6 Volts?... What do you mean, 6 Volts?

Varying Signals.

Figures 4.1a, b, and c illustrate three different ways in which the signal voltage somewhere in an electronic circuit might vary with time. In each case one of the most basic things we'd like to know is, “How big is this signal?”

fig1a.gif - 4Kb
When the voltage is a ‘d.c.’ level as shown in 4.1a answering this question is fairly obvious. Whenever we examine the voltage (with a voltmeter, oscilloscope, or whatever) we find it equals volts. Hence we can quote this value as a measure of the signal's size. Unfortunately, this approach isn't any good for a signal whose voltage varies from moment to moment.

fig1b.gif - 6Kb
Consider the example of a voltage which varies sinusoidally with time. Given an oscilloscope we can display how the signal level varies with time and will see the pattern shown in figure 4.1b. We can then use the scales on the 'scope to read off the screen the signal's peak-to-peak voltage, . This represents the range of voltages over which the signal swings.
For a sinewave we can expect the voltage at any moment to obey an expression of the form


where is the amplitude of the sinusoidal wave, f is the frequency of the wave, and t is the time. [By the way, note that in these pages I will use curly brackets, { }, to indicate when one quantity depends upon another. In this case the voltage depends upon the time.]

Looking at equation 1 we can see that will swing back and forth between a maximum voltage, , and a minimum voltage, — i.e. over a total range . This equals the peak-to-peak range we can see on a 'scope, so we can say that the amplitude of the sinewave signal voltage is . Hence we can determine the size (amplitude) of the sinewave by measuring the peak-to-peak size and halving its value. This technique of measuring peak-to-peak voltages is a convenient way to determine the size of sinewaves and other simple waveforms (squarewaves, triangle-waves, etc). As a result many electronics measurements can be made using an oscilloscope in this way to obtain peak-to-peak quantities. Alas, as with simple d.c. measurements, this method has its limitations.

fig1c.gif - 7Kb
Consider now the signal waveform shown in figure 4.1c. This represents the kind of fairly complex signal level variations we tend to get with ‘real’ signals which carry information — for example, the voltage pattern we'd get from a microphone an orchestra or rock band were playing into. Here the level fluctuates from moment to moment in a way that doesn't have a single obvious peak-to-peak size.
To deal with signals like this we need to take a different approach. It's a general rule in science that you can only know what you can measure. (This fact forms the basis of Information Theory.) Long before transistors — or even valves — were invented scientists & engineers had the problem, how were they to measure voltages that varied with time? They did have resistors, whose properties were known from ‘d.c.’ measurements. They also had accurate thermometers. They therefore connected the signal to a resistor and measured how much the signal’s power warmed it up.

From part 2 we know that the amount of power dissipated in a resistance, R, by an applied voltage, V, is . Since (Ohm's Law) this means that


It takes a moderate time to warm up a resistor, so we can use the change in resistor temperature when a signal's applied to indicate the time-averaged (or mean) level of the squared voltage, . [In these pages I'll use a pair of angle brackets, , to indicate that a quantity has been averaged over a period of time to produce a mean level. Some books use a little ‘hat’ symbol, for the same thing.]

For a steady d.c. voltage we can expect , and the resulting power level will be


For the sinewave


If we look at a helpful maths book (assuming that's not a self-contradiction!) we can discover that so we can rewrite equation 4 as


The way this varies with time is illustrated in figure 4.2. Since spends as much time being +ve as it does being -ve, the value of for the sine wave oscillates above and below an average level . As a result, for a sinewave we can expect that . The heating power of the sinewave signal will therefore be


Content and pages maintained by: Jim Lesurf (jcgl@st-and.ac.uk)
using TechWriter Pro and HTMLEdit on a StrongARM powered RISCOS machine.
University of St. Andrews, St Andrews, Fife KY16 9SS, Scotland.