ELE-280 · Lesson 7 of 12 · 27 min

Oscilloscopes II: bandwidth, sampling and grounding

An oscilloscope shows a picture of a signal, but the picture is shaped by the instrument. Its bandwidth rounds fast edges, its sampling can invent frequencies that are not there, its probe loads the circuit, and its ground connection can short a circuit or add noise. This lesson covers the limits that decide whether what you see is the signal or an artifact of the measurement, and it ends with the frequency-domain view that most digital scopes offer through a built-in FFT.

Leo has started to notice odd things. The shift-register clock looks rounder on one scope than on another. A slow sensor signal shows a fast wiggle that the sensor cannot produce. And when he clipped a probe ground to the wrong pin on a mains-powered test fixture, a breaker tripped. Amara uses each incident to explain one limit.

Bandwidth and rise time

A scope's analog front end, probe and amplifier together, behaves like a low-pass filter. Its bandwidth is the frequency at which a sine wave's displayed amplitude has fallen by 3 dB, to about 70.7 % of its true value. A 100 MHz scope does not show 100 MHz sines at full height; it shows them about 30 % low. Even well below the bandwidth there is some loss: for a simple first-order response, a sine at half the bandwidth is displayed about 11 % low, and at one-fifth of the bandwidth about 2 % low. The common rule of thumb is to use a scope whose bandwidth is at least three to five times the highest frequency you need to measure accurately.

A graph of displayed amplitude in percent against frequency as a fraction of scope bandwidth, on a logarithmic frequency axis from 0.05 to 3. The curve is flat near 100 percent at low frequency, falls to about 98 percent at 0.2, about 89 percent at 0.5 and 70.7 percent at 1.0, labeled the minus 3 dB bandwidth, and continues to fall above it.
Amplitude error of a first-order front end. At the rated bandwidth the scope shows only 70.7 % of a sine's true amplitude.

For edges, the useful relation links bandwidth to rise time. A system with a first-order or Gaussian-like response has a 10–90 % rise time of about 0.35 divided by its bandwidth: a 100 MHz scope has an intrinsic rise time of about 3.5 ns. When a signal with its own rise time passes through a scope, and through a probe with its own rise time, the displayed rise time is approximately the root-sum-square of all of them. That tells you how fast a scope must be to measure a given edge, and how to correct a measurement made with one that is not quite fast enough.

\[t_r \approx \frac{0.35}{BW}, \qquad t_{r,\text{displayed}} \approx \sqrt{t_{r,\text{signal}}^2 + t_{r,\text{scope}}^2 + t_{r,\text{probe}}^2}\]

Worked problem

How fast is the clock edge really?

Leo measures the shift-register clock's rise time as 8.0 ns on a 50 MHz scope and as 4.3 ns on a 200 MHz scope, both with probes rated well above the scope bandwidth. Estimate the true rise time from each and decide which measurement to trust.

Show solution
  1. The 50 MHz scope's own rise time is about 0.35/50 MHz = 7.0 ns. The 200 MHz scope's is about 1.75 ns.
  2. \[t_{r,\text{signal}} \approx \sqrt{8.0^2 - 7.0^2} \approx 3.9\ \text{ns} \quad (50\ \text{MHz scope})\]
  3. \[t_{r,\text{signal}} \approx \sqrt{4.3^2 - 1.75^2} \approx 3.9\ \text{ns} \quad (200\ \text{MHz scope})\]
  4. Both estimates agree, but the first relies on subtracting two nearly equal numbers, so any error in the 7.0 ns assumption changes the answer a lot.

Answer: The edge rises in about 3.9 ns. Trust the 200 MHz measurement: its correction is small, while the 50 MHz scope's displayed value is mostly its own rise time. A rule of thumb is that the scope's rise time should be at most a third to a fifth of the edge you want to measure.

Sampling and aliasing

A digital scope's converter takes samples at a finite rate, and the sampling theorem from Signals and Systems applies: a signal can be represented faithfully only if it contains no frequencies above half the sample rate, the Nyquist frequency. Content above it folds back and appears at a false, lower frequency, an alias. A real-time scope samples fast enough that its own front-end bandwidth keeps the input below Nyquist, typically sampling at least 2.5 to 5 times its bandwidth. The danger comes when the timebase is slow: with shallow memory, a long time window forces a low sample rate, and a fast signal can alias into a slow-looking waveform that seems stable and plausible.

