Question

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.

• A. 0.5
• B. 0.707
• C. 1.0
• D. 1.414

Hint:

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.

Answer 1

The Correct Option is C

Step 1: Identify the circuit and mark the virtual-ground nodes.
The circuit has three ideal op-amp stages in cascade. Because each op-amp works with negative feedback and its non-inverting input is grounded, the inverting input of every stage sits at virtual ground (0 V). Call the inverting-input node of the first op-amp A, of the second op-amp B, and of the third op-amp C. Let \(V_1\) be the output of the first op-amp, \(V_2\) the output of the second op-amp, and \(V_3 = V_{out}\) the output of the third (and last) op-amp.

Step 2: Write the current equation at node A.
Node A has three branches meeting at it: the input resistor \(R\) from \(V_{in}\), the parallel \(R \| C\) feedback branch to \(V_1\) (the first op-amp's own output), and a long resistor \(R\) that runs across the top of the diagram straight from \(V_{out}\) back to this same node, an overall feedback path around the whole three-stage loop. Since no current enters an ideal op-amp input, the branch currents into node A must add to zero:
\[ \frac{V_{in}-0}{R} + \frac{V_1-0}{R\|(1/j\omega C)} + \frac{V_3-0}{R} = 0 \]
The parallel combination of \(R\) and \(C\) has impedance \(Z_f = \dfrac{R}{1+j\omega RC}\), so \(1/Z_f = (1+j\omega RC)/R\). Substituting and multiplying through by \(R\) gives
\[ V_{in} + V_1(1+j\omega RC) + V_3 = 0 \]
so
\[ V_1 = -\frac{V_{in}+V_3}{1+j\omega RC} \quad \text{...(1)} \]

Step 3: Write the current equation at node B (pure integrator stage).
The second stage has only an input resistor \(R\) (from \(V_1\)) and a feedback capacitor \(C\) (to \(V_2\)), with no parallel resistor, so it is a pure inverting integrator:
\[ \frac{V_1-0}{R} + \frac{V_2-0}{1/(j\omega C)} = 0 \]
\[ \frac{V_1}{R} + V_2\, j\omega C = 0 \ \Rightarrow\ V_2 = -\frac{V_1}{j\omega RC} \quad \text{...(2)} \]

Step 4: Write the current equation at node C (unity-gain inverter stage).
The third stage has an input resistor \(R\) (from \(V_2\)) and a feedback resistor \(R\) (to \(V_3=V_{out}\)), both of the same value, so it is a plain inverting amplifier of gain \(-1\):
\[ \frac{V_2}{R} + \frac{V_3}{R} = 0 \ \Rightarrow\ V_3 = -V_2 \quad \text{...(3)} \]

Step 5: Combine the three equations.
From (3), \(V_2 = -V_3\). Put this in (2):
\[ -V_3 = -\frac{V_1}{j\omega RC} \ \Rightarrow\ V_3 = \frac{V_1}{j\omega RC} \quad \text{...(4)} \]
Substitute (4) into (1) to eliminate \(V_3\):
\[ V_1 = -\frac{V_{in} + V_1/(j\omega RC)}{1+j\omega RC} \]
\[ V_1(1+j\omega RC) + \frac{V_1}{j\omega RC} = -V_{in} \]
\[ V_1\left[(1+j\omega RC) + \frac{1}{j\omega RC}\right] = -V_{in} \quad \text{...(5)} \]

Step 6: Substitute the numbers.
\(R = 1\ \text{k}\Omega = 1000\ \Omega\), \(C = 1\ \mu\text{F} = 10^{-6}\) F, \(\omega = 1000\) rad/s, so
\[ \omega RC = 1000 \times 1000 \times 10^{-6} = 1 \]
This is the key simplification the numbers were chosen to give. With \(\omega RC = 1\), equation (5) becomes
\[ V_1\big[(1+j) + \tfrac{1}{j}\big] = -V_{in} \]
Since \(1/j = -j\), the bracket is \((1+j) + (-j) = 1\), so
\[ V_1 = -V_{in} \]
Because \(V_{in}\) has amplitude 1 V, \(|V_1| = 1\) V.

Step 7: Find Vout from V1.
From (4), with \(j\omega RC = j\):
\[ V_3 = \frac{V_1}{j} = -j\,V_1 = -j(-V_{in}) = j\,V_{in} \]
Multiplying by \(j\) only rotates the phasor by 90 degrees, it does not change its magnitude, so
\[ |V_{out}| = |V_3| = |V_{in}| = 1\ \text{V} \]

Step 8: Why the other options are wrong.
0.707 V (option B) is what you get if you treat the RC combination as a simple first-order low-pass and use \(1/\sqrt{1+(\omega RC)^2} = 1/\sqrt{2}\) without carrying the extra feedback and integrator stages through the loop. 1.414 V (option D) is the reciprocal mistake, using \(\sqrt{1+(\omega RC)^2}=\sqrt2\) as a gain instead of properly closing the three-stage loop. 0.5 V (option A) would come from assuming a simple two-stage attenuation of \(1/2\) without accounting for the 90 degree phase rotation that leaves the magnitude unchanged. Solving the full loop equation, as above, is required, and it gives exactly unity magnitude.

Final Answer:
The amplitude of \(V_{out}\) is 1.0 V. \[ \boxed{|V_{out}| = 1.0\ \text{V}} \]