# SNR when calculating noise

I am trying to solve a few example questions relating to input voltage and input noise for the following questions. and I understand that I need to calculate the maximum noise voltage acceptable for the required SNR using the equation, SNR = 20log(Signal/Noise).

The thing I am having trouble understanding is for the first question, in the solutions the input noise voltage is calculated using: Whilst in the solution for the second problem it uses: I cannot see any difference in the question as to why one would be squared and the other not. Thanks in advance.

## 1 Answer

Let's assume that 500 uV input is 500 uV RMS. We are allowed 40 dB SNR, meaning the ratio of Vinsignal/Vinnoise is 100:1. We now know the allowed input noise is 5 uVrms. This 5 uV input noise must be from all the sources attached to that input node.

To make the math easy, assume the bandwidth is 1,000,000 Hz. (the 1.2 MHz is about 2 dB error; I'll let you correct for that).

We need the input noise density for 1Hertz bandwidth. Power of noise changes linearly with bandwidth. Voltage of noise changes with square-root of bandwidth.

The 1 hertz noise density is 5uV/sqrt(1,000,000) = 5uV/1,000 = 5 nanovolts.

The noise floor of the signal source must be included in this; that 1 kΩ resistor provides 4 nanovolts RMS in a 1 hertz bandwidth.

Your budget allows only 5 nanovolts. Will the extra factor of 1.2x bandwidth cause this to be an impossible design task, intended as a learning experience by your teacher?