In the following active filter circuit with ideal op-amps, \(R\) is \(1\ \text{k}\Omega\), \(C\) is \(1\ \mu\text{F}\), and \(V_{in}\) has an amplitude of \(1\) V at \(1000\) rad/s. The amplitude of \(V_{out}\) is __________ V.
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Note that \(\omega\) is chosen so that \(\omega RC = 1\) exactly, which greatly simplifies the loop algebra. Treat each op-amp stage as an inverting amplifier or integrator with virtual ground at its inverting input, and remember the long resistor at the top closes a feedback loop from \(V_{out}\) back to the first stage.
A different route: build one overall transfer function H(s) = Vout/Vin first, then plug in the numbers. Setting up the stage gains.
Let $x = j\omega RC$. Stage 1 is a summing amplifier with two inputs, $V_{in}$ (through $R$) and the global feedback $V_{out}$ (through the top $R$), and a feedback impedance $Z_f=R/(1+x)$ combining $R$ and $C$ in parallel. For a summing inverting amplifier, each input current adds independently:
\[
V_1 = -\frac{Z_f}{R}\big(V_{in}+V_{out}\big) = -\frac{1}{1+x}\big(V_{in}+V_{out}\big)
\]
Stage 2 is a plain integrator.
Its gain is $-\dfrac{1}{j\omega RC}=-\dfrac{1}{x}$, so
\[
V_2 = -\frac{V_1}{x}
\]
Stage 3 is a unity inverter.
\[
V_{out} = -V_2 = \frac{V_1}{x}
\]
Closing the loop algebraically.
Put $V_1 = x V_{out}$ (from the last line) into the stage-1 equation:
\[
x V_{out} = -\frac{1}{1+x}\big(V_{in}+V_{out}\big)
\]
Multiply both sides by $(1+x)$:
\[
x(1+x) V_{out} = -V_{in} - V_{out}
\]
\[
V_{out}\big[x(1+x) + 1\big] = -V_{in}
\]
\[
\frac{V_{out}}{V_{in}} = \frac{-1}{x^2+x+1}
\]
This is the overall transfer function of the loop, valid for any $\omega$.
Plugging in the numbers.
$R=1000\ \Omega$, $C=10^{-6}$ F, $\omega=1000$ rad/s give $\omega RC = 1$, so $x = j$. Then
\[
x^2+x+1 = (j)^2+j+1 = -1+j+1 = j
\]
\[
\frac{V_{out}}{V_{in}} = \frac{-1}{j} = j
\]
using $-1/j = j$, since $1/j=-j$.
Reading off the magnitude.
$|V_{out}/V_{in}| = |j| = 1$. Since $|V_{in}| = 1$ V,
\[
|V_{out}| = 1 \times 1 = 1\ \text{V}
\]
This matches the node-by-node method exactly, confirming the answer independently: the magnitude is unity, only the phase shifts by 90 degrees. The distractor values 0.5 V, 0.707 V and 1.414 V all come from stopping the algebra early, using a single stage's gain instead of solving the full closed loop, so they do not satisfy the complete transfer function above.
\[ \boxed{|V_{out}| = 1.0\ \text{V}} \]
