= Solution
For an <ideal operational amplifier> under <negative feedback>, the grounded non-inverting terminal makes the inverting terminal a <virtual ground>, and no <electric current> enters either input. The feedback <resistor> and <capacitor> have parallel <electrical impedance>
$$
Z_f=\left(\frac1{R_2}+i\omega C\right)^{-1}=\frac{R_2}{1+i\omega R_2C}.
$$
<Kirchhoff's current law> at the inverting node gives $\widetilde V_i/R_1+\widetilde V_o/Z_f=0$, and hence
$$
\boxed{\frac{\widetilde V_o}{\widetilde V_i}=-\frac{R_2/R_1}{1+i\omega R_2C}.}
$$
With the stated components, $R_2C=16.0\,\mu\mathrm s$ and $R_2/R_1=5$. Thus this <inverting low-pass amplifier> has exactly the same ideal <cutoff frequency> and normalized <frequency response> as the passive <RC low-pass filter>, with output multiplied by $-5$:
$$
\boxed{V_o(t)=-\frac{5V_{i0}}{\sqrt{1+(\omega\tau)^2}}\cos\!\left(\omega t-\tan^{-1}(\omega\tau)\right).}
$$
Its <amplitude> ratios at the three stated <frequencies> are $4.99975$, $3.52616$, and $0.97560$. One principal choice of <phase shift> is $180^\circ-\tan^{-1}(\omega\tau)$: respectively $179.424^\circ$, $134.848^\circ$, and $101.252^\circ$. Adding or subtracting $360^\circ$ describes the same output; the extra inversion must be retained.
Back to article page
