Basic Electronics, Chapter 4: Challenge 2 solution (RC low-pass filter)

Formulas: fc = 1 / (2 pi R C)    gain = 1 / sqrt(1 + (f/fc)^2)    tau = R C
          time to a fraction f of final value: t = -tau ln(1 - f)

(a) Resistance for fc = 500 Hz with C = 47 nF
  R = 1 / (2 pi x fc x C) = 1 / (2 x 3.1416 x 500 x 47e-9)
    = 1 / (1.4765e-4) = 6,773 ohm
  Nearest E24 value: 6.8 k ohm
  Actual cut-off:  fc = 1 / (2 pi x 6800 x 47e-9) = 1 / (2.0081e-3) = 497.98 Hz  (about 498 Hz)

(b) Gain at 50 Hz, 500 Hz, 5 kHz (fc = 498 Hz)
  50 Hz:    f/fc = 0.1004   gain = 1 / sqrt(1 + 0.01008) = 1 / 1.00503 = 0.995
  500 Hz:   f/fc = 1.0040   gain = 1 / sqrt(1 + 1.00802) = 1 / 1.41705 = 0.706
            (at fc itself it is exactly 0.707, -3 dB; 500 Hz is just above fc)
  5 kHz:    f/fc = 10.04    gain = 1 / sqrt(1 + 100.8)    = 1 / 10.09  = 0.099  (about -20 dB)
  Check by reactance at 5 kHz: Xc = 1 / (2 pi x 5000 x 47e-9) = 677 ohm;
    gain = 677 / sqrt(6800^2 + 677^2) = 677 / 6833 = 0.099  (agrees)

(c) Step response to 5 V
  tau = R C = 6800 x 47e-9 = 319.6e-6 s = 320 us
  90 % point:  t = -tau x ln(1 - 0.9) = tau x 2.3026 = 319.6 us x 2.3026 = 736 us
  The 10 %-to-90 % rise time is tau x ln(9) = 2.197 x tau = 702 us.
  Rule of thumb check: 0.35 / fc = 0.35 / 498 = 703 us  (agrees with 702 us)
  Final value 5 V; at 1 tau the output is 3.16 V (at 320 us), at 3 tau (0.96 ms) 4.75 V.

Simulator check (Falstad)
  AC source 1 V amplitude, 6.8 k resistor, 47 nF capacitor to ground, scope the capacitor node:
  at 50 Hz the amplitude is about 0.995 V, at 500 Hz about 0.706 V (output lags the input by about 45 degrees),
  at 5 kHz about 0.099 V.

Checklist
  - fc is where reactance equals resistance, Xc = R.
  - Gain is 0.707 at fc and falls roughly 10 x per 10 x frequency above it.
  - Choose E24 value, then recalculate with the real value.