First-Order Bode Phase
φ = −arctan(f/fc)
Quick answer
This calculator solves φ = −arctan(f / fc) for Phase (degrees). With Signal Frequency 1 kHz, Cutoff Frequency 2.2 kHz, the result is -24.44. It solves in 3 directions, so you can enter the values you have and get the one you need.
Calculator
Formula
Description
A first-order low-pass filter introduces a phase shift that varies from 0° at DC to −90° at high frequencies, passing through −45° exactly at the cutoff frequency. The phase response is a smooth arctan curve on a logarithmic frequency scale. Phase shift is critical in feedback control systems because excessive phase lag at the unity gain frequency causes oscillation. In audio systems, phase differences between drivers in a crossover network affect the acoustic summing and can create lobes or nulls in the radiation pattern.
Variables
- φ — Phase shift in degrees (negative for low-pass)
- f — Signal frequency (Hz)
- fc — Filter cutoff (−3 dB) frequency (Hz)
Practical Notes
At f = 0.1×fc, phase ≈ −5.7°. At f = fc, phase = −45°. At f = 10×fc, phase ≈ −84.3°. A second-order filter reaches −180° at high frequencies (−90° at fc). Phase margin in feedback systems requires the total loop phase to be less than −180° at the gain crossover frequency. A first-order filter contributes at most −90° of phase, so a single-pole system is inherently stable.
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| phi | Phase (degrees) | — | — |
| f | Signal Frequency | Hz | Hz, kHz, MHz, GHz |
| fc | Cutoff Frequency | Hz | Hz, kHz, MHz, GHz |
Step by step
- 1. What you know: Signal Frequency 1 kHz, Cutoff Frequency 2.2 kHz
- 2. The relation: φ = −arctan(f / fc)
- 3. Substitute: f = 1 kHz, fc = 2.2 kHz
- 4. Result: Phase (degrees) = -24.44
Worked examples
Minimum range
Given: Signal Frequency 5 Hz, Cutoff Frequency 11 Hz
Result: Phase (degrees) = -24.44
Low range
Given: Signal Frequency 50 Hz, Cutoff Frequency 110 Hz
Result: Phase (degrees) = -24.44
Typical
Given: Signal Frequency 1 kHz, Cutoff Frequency 2.2 kHz
Result: Phase (degrees) = -24.44
High range
Given: Signal Frequency 20 kHz, Cutoff Frequency 44 kHz
Result: Phase (degrees) = -24.44
Maximum range
Given: Signal Frequency 0.2 MHz, Cutoff Frequency 0.44 MHz
Result: Phase (degrees) = -24.44
Solving for each variable
Solve for Phase (degrees)
Signal Frequency 5 Hz, Cutoff Frequency 11 Hz → -24.44
Solve for Signal Frequency
Phase (degrees) 25000, Cutoff Frequency 11 MHz → 4.004 MHz
Solve for Cutoff Frequency
Phase (degrees) 25000, Signal Frequency 5 MHz → 13.74 MHz
How Phase (degrees) changes with Signal Frequency
| Signal Frequency (input) | Phase (degrees) (result) |
|---|---|
| 100 Hz | -2.603 |
| 0.25 kHz | -6.483 |
| 0.5 kHz | -12.8 |
| 0.75 kHz | -18.82 |
| 1 kHz | -24.44 |
| 1.5 kHz | -34.29 |
| 2 kHz | -42.27 |
| 4 kHz | -61.19 |
| 7 kHz | -72.55 |
| 10 kHz | -77.59 |
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<p><a href="https://vbatpower.com/formulas/bode-phase-1st-order/">First-Order Bode Phase</a> Calculator by vbatpower.com</p>Frequently asked questions
How do you calculate First-Order Bode Phase?
Use φ = −arctan(f / fc). With Signal Frequency 1 kHz, Cutoff Frequency 2.2 kHz, Phase (degrees) works out to -24.44.
What is the formula for First-Order Bode Phase?
The formula is φ = −arctan(f / fc). A first-order low-pass filter introduces a phase shift that varies from 0° at DC to −90° at high frequencies, passing through −45° exactly at the cutoff frequency.
How do you find Signal Frequency when you know the other values?
Rearrange for Signal Frequency: with Phase (degrees) 25000, Cutoff Frequency 11 MHz, Signal Frequency is 4.004 MHz.
How do you find Cutoff Frequency when you know the other values?
Rearrange for Cutoff Frequency: with Phase (degrees) 25000, Signal Frequency 5 MHz, Cutoff Frequency is 13.74 MHz.
Related Concepts
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