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From: Jon Hagen

Date: 6-19-01

Subject: Johnson Noise

A resistor at temperature T is a noise generator. If connected to a second resistor of equal resistance, it will supply that resistor with a power kTB. One way to derive this formula is to consider a transmission line of length L. Let the ends be shorted so that the line is a one- dimenstional cavity, capable of storing energy in each of the possible standing wave modes (modes that have V=0 at x=0 and at L=0, to satisfy the boundary conditions). If the line or the terminations have even the slightest bit of loss, thermal energy will eventually excite each mode to an average energy kT. These standing waves consist of equal power flow in both directions. The number of modes per Hertz is proportional to the length of the line, so as L is increased, the energy density per unit length becomes a constant. If the line is made longer,the energy density in each mode is reduced, but there will be more modes per Hz, so the energy density per Hertz becomes constant. Let us calculate that constant:

The boundary conditions require that f = nc/2L where n= 1,2,3.... and c is the speed of light. Therefore the number of modes n in a frequency range, f= B, is given by n = 2L B /c. The energy density in Watts per Hz per unit length is therefore u= 2BkT/c, where we have assigned an energy kT to each more. The resulting power flow in each of the two directions is therefore

P= uc/2 = ktB Watts/Hz.

Now suppose that the terminations are not perfect shorts, but are resistors