# Additive Gaussian white noise - bandwidth

The average power at output when additive Gaussian white noise is present is given by the formula below:

$$P_0 = \int\limits_{-B}^{B} \frac N2\ df ,\quad \text{with B being the bandwidth.}$$

My question is when the bandwidth is in terms of $\textrm{rad/s}$, what would be the formula? The one stated divided by $2\pi$ or another formula?

• Ok, you integrate from $-B$ to $B$, so your overall bandwidth for noise observation is definitely $2B$, not $B$. The formula doesn't care what units your $B$ or $N$ has. So your question doesn't mean anything. – Marcus Müller May 16 '16 at 10:04
• Why should the power at the output be different for you who are using Hertz and your friend who is analyzing the same system but using rad/s? The answer is the same in both cases! Read my comment on Jan Kruger's answer. – Dilip Sarwate May 16 '16 at 13:20

Just to be clear about the units:

\begin{align} \Omega &= 2\pi F, \quad \text{with $\Omega$ in } \rm\underline{radians/sec}\text{ and $F$ in } \rm\underline{cycles/sec} \text{ or }\rm\underline{Hz}\tag{1}\\ \omega &= 2\pi f, \quad \text{with $\omega$ in } \rm\underline{radians/samples}\text{ and $f$ in } \rm\underline{cycle/sample}\tag{2}\\ \omega&=\Omega T_s\quad \text{and}\quad f=F/F_s\tag{3} \end{align} Read @Dilip Sarwate's comments; what should happen there is a change of variables. \begin{align} \omega &= 2\pi f\Longrightarrow d\omega = 2\pi df\iff df=\frac{d\omega}{2\pi}\tag{4}\\ f&=\pm B \Longrightarrow \omega= \pm 2\pi B\tag{5}\\ \end{align} Using $(4)$ and $(5)$ you then have the following results: $$P_0=\int_{-B}^{B} \frac N2 df = NB \Longrightarrow P_0= \int_{-2\pi B}^{2\pi B} \frac N2 \cdot \frac{1}{2\pi}d\omega=\frac{1}{2\pi}\int_{-2\pi B}^{2\pi B} \frac N2 d\omega = NB\tag{6}\\$$

you integrate from $-B$ to $B$. If we do integrate the term we get:

$$P_o = \int_{-B}^B \frac{N}{2} df = \left. \left[ \frac{N}{2}f \right] \right\rvert_{-B}^{B} = \frac{N}{2} \left[ B + B \right] = NB$$

So your noise scales linear with the bandwidth. Assuming our noise term is defined as $$[P_{gauss}] = \left[\frac{V^2}{Hz}\right]$$ we need to use the frequency and not the angular frequency for the bandwidth. You can obtain it from the following formula: $$\omega = 2\pi f$$ with the units $$\left[\frac{rad}{s}\right] = 2\pi [Hz] = \underbrace{2\pi}_{"rad"} \left[\frac{1}{s}\right]$$

Therefore, if you have$P_{gauss}$ defined in the unit $Hz$ you need to correct for $2\pi$ like so $$P_{gauss, \omega} = \frac{NB}{2\pi}$$

Please keep in mind that $2\pi$ does not have unitx. The term $rad$ means radians and is only added to be able to distinguish between the frequency in Hz and the angular frequency. In signal processing we often deal with sines, cosines and thelike which have a periodicity of $2\pi$, so $\omega$ is often used to simplify the calculations.

• This is incorrect: the power at the output does not depend on whether the engineer is using Hertz or angular frequency! The power spectral density of white noise has the same numerical value regardless of whether one uses frequency in Hz or angular frequency $\omega$. With angular frequency, the formula is $$P = \frac{1}{2\pi}\int_{-2\pi B}^{2\pi B} \frac{N}{2}\,\mathrm d\omega = NB$$ just as before. – Dilip Sarwate May 16 '16 at 13:16
• I tried to make clear that if the power of the gaussian noise is defined as V^2/Hz, he needs to add the correction factor to get the right answer. The formula is correct as is, but if you try to replace B with the according value in radians/s, you will get the wrong result. If I didn't make that clear enough, please improve my answer! Thanks! – Jan Krüger May 16 '16 at 13:26
• If the noise is defined as $$\left[P_{gauss}\right] = \left[\frac{V^2}{2\pi Hz}\right]$$ Then the engineer can use $\omega$ at instant. It is a question of scaling the frequency axis with 2*Pi or not. Our professor likes to do this alot to test us. – Jan Krüger May 16 '16 at 13:31