This came up during an assignment question, but is a tangent.

Thermal Noise is given by; N=k T B where;

k= Boltzmann's Constant (J/K) ir (W-s/K)

T = Temperature (K)

B = Bandwidth (Hz)

That would mean that the units of thermal noise can be expressed as Watt-cycles? That doesn't make sense to me.

Normally noise is expressed in dB or dBm. How does that relate to the units here?


2 Answers 2


Hertz is 1/second. So N in your formula is Watts. The formula is not for noise voltage figure but for noise power. Power is watts, so everything computes OK.

To fully answer your question, noise is measured either in absolute units or as a ratio. Relative unitless ratio is just a number. For each untiless number scale there is always some absolute value in physical units which is agreed upon. Say if reference number is milliwatt, then microwatt of noise is 0.001 unitless on milliwatt scale. Or logarithmically -30dBm or -60dBW, etc.

Side note: non zero thermal noise strictly speaking has no absolute maximum voltage amplitude for infinite period of observation. Every observed momentary value can always be exceeded in any of next observations. Saying that noise is 0.01V p-p (peak to peak) is oversimplification or misuse of terminology. Preferred quantification of noise is in any units but with note of the method of observation, like 0.5A rms (root mean square).

  • \$\begingroup\$ Ha ha. No problem. Good luck with assignment \$\endgroup\$
    – user924
    May 31, 2012 at 23:37
  • \$\begingroup\$ Note that the "d" in "dB" is a "d", not a "D". \$\endgroup\$ Jun 1, 2012 at 0:11

The dimensions:

\$ \dfrac{W \cdot s}{K} \cdot K \cdot Hz = \dfrac{W \cdot s}{K} \cdot K \cdot \dfrac{1}{s} = W \$

So your noise is a power quantity. The expression gives power as an absolute value. We're often more interested in power compared to a reference level, and to keep it practical, because the ratios can vary from very low to very high, a logarithmic scale is used:

\$ dB = 10 \cdot log\left(\dfrac{P}{P_{REF}}\right) \$

That's the 10-based logarithm. There exist several different power references, and each will give a different dB number, so it's important to state clearly which scale you're using. dBm for instance has a 1mW reference, that's for instance 775mV in 600\$\Omega\$. If your mobile phone transmits 500mW of power that's

\$ 10 \cdot log\left(\dfrac{500mW}{1mW}\right) = 27 dBm \$

A 3dB difference is a factor two; 3dB higher is twice the power.

You can also express voltage levels in dB. Then

\$ 10 \cdot log\left(\dfrac{P}{P_{REF}}\right) = 10 \cdot log\left(\dfrac{\dfrac{V^2}{R}}{\dfrac{V_{REF}^2}{R}}\right) = 10 \cdot log\left(\dfrac{V^2}{V_{REF}^2}\right) = 20 \cdot log\left(\dfrac{V}{V_{REF}}\right) \$

So, while for power you have to multiply by 10, for voltages it's by 20.
6dB lower is half the voltage.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.