= Solution
An <ideal operational amplifier> has infinite <input impedance> and <open-loop gain>, zero <output impedance> and no offsets or response limits. Its input currents vanish. Stable <operational-amplifier negative feedback>, a finite output and unsaturated linear operation then imply a <virtual short>, $v_+=v_-$. Real amplifiers have finite frequency-dependent gain, <input bias currents>, offset <voltage>, output loading, finite <bandwidth>, <slew rate> and supply-rail limits. Equal input <voltages> cannot be assumed in an open-loop or saturated circuit.
Here the noninverting input is grounded. In the ideal-feedback limit the inverting node is a <virtual ground>, so the <resistor> current charges the feedback <capacitor>:
$$
\frac{V_{\rm in}}R=C\frac{d(0-V_{\rm out})}{dt},\qquad
\boxed{V_{\rm out}(t)=V_{\rm out}(t_0)-\frac1{RC}\int_{t_0}^t V_{\rm in}(u)\,du}.
$$
The initial <capacitor> charge fixes the integration constant. This is an <op amp integrator>, with $H(i\omega)=-1/(i\omega RC)$.
A tiny direct-current input, <input bias current> or offset can drive the output towards saturation. A parallel feedback <resistor> $R_f$ provides a direct-current return path, giving $H(s)=-(R_f/R)/(1+sR_fC)$; choose $\omega R_fC\gg1$ over the intended integration band. A reset switch sets initial charge; low-offset components and suitable <signal> amplitude reduce drift and saturation. The useful band must also remain below the amplifier's <bandwidth> and slew-rate limits.
For the stipulated finite but constant real gain $A=100$, infinite <input impedance> and negligible <output impedance>, write the inverting-node <voltage> as $v=-V_{\rm out}/A$. With <phasors> proportional to $e^{i\omega t}$,
$$
\frac{V_{\rm in}-v}{R}=i\omega C(v-V_{\rm out}),\qquad
\boxed{H(i\omega)=\frac{V_{\rm out}}{V_{\rm in}}=-\frac{A}{1+i(1+A)\omega RC}}.
$$
Therefore
$$
\boxed{|H|=\frac{100}{\sqrt{1+(101\omega RC)^2}},\qquad
\arg H=\pi-\tan^{-1}(101\omega RC)\quad(\omega>0)}.
$$
The low-frequency gain is $-100$. The corner is $\omega_c=1/(101RC)$; above it the amplitude falls as $100/(101\omega RC)$ and the <phase> approaches $+\pi/2$. <phase> $+\pi$ and $-\pi$ describe the same low-frequency inversion. This finite-gain circuit is a low-pass response with an integrating high-frequency regime, not an ideal integrator down to zero <frequency>. A genuinely frequency-dependent $A(i\omega)$ would introduce additional high-frequency limitations beyond this specified model.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2010/paper-2-integrator-response.png]
{title=Finite-gain integrator response}
Back to article page
