Average Voltage
San Luis Obispo, Calif. In the last article in this series, we noticed
that the instantaneous voltage for an AC voltage is changing continuously as a
function of time. The most common AC waveform is a sine wave. The voltage at any
instant in time for a sine wave can be determined as:
V(t)=V_{P}*sin(wt)
where V_{P} is the peak voltage, w (omega) is the frequency in
radians per second (equivalent to 2*pi*f where f is the frequency in Hz), and t
is the time since the waveform started. This equation assumes the sine function
is based on an angle in radians (instead of degrees). We'll be switching back
and forth between using degrees and radians, depending upon which gives us the
simplest math. Find that radian/degree switch on your calculator! We can also
add a constant for phase shift, but we'll leave that for later.
Quite often we are interested in where we are in a waveform in terms
of how much of the cycle we have completed instead of how many seconds,
microseconds, or whatever, since the waveform began. Again, we can use either
degrees or radians. If we want to know the instantaneous voltage of a sine
waveform at any point in a cycle, independent of frequency and time (since we'll
use phase instead), the above equation becomes:
V(a)=V_{P}*sin(a)
where V_{P} is the peak voltage and a is how far we are in to
the waveform in degrees or radians (depending upon which sine function we're
using). Table 1 shows the voltage at various points through a single cycle of a
1 volt peak sine wave.
Radians 
Degrees 
Volts 
0 
0 
0 
0.392699 
22.5 
0.382683 
0.785398 
45 
0.707107 
1.178097 
67.5 
0.92388 
1.570796 
90 
1 
1.963495 
112.5 
0.92388 
2.356194 
135 
0.707107 
2.748894 
157.5 
0.382683 
3.141593 
180 
0 
3.534292 
202.5 
0.382683 
3.926991 
225 
0.707107 
4.31969 
247.5 
0.92388 
4.712389 
270 
1 
5.105088 
292.5 
0.92388 
5.497787 
315 
0.707107 
5.890486 
337.5 
0.382683 
Average Voltage
Now that we can determine the instantaneous voltage at any point in
the waveform, let's determine the average voltage through one cycle. The
average is also known as the arithmetic mean. We can simply add up the
instaneous voltages and divide by the number of voltages we've added. Notice
that the sum is zero? For every positive voltage during the first half cycle, we
have an equal negative voltage during the second half cycle. If you connect a DC
meter (which mechanically averages the instantaneous voltages) to an AC source,
it will read 0 volts, since the average is indeed 0 volts. However, the average
for a half cycle is not zero volts. If we average the first half cycle (the
first 8 voltages in the table above), we get about 0.615 volts. This is
approximately the average voltage during a half cycle.
For a more exact determination of the average voltage of a half cycle, let's
reconsider how we determine the average. If, for example, you travel 10 miles
per hour for 1 hour and 20 miles per hour for two hours, you might be tempted to
say the average speed is 15 miles per hour (add the speeds and divide by the
number of speeds). However, we know that what we must really do is determine the
total distance traveled (10 miles + 40 miles), then divide by 3 hours,
yielding 16.667 miles per hour. Our new definition of average is the sum of
the products of the value times the amount of time we spent at that value,
divided by the total time. Applying this to the voltages in a sine wave, we
can approximate the average voltage as shown in figure 1. In equation 1,
sigma indicates we're taking the sum of several terms, V(t) is the
instantaneous voltage, and delta t is the change in time from the this
time to the next time we determine the voltage. Since the voltage is changing
continuously, it really doesn't spend any time at a particular voltage.
Therefore, we keep increasing the number of samples and multiplying them by a
smaller and smaller delta t. The limit of this smaller and smaller
approach turns the summation into an integral and turns delta t into dt.
So, if we say the average voltage can be computed using the integral in equation
2, you know where it comes from!
Equation 2 is a definite integral. We are to do this infinite number
of multiplications and additions for t values between 0 and pi (radians).
Substituting the function for a 1 volt peak sine wave, we get the equation 3.
This value of this definite integral is determined by taking the definite
integral of sin(t) dt, which is cos(t) (we'll leave the calculus details for
another time!), and subtracting the integral evaluated at the lower limit from
that evaluated at the upper limit. This is shown in equation 4 where the
evaluation limits are shown to the right of the cos term. These evaluation
limits are inserted in the equation in equation 5. Finally, the values of cos(pi)
and cos(0) are substituted in equation 6. This yields an exact value for the
average of a half cycle of a sine wave as 2/pi, or about 0.636 times the peak
voltage.
Next time, we'll evaluate the Root Mean Square of a sine wave. To start you
thinking, we're going to take the square root of the average of the squares
of the voltages. We'll also discuss why we do that.
