Let's say there's some 50 ohm resistor connect across the input of some 2 port network that has an input impedance of 50 ohm, so like this:


simulate this circuit – Schematic created using CircuitLab

Does the input resistor (in the two port box) get counted in the thermal noise calculations? If it does, then it would seem as though the voltage across it would then be $$ V_\mathrm{Two\ port\ resistor} = \sqrt{4kTBR} $$ and so the power would be $$ P_\mathrm{Two\ port\ resistor} = \frac{\sqrt{4kTBR}^2}{R} = 4kTB $$ however, wherever I've read seems to say that the noise power delivered to a matched impedance is \$kTB\$. What I don't understand is, this only seems to take into account one of the two impedances.

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    \$\begingroup\$ The voltage sum of two uncorrelated noise sources is \$\sqrt{V_1^2+V_2^2}\$ and not \$V_1 + V_2\$ as your third diagram shows. \$\endgroup\$ – MikeJ-UK Jun 6 '13 at 14:57
  • \$\begingroup\$ Not too sure what you mean by that; could you please elaborate? Both sources are in volts so not sure why I cannot simply add them. If the square root were to not be there, then that'd make sense to me! Also, doing what you propose would not reach an accurate answer, it'd end up being \$\sqrt(8)kTB\$ I think. \$\endgroup\$ – user968243 Jun 6 '13 at 15:06
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    \$\begingroup\$ Think of two sine waves. If they are in-phase you can just add the amplitudes. If they are 180 deg out of phase, the sum is zero. If they are 90 deg out of phase, the result is as per the root-sum-square formula and statistically, two uncorrelated random noise sources will behave the same way. \$\endgroup\$ – MikeJ-UK Jun 6 '13 at 15:10
  • \$\begingroup\$ My two books say that if you have two resistors in series, R1 and R2, you can say that the thermal noise of the two is \$\sqrt{4kTB(R1 + R2)}\$. Oh, I guess that actually probes your point! :D \$\endgroup\$ – user968243 Jun 6 '13 at 15:14
  • \$\begingroup\$ You should also tell us what is B. Looking at other answers, it seem to be bandwith = \$\Delta f\$. \$\endgroup\$ – not2qubit May 15 '19 at 17:03

Your problem is combining the voltage sources. This is incorrect, first because you can't add uncorrelated noise to each other, second because we don't even need to do worry about the other resistor's power generation for this problem.

Since we are only looking at the power that one resistor transfers to another, we look only at the voltage it generates and transfers to the other.


simulate this circuit – Schematic created using CircuitLab

Now, we look at the voltage that would appear on the transferred resistor, which would be exactly half.

$$ V_\mathrm{transferred} = \frac{\sqrt{4k_BT \Delta FR}}{2} $$

Now with power: $$ P_\mathrm{transferred} = \frac{V^2}{R} $$ $$ P_\mathrm{transferred} = \frac{4k_BT \Delta FR}{4R} $$ $$ P_\mathrm{transferred} = k_BT \Delta F $$

Hope this helps!

  • \$\begingroup\$ Thanks, that makes sense I guess; however, is what I've done correct though, correct in that you'd have to include the noise source from both resistance? \$\endgroup\$ – user968243 Jun 6 '13 at 15:08
  • \$\begingroup\$ Since we're only looking at the power from one resistor to another, we don't need the power the other resistor is generating from noise. If you wanted to combine their voltages, you have to actually add them together via Parseval's theorem as they are RMS values and uncorrelated (they don't depend on each other). That would give a total voltage of $$ \sqrt{V_{R1}^2 + V_{R2}^2} $$ which works out to be $$ \sqrt{8k_BT \Delta FR} $$ when the two resistors are matched. \$\endgroup\$ – Steven Goldade Jun 6 '13 at 15:27
  • \$\begingroup\$ If you then used that voltage, you would be treating the power that is transferred from both resistors to one, not just from the one to the other. That's where your second mistake is. \$\endgroup\$ – Steven Goldade Jun 6 '13 at 15:31
  • \$\begingroup\$ Okay, I kind of see that. I guess what I'm not clear on is: what is the total thermal noise across one of the two resistors? Do I have to include the thermal noise the the resistor itself generates? You calculate the noise power that is transferred from one resistor to another, what I'm wondering is, on top of that transfer, does the resistor itself also generate noise that affects itself (thus doubling the noise voltage across it if they're matched)? Thanks for you help! \$\endgroup\$ – user968243 Jun 6 '13 at 15:40
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    \$\begingroup\$ I'd have to think further about it, but noise voltages can be tricky as noise is random. This means everything is actually done with probability density functions, in this case with the spectral power density. That noise would dissipate into itself, and transfer $$k_BT \Delta F $$ into the other resistor, as is normal. The other resistor would do the same, but that noise is not necessarily correlated, so they would statistically add like mentioned above, leaving $$\sqrt{2} k_BT \Delta F $$. I may however be making an incorrect assumption about the correlation. \$\endgroup\$ – Steven Goldade Jun 6 '13 at 16:25

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