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Signal Parameters & Measuring Signals

Why Measure Signals?

What Can You Measure In A Signal? 

Peak-To-Peak Value Average Value RