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1. Ground the excitation voltage of the circuit and measure the white noise voltage produced by connecting the output of the circuit to a 8.5DVM

2. Subtract the Avg DC offset of from each white noise voltage measurement taken. This will leave me with the AC content of the measurements.

3. Perform FFT on the AC content.

4. Divide the Spectrum magnitude produced by the FFT by 2^0.5, in order to obtain the spectrum magnitude in terms of Vrms/root Hz

6.Replace the resistor in question with a short circuit (copper wire)

7. Repeat the measurement using the 8.5 DVM, measuring the output of the circuit for white noise voltage for when the resistor is replaced short circuit.

8. Subtract the Avg DC offset of from each white noise voltage measurement taken. This will leave me with the AC content of the measurements.

9. Perform FFT on the AC content.

10.Divide the Spectrum magnitude produced by the FFT by 2^0.5, in order to obtain the spectrum magnitude in terms of Vrms/root Hz

11. From the 2 FFT plots obtained, choose a frequency and read off the magnitude.

12. Use the following equation to calculate the Noise Voltage Rms of the resistor:

Ediff_rms= [(Eres_in_circuit)^2 - (Eres_replaced_with_short)^2]^0.5 * freq

Can anyone verify if this method of measurement is correct?

I thank you all in advanced.