Consider a simple first-order RC filter with AWGN input noise with single-sided density of No. We know the output noise power is No/(4RC). What is rarely noted is that this is true after steady-state has been reached. I simulated the noise from such a filter starting from rest at t=0 and found that the ensemble average of the instantaneous noise power seems to increase from 0 to No/(4RC) exponentially with a time constant of RC/2, not the circuit time constant of RC. Can anyone derive the equation for this?

Just to clarify this is input noise that is switched on by a unit-step at t=0 at the input to the filter. So I am guessing the usual analysis based on the assumption of stationarity probably does not apply.

Thanks, John

  • $\begingroup$ Is the time constant $RC$ or $1/RC$? Dimensional analysis suggests that it is $RC$, the time required for an exponential to decay by a factor of $1/e$ from its initial value. $\endgroup$ Feb 24, 2015 at 17:20

1 Answer 1


Consider a simple RC filter whose step response is $1-e^{-t/RC}$ for $t > 0$, and impulse response is $$h(t) = \begin{cases} \frac{1}{RC}e^{-t/RC}, & t \ge 0,\\ 0, &t < 0.\end{cases}$$ Then, when the input is white noise, the steady-state output variance is $$\sigma^2 = \frac{N_0}{2} \int_{-\infty}^\infty h^2(t)\,\mathrm dt = \frac{N_0}{2}\int_0^\infty \frac{1}{(RC)^2}e^{-2t/RC}\,\mathrm dt = \frac{N_0}{4RC}.$$ However, if the input noise is also $0$ for $t < 0$, then at time $T$, the noise variance at the output is $$\begin{align} \sigma_T^2 &= E\left[\int_0^T N(t)\frac{1}{RC}e^{-(T-t)/RC}\,\mathrm dt \int_0^T N(s)\frac{1}{RC}e^{-(T-s)/RC}\,\mathrm ds\right]\\ &= \int_0^T\int_0^T E[N(t)N(s)]\frac{1}{(RC)^2}e^{-(2T-t-s)/RC} \,\mathrm dt\,\mathrm ds\\ &= \int_0^T\int_0^T \frac{N_0}{2}\delta(t-s)\frac{1}{(RC)^2}e^{-(2T-t-s)/RC} \,\mathrm dt\,\mathrm ds\\ &= \int_0^T \frac{N_0}{2}\frac{1}{(RC)^2}e^{-(2T-2s)/RC} \,\mathrm ds\\ &= \frac{N_0}{4RC}e^{-2T/RC}\int_0^T \frac{2}{RC}e^{2s/RC}\,\mathrm ds\\ &= \frac{N_0}{4RC}e^{-2T/RC}[e^{2T/RC}-1]\\ &= \left.\left.\frac{N_0}{4RC}\right[1 - e^{-2T/RC}\right] \end{align}$$ showing that $\sigma_T^2$ increases to its asymptotic value $\displaystyle \sigma^2 = \frac{N_0}{4RC}$ more rapidly than the unit step response increases to its asymptotic value $1$.

The time constant of the RC circuit is $RC$, not $1/RC$ as the OP claims, and the time constant of the increase in the noise variance is $RC/2$, half that of the RC circuit. The smaller the time constant, the more rapid the decay to $0$ (or rise to the asymptotic value).

  • $\begingroup$ @MattL Thanks for making the correction. That's what happens when one copies and pastes the previous line and forgets to make all the needed changes. $\endgroup$ Feb 25, 2015 at 14:00
  • $\begingroup$ You're welcome, I think one error in all those formulas is not such a bad average ... $\endgroup$
    – Matt L.
    Feb 25, 2015 at 14:03

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