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Natural Frequency (Electrical Analog)

The frequency at which a second-order LC system would oscillate with no damping present.

Control SystemsSecond-Order Systems

Natural Frequency Calculator ωn = 1 / √(LC)

ωn = 1 / √(L · C)
ωn = natural angular frequency (rad/s)  ·  L = inductance (H)  ·  C = capacitance (F)
⟹ Solve ωn, L, C
H
F
rad/s
Please fix the errors above.
Solve for:
Presets:
Angular Frequency
L: C: ωn: f:
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Angular Frequency (ωn)
Low (< 1e3 rad/s) Moderate (1e3–1e6) High (> 1e6)
ωn = 1 / √(LC)  ·  The natural frequency of an LC circuit determines its resonant behavior.

Interpretation

Natural frequency (electrical analogue) ω_n = 1/√(LC) is the frequency at which an RLC circuit would oscillate if there were no damping.
It is the resonant frequency of the circuit.
Example: L=0.1H, C=100µF → ω_n = 1/√(0.1 × 100e-6) = 1/√(1e-5) = 316.2 rad/s.

ω_n = 1 / √(LC)
Natural Frequency (Electrical Analog)

Variables

SymbolQuantityUnit
ω_nNatural (undamped) frequencyrad/s
LInductanceH
CCapacitanceF

What it means

The natural frequency ω_n of a series RLC circuit is the frequency at which it would oscillate if there were no damping. It is given by ω_n = 1/√(LC). This frequency is also the resonant frequency of the circuit. The natural frequency is a key parameter in second‑order systems, determining the speed of response. It is used to design filters and oscillators. Example: For L=0.1H and C=100µF, ω_n = 1/√(0.1 * 100e-6) = 1/√(1e-5) = 316.2 rad/s. The corresponding natural frequency in Hz is f_n = ω_n/(2π) ≈ 50.3Hz. This is the frequency at which the circuit would oscillate if undamped, and it is the centre frequency of the band‑pass response.

Worked example

Natural Frequency – Practical Example

Real‑World
Scenario: Using the same RLC (L=10 mH, C=1 µF), find the natural frequency.
ParameterValue
L0.01 H
C1×10⁻⁶ F
Formulaωn = 1 / √(LC)
1LC = 0.01 × 1e-6 = 1e-8
2ωn = 1 / √(1e-8) = 1 / 1e-4 = 10,000 rad/s
3fn = 10000/(2π) ≈ 1.59 kHz
Final Design fn ≈ 1.59 kHz ✓ Natural frequency
Why: Natural frequency is the frequency at which the circuit would oscillate if undamped – it depends on L and C.

Common mistakes

Watch unit consistency and the assumptions behind the formula; misapplying it outside its valid conditions is the most frequent error.

Applications

Natural frequency (electrical analogue) ω_n = 1/√(LC) is the undamped resonant frequency of a circuit. It determines the centre frequency of filters and the natural response of systems. Engineers use it to design tuned circuits, to select component values for desired frequency, and to analyse resonant behaviour. This is a key parameter for all resonant circuits.

  • Tuned amplifier and oscillator design
  • Filter centre frequency selection
  • Resonant converter design (LLC, series resonant)
  • Antenna and matching network design
  • Educational understanding of resonance

Frequently Asked Questions

Q01What is the natural frequency of an RLC circuit?
A01

The natural frequency is ω_n = 1 / √(LC). It is the frequency at which the circuit would oscillate if there were no damping.

Q02What is the difference between natural frequency and resonant frequency?
A02

For an undamped system, they are the same. For a damped system, the resonant frequency is slightly lower: ω_d = ω_n√(1−ζ²).

Q03How does ω_n affect the circuit response?
A03

Higher ω_n means faster response (shorter rise time and settling time), but it also affects the resonant peak.

Q04What are common mistakes when using ω_n?
A04

Common errors: 1) using the wrong formula for parallel RLC, 2) confusing with damped frequency, 3) applying to non‑RLC circuits, 4) forgetting the square root.

Q05What are practical applications?
A05

Designing filters, oscillators, and understanding transient response.

Q06What is the relationship between ω_n and the bandwidth?
A06

The bandwidth is approximately ω_n / Q for high Q circuits.

Q07How does the natural frequency relate to the pole locations?
A07

The poles of a second‑order system are at s = −ζω_n ± jω_n√(1−ζ²).

Q08What is the natural frequency of a circuit with L=10 mH and C=100 µF?
A08

ω_n = 1 / √(10e−3 × 100e−6) = 1 / √(1e−6) = 1000 rad/s.