AoNArt of Nanoelectronics
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05 · Intermediate

Bandwidth, rise time & reproducible waveform measurement

Close the measurement chain by treating oscilloscope/probe bandwidth as part of the transfer function. Derive the exact 10–90% rise-time relation of a first-order low-pass, show why the familiar BW·tr≈0.35 constant is model-dependent rather than universal, quantify amplitude error versus bandwidth, combine independent measurement-chain rise times with stated approximations, and finish with a reproducible waveform-measurement record that separates analog bandwidth, probing, acquisition, triggering and numerical extraction.

160–205 minIntermediate
In this lesson
  • Define oscilloscope bandwidth as a property of the analog measurement transfer function rather than sample rate.
  • Derive the first-order low-pass step response and the exact 10–90% relation tr=τ ln 9.
  • Combine f3dB=1/(2πτ) with tr=τ ln 9 to derive f3dB tr=ln 9/(2π)=0.3497 for a single-pole model.
  • Explain why real oscilloscopes can use different bandwidth–rise-time constants depending on frequency-response shape.
  • Derive sinusoidal amplitude error at a finite bandwidth and calculate the bandwidth ratio needed for a target amplitude-error limit in a one-pole model.
  • Treat probe, oscilloscope and DUT as a cascaded measurement system and state when root-sum-square rise-time approximations are valid.
  • Estimate DUT rise time from measured and measurement-system rise times only with an explicit response-shape approximation.
  • Distinguish analog bandwidth, sample rate, record length, trigger and numerical measurement algorithms in a final uncertainty/error budget.
  • Create a reproducible waveform-measurement evidence record containing topology, probe, loading, bandwidth, acquisition and extraction settings.
Starting question

A fast edge enters the probe. The oscilloscope displays a slower edge. Which part belongs to the DUT, which part belongs to the measurement system, and when is a correction mathematically justified?

The final MS-02 chapter closes the chain by treating bandwidth, rise time and the measurement record as one coherent estimation problem.

05.1Bandwidth is an analog transfer-function property

For a voltage measurement chain, define a small-signal transfer function from the actual DUT voltage at the measurement point to the analog signal presented to the digitizer:

Hmeas(jω)=Vanalog,out(jω)/VDUT(jω)

Oscilloscope bandwidth refers to the frequency response of this analog path, not to the sample rate. For a low-pass response normalized to unity at low frequency, the conventional bandwidth edge is commonly the frequency where

|Hmeas(jωB)|=1/√2
20log10|H|=−3.0103 dB

Sample rate, record length and trigger were separated in chapter 03 because they solve different problems.

Figure 5.1 — Where does measurement bandwidth actually live?

The one question is: does the displayed waveform belong to the DUT alone, or to the whole chain?

sourceDUTprobeanalog front endacquisition

Every block can alter amplitude, phase or rise time. The displayed trace is the output of the complete measurement chain, not an ideal window directly onto the DUT.

05.2Derive the 10–90% rise time of a one-pole measurement system

Take the canonical first-order low-pass

H(s)=1/(1+sτ)

Its unit-step response is

y(t)=1−e−t/τ, t≥0

At 10%:

0.1=1−e−t10/τ ⇒ t10=−τ ln(0.9)

At 90%:

0.9=1−e−t90/τ ⇒ t90=−τ ln(0.1)

Therefore the 10–90% rise time is

tr=t90−t10=τ ln 9≈2.19722τ

This is an exact result for the one-pole model and this 10–90% definition.

05.3Where BW·tr≈0.35 actually comes from

For the same one-pole system,

f3dB=1/(2πτ)

Multiplying by the exact 10–90% rise time gives

f3dBtr=[1/(2πτ)]·[τ ln9]=ln9/(2π)=0.34970...

The familiar 0.35 relation is therefore not a universal law of oscilloscopes. It is the consequence of a particular response shape and rise-time definition.

Modern oscilloscopes can have deliberately shaped multi-pole responses; vendor guidance therefore often uses a constant K somewhere around 0.35–0.45:

BW·tr≈K

05.4Derive sinusoidal amplitude error from finite bandwidth

For the same normalized one-pole response,

|H(jf)|=1/√[1+(f/fB)2]

If the allowed fractional amplitude error is ε, require

|H|≥1−ε

Solving for the required bandwidth ratio gives

fB/f ≥ 1/√[1/(1−ε)2−1]

Examples for the one-pole model:

Allowed amplitude errorRequired fB/f
3%≈3.99
1%≈7.02

This derivation explains why “use several times more bandwidth than the sine frequency” appears as a rule of thumb: the multiplier encodes an amplitude-error target and a response-shape assumption.

05.5Fast edges contain many frequencies

A square-looking digital waveform is not characterized by clock frequency alone. Its transition shape requires higher-frequency components. A bandwidth adequate for the fundamental can still round the edge substantially.

