SEAB Original-style practice: A sine wave displayed on an oscilloscope has a period of 5.0\ ms and a peak voltage of 3.0\ V. Calculate (a) the frequency and (b) the peak-to-peak voltage.

(a) Frequency is one divided by the period: f = 1T = 15.0 × 10^-3 = 200\ Hz. (b) The peak-to-peak voltage is twice the peak (amplitude): V_pp = 2 × 3.0 = 6.0\ V. What markers reward: converting the period to seconds and using f = 1/T for 200\ Hz, and doubling the peak for the peak-to-peak value of 6.0\ V.

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What makes a signal analogue, and how do we describe a repeating waveform using its amplitude, period and frequency?

Describe an analogue signal, read a waveform on an oscilloscope, and calculate amplitude, period and frequency

A focused answer to the O-Level Electronics outcome on analogue signals. What makes a signal analogue, reading a waveform, and calculating amplitude, period and frequency from an oscilloscope trace.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to describe what makes a signal analogue, to read a repeating waveform from an oscilloscope, and to calculate its amplitude, period and frequency. The central insight is that an analogue signal varies smoothly and can take any value in a range, and that a repeating waveform is fully described by how big it is (amplitude), how long one cycle lasts (period) and how many cycles happen each second (frequency).

The answer

What makes a signal analogue

An analogue signal is one that varies continuously and can take any value within its range, rather than jumping between fixed levels. The voltage from a microphone, the output of a temperature sensor, and the mains supply are all analogue: they change smoothly over time. This is the opposite of a digital signal, which is allowed only two levels.

Describing a waveform

A repeating signal, such as a sine wave, is described by three quantities:

Amplitude is the maximum displacement from the centre (zero) line, measured in volts. It tells you how large the signal is.
Period ( $T$ ) is the time for one complete cycle, measured in seconds.
Frequency ( $f$ ) is the number of complete cycles per second, measured in hertz ( $\text{Hz}$ ).

Frequency and period are reciprocals:

f = \frac{1}{T}

So a short period means a high frequency. The peak-to-peak voltage is the full height of the wave, from the lowest point to the highest, which is twice the amplitude.

Reading an oscilloscope

An oscilloscope draws a graph of voltage (vertical) against time (horizontal). Two controls set the scale:

The timebase sets the time per horizontal division (for example, $2\ \text{ms}$ per division).
The vertical sensitivity sets the volts per vertical division (for example, $1\ \text{V}$ per division).

To find the period, count the horizontal divisions for one full cycle and multiply by the timebase. To find the amplitude, count the vertical divisions from the centre line to a peak and multiply by the sensitivity. Then use $f = 1/T$ for the frequency.

Worked example

An oscilloscope shows a sine wave. The timebase is $5.0\ \text{ms}$ per division and the vertical sensitivity is $2.0\ \text{V}$ per division. One full cycle spans $4.0$ horizontal divisions, and the wave rises $3.0$ divisions above the centre line. Find the period, frequency, amplitude and peak-to-peak voltage.

Step 1: Find the period from the timebase

One cycle spans $4.0$ divisions at $5.0\ \text{ms}$ each: $T = 4.0 \times 5.0 = 20\ \text{ms} = 0.020\ \text{s}$ .

Step 2: Find the frequency

Frequency is the reciprocal of the period: $f = \dfrac{1}{T} = \dfrac{1}{0.020} = 50\ \text{Hz}$ .

Step 3: Find the amplitude from the sensitivity

The peak is $3.0$ divisions at $2.0\ \text{V}$ each: amplitude $= 3.0 \times 2.0 = 6.0\ \text{V}$ .

Step 4: Find the peak-to-peak voltage

The peak-to-peak value is twice the amplitude: $V_{pp} = 2 \times 6.0 = 12\ \text{V}$ .

Common traps

Confusing amplitude and peak-to-peak: Amplitude is from the centre to one peak; peak-to-peak is the full height, twice the amplitude.
Forgetting to convert milliseconds to seconds: Frequency in hertz needs the period in seconds, so $20\ \text{ms}$ must become $0.020\ \text{s}$ before using $f = 1/T$ .
Reading the wrong number of divisions: Count divisions for exactly one complete cycle for the period, and from the centre line to a peak for the amplitude, not from peak to peak.
Mixing up the timebase and the sensitivity: The timebase is horizontal (time per division); the sensitivity is vertical (volts per division). Using one for the other gives nonsense.
Calling every changing signal digital: An analogue signal varies smoothly through any value; only a two-level signal is digital.

Examples in context

Example 1. A microphone signal. When you speak into a microphone, the sound pressure is turned into a smoothly varying voltage. Displayed on an oscilloscope, a steady note shows a repeating waveform whose frequency matches the pitch and whose amplitude matches the loudness. This is a real analogue signal you can measure directly.

Example 2. Checking a power supply. An oscilloscope on the output of a rectifier-and-smoothing circuit shows whether the direct voltage is steady or still has ripple. The size of the ripple (its amplitude) and how often it repeats (its frequency) tell you how well the smoothing is working, which is a routine practical use of waveform reading.

Try this

Cue. A waveform has a frequency of $250\ \text{Hz}$ . Find its period. Use $T = 1/f = 1/250 = 4.0\ \text{ms}$ .
Cue. A sine wave has a peak-to-peak voltage of $8.0\ \text{V}$ . State its amplitude. Amplitude is half the peak-to-peak value, so $4.0\ \text{V}$ .
Cue. State one difference between an analogue and a digital signal. An analogue signal varies smoothly and can take any value, while a digital signal is allowed only two fixed levels.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksA sine wave displayed on an oscilloscope has a period of

5.0\ \text{ms}

and a peak voltage of

3.0\ \text{V}

. Calculate (a) the frequency and (b) the peak-to-peak voltage.

Show worked answer →

(a) Frequency is one divided by the period: $f = \dfrac{1}{T} = \dfrac{1}{5.0 \times 10^{-3}} = 200\ \text{Hz}$ .

(b) The peak-to-peak voltage is twice the peak (amplitude): $V_{pp} = 2 \times 3.0 = 6.0\ \text{V}$ .

What markers reward: converting the period to seconds and using $f = 1/T$ for $200\ \text{Hz}$ , and doubling the peak for the peak-to-peak value of $6.0\ \text{V}$ .

Original4 marksOn an oscilloscope the timebase is set to

2.0\ \text{ms}

per division and the vertical sensitivity to

1.0\ \text{V}

per division. A sine wave occupies

4.0

divisions for one complete cycle and has a peak height of

2.5

divisions from the centre line. Find the frequency and the amplitude of the signal.

Show worked answer →

The period is the cycle length times the timebase: $T = 4.0 \times 2.0\ \text{ms} = 8.0\ \text{ms} = 8.0 \times 10^{-3}\ \text{s}$ . So $f = 1/T = 125\ \text{Hz}$ .

The amplitude is the peak height times the sensitivity: $V = 2.5 \times 1.0 = 2.5\ \text{V}$ .

What markers reward: multiplying divisions by the timebase to get the period then inverting it, and multiplying divisions by the volts-per-division to get the amplitude. Reading the screen settings correctly is the key skill.