Analog and electronic circuits

For a circuit with first-order low-pass behavior, which approximately relates its bandwidth BW to its rise time tr for a step input? (10% to 90% rise time)

  1. Atr ≈ 0.35 / BW
  2. Btr ≈ BW / 0.35
  3. Ctr ≈ 0.35 × BW
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Answer: A. tr ≈ 0.35 / BW

For a first-order low-pass circuit, the 10% to 90% rise time is about 2.2RC and the bandwidth is 1/(2πRC), so tr ≈ 2.2/(2π BW) ≈ 0.35/BW. The wider the bandwidth, the faster the rise. This is also used as a guideline when choosing oscilloscope bandwidth.

  • A. Correct. tr ≈ 0.35/BW.
  • B. Time and frequency are reciprocals, so BW should not be in the numerator.
  • C. A wider bandwidth gives a faster rise, so tr is not proportional to BW.
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