Chapter 5, Problem 5.9QAP

Interpretation Introduction

Interpretation:

Root-mean-square thermal noise associated with a resister should be calculated. The fate of the thermal noise when the bandwidth is reduced to $100\text{ Hz}$ should be predicted.

Concept introduction:

Data given:

Temperature (T) = Room temperature = $298\text{ K}$

Bandwidth ( $\Delta f$ ) = $1\text{ MHz}$

Resistance of the resistor (R) = $1.0\text{ M}\Omega$

Thermal noise for a resistive circuit element can be given as

$\begin{array}{l}{\overline{\upsilon }}_{rms}=\sqrt{4kTR\Delta f}\\ \text{Where,}\\ k\text{ = Boltzmann's constant = 1}\text{.38}\times {\text{10}}^{-23}\text{ J/K}\\ T\text{ = Temperature in K}\\ R\text{ = Resistance in }\Omega \\ \Delta f\text{ = Bandwidth}\end{array}$