For a band-pass filter, the center (resonant) frequency f0 is always equal to which of the following, in terms of the lower and upper −3 dB cutoff frequencies f1 and f2?

Introduction:In narrow-band band-pass design and analysis, relating the center frequency to the cutoff points is essential for tuning, measuring, and specifying filters. The correct relationship uses a geometric mean, not an arithmetic mean, reflecting how frequency responses scale multiplicatively.

Given Data / Assumptions:

• Lower cutoff (−3 dB) frequency: f1.
• Upper cutoff (−3 dB) frequency: f2.
• Narrow-band assumption generally applies, but the identity holds broadly for standard definitions.

Concept / Approach:The center (resonant) frequency of a band-pass filter satisfies f0 = sqrt(f1 * f2). This arises from the symmetry of logarithmic frequency scaling and is consistent with constant-Q filter behavior where the passband is centered on a logarithmic axis.

Step-by-Step Solution:Define bandwidth: BW = f2 − f1 and quality factor Q = f0 / BW.For many common band-pass implementations, the midband gain occurs near f0 where the reactive components balance.Using filter theory, the −3 dB points satisfy f0^2 = f1 * f2 ⇒ f0 = sqrt(f1 * f2).Therefore, the correct relationship is the geometric mean of the cutoff frequencies.

Verification / Alternative check:On a Bode magnitude plot versus log frequency, f0 lies midway between f1 and f2 on the log axis, which mathematically corresponds to the geometric mean. Measured data from RLC band-pass circuits confirms the identity within approximation errors for real components.

Why Other Options Are Wrong:

• Bandwidth or bandwidth/Q: Related parameters, but not equal to f0.
• −3 dB frequency: There are two such frequencies (f1 and f2); f0 is not itself a −3 dB point.
• Arithmetic average: Only correct for perfectly symmetric responses on a linear frequency axis, which standard band-pass filters are not.

Common Pitfalls:

• Using arithmetic mean, which misplaces f0 for wide or even moderate bandwidths.
• Confusing midband gain frequency with one of the cutoff points.

Final Answer:geometric average of the critical frequencies