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Circuit Analysis: AC, Linear Circuits and Transients

A sine wave of angular frequency ω = 10/(RC) is applied to the RC low-pass filter in figure (a). Which of the following is closest to the ratio of the output amplitude to the input amplitude (the gain)?

RC low-pass filter and RC high-pass filter (a) vin vout R C (b) vin vout C R H(s) = 1/(1 + sRC) H(s) = sRC/(1 + sRC)
RC low-pass filter and RC high-pass filter · View in diagrams
  1. A−20 dB
  2. B−10 dB
  3. C−3 dB
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Answer: A. −20 dB

Method 1 (calculate with complex numbers): the transfer function is H(jω) = 1/(1 + jωRC), and for ωRC = 10, |H| = 1/√(1 + 10²) = 1/√101 ≈ 0.0995. In decibels this is 20 log10 0.0995 ≈ −20.0 dB, and the phase is −tan⁻¹ 10 ≈ −84.3°. Method 2 (straight-line Bode approximation): above the cutoff angular frequency ωc = 1/(RC) the gain falls at −20 dB/dec. ω = 10ωc is exactly one decade above the cutoff, so the gain reads about −20 dB. Near the cutoff the straight-line approximation is off by up to 3 dB, but the farther away, the better it agrees.

  • A. Correct. |H| ≈ 0.0995, about −20 dB.
  • B. This finds the amplitude ratio of about 1/10 but converts the voltage ratio to dB with 10 log10. A voltage ratio is expressed with 20 log10.
  • C. −3 dB is the value at the cutoff angular frequency ω = 1/(RC).
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