Signal Parameters & Measuring Signals
Why Measure Signals?
What Can You Measure In A Signal?
Peak-To-Peak Value
Average Value
RMS Value
Problems
You are at:  Basic Concepts - Signals - Signal Features & Measurement
Why Measure Signals?

Electrical signals - time varying voltages and currents - in many cases have important properties that you may have to measure.  Sometime in the future you might have to make any of these kinds of measurements.

• Power in an audio signal - as you test an audio amplifier's output capability
• Frequency - as you use an AC tachometer to measure a motor's rpm.
• Amplitude - as you measure signal strength in a communication system.

Goals For This Lesson

Goals for this lesson are simple.

Given a signal,
Be able to determine signal parameters including RMS voltage, peak-to-peak voltage and average voltage.

What Can You Measure In Signals?

There are lots of different properties of signals that you can measure.  If we examine sinusoidal signals we can note several properties of a sinusoidal signal we might want to measure.  Those are the three parameters you need to specify to describe a sinusoidal signal completely.

The mathematical function we use to describe a sinusoidal signal is a general sine function.  Let's say that we have a sinusoidal voltage signal, V(t).  Then we must have:

V(t) = Vmaxsin(wt + f)

There are three parameters here.

• Vmax = amplitude,
• w = angular frequency,
• And, w = 2pf.
• f = phase.
Now, over the years various kinds of instruments have been designed to measure amplitude, frequency and phase.  However, there are instruments that also measure other asspects of signals - even other aspects of sinusoidal signals.  There may be aspects of signals that you haven't thought about.  Consider the signal in this figure.

Closely examining this signal will let you see that the signal runs from -10 to +14 volts.  It can't be represented as a pure sinusoid with an expression like:

V(t) = Vmaxsin(wt + f)

It can't be represented that way because a pure sinusoid has positive and negative extremes of the same absolute value.  One way to characterize this signal is to give the peak-to-peak value of the signal.  The peak-to-peak (or just P-P, and we might represent a voltage as Vpp.) value is the algebraic difference between the largest voltage in the signal and the lowest voltage in the signal.  Here we would have:

Vpp = 14 - (-10) = 24v

Here are some other things you might want to measure for a sinusoidal signal.

• The peak-to-peak voltage of the signal.
• The RMS value of the signal.

Question

Q1   Here is a simulator that will let you add a DC component to a cosine wave.

Using this simulator set the amplitude to 120v, run the simulator and determine the peak-to-peak voltage.  Record that value for later use.

Then, add a DC component of -50v, clear the plot and run the simulator again.  Again, determine the peak-to-peak voltage, then answer this question.  Is the peak-to-peak smaller, larger or the same?

Peak-to-peak Voltage

Peak-to-peak voltage is a pretty simple concept.  If you have a signal, the peak-to-peak value of the volage is simply the difference between the largest voltage (usually positive)  and the smallest voltage (usually negative).  Here is the example signal from above.

As we found earlier, the peak-to-peak voltage is given by:

Vpp = 14 - (-10) = 24v

Looking at the signal a little closer we might suspect that the signal plotted above has a mathematical description something like the following:

V(t) = 2 + 12sin(2p500t)

That's the function we used to generate the signal.  It illustrates two aspects of signals.

• More complex signals can often be expressed as sums of simpler signals.  That will be important when you need to use Fourier Series.
• Signals do not have to be symmetric around zero.  They don't have to be symmetric at all.
• A constant component (DC component) will shift the signal.

Average

Let's examine another aspect of a signal that is often important.  Signals can have an average value.  The average value - denoted by Vavg here - is given by:

Let's look at an example - the signal we saw earlier.

Example

If we have a signal that is represented by:

V(t) = 2 + 12sin(2p500t)

This is a periodic signal, and we can compute the average of this signal.  Since the average involves an integral over one period of the periodic signal, we are free to choose where the interval starts.  In this example, the simplest thing is to integrate from t = 0 to t= T, noticing that the signal repeats 500 times a second so the period is .002sec.

