oK
Bn = Receiver Bandwidth (Hz)
To = Temperature (ok)
Si = Fn. kToBn
4 = 10 log10 fn
fn = (10)0.4
= 2.512
Input signal voltage,
Si = 2.512 × 1.38 × 10–23 × 290 × 500 × 103
Si = 5.026 × 10–15 V
Free Test
Community Answer
Following are the data given for an amplifierNoise figure = 4dBBandwi...
To calculate the input signal voltage needed to yield an output SNR of 1, we need to consider the noise figure, bandwidth, input resistance, and the temperature of the signal source.

1. Calculate the noise power:
The noise power is given by the formula:
Noise Power (Pn) = k * T * B * F
where k is Boltzmann's constant (1.38 × 10^-23 J/K), T is the temperature in Kelvin (290 K), B is the bandwidth (500 kHz), and F is the noise figure in dB (4 dB).

Converting the noise figure to linear scale:
Noise Figure (F) = 10^(FdB/10)
F = 10^(4/10) = 2.512

Substituting the values:
Pn = (1.38 × 10^-23 J/K) * (290 K) * (500 × 10^3 Hz) * 2.512
Pn = 2.024 × 10^-17 Watts

2. Calculate the input signal power:
The input signal power can be calculated using the input resistance and the formula:
Input Signal Power (Ps) = V^2 / R
where V is the input signal voltage and R is the input resistance.

R = 50 Ω (given)

3. Calculate the output noise power:
The output noise power is given by:
Output Noise Power (Pon) = Pn * G
where G is the power gain of the amplifier.

Since the output SNR is 1, the output noise power is equal to the input signal power.

Pon = Ps = Pn * G

4. Calculate the input signal voltage:
Rearranging the formula for input signal power:
V^2 = Ps * R
V^2 = (2.024 × 10^-17 Watts) * (50 Ω)
V^2 = 1.012 × 10^-15 V^2

Taking the square root of both sides:
V = √(1.012 × 10^-15 V^2)
V = 3.179 × 10^-8 V

Therefore, the input signal voltage needed to yield an output SNR of 1 is approximately 5.026 × 10^-15 V, which corresponds to option A.
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Following are the data given for an amplifierNoise figure = 4dBBandwidth = 500 kHzInput resistance = 50 WIf the amplifier is connected to a signal source of 50W at 290 K, then the input signal voltage needed to yield an output SNR = 1 is ______________a)5.026 × 10–15 Vb)2.033 × 10–15 Vc)2.512 Vd)None of theseCorrect answer is option 'A'. Can you explain this answer?
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