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Capacitors Meet Geometric Series
We'll analyze the charge present in a capacitor under the influence of a square wave with symmetric semi-period and positive average value. The step function takes the values: The charge function $q(t)$ during charging (when voltage = $E$) is: $$q(t) = \text{E}\text{C} + (q_{in} - \text{E}\text{C}) \ e^{\frac{-t}{\tau}}$$ During discharge (when voltage = 0): $$q(t) = q_{in} \ e^{\frac{-t}{\tau}}$$ where: From $q_{2n}$ to $q_{2n+1}$, we can express: $$q_{2n+1} = E C + (q_{2n} - E C) e^{\frac{-T}{2 \tau}} = E C (1 - e^{\frac{-T}{2 \tau}})+ q_{2n} e^{\frac{-T}{2 \tau}}$$ where: For $q_{2n+2}$: $$q_{2n+2} = q_{2n+1} e^{\frac{-T}{2\tau}} = \alpha q_{2n+1}$$ The geometric series converges to: For odd-numbered steps: $$\lim_{n\rightarrow \infty}q_{2n+1}=\frac{EC}{1+\alpha}$$ For even-numbered steps: $$\lim_{n\rightarrow \infty}q_{2n}=\frac{EC\alpha}{1+\alpha}$$ The capacitor has sufficient time to fully charge and discharge.
We'll analyze the charge present in a capacitor under the influence of a square wave with symmetric semi-period and positive average value. $\text{E}$ is the voltage source value$\text{C}$ is capacitance$\tau$ is the time constant$q_{in}$ is initial charge The circuit acts as a low-pass filter, averaging the input voltage.
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