Worked problem

The wiggle that was not there

Leo watches a slow temperature-sensor signal with the timebase at 200 ms per division on a scope with 20 kpoints of memory. He sees a small 4 Hz wiggle that the sensor cannot produce. Nearby, a switching regulator runs at 100.004 kHz and couples a little noise onto the trace. What sample rate is the scope using, and can the regulator explain the wiggle?

Show solution
  1. The window is 10 × 200 ms = 2 s, so the sample rate is 20,000 / 2 s = 10 kSa/s, with a Nyquist frequency of 5 kHz.
  2. The regulator's 100.004 kHz is far above Nyquist. Its alias is its distance from the nearest multiple of the sample rate.
  3. \[f_{alias} = |100\,004 - 10 \times 10\,000|\ \text{Hz} = 4\ \text{Hz}\]

Answer: The 4 Hz wiggle is the regulator noise aliased by the slow sample rate. Speeding up the timebase raises the sample rate and the wiggle changes or vanishes, which is the standard test for aliasing. Using a deeper memory setting, a peak-detect acquisition mode, or the scope's bandwidth-limit filter avoids the problem.

Two acquisition modes help. Peak-detect mode samples at the maximum rate internally and keeps the highest and lowest value in each display interval, so narrow glitches and high-frequency noise show up as a thick trace rather than as a misleading alias. High-resolution or averaging modes do the opposite: they combine adjacent fast samples, or successive triggered captures, to reduce noise and increase effective resolution at the cost of bandwidth. The bandwidth-limit switch, usually 20 MHz, inserts a filter that removes high-frequency noise you do not care about, which cleans up power-rail ripple measurements considerably.

Probe loading at high frequency

Lesson 6 described a 10X probe as 10 MΩ in parallel with about 10 to 15 pF. At DC the 10 MΩ dominates and the loading is tiny. At high frequency the capacitance dominates: 12 pF has a reactance of about 1.3 kΩ at 10 MHz and 130 Ω at 100 MHz. Probing a node with a source resistance of a few kilohms at those frequencies forms a low-pass filter that slows its edges, sometimes enough to make a marginal circuit fail only while you are probing it, or to make it work only while you are probing it. For fast, high-impedance signals, active probes with input capacitance below 1 pF, or low-impedance probes for 50 Ω systems, reduce the loading.

Amplitude loss versus frequency, rise-time correction, probe-loading corner frequency, and alias frequency at four sample rates.
# Bandwidth, rise time, probe loading and alias frequency calculations
import math

def displayed_fraction(f, bw):          # first-order front end
    return 1 / math.sqrt(1 + (f / bw) ** 2)

for frac in (0.2, 0.33, 0.5, 1.0):
    print(f"sine at {frac:4.2f} x bandwidth displayed at {100*displayed_fraction(frac, 1):5.1f} %")

def true_rise(displayed_ns, scope_bw_mhz, probe_bw_mhz=None):
    parts = [(0.35 / scope_bw_mhz) * 1e3]
    if probe_bw_mhz: parts.append((0.35 / probe_bw_mhz) * 1e3)
    return math.sqrt(max(displayed_ns**2 - sum(p * p for p in parts), 0))

print(f"edge on 50 MHz scope:  {true_rise(8.0, 50):.2f} ns")
print(f"edge on 200 MHz scope: {true_rise(4.3, 200):.2f} ns")

