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How can the channel noise power be below thermal noise?

Question asked by rulonv Employee on Aug 16, 2004
We know that a noisy resistor has an equivalent noise voltage of sqrt( 4kTBR) into a noiseless resistor.  Consequently, a 1 ohm resistor will have thermal noise at its output.

However, if we follow this resistor with a filter made up of ideal capacitors and inductors we know that the output noise must be below thermal noise since the inductors and capacitors can't add noise ... assuming we are down the skirt of the filter.

This is just a simple illustration ... obviously in real life the filter components have Q which is a result of component resistances.  These resistances do add noise (even though this resistance is generally small) and appears as the dissipative loss of the filter (or insertion loss).  Spectrasys correctly models all of these effects.

In the book, Radio Receiver Design, by Kevin McClaning and Tom Vito, Noble Publishing, 2000, pg 465 (this is an excellent book ... it can be purchased at "")

In Chapter 5 titled "Amplifiers and Noise" they make the following statements in the section called

"Miscellaneous Comments"
"Although some of the following points may seem obvious, we have often found it necessary to include them."
    "- Filters will filter noise.  Noise is just like any other signal."
    "- Attenuators will attenuate noise."
    "- Noise can and does exist at levels below kTB, or -174 dBm."
We could never measure any noise below thermal because the output port or measurement device adds noise which will be greater than or equal to the thermal noise floor.