Physics Asked by Vibhor Sharma on June 19, 2021
"The RC circuit averages past history at the input with a weighting factor of e−Δt/RC."
I can see why you would find this statement opaque. Here is my take on it.
When we perform an integral, we are summing up a time series of events for infinitesimal slices or steps of time. this means that the integral contains (across its start and stop limits) an accounting of what happened right at the start of the summation, the middle, and the final point of that summation- in other words, it represents a cumulative history of what happened between the integration limits in time.
The system response (the specific voltage at any time point t) is scaled by the details of the dynamics of the system (how big the capacitance and resistance are and how they are connected).
Awkwardly, then, these facts can be expressed in the form given in your textbook.
Answered by niels nielsen on June 19, 2021
Rewrite that integral (which defines a mathematical operation called convolution) by performing the substitution $taumapsto t-Delta t$, $mathrm{d}taumapsto -mathrm{d}(Delta t)$:
$$V(t) = frac{1}{RC}int_0^infty V_mathrm{i}(t-Delta t)mathrm{e}^{-Delta t/RC}mathrm{d}(Delta t).$$
You can now see that $V_mathrm{i}(t-Delta t)$, with $Delta t$ running from $0$ to $infty$, represents all the past (up to time $t$) values of $V_mathrm{i}$ (that is, its history) and each past value is weighted by $mathrm{e}^{-Delta t/RC}$ (actually by $(RC)^{-1}mathrm{e}^{-Delta t/RC}$).
Answered by Massimo Ortolano on June 19, 2021
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