Clock Frequency & Period Calculator
The simplest relationship in timing: f = 1 / T. Solve for frequency or period, then read the duty cycle, high time, Nyquist limit and free-space wavelength.
Inputs
Choose what to solve for, then fill in the other box. Scientific notation such as 1e9 or 2.5e-9 is accepted. Units convert automatically.
Results
Enter a value and press Calculate. Your answer will appear here.
Clock waveform
Three periods of the clock drawn to scale. T is the full period, tH the high time and tL the low time. At 50 % duty the high and low times are equal.
Frequency vs. period (log–log)
On log–log axes f = 1/T is a straight line of slope −1. One decade more frequency is exactly one decade less period. The dashed line marks the Nyquist limit, f / 2.
What is being calculated
Common clock frequencies
| Source | Frequency | Period | Half-period |
|---|
Computed by the same engine at 50 % duty. Illustrative values, not manufacturer data.
Frequency unit ladder
| Prefix | Value | Period when f = 1 of that unit |
|---|---|---|
| 1 Hz | 1 × 10⁰ Hz | 1 s |
| 1 kHz | 1 × 10³ Hz | 1 ms |
| 1 MHz | 1 × 10⁶ Hz | 1 µs |
| 1 GHz | 1 × 10⁹ Hz | 1 ns |
| 1 THz | 1 × 10¹² Hz | 1 ps |
The prefix steps by 1000 while the period steps by 1/1000 — that is the f = 1/T relationship read straight off the SI ladder.
Key relations and unit conversions
| Quantity | Relation |
|---|---|
| Frequency and period | f = 1 / T · T = 1 / f · f · T = 1 |
| Angular frequency | ω = 2πf = 2π / T (rad/s) |
| Duty cycle | D = t_H / T · t_H = D·T · t_L = (1 − D)·T |
| Time for N cycles | t = N · T = N / f |
| Cycles in a time t | N = f · t |
| Nyquist limit | fmax = fsample / 2 |
| Wavelength (free space) | λ = c / f · c = 299 792 458 m/s |
| Frequency units | 1 Hz = 10⁻³ kHz = 10⁻⁶ MHz = 10⁻⁹ GHz = 10⁻¹² THz |
| Period units | 1 s = 10³ ms = 10⁶ µs = 10⁹ ns = 10¹² ps |
| Period ↔ frequency pairs | 1 Hz ↔ 1 s · 1 kHz ↔ 1 ms · 1 MHz ↔ 1 µs · 1 GHz ↔ 1 ns |
| RPM to hertz | 1 rpm = 1/60 Hz ≈ 16.667 mHz |
| Jitter / ppm | 1 ppm of T = T × 10⁻⁶ s of period error |
Formula
- f: frequency, cycles per second (Hz)
- T: period, seconds per cycle (s)
- f · T = 1 for any periodic signal
Example: a 100 MHz clock has T = 1 / (100 × 10⁶ Hz) = 10 × 10⁻⁹ s = 10 ns. One cycle lasts ten nanoseconds; a 50 % duty square wave is high for 5 ns and low for 5 ns.
The relationship is exact and unit-free in the sense that any frequency unit paired with its reciprocal time unit works out — 1 kHz is 1 ms, 1 MHz is 1 µs, 1 GHz is 1 ns. That ladder is why the prefixes kHz, MHz and GHz line up so neatly with ms, µs and ns.
Real clocks are not perfect square waves. The edges take a finite time to rise and fall, the duty cycle is rarely exactly 50 %, and the period wanders slightly from cycle to cycle — jitter — or drifts slowly with temperature — wander. A crystal specified at ±20 ppm, for instance, has a period that can be off by 20 parts per million, which is 200 ps on a 10 ns period. When a clock drives a data bus, the period sets the setup and hold window, so the achievable data rate is limited by the period minus the edge and skew budgets.
Advanced mode adds the duty cycle so the high and low times are reported separately, an optional cycle count so the total time to count N cycles is shown, plus the angular frequency ω = 2πf, the free-space wavelength λ = c/f, the Nyquist limit f/2 and the number of cycles per day.
References
- Wikipedia, Frequency: definition and the f = 1/T relation
- Wikipedia, Clock signal: duty cycle, rise time and clock distribution
- Wikipedia, Duty cycle: high time and low time
- Wikipedia, Nyquist frequency: the f/2 sampling limit
- NIST SP 811, Guide for the Use of the SI: unit conversion factors
How to Use This Tool
- Choose whether to solve for frequency f or period T.
- Pick a preset, or type the known value and choose its unit.
- In Advanced, set the duty cycle and an optional cycle count, or change the significant figures.
- Click Calculate.
- Read the highlighted answer, then check the waveform and the log–log chart.
What This Tool Offers
- Solves f = 1 / T for either frequency or period
- Automatic unit conversion from hertz to terahertz and seconds to picoseconds
- Duty cycle with high time and low time (Advanced)
- Angular frequency, free-space wavelength and Nyquist limit
- Total time for an optional number of cycles
- Scaled clock waveform and a log–log chart with slope −1
- Ready-made presets and a mobile-friendly layout
Variables
| Symbol | Quantity | Unit |
|---|---|---|
| f | Frequency | Hz |
| T | Time period | s |
What it means
The clock frequency f (in Hz) and period T (in seconds) are reciprocals: f = 1/T. The period is the time between successive rising (or falling) edges of the clock signal. This relationship is fundamental in digital design, as it determines how often operations can occur. A higher frequency means more operations per second but also tighter timing constraints. The period must be long enough to accommodate the propagation delays of the logic path. Example: A clock with a period of 10ns corresponds to a frequency of 1/(10×10⁻⁹) = 100 MHz. In a processor, the clock speed is often quoted in GHz; a 3GHz processor has a period of 0.333 ns.
Worked example
Clock Frequency–Period – Practical Example
Real‑World| Parameter | Value |
|---|---|
| f | 2.5 GHz = 2.5×10⁹ Hz |
| Formula | T = 1 / f |
Common mistakes
Watch unit consistency and the assumptions behind the formula; misapplying it outside its valid conditions is the most frequent error.Applications
Clock frequency and period are reciprocals: f = 1/T. This simple relation is used to determine the speed of digital systems. Engineers use it to design clock generators, to set operating frequencies of microcontrollers and processors, and to meet performance specifications. A higher frequency enables faster processing but may increase power consumption and signal integrity challenges. This formula is essential for timing design in digital electronics.
- Clock generation and distribution design
- Microprocessor and FPGA speed specification
- Timing budget and system performance analysis
- Power‑performance trade‑offs in digital circuits
- Educational understanding of clock signals
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