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Ex 5 : For a non-ideal amplifier with a following parameters, determinea. input S/N ratio (dB)b. output S/N ratio (dB)c. noise factor and noise figureInput signal power = 2 x 10 -10 WInput noise power = 2 x 10 -18 WPower gain = 1000000Internal noise N d = 6 x 10 -12 W

Question

Ex 5 : For a non-ideal amplifier with a following parameters, determinea. input S/N ratio (dB)b. output S/N ratio (dB)c. noise factor and noise figureInput signal power = 2 x 10 -10 WInput noise power = 2 x 10 -18 WPower gain = 1000000Internal noise N d = 6 x 10 -12 W

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Solution

To determine the input S/N ratio (dB), we need to calculate the ratio of the input signal power to the input noise power.

Given: Input signal power = 2 x 10^-10 W Input noise power = 2 x 10^-18 W

The input S/N ratio (dB) can be calculated using the formula: Input S/N ratio (dB) = 10 * log10(input signal power / input noise power)

Substituting the given values: Input S/N ratio (dB) = 10 * log10(2 x 10^-10 / 2 x 10^-18)

Simplifying the expression: Input S/N ratio (dB) = 10 * log10(10^8) Input S/N ratio (dB) = 10 * 8 Input S/N ratio (dB) = 80 dB

To determine the output S/N ratio (dB), we need to calculate the ratio of the output signal power to the output noise power.

Given: Power gain = 1000000 Internal noise Nd = 6 x 10^-12 W

The output signal power can be calculated using the formula: Output signal power = input signal power * power gain

Substituting the given values: Output signal power = 2 x 10^-10 * 1000000 Output signal power = 2 x 10^-4 W

The output noise power can be calculated using the formula: Output noise power = input noise power * power gain + internal noise Nd

Substituting the given values: Output noise power = 2 x 10^-18 * 1000000 + 6 x 10^-12 Output noise power = 2 x 10^-12 + 6 x 10^-12 Output noise power = 8 x 10^-12 W

The output S/N ratio (dB) can be calculated using the formula: Output S/N ratio (dB) = 10 * log10(output signal power / output noise power)

Substituting the calculated values: Output S/N ratio (dB) = 10 * log10((2 x 10^-4) / (8 x 10^-12))

Simplifying the expression: Output S/N ratio (dB) = 10 * log10(2.5 x 10^7) Output S/N ratio (dB) = 10 * 7.39794 Output S/N ratio (dB) = 73.9794 dB

To determine the noise factor, we need to calculate the ratio of the output noise power to the input noise power.

The noise factor can be calculated using the formula: Noise factor = (output noise power / input noise power)

Substituting the calculated values: Noise factor = (8 x 10^-12) / (2 x 10^-18) Noise factor = 4 x 10^6

To determine the noise figure, we need to calculate the ratio of the noise factor to the power gain.

The noise figure can be calculated using the formula: Noise figure = 10 * log10(noise factor)

Substituting the calculated values: Noise figure = 10 * log10(4 x 10^6) Noise figure = 10 * 6.60206 Noise figure = 66.0206 dB

Therefore, the answers are: a. Input S/N ratio (dB) = 80 dB b. Output S/N ratio (dB) = 73.9794 dB c. Noise factor = 4 x 10^6 Noise figure = 66.0206 dB

This problem has been solved

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