Bode Plotter

Quick experiments

• Prove the −3 dB rule. On RC low-pass with R = 1 kΩ and C = 100 nF, the cutoff is ~1.59 kHz. The curve crosses the orange dashed line exactly there, and the phase reads −45°.
• Mirror LP vs HP. Same R and C — flip between low-pass and high-pass. Cutoff is identical; the pass-band is on the opposite side and the phase runs +90° → 0° instead of 0° → −90°.
• Decade-per-decade slope. Far from f_c, first-order filters roll off 20 dB per decade. Count gridlines: from 10× the cutoff to 100× the cutoff the magnitude drops 20 dB — the same thing as 6 dB per octave.
• Band-pass Q. Switch to RLC band-pass. Drop R → the peak narrows; raise R → the peak broadens. BW ≈ f₀ / Q is visible right in the plot.
• Trade R against C for the same corner. Halve C to 50 nF and double R to 2 kΩ. The RC product is unchanged, so the curve does not move — which is why designers pick R for impedance and let C follow.
• See why phase margin matters. Stack two identical stages. Phase now reaches −180° while there is still gain left — exactly the condition that turns a feedback amplifier into an oscillator.

Formula reference

Cutoff (corner) frequency of an RC filter

$$f_c = \frac{1}{2\pi R C}$$

R = 1 kΩ with C = 100 nF gives a cutoff near 1.59 kHz.

Magnitude in decibels

$$A_{dB} = 20 \log_{10}\!\left(\frac{V_{out}}{V_{in}}\right)$$

A ratio of 0.707 gives −3 dB, which is half the power.

First-order low-pass response

$$\left|H(f)\right| = \frac{1}{\sqrt{1 + (f/f_c)^2}}$$

At the cutoff frequency this equals 1/√2 = 0.707.

Phase shift of a first-order low-pass

$$\phi(f) = -\arctan\!\left(\frac{f}{f_c}\right)$$

0° well below cutoff, −45° at cutoff, approaching −90° well above.

Common mistakes

• ✗Reading the cutoff as the point where output stops.

  Cutoff is where output has fallen to 70.7 % of input, not to zero. A first-order filter still passes a substantial signal an octave past cutoff — it rolls off gradually, it does not switch off.

• ✗Forgetting to convert to radians per second.

  ω = 2πf. Using f directly where the formula expects ω puts the answer out by a factor of 6.28, which looks plausible enough to slip through unnoticed.

• ✗Plotting frequency on a linear axis.

  Bode plots use a logarithmic frequency axis. On a linear axis the decades are crushed together, the constant-slope roll-off curves, and the whole point of the plot is lost.

• ✗Assuming 6 dB/octave and 20 dB/decade are different slopes.

  They are the same slope in different units. A decade is a factor of 10 and an octave a factor of 2, so 20 dB per decade works out to about 6 dB per octave.

• ✗Ignoring source and load impedance.

  The formula assumes an ideal source and no loading. A source resistance adds to R and shifts the corner down; a load resistance across the output lowers the passband gain.

Frequently asked questions