# RC corner when a 12 pF probe loads a node with 4.7 kohm source resistance
R, C = 4.7e3, 12e-12
print(f"probe-loaded corner frequency: {1/(2*math.pi*R*C)/1e6:.2f} MHz, added rise time {2.2*R*C*1e9:.1f} ns")

def alias(f, fs):
    k = round(f / fs)
    return abs(f - k * fs)

for fs in (10e3, 100e3, 1e6, 10e6):
    print(f"fs = {fs/1e3:7.0f} kSa/s: 100.004 kHz appears at {alias(100004, fs):10.1f} Hz")
Output
sine at 0.20 x bandwidth displayed at  98.1 %
sine at 0.33 x bandwidth displayed at  95.0 %
sine at 0.50 x bandwidth displayed at  89.4 %
sine at 1.00 x bandwidth displayed at  70.7 %
edge on 50 MHz scope:  3.87 ns
edge on 200 MHz scope: 3.93 ns
probe-loaded corner frequency: 2.82 MHz, added rise time 124.1 ns
fs =      10 kSa/s: 100.004 kHz appears at        4.0 Hz
fs =     100 kSa/s: 100.004 kHz appears at        4.0 Hz
fs =    1000 kSa/s: 100.004 kHz appears at   100004.0 Hz
fs =   10000 kSa/s: 100.004 kHz appears at   100004.0 Hz

The last section of the output is a practical test. At 10 kSa/s and at 100 kSa/s the regulator noise appears at 4 Hz, because 100.004 kHz sits just above a multiple of both sample rates; once the sample rate rises above twice 100 kHz, it appears at its true frequency. Real signals do not change frequency when you change the timebase; aliases do.

Grounding and the mains-referenced scope

On almost all bench oscilloscopes, the outer conductor of every input connector, and therefore every probe ground clip, is connected to the scope's chassis and through the power cord to the building's protective earth. A probe measures the voltage between its tip and earth, not between its tip and its ground clip. That is harmless when the circuit's ground is also at earth potential or floating, as with a battery-powered board or a board powered from an isolated bench supply. It is dangerous when you clip the ground lead to a point that is not at earth potential, such as the high side of a resistor, the switch node of a converter, or anything in a mains-connected, non-isolated circuit. The ground clip then shorts that point to earth through the probe, and the fault current flows through the probe's thin shield.

A diagram of a mains-powered circuit on the left with line and neutral from the wall, a rectifier and a capacitor. On the right an oscilloscope whose input connector shield is connected through its power cord to protective earth. A probe ground clip attached to a point on the rectifier circuit creates a short-circuit path, drawn in red, from that point through the probe shield, the scope chassis and the earth wire back to the neutral of the supply.
Clipping a mains-referenced probe ground to a point that is not at earth potential creates a short circuit through the scope. The fault current can vaporize the ground lead or damage the circuit and the scope.

That is what happened on Leo's test fixture. The fixture's power supply rectifies mains directly, so its internal ground sits near line potential for half of each cycle. When he clipped the probe's ground there, the fault current tripped the breaker. He was lucky: the ground lead could have burned open and the scope's input could have been damaged. The correct ways to measure such points are a differential probe, which measures the voltage between two points without connecting either to earth; an isolated-input scope or a battery-powered handheld scope rated for the measurement category; or two channels in the scope's math mode, subtracting one from the other, which works for slow signals with small common-mode voltages. Defeating the scope's earth with a cheater plug is never acceptable; it puts the chassis and every exposed metal part at the circuit's voltage.

Grounding also causes noise even when nothing is shorted. If the circuit under test is earthed at one point, through its supply, and the scope is earthed at another, through its cord, the probe ground lead closes a ground loop. Small voltage differences between the two earth points drive current around the loop, and the loop picks up magnetic fields from nearby wiring. The result is often a 60 Hz hum or spiky noise on the trace that is not in the circuit. Keeping the probe ground lead short, using a single ground point for the setup, and powering the circuit from a floating supply all reduce it.

The frequency domain: the scope's FFT

Most digital scopes can display a fast Fourier transform of the captured waveform, showing amplitude against frequency instead of against time. It is the quickest way to answer questions such as which harmonics are in the LED driver's switching current, where the noise on a sensor line comes from, or whether a filter is removing what it should. The FFT's frequency resolution is the sample rate divided by the number of points, so a longer capture gives finer resolution. Its span reaches half the sample rate. A window function, such as Hann or flat-top, reduces the leakage of a strong component into neighboring bins at the cost of slightly wider peaks. The vertical axis is usually in dBV or dBm, which makes small components visible next to large ones.

