I have been reading through "Johnson, J. B.: Thermal Agitation of Electricity in Conductors" from 1928 and he keeps on talking about the "real resistance component" \$R(\omega)\$ of a resistor shunted with its own shunt capacity. I think these are questions of historical perception or change in usage of technical language:

  1. For me a general impedance is defined as \$Z(\omega)=R+jX(w)\$, with \$R\$ not being depended on \$\omega\$. For an RC circuit this would be $$ Z(\omega)=R+jX(w)=R+\frac{1}{j\omega C}.~~~~~~~~~~~~~~(1) $$ So why should \$R\$ depended on \$\omega\$?
  2. Nyquist talks about a "pure" resistance \$R_0\$. Is he talking about the purely resistive part of his RC circuit, namely \$R\$ in the schematic given below?


simulate this circuit – Schematic created using CircuitLab

  1. How does he deduce eq. (2) in the paper: $$ R(\omega)=\frac{R_0}{1+\omega^2C^2R_0^2}~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2) $$ If I talk about real part, I solely talk about \$R\$. If I talk noise, I usually go with the absolute-square. The absoulte yields $$ \left|\frac{R\cdot\frac{1}{j\omega C}}{R+\frac{1}{j\omega C}}\right|=\frac{R}{\sqrt{1+\omega^2C^2R^2}}.~~~~~~~~~~~~~~~~(3) $$ Squaring eq. (3) does not equal eq. (2). Where is my error?
  • 1
    \$\begingroup\$ In a purely DC world resistance is resistance. In the real world resistors have both capacitive and inductive effects. At low frequencies the capacitive effects can be neglected as minor. At high frequencies the capacitive effects of the resistor will dominate over the resistive effects. This way the resistor becomes frequency dependent. \$\endgroup\$
    – vini_i
    Commented Sep 14, 2016 at 16:51
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    \$\begingroup\$ For an example of a case where resistance is frequency-dependent, look up skin effect. \$\endgroup\$
    – The Photon
    Commented Sep 15, 2016 at 4:48
  • \$\begingroup\$ @ vini_i: I totally agree about its physical meaning, but this does not change the way of mathematical description used today separating real and imaginary parts of an Impedance - that does not fit an \$R(\omega)\$. This is not to be find in any modern textbook I have looked through. And Johnson does not differentiate frequency regimes, probably because back then he was very restricted in frequency. \$\endgroup\$
    – Irenaius
    Commented Sep 15, 2016 at 6:17
  • \$\begingroup\$ I have a strong feeling that the R(w) being referred to, did have some frequency dependence, that is why he used the term "real resistance component". This term would be the equivalent of what today we call impedance. The reason your equation and equation (2) do not match, is that the person that printed the book, made the mistake of leaving the root symbol out. Yours is the correct one. \$\endgroup\$
    – Guill
    Commented Sep 16, 2016 at 22:48
  • \$\begingroup\$ @ Guill: I figured out why Johnson used R(w), it is all correct. And he did not use the absolute square. Please see my answer. \$\endgroup\$
    – Irenaius
    Commented Sep 21, 2016 at 10:52

1 Answer 1


It is a fact that \$R(\omega)\$ depends on \$\omega\$, the proof is as follows:

How do you model a real, physical resistor? Every resistor has a certain, sometimes very small capacitance. How to include it in an equivalent circuit, in parallel or in series? If you put it in series it becomes a highpass, and in reality the noise of any resistor does not show highpass behavior. Thus, it needs to go in parallel, as I drew earlier (see question).

Now it needs to be put into mathematical equations. In general a parallel circuit is most easily treated with admittances, thus $$ Y(\omega)=G(w)+jB(w)=G+jB(w).~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(4) $$ with \$G(\omega)\$ being the inverse of the so called "real resistance component" \$R_0\$. And since \$R_0\$ does not depend on \$\omega\$, \$G(\omega)\$ does not either: \$G(\omega)=G=const\$.

Now Johnson, in historical perspective, preferred to think in impedances. So he converted the admittance into an impedance by taking its inverse or simply considering the parallel circuit of \$R\$ and \$C\$: $$ Z(\omega)=\frac{1}{Y(\omega)}=R||C=\frac{R\cdot\frac{1}{j\omega C}}{R+\frac{1}{j\omega C}}=\frac{R}{1+j\omega RC}.~~~~~~~~~~~~~~~(5) $$ Now a little conversion, starting with the multiplication of (5) with \$1=(1-j\omega RC)/(1-j\omega RC)\$: $$ Z(\omega)=\frac{R\cdot(1-j\omega RC)}{(1+j\omega RC)(1-j\omega RC)}=\frac{R-j\omega R^2C}{1+j(\omega RC)^2}~~~~~~~~~~~~~~~(6), $$ utilizing \$(a-b)(a+b)=a^2-b^2\$. Now we separate the real part and imaginary part which yields $$ Z(\omega)=\frac{R}{1+j(\omega RC)^2}-\frac{j\omega RC}{1+j(\omega RC)^2}=R(\omega)+jX(\omega).~~(7) $$ Thus, the resistive part is frequency dependend, which answers question 1). The answer to question 2) is that the \$R\$ in eq. (7) is really \$R_0\$, so the assumption was correct. The above deduction also answers question 3).


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