@Muse - An RC low-pass filter acts as an approximate passive integrator for signals whose frequencies lie well above its cutoff frequency:
$$f_c = \frac{1}{2\pi R C} = \frac{1}{2\pi \times 1\text{ k}\Omega \times 1\,\mu\text{F}} \approx 159.15\text{ Hz}$$
### Circuit Operation
- **Input (#in):** A $1\text{ kHz}$ sinusoidal voltage source #Vin with $1\text{ V}$ peak amplitude. Since $1\text{ kHz} \gg 159\text{ Hz}$ (about $6.3\times$ higher), the capacitor impedance is significantly smaller than the resistor value ($X_C \approx 159\,\Omega \ll R_1 = 1\text{ k}\Omega$).
- **Integration Behavior:** Because nearly all the input voltage drops across resistor #R1, the current charging capacitor #C1 is proportional to the input voltage:
$$i(t) \approx \frac{v_{\text{in}}(t)}{R}$$
The capacitor voltage at node #out is therefore proportional to the time integral of the input:
$$v_{\text{out}}(t) = \frac{1}{C}\int i(t)\,dt \approx \frac{1}{RC}\int v_{\text{in}}(t)\,dt$$
- **Phase and Amplitude:** Integrating a sine wave yields a negative cosine wave, resulting in a phase lag of approximately $-90^\circ$ (specifically $\approx -81^\circ$). The peak output amplitude is attenuated to:
$$V_{\text{out, peak}} \approx \frac{V_{\text{in, peak}}}{\sqrt{1 + (2\pi f R C)^2}} \approx \frac{1\text{ V}}{\sqrt{1 + (6.28)^2}} \approx 157\text{ mV}$$
