PhysicsCore16 min read

Uses of an Oscilloscope

Reading a trace: time-base, amplitude, period and frequency

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01

What the screen is actually showing

A cathode-ray oscilloscope draws a graph. That is the whole idea, and holding on to it makes every reading straightforward. The horizontal axis is time, and the vertical axis is voltage. A beam of electrons is swept steadily across the screen from left to right while the input voltage pushes it up and down.

Because the sweep is steady, horizontal distance is proportional to time. Because the deflection is proportional to the input, vertical distance is proportional to voltage. The trace you see is a voltage–time graph being drawn thousands of times a second.

Two controls set the scales, and neither changes the signal — only how it is displayed. The time-base sets how much time each horizontal division represents. The Y-gain sets how many volts each vertical division represents.

ControlAxisWhat it setsTypical units
Time-basehorizontaltime per divisionms/div or μs/div
Y-gainverticalvolts per divisionV/div or mV/div

Neither control changes the signal

Turning the Y-gain up makes the trace taller, not the voltage larger. Turning the time-base down fits more waves on the screen, not a higher frequency. Questions frequently test whether you have separated the display from the thing being displayed.

02

Reading a trace

Every oscilloscope reading is the same two-step process: count divisions, then multiply by the setting. Count carefully — half and quarter divisions matter, and most marks lost here are lost in the counting rather than the arithmetic.

For time, count the divisions occupied by exactly one complete cycle. One complete cycle runs from a point on the wave to the next point in the same position moving the same way — crest to crest is easiest. Multiply by the time-base setting to get the period T, then take the reciprocal for the frequency.

For voltage, count from the centre line up to a crest. That gives the amplitude, sometimes called the peak voltage. The distance from the bottom of a trough to the top of a crest is the peak-to-peak voltage, which is twice the amplitude. Read which one the question asks for.

T = (divisions per cycle) × (time-base setting)f = 1 / TV = (divisions from centre) × (Y-gain setting)peak-to-peak voltage is twice the amplitude
T
periods
f
frequencyHz
V
amplitudeV
Worked example 15 marks

A trace shows one complete cycle across 5.0 divisions with the time-base at 2.0 ms/div. The trace reaches 3.0 divisions above the centre line with the Y-gain at 0.50 V/div. Find the frequency and the peak-to-peak voltage.

  1. T = 5.0 × 2.0 ms = 10 ms.Divisions for one full cycle × time per division.
  2. Convert: 10 ms = 0.010 s.Frequency in hertz needs the period in seconds.
  3. f = 1/T = 1 / 0.010 = 100 Hz.
  4. Amplitude = 3.0 × 0.50 = 1.5 V.Divisions from the centre line × volts per division.
  5. Peak-to-peak = 2 × 1.5 = 3.0 V.Crest to trough is twice the amplitude.

f = 100 Hz, V_pp = 3.0 V

The lower panel is exactly what an oscilloscope draws: displacement against time. Change the mass and the waves stretch out — a longer period, a lower frequency. Change the amplitude and the trace gets taller while the spacing stays identical, which is the distinction between amplitude and frequency in one picture.

03

Telling a.c. from d.c. on the screen

The oscilloscope is the quickest way to see the difference between direct and alternating current, and the comparison is a standard exam question.

A d.c. supply gives a straight horizontal line, displaced from the centre by an amount that represents its voltage. The line is flat because the voltage does not change with time. Reverse the connections and the line moves to the other side of the centre.

An a.c. supply gives a repeating wave, rising as far above the centre line as it falls below it. The voltage is changing continuously and reverses direction twice every cycle. The mains supply in Pakistan alternates at 50 Hz, so its trace completes fifty full cycles every second.

If the time-base is switched off entirely, the beam no longer sweeps sideways. A d.c. input then gives a single stationary dot, and an a.c. input gives a vertical line, because the beam is moving up and down with nowhere to travel horizontally.

Key points

  1. d.c. — a flat horizontal line, displaced from the centre.
  2. a.c. — a repeating wave, equal distances above and below the centre.
  3. Time-base off: d.c. gives a dot, a.c. gives a vertical line.
  4. The height gives voltage; the horizontal spacing gives period.
  5. Doubling the frequency doubles the number of waves and leaves the height alone.
04

What an oscilloscope is used for

Beyond the classroom the instrument earns its place because it shows the shape of a signal, not just a number. A voltmeter connected to a distorted mains supply reads a single value; an oscilloscope shows you the distortion.

It is used to measure short time intervals directly — the delay between an ultrasound pulse being sent and its echo returning, for instance, which is how depth sounding and medical scanning work. Two traces displayed together let you compare the timing of two signals.

It also measures frequencies far too high to count by hand. Anything from an audio tone at a few hundred hertz to a radio signal at megahertz is read the same way: count the divisions, multiply by the time-base, take the reciprocal.

