Finding Frequency with an Oscilloscope (Paper 2)

Explained

Finding frequency with an oscilloscope

An oscilloscope draws a graph of voltage against time. Sound arrives through a microphone, which turns it into a changing voltage, and the screen shows that voltage as a wave. What you measure directly is the period, and frequency comes from it.

Setting it up

Connect a microphone, or a signal generator if you want a controlled test, to the oscilloscope input. Adjust the volts per division control until the waveform fills a useful part of the screen without going off the top or bottom, which is called clipping.

Then adjust the time base until one or two complete waves are visible. Too many waves squeezed on screen makes measurement imprecise, and too few makes it impossible to see a full cycle.

Reading the period from the screen

The time base tells you how much time each horizontal division represents, for example 2 milliseconds per division.

Count the horizontal divisions for one complete wave, from one point on the wave to the same point on the next. Multiply the number of divisions by the time base setting, and that is the period.

Measure from peak to peak, or from one zero crossing going upwards to the next going upwards. Measuring peak to trough gives half a wave, and it is the commonest way to halve an answer without noticing.

Converting period to frequency

Frequency equals one divided by the period. If the period is in seconds, the frequency comes out in hertz.

Time base settings are usually in milliseconds or microseconds, so convert before dividing. A period of 2 milliseconds is 0.002 seconds, which gives a frequency of 500 Hz. Leaving it as 2 gives 0.5 Hz, which is wrong by a factor of a thousand and looks obviously wrong once you know what to expect: audible sound runs from about 20 Hz to 20,000 Hz.

Improving the measurement

Measuring across several complete waves and dividing by the number of waves gives a better value for the period than measuring one, because any error in reading the screen is spread across all of them.

That is the standard improvement answer for this practical, and it is the same reasoning as timing twenty swings of a pendulum rather than one.