= RC low-pass frequency response
{c}
{title2=$H(\omega)=1/(1+\mathrm i\omega RC)$}
Use the <complex amplitude> convention $v(t)=\operatorname{Re}(\widehat V e^{\mathrm i\omega t})$, where $\omega=2\pi f$ is the <angular frequency> and $f$ is the <frequency> in cycles per second. Substitution into the <RC low-pass filter> equation replaces the time <derivative> with multiplication by $\mathrm i\omega$, giving its <continuous-time frequency response>
$$
H(\omega)=\frac{\widehat V_{\rm out}}{\widehat V_{\rm in}}
=\frac1{1+\mathrm i\omega\tau},\qquad
|H|=\frac1{\sqrt{1+(\omega\tau)^2}},\qquad
\phi=\arg H=-\arctan(\omega\tau).
$$
Here $\phi$ is the output-minus-input <phase difference>; a negative value means the output lags. Equivalently, the <voltage divider> formed from the <electrical impedances> $R$ and $1/(\mathrm i\omega C)$ gives the same ratio. For an input $A\cos(\omega t)$, the sustained output is $A|H|\cos(\omega t+\phi)$ after the initial <exponential decay> has died away.
Define the <cutoff frequency> by a voltage <amplitude> ratio $1/\sqrt2$ relative to the low-<frequency> value:
$$
\omega_c=\frac1\tau,\qquad f_c=\frac1{2\pi\tau},\qquad
|H(\omega_c)|=\frac1{\sqrt2},\qquad \phi(\omega_c)=-\frac\pi4.
$$
On a <Bode plot>, the voltage gain in <decibels> is
$$
G_{\rm dB}=20\log_{10}|H|=-10\log_{10}\left(1+(f/f_c)^2\right).
$$
At the <cutoff frequency> it is $-3.0103\,\mathrm{dB}$, and far above that <frequency> it approaches a decrease of $20\,\mathrm{dB}$ per factor of ten in $f$. There is a smooth transition rather than an abrupt wall. The squared voltage ratio is one half at the corner; converting that statement to a <power> ratio requires the same real <electrical resistance> for the compared voltages. It does not mean that an unloaded <capacitor> dissipates half the source <power>.
For $R=2.2\,\mathrm{k}\Omega$ and $C=47\,\mathrm{nF}$, the <cutoff frequency> is $f_c\approx1.5392\,\mathrm{kHz}$. A $1\,\mathrm V$ peak input at $0.1f_c$, $f_c$, and $10f_c$ produces peak output <amplitudes> of approximately $0.9950\,\mathrm V$, $0.7071\,\mathrm V$, and $0.09950\,\mathrm V$, with <phase differences> $-5.711^\circ$, $-45^\circ$, and $-84.289^\circ$. Thus a wanted oscillation at $0.1f_c$ is almost preserved while an unwanted one at $10f_c$ is reduced by about a factor of ten in voltage. The <superposition principle> lets these two responses be added when both oscillations occur in the same <signal>.
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