Worked example 24 marks

An ultrasound pulse and its echo appear 6.0 divisions apart with the time-base at 0.10 ms/div. The speed of sound in the material is 1500 m s⁻¹. Calculate the depth of the reflecting surface.

  1. Time between pulses = 6.0 × 0.10 ms = 0.60 ms = 6.0 × 10⁻⁴ s.Read the separation the same way as any time measurement.
  2. Total distance = v t = 1500 × 6.0 × 10⁻⁴ = 0.90 m.
  3. That distance is there and back.The pulse travels down to the surface and returns.
  4. Depth = 0.90 / 2.Halving is a mark in its own right.

0.45 m

Practice questions

6 questions · 25 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

2 · 4 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
State what the time-base control and the Y-gain control each adjust.
Model answer

The time-base sets the time represented by each horizontal division. The Y-gain sets the voltage represented by each vertical division.

Examiner tip. One mark each, and both must name the axis they affect.

SQ2[2 marks]
Describe how the trace differs between a d.c. and an a.c. input of the same peak voltage.
Model answer

D.C. gives a straight horizontal line displaced from the centre. A.C. gives a repeating wave that rises the same distance above and below the centre line.

Examiner tip. Describe both traces. Naming only one costs a mark.

Long questions

1 · 8 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[8 marks]
An oscilloscope is connected to a signal generator. The time-base is 0.50 ms/div and the Y-gain is 0.20 V/div. One complete cycle occupies 8.0 divisions and the trace reaches 3.5 divisions above the centre line.
  1. Calculate the period of the signal. [2]
  2. Calculate its frequency. [2]
  3. State the amplitude and the peak-to-peak voltage. [2]
  4. The frequency is doubled with no other change. Describe how the trace changes. [2]
Mark scheme
  1. T = 8.0 × 0.50 ms = 4.0 ms[1]
  2. = 4.0 × 10⁻³ s[1]
  3. Uses f = 1/T[1]
  4. f = 250 Hz[1]
  5. Amplitude = 3.5 × 0.20 = 0.70 V[1]
  6. Peak-to-peak = 1.4 V[1]
  7. Twice as many complete waves appear across the screeneach cycle now takes 4 divisions[1]
  8. The height of the trace is unchanged, since the voltage has not alteredthe mark most often missed[1]

(a) 4.0 ms (b) 250 Hz (c) 0.70 V and 1.4 V (d) twice as many waves, same height

Examiner tip. In part (d) say explicitly that the height does not change. Examiners award that separately, and it tests whether you have really separated frequency from amplitude.

Exam questions

3 · 13 marks

Multi-part questions with a full mark scheme.

Q1[6 marks]
An oscilloscope displays a sinusoidal signal. One complete cycle occupies 6.0 horizontal divisions, and the trace extends 2.5 divisions above and 2.5 divisions below the centre line. The time-base is set to 2.0 ms/div and the Y-gain to 4.0 V/div.
  1. Calculate the period of the signal.
  2. Calculate its frequency.
  3. State the amplitude and the peak-to-peak voltage.
Mark scheme
  1. T = 6.0 × 2.0 ms = 12 ms[1]
  2. = 0.012 sconversion needed before finding f[1]
  3. Uses f = 1/T[1]
  4. f = 83 Hzaccept 83.3[1]
  5. Amplitude = 2.5 × 4.0 = 10 Vcentre to peak[1]
  6. Peak-to-peak = 20 Vtwice the amplitude[1]

T = 0.012 s, f = 83 Hz, amplitude 10 V, peak-to-peak 20 V

Examiner tip. Convert milliseconds to seconds before inverting. Working in ms gives 0.083 Hz — a thousand times out, and it looks plausible enough to go unnoticed.

Q2[4 marks]
A microphone is connected to an oscilloscope and two notes are played in turn. Note B produces a trace with waves that are half as far apart and twice as tall as note A.
  1. Compare the frequencies of the two notes.
  2. Compare their loudness.
Mark scheme
  1. B has twice the frequency of A[1]
  2. Because the period is halved, and f = 1/T[1]
  3. B is louder than A[1]
  4. Because its amplitude is twice as large[1]

Examiner tip. Waves "half as far apart" means half the period, which means double the frequency. Say the intermediate step — it is a mark.

Q3[3 marks]
Describe how the trace on an oscilloscope differs when it is connected to a d.c. supply rather than an a.c. supply, and explain the difference.
Mark scheme
  1. D.C. gives a straight horizontal linedisplaced from the centre by an amount showing the voltage[1]
  2. A.C. gives a repeating wave[1]
  3. Because d.c. has a constant voltage while a.c. reverses direction and varies continuously with time[1]