Doing the integration, we have:

We will leave the details to you.  In particular:
• The integral of the sine - over a single full period - is zero.
• Check it by doing the integration.
• Consider a sine wave plot for a single full period, and notice that the area above the axis is the same as the area below the axis.  Remember, it's just the sine wave, and it doesn't include the 2.
• The integral of the constant part becomes:
• (1/T)*(2T) = 2
• So, the average of 2 is 2.  Seems right.
The net result is that the average of this signal is simply 2v.  Note the following.
• Sinusoidal signals have no average value.
• Constant terms have an average value equal to their value.
• The average value is very often referred to as the DC Component of the signal.

RMS Voltage

RMS voltage is a more sophisticated concept - even though it is a concept that has a longer history than most electrical concepts.

When AC and DC power distribution systems were both in existence, there was a need for some standardization between the two different systems.  If you bought a 100 watt light bulb it would only consume 100 watts if the voltages were the same.  But, how do you measure AC voltage in a way that assures that it will work to produce the same amount of light as the same DC voltage?  After all, the AC voltage is changing value all of the time, so you can't way what voltage it is because it's going to change.  That won't help you with your light bulb.  What was needed was a measure of AC voltage that allowed you to use the AC voltage value the same way you used the DC voltage value when you computed power.

The problem here is that power is computed as V2/R.  If you have 100 volts and you have a DC voltage there's no problem.  If you have an AC voltage you might have a signal that looks like this one.

Is this AC voltage 100 v?  Well, actually, sometimes it is much bigger than 100v, and it looks like the positive peak is around +140 v and the negative peak is around -140 v.  That means that sometimes it is much larger than 100v.  However,  there are other times when the voltage is much less than 100v, and it even passes through 0v every so often!  It begins to look like you can't say what this voltage is because it changes constantly.

Now, let's think about that light bulb again.  In a light bulb, the thing that determines the amount of light is the power that is put into the filament.  It's the power that we need to be concerned about.

In the early days of electricity, one of the first products to become widely used was the incandescent light bulb.  In the light bulb, the amount of light produced depends upon the temperature of the filament, which in turn depends upon the power dissipated as heat.  Also, you don't get the perception that the light bulb lights up and goes off 120 times a second.  (The signal above has a peak - either positive or negative - 120 times a second because it is a 60 Hz. signal.)  When the voltage goes through zero and the power goes to zero when the voltage goes to zero.  Even though the power goes to zero, the light bulb does not heat up or cool off instantaneously.  What really counts in the light bulb is average power.  Let's compute the average power.  We'll do that by assuming that the voltage appears across a resistor, R.

• First, note that the instantaneous power is v2(t)/R.
• Second, note that we can then compute the average power.  To get the average power we must compute the value of the integral below - which is not difficult to get to once you have computed the average above.

Now, let's look at the evaluation of this integral.  We can substitute a general expression for the voltage as a function of time.  Here is the result.

The easiest way to evaluate this integral is to expand the squared-sine function.

Now, we know that the cosine function (at twice the frequency) will have no net area, and the integral becomes:

Now, if we had a DC voltage, V, the power in a resistor, R, would be V2/R.  If we wanted to adjust the voltage, V, so that it produced the same average power as the AC voltage (and call that value that produces the same average power Veq), we would have to have.

Actually, this value of voltage is only sometimes called the equivalent voltage.  If you trace the derivation backwards you will see that the equivalent voltage is the root (the square root) of the mean (i.e. the average) of the square (the square of the voltage as a function of time).  It's usually called the RMS (for Root-Mean-Square) voltage.  That's why we gave that title to this section after all.

Example

We all know that the AC line voltage is 115 volts or thereabouts.  What that means is that the RMS voltage is 115 volts.  If you wanted to write a time function for the voltage we would write:

V(t) = 162.6sin(2p60t + f)

Where did we get 162.6?  We got that by multiplying 115 by the square root of 2.  And, we included an arbitrary phase angle when we wrote the function.

Problems

Here is a sinusoidal signal.  Willy Nilly has measured this signal, and acquired it in a computer file and plotted it for you.  Determine the amplitude of this signal.

P1.   What is the Peak-to-Peak voltage for this signal?

Enter your answer in the box below, then click the button to submit your answer.  You will get a grade on a 0 (completely wrong) to 100 (perfectly accurate answer) scale.

P2.   What is the RMS voltage for this signal?

Enter your answer in the box below, then click the button to submit your answer.