The upper unity inverter gives −Vᵢ; the bottom unity inverter gives −Vₒ. With τ=(1 kΩ)(100 pF), the integrator output is U=−(Vᵢ−Vₒ)/(sτ). The final summer gives Vₒ=(10 kΩ/R₂)Vᵢ−2U. Therefore H=[a sτ+2]/[sτ+2], a=10 kΩ/R₂. Its magnitude is constant when a=1, giving R₂=10 kΩ, B. In the printed ideal topology this reduces to the constant unity all-pass response.
More Operational Amplifiers questions
GATE 2022 BM Q57– The shown ideal op-amp has infinite gain and bandwidth and ±10 V supplies. Inputs 0.1 V…
GATE 2021 BM Q43– All op-amps in the shown circuit are ideal with infinite gain and bandwidth. Find…
GATE 2021 BM Q61– The shown op-amp has ±10 V supplies. Its non-inverting node connects to ground through 2R…
GATE 2020 BM Q38– For Vᵢₙ=10sin(2πt) applied to the circuit shown, which output waveform results? Both…
GATE 2026 BM Q48– For the ideal op-amp circuit shown, find input impedance looking into Vin.
GATE 2026 BM Q55– In the ideal instrumentation amplifier shown, one input-stage feedback resistor R* differs…