For edge measurements, work from the fastest expected rise/fall time and the measurement-system rise time rather than only from repetition rate. Conversely, if only a sine amplitude matters, a direct frequency-response error calculation is more meaningful than a rise-time rule.

This separates two different measurands: sinusoidal amplitude accuracy and transient edge fidelity.

05.6Cascaded bandwidths and rise times: state the approximation

The complete measurement chain may include DUT output dynamics, probe, oscilloscope front end and other filtering:

Htotal(s)=HDUT(s)Hprobe(s)Hscope(s)…

For approximately Gaussian independent responses, a common engineering approximation is root-sum-square rise time:

tr,total2≈tr,12+tr,22+…

Then, if the measurement-system rise time is known and the approximation is justified,

tr,DUT≈√(tr,meas2−tr,system2)

This is not exact for arbitrary pole/zero structures, ringing systems or strongly non-Gaussian responses. In those cases use the actual transfer functions or a validated de-embedding model.

05.7Example — when measurement-system speed is not negligible

Suppose a measured 10–90% edge is 5.00 ns and an independently characterized approximately Gaussian probe+scope system has tr,system=3.00 ns. The RSS estimate gives

tr,DUT≈√(5.002−3.002) ns=4.00 ns

The correction is substantial. By contrast, if the measurement system were 1.00 ns,

√(5.002−1.002)=4.90 ns

so the correction would be small. This motivates designing the measurement system to be materially faster than the edge being characterized.

05.8Reproducibility requires more than a screenshot

A waveform image alone cannot reconstruct the measurement. Record at least:

  1. DUT state, bias and measurement nodes/reference.
  2. Source model/configuration, output impedance assumption, load and amplitude convention.
  3. Probe type/attenuation, compensation status, connection geometry and relevant bandwidth/loading data.
  4. Oscilloscope input mode, analog bandwidth limit, coupling and channel scaling.
  5. Sample rate, record length, time span and acquisition mode.
  6. Trigger source, threshold, slope and position.
  7. Measured quantity definition: Vpp, RMS, period, 10–90% rise time, overshoot, etc.
  8. Algorithm/threshold/interpolation method used by software or cursor extraction.
  9. Repeated measurements or uncertainty/resolution information when the result is reported quantitatively.

This extends the MS-01 evidence-record concept from scalar DC measurements to time-varying signals.

05.9Final diagnostic workflow

  1. Define the measurand. Amplitude, frequency, phase, rise time and overshoot require different bandwidth/acquisition choices.
  2. Model loading and references. Reuse chapters 01–02.
  3. Model the source. Reuse chapter 04.
  4. Choose analog bandwidth intentionally. Avoid both insufficient bandwidth and unnecessary noise bandwidth.
  5. Choose sample rate and record length separately. Reuse chapter 03.
  6. Establish a stable trigger.
  7. Check invariance. Change probe, bandwidth limit, sample rate or vertical scale one factor at a time; a real DUT feature should survive justified measurement changes.
  8. Report the chain. Do not present a corrected DUT result without the correction model and assumptions.

05.10Mastery exercises

  1. Derive tr=τln9 for a one-pole step response.
  2. Derive f3dBtr=ln9/(2π).
  3. A one-pole measurement front end has fB=100 MHz. Estimate its 10–90% rise time.
  4. For the same one-pole system, what amplitude ratio is measured for a 50 MHz sine?
  5. Derive the bandwidth multiplier required for ≤3% sine-amplitude error.
  6. A measured rise time is 4 ns and an approximately Gaussian system rise time is 2 ns. Estimate the DUT rise time.
  7. Why is the RSS correction not valid for every measurement chain?
  8. Name six items that must accompany a reproducible rise-time measurement.
Solutions

1. t10=−τln0.9 and t90=−τln0.1; subtraction gives τln9.

2. Multiply 1/(2πτ) by τln9 to obtain 0.34970.

3. tr≈0.3497/100 MHz≈3.50 ns for the one-pole model.

4. |H|=1/√(1+0.5²)=0.8944, about −0.969 dB.

5. Set 1/√(1+r²)≥0.97 with r=f/fB; solving gives fB/f≈3.99.

6. √(4²−2²)=3.46 ns.

7. It assumes compatible approximately Gaussian/independent response shapes; arbitrary poles, zeros and ringing need a fuller model.

8. Examples: nodes/reference, probe and attenuation, connection geometry, analog bandwidth, sample rate, record length, trigger, rise-time thresholds/algorithm, repetitions/uncertainty.

MS-02 is now complete as a general measurement course: reference topology → probe loading → sampling/trigger → source impedance → bandwidth/rise-time/reproducibility. The next curriculum step can proceed without depending on any personal oscilloscope or thesis-specific context.

Volume ITable of contents