Why does a slow clock cause high-frequency emissions?

Because the clock frequency sets where the harmonics fall, while the rise time sets how far up they persist. A trapezoidal waveform has an envelope with two corner frequencies. The first, at one over pi times the pulse width, is where the envelope starts falling at 20 dB per decade. The second, at one over pi times the rise time, is where it steepens to 40 dB per decade. A 10 MHz clock with 1 nanosecond edges has its second corner at 318 MHz, so it still has significant content in the FM band and above. The fundamental tells you almost nothing about the emission profile; the edge does.

What does slowing an edge actually buy?

Doubling the rise time moves the second corner down an octave, and asymptotically that is worth 6.02 dB at every frequency well above the new corner. It is worth nothing at all below it. Modelling a 100 MHz clock with 50 percent duty and comparing 1 nanosecond against 2 nanosecond edges gives exactly 0.00 dB of improvement at 100 MHz, 5.51 dB at 300 MHz and 6.02 dB at 1 GHz and above. So slowing edges is a targeted fix for a high-frequency problem and does nothing for a low-frequency one, which is worth knowing before spending timing margin on it.

What is the 0.35 over rise time rule?

An estimate of the bandwidth a signal occupies, from the relation between a first-order system's 10 to 90 percent rise time and its 3 dB bandwidth. For a 1 nanosecond edge it gives 350 MHz. It is a different quantity from the spectral corner at one over pi times the rise time, which is 318 MHz for the same edge, and the two are often confused. They agree closely enough that either works as a rough guide to how far up the spectrum a signal reaches, but the corner frequency is the one that belongs on an envelope plot.

Why do only odd harmonics appear on a square wave?

Because a waveform with exactly 50 percent duty cycle has a half-wave symmetry that cancels the even terms. In the envelope expression that shows up as the first cardinal sine landing on a null at every even harmonic. In practice duty cycle is never exactly 50 percent, so even harmonics appear at a reduced level rather than vanishing, and a scan showing strong even harmonics on a nominally square clock is telling you the duty cycle is off.

Does spread spectrum reduce total emitted energy?

No. It redistributes the same energy over a wider band, so the peak measured in any one resolution bandwidth falls while the total is unchanged. That is genuinely useful against a limit expressed as a level in a defined bandwidth, which is how emission limits are written, and it is why the technique works. It is not a reduction in interference in every sense: a victim integrating over a wide band sees no benefit, and the technique can turn a narrow overshoot into a broad elevation that is harder to diagnose.

Time and frequency domain: why edges set emissions

Author Hugues Orgitello Last updated 30 August 2026 ~5 min read

Guide, EMC fundamentals

A designer works in the time domain and a test house works in the frequency domain, and the translation between them is where most EMC surprises live. The most useful consequence is counter-intuitive: the clock frequency barely matters to an emission profile, while the rise time almost entirely determines it. This page derives why, and quantifies what slowing an edge is actually worth.

The trapezoid, and its two corners

Real digital signals are trapezoids: a rise, a flat top, a fall, a flat bottom. Their Fourier coefficients are

|cn| = 2A (tau/T) x |sin(pi n tau/T)/(pi n tau/T)| x |sin(pi n tr/T)/(pi n tr/T)|

with A the amplitude, T the period, tau the pulse width and tr the rise time. The two cardinal sine terms give the envelope two corner frequencies:

Corner	Set by	Behaviour above it
f1 = 1 / (pi tau)	Pulse width	Envelope falls at 20 dB/decade
f2 = 1 / (pi tr)	Rise time	Envelope falls at 40 dB/decade

Below f1 the envelope is flat. Between f1 and f2 it falls at 20 dB per decade. Above f2 it falls at 40.

The rise time sets the second corner, so the rise time decides how far up the spectrum a signal still has energy. The clock frequency only decides where the harmonic lines sit within that envelope.

What that means in numbers

Rise time	f2 = 1/(pi tr)	0.35/tr bandwidth
10 ns	31.8 MHz	35 MHz
5 ns	63.7 MHz	70 MHz
2 ns	159 MHz	175 MHz
1 ns	318 MHz	350 MHz
0.5 ns	637 MHz	700 MHz
0.2 ns	1592 MHz	1750 MHz

A 10 MHz clock with 1 ns edges reaches its second corner at 318 MHz. Slowing the clock to 5 MHz changes nothing about that; only the edge does.

The two columns are different quantities that are often confused. The corner frequency is where the envelope changes slope. The 0.35 over rise time figure estimates the bandwidth a signal occupies, from the relation between a first-order system's 10 to 90 percent rise time and its 3 dB bandwidth. They are close enough to serve as the same rough guide, but only the corner belongs on an envelope plot.

Worked example

A 100 MHz clock, 50 percent duty, 1 ns edges:

period 10 ns, pulse width 5 ns, rise time 1 ns
f1 = 63.7 MHz, f2 = 318 MHz

So between 64 and 318 MHz the envelope falls at only 20 dB per decade, and the 40 dB per decade roll-off does not begin until 318 MHz. That is why this clock is a nuisance across the whole 30 MHz to 1 GHz scan rather than only near its fundamental.

