Final answer:
The frequency at which the gain of an op-amp-based inverting integrator is reduced to -1V/V can be calculated using the gain formula as 1000 kHz. The time constant can be determined from the gain at a known frequency using the impedance of the capacitor.
Step-by-step explanation:
When we look at an op-amp-based inverting integrator, we know that it functions as a frequency-dependent amplifier where the gain decreases as the frequency increases. The gain at a specific frequency for an integrator can be calculated using the formula A(-j\omega) = -1/(j\omega RC), where \(A(-j\omega)\) is the gain at angular frequency \(\omega\), \(R\) is the resistor value, and \(C\) is the capacitor value. The gain in decibels is given by \(20 \log |A(-j\omega)|.
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The integrator's time constant (\(\tau\)) is equivalent to the product of the resistance (\(R\)) and the capacitance (\(C\)). We can find the time constant by analyzing the given gain at 10 kHz (10,000 Hz) and when the gain is -1V/V. Given a gain of -10\(^4\)V/V at 10 kHz, if the gain is -1V/V, the frequency at which this occurs can be found by solving for \(\omega\) in the gain formula, equating the magnitude of the gain with 1. Based on the integrator's property of having a gain inversely proportional to frequency, we can see that the gain would be -1V/V when the frequency is 10^6 Hz (or 1000 kHz).
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The time constant of the integrator can be determined by recognizing that the gain at 10 kHz is the result of the time constant determining the capacitor's impedance at this frequency. Since the gain at 10 kHz is given as -10\(^4\)V/V, the time constant \(\tau = RC\) can be calculated by finding the impedance of the capacitor and equating the gain formula \(A = -1/(\omega C)\) to -10\(^4\). The frequency at which the gain is -1V/V gives us another point of reference to determine the time constant more precisely.