Worked problem

Finding the source of noise

An FFT of the microphone amplifier output, captured over 100 ms at 1 MSa/s with a Hann window, shows peaks at 120 Hz, 240 Hz and 360 Hz about 40 dB below the 1 kHz test tone, and a group of peaks around 100 kHz. What is the frequency resolution, and what are the likely sources?

Show solution
  1. Frequency resolution = sample rate / number of points = 1 MSa/s / 100,000 = 10 Hz, so 120 Hz harmonics are clearly resolved.
  2. The 120 Hz family matches full-wave rectifier ripple on the 24 V rail coupling into the amplifier's supply.
  3. The 100 kHz group matches the switching regulator; its sidebands come from load changes modulating the switching.

Answer: Resolution 10 Hz. The 120 Hz harmonics, at −40 dB relative to the tone, point to supply ripple; better decoupling or a quieter supply for the amplifier is the fix. The 100 kHz components are above the audio band and will be removed by the amplifier's output filter.

Common mistakes

  • MistakeAssuming a scope shows full amplitude up to its rated bandwidth.

    InsteadAt the −3 dB bandwidth it shows 70.7 %; choose a bandwidth three to five times the highest frequency of interest.

  • MistakeTrusting a stable-looking waveform at a slow timebase.

    InsteadCheck for aliasing by changing the timebase or using peak-detect; real frequencies do not change.

  • MistakeIgnoring probe capacitance at high frequencies.

    InsteadAbout 12 pF loads high-impedance nodes heavily above a few megahertz; use an active probe if needed.

  • MistakeClipping the ground lead to a point not at earth potential.

    InsteadUse a differential probe or an isolated instrument; never float a scope by defeating its earth.

  • MistakeRunning an FFT on too short a capture.

    InsteadResolution is sample rate divided by points; capture enough time to separate the frequencies you care about.

Summary and key terms

  • Scope bandwidth is the −3 dB point; use three to five times the highest frequency of interest.
  • Rise time is about 0.35/BW, and displayed rise times combine as a root-sum-square of signal, scope and probe.
  • Slow timebases with shallow memory lower the sample rate and can alias fast signals into false slow ones.
  • Probe capacitance dominates loading at high frequency; active probes reduce it.
  • Bench scope grounds are tied to earth; clipping them to non-earth points causes shorts, and ground loops cause hum.
  • The scope's FFT shows spectra; resolution is f_s/N and windows reduce leakage.
Bandwidth
The frequency at which an instrument's response has fallen by 3 dB, to 70.7 % of the true amplitude.
Aliasing
The appearance of a signal component above half the sample rate at a false, lower frequency.
Peak detect
An acquisition mode that records the maximum and minimum within each display interval so narrow events are not missed.
Differential probe
A probe that measures the voltage between two points without connecting either to earth.
Ground loop
A closed path formed by multiple ground connections, in which stray currents induce noise.
Window function
A weighting applied to a captured record before an FFT to reduce spectral leakage.
Bandwidth limit
A selectable low-pass filter, often 20 MHz, that removes high-frequency noise from the display.

Practice problems

Work each problem on paper first, then reveal the answer to check.

  1. What is the intrinsic rise time of a 350 MHz scope?

    Show answer

    About 1.0 ns.

  2. A 20 MHz sine is viewed on a 100 MHz scope with a first-order response. What percentage of its amplitude is displayed?

    Show answer

    1/√(1 + 0.04) ≈ 98 %.

  3. A scope samples at 50 kSa/s. Where does a 49 kHz signal appear?

    Show answer

    At 1 kHz.

  4. What FFT resolution do 1 Mpoints at 100 MSa/s give?

    Show answer

    100 Hz.

  5. Why can't you measure the voltage across a high-side current-sense resistor in a mains-connected circuit with a standard probe's ground clip?

    Show answer

    The clip is earthed; attaching it to a non-earth point creates a short. Use a differential probe.

  6. A 15 pF probe loads a node with 10 kΩ source resistance. What corner frequency does it form?

    Show answer

    About 1.06 MHz.

Lesson quiz · 5 questions

Score 60% or more to complete this lesson.

Take the quiz