What slowing the edge is worth

Doubling the rise time halves f2, so one octave that used to fall at 20 dB per decade now falls at 40. Modelling the same 100 MHz clock with 1 ns and 2 ns edges:

Frequency	1 ns edge	2 ns edge	Gain
100 MHz	-3.92 dB	-3.92 dB	0.00 dB
300 MHz	-13.46 dB	-18.97 dB	5.51 dB
1 GHz	-33.87 dB	-39.89 dB	6.02 dB
3 GHz	-52.95 dB	-58.97 dB	6.02 dB

Levels are relative to the flat part of the envelope.

Two conclusions, and the second is the one usually missed. Doubling the rise time is worth 6.02 dB, asymptotically, which is 20 log10(2) and the same factor of two in voltage that appears throughout decibels for EMC. And it is worth exactly nothing below the corner: at 100 MHz the improvement is 0.00 dB.

Slowing edges is therefore a targeted fix for a high-frequency overshoot, not a general-purpose quietening measure. Spending timing margin on it to solve a 50 MHz problem buys nothing.

Odd and even harmonics

A waveform with exactly 50 percent duty has half-wave symmetry, which cancels the even harmonics: the first cardinal sine term lands on a null at every even multiple. Real duty cycles are never exactly 50 percent, so even harmonics appear at reduced level rather than vanishing.

That is a useful diagnostic. Strong even harmonics on a nominally square clock indicate the duty cycle is off, which is often easier to fix than the emission itself.

Detectors: the other translation

The receiver performs its own time-to-frequency translation, and which detector it uses changes the number.

Peak responds to the highest instantaneous value and is fastest, which is why pre-scans use it: a peak result that passes guarantees the others will.

Quasi-peak applies charge and discharge time constants defined in CISPR 16-1-1, weighting by repetition rate so that infrequent impulses read lower. It approximates the annoyance of interference to a listener, and formal limits below 1 GHz are mostly written against it.

Average reads lower still for pulsed signals and is used alongside quasi-peak in some limit pairs.

A broadband impulsive source can read many decibels lower in quasi-peak than in peak, while a continuous carrier reads almost the same in both. The gap between the two detectors is itself information about whether a source is continuous or impulsive.

Spread spectrum, honestly

Spread-spectrum clocking dithers the clock so its energy is smeared across a band instead of concentrated in a line. The peak in any one resolution bandwidth falls; the total emitted energy does not change.

Against a limit expressed as a level in a defined bandwidth, which is how emission limits are written, that is a real and legitimate improvement. But it is not a reduction in interference in every sense: a victim integrating across a wide band sees no benefit, and the technique can convert a narrow overshoot into a broad elevation that is harder to diagnose later.

Key takeaways

Rise time sets the spectrum, clock frequency sets the lines. f2 = 1/(pi tr) is where roll-off steepens to 40 dB/decade.
1 ns edges reach 318 MHz regardless of whether the clock is 10 MHz or 100 MHz.
Doubling rise time buys 6.02 dB above the corner and 0.00 dB below it.
Strong even harmonics mean the duty cycle is not 50 percent.
Peak, quasi-peak and average are three different translations. The gap between them tells you whether a source is impulsive.
Spread spectrum redistributes energy, it does not remove it.

Frequently asked questions

Why does a slow clock cause high-frequency emissions?: Because the clock frequency sets where the harmonics fall, while the rise time sets how far up they persist. A trapezoidal waveform has an envelope with two corner frequencies. The first, at one over pi times the pulse width, is where the envelope starts falling at 20 dB per decade. The second, at one over pi times the rise time, is where it steepens to 40 dB per decade. A 10 MHz clock with 1 nanosecond edges has its second corner at 318 MHz, so it still has significant content in the FM band and above. The fundamental tells you almost nothing about the emission profile; the edge does.
What does slowing an edge actually buy?: Doubling the rise time moves the second corner down an octave, and asymptotically that is worth 6.02 dB at every frequency well above the new corner. It is worth nothing at all below it. Modelling a 100 MHz clock with 50 percent duty and comparing 1 nanosecond against 2 nanosecond edges gives exactly 0.00 dB of improvement at 100 MHz, 5.51 dB at 300 MHz and 6.02 dB at 1 GHz and above. So slowing edges is a targeted fix for a high-frequency problem and does nothing for a low-frequency one, which is worth knowing before spending timing margin on it.
What is the 0.35 over rise time rule?: An estimate of the bandwidth a signal occupies, from the relation between a first-order system's 10 to 90 percent rise time and its 3 dB bandwidth. For a 1 nanosecond edge it gives 350 MHz. It is a different quantity from the spectral corner at one over pi times the rise time, which is 318 MHz for the same edge, and the two are often confused. They agree closely enough that either works as a rough guide to how far up the spectrum a signal reaches, but the corner frequency is the one that belongs on an envelope plot.
Why do only odd harmonics appear on a square wave?: Because a waveform with exactly 50 percent duty cycle has a half-wave symmetry that cancels the even terms. In the envelope expression that shows up as the first cardinal sine landing on a null at every even harmonic. In practice duty cycle is never exactly 50 percent, so even harmonics appear at a reduced level rather than vanishing, and a scan showing strong even harmonics on a nominally square clock is telling you the duty cycle is off.
Does spread spectrum reduce total emitted energy?: No. It redistributes the same energy over a wider band, so the peak measured in any one resolution bandwidth falls while the total is unchanged. That is genuinely useful against a limit expressed as a level in a defined bandwidth, which is how emission limits are written, and it is why the technique works. It is not a reduction in interference in every sense: a victim integrating over a wide band sees no benefit, and the technique can turn a narrow overshoot into a broad elevation that is harder to diagnose.