# AC Waveform Characteristics

Part 1:  AC Voltages and Current: AC Waveform Characteristics

It is very common, particularly in power distribution, for an AC waveform to be sinusoidal.  Figure 1 depicts a sine wave.

Figure 1.  Sine wave.

The peak amplitude, both negative and positive, are the same for a sine wave.  It is centered about 0V.  A sine wave is composed of one shape, or one cycle, repeated over and over.  Figure 1 depicts the amount of time, shown as ∆t, it takes for one cycle to occur.

When quantifying a sine wave, there are two values needed to describe it.  They are the voltage of the waveform and the number of times the cycle repeats in one second.  The number of cycles that occur per second is called the frequency of the waveform.  The frequency, abbreviated as “f”, carries the unit Hertz, which is abbreviated as Hz.  A waveform with a frequency of 1,000Hz has one cycle repeated 1000 times per second.  Although it is somewhat rare, you will occasionally see “cps” used in lieu of Hz.  The abbreviation cps stands for “cycles per second”.

In Figure 2, the time it takes for one cycle to repeat is shown.  The name given to the amount of time it takes to complete one cycle is called the period.  In Figure 1, the period is found by measuring the amount of time it takes between the zero crossing points of one cycle.  In Figure 2, it is found by looking at two sequential peaks.  This yields the same time because of the regularity of the waveform.

Figure 2.  Sine wave with a period of 0.002 seconds.

Using the value of the period, which is measured in seconds per cycle, it is possible to find the value for the frequency, which is measured in cycles per second.  They have a reciprocal relationship as shown in the following equation.  In Figure 2, the period is given as 0.002 second, therefore the frequency is found by f = 1/0.002 or f = 500Hz.

Equation 1.  Frequency and period.

With a DC voltage, it is easy to understand what is meant when its voltage is specified.  There is only one unchanging value.  In the case of an AC voltage, what voltage is being referred to?  There are actually three common ways of referring to the voltage of an AC waveform.  These are illustrated in the following figure.

Figure 3.  Different ways to indicate an AC waveform voltage.

The thee ways to typically refer to magnitude of an AC waveform are the peak voltage, the peak to peak voltage (abbreviated as P-P), and the RMS voltage.  Which one is used, is based on what is most convenient in a particular circumstance.  The left hand portion of Figure 3 illustrates the peak, and peak to peak voltage.  They are what you would expect from the names given to them.  The peak value is simply the highest voltage measured.  The peak to peak voltage is the difference between the highest and lowest value.  In the case of the sine wave shown in Figure 3, the positive peak voltage and the negative peak voltage are the same except for the polarity and the peak to peak voltage is simply twice that of the peak voltage.

The RMS value requires a bit more effort to find.  The RMS value is the voltage that can do the same amount of work as an equivalent DC voltage with a resistive load.  In the previous chapter, we discussed the fact that a resistor converts the potential energy provided by a voltage source to heat.  The RMS value for an AC voltage is therefore value of a DC voltage that would produce the same amount of heat in a resistor.

Graphically, you can imagine that the RMS value is found first by making the negative part of the waveform positive (called full wave rectification) and then finding the equivalent DC value in terms of the power delivered to a load.  The math used to find the RMS value of a sine wave is straight forward as shown in the following equation.

Equation 2.  RMS and peak voltage relationship.

Note that, if the AC waveform is not a sine wave, the equation above will not work because the math used to derive it relies on the shape of the waveform.  Rather than calculating the square root of two each time, remembering that Vrms = Vpeak * 0.707 (1/√2 is about 0.707) provides a value that is close enough in most cases.  Alternatively, √2 is about 1.414, so you can also find the RMS value by using Vrms = Vpeak / 1.414 instead.

When you use a typical voltmeter to measure an AC voltage it displays the RMS value.  Some high end voltmeters are able to display the RMS value of a non sinusoidal waveform but most display a value assuming that what is being measured is a sine wave.  As a side note, if you averaged a rectified AC sine wave, you would find that the average voltage is 0.637 x Vpeak.  The average voltage value, and the voltage for the equivalent power delivered to a load (the RMS value), are not the same.  Unless otherwise stated, always assume that, when the voltage of an AC waveform is given, it is the RMS voltage.

To convert from RMS to Vpeak use:

Equation 3.  RMS and peak voltage relationship.

To quickly convert from an RMS voltage to a peak voltage, use Vpeak = 1.414 x Vrms.  As seen in !Fig(peakandRMS), the peak to peak voltage can be found by doubling the peak voltage.  Again, remember that these equations only apply to sine waves.

So far, we have dealt with the common characteristics, the voltage and frequency, of a sine wave.  The peak voltage and the RMS value do not give you any information about what the voltage is at any specific point in time — they tell us about voltage values that are common to the waveform as a whole.  What if you wanted to know the instantaneous value of the voltage at some point in time?  The voltage at any point in time for a sine wave is given by the following equation.

Equation 4. The instantaneous voltage at a given time.

In Equation 4, V is the instantaneous voltage, f is the frequency, t is the time (in seconds) from the beginning of the sine wave, and Vpeak is the peak voltage.

Suppose you had a 10V peak sine wave with a frequency of 5,000 Hz and wanted to find the voltage 2.3 seconds after it began.  You can do this by substituting the values given into Equation 4 and solving as shown below.  In this case, the instantaneous voltage is -9.729V.

V = 10V * sin(2*π*5000*2.3)

V = 10V * sin(72,256.631)

V= 10 * -0.9729

V = -9.729V

As a side note, it is common to work in radians when dealing with trigonometric functions.  For that reason you will most likely see Equation 4 written as shown in the following equation.

Equation 5.  The instantaneous voltage at a given time in radians.

In Equation 5, ω, the lower case Greek letter omega, is the radian frequency which has units of radians per second.  It has the relationship to f, the frequency in cycles per second, shown in the following equation.

Equation 6.  The relationship between ω and f.

As you can see from looking at Equation 4 and Equation 6, Equation 5 was arrived at by simple substitution.  In any equation you encounter, the radian frequency ω can be replaced by 2πf.

Key Concepts

• AC voltages are those whose polarity vary over time.

• At a minimum it take two values, the voltage and frequency (or period), to describe an AC voltage.

• There are three common methods to describe the magnitude of an AC voltage: peak, peak to peak, and RMS.  The RMS value of an AC waveform is the voltage that can do the same amount of work as an equivalent DC voltage.

• The average value of a sine wave is 0V.  The average value of a rectified sine wave is not the same as its RMS value.