Formula
A first-order RC low-pass and a first-order RC high-pass share the cutoff 1 / (2πRC). At that frequency the magnitude is 1/√2. The low-pass magnitude is 1 / √(1 + (f/fc)²). The high-pass magnitude is (f/fc) over that same square root. Decibels are 20 log10 of the magnitude.
fc = 1 / (2π R C)
f0 = 1 / (2π √(L C))
Q = (1/R) √(L/C)
The resonant frequency calculator on this page is the LC mode: f0 = 1 / (2π √(LC)). Series RLC band-pass uses that f0, a bandwidth R / (2πL), and Q = f0 / bandwidth. The plotted band-pass magnitude is the voltage across the resistor, 1 / √(1 + Q² (f/f0 − f0/f)²). The LC curve is series reactance XL − XC, in ohms, not a filter gain. The time constant of the same R and C is RC, and the impedance calculator evaluates XL and XC at one frequency.
Worked examples
1 kΩ and 1 µF, low-pass. Cutoff is 159.155 Hz. Magnitude at the cutoff is 0.7071, −3.010 dB.
The same parts, high-pass. The cutoff frequency does not move. The curve rises instead of falling.
10 µH and 100 pF. Resonant frequency is 5.032921 MHz. With 50 Ω in series, Q is 6.325 and the 3 dB bandwidth is 795.8 kHz.
Reference
These are ideal lumped parts. A real inductor’s self-resonance, and a capacitor’s series resistance, move the peak.
Questions
What is the cutoff of 1 kΩ and 1 µF?
159.15 Hz, which is 1 / (2πRC). At that frequency a first-order low-pass or high-pass is at 1/√2, about −3.01 dB.
Where is the resonant frequency?
On this page, in the LC mode. 10 µH and 100 pF resonate at 5.033 MHz. The same LC with a series R is the band-pass mode.
Is the plot the only result?
No. The table under the plot lists frequency and magnitude. The curve repeats that table.
Sources
- First-order RC cutoff fc = 1 / (2πRC), and the magnitude responses above.
- LC resonance f0 = 1 / (2π √(LC)). Series RLC Q = (1/R) √(L/C), bandwidth R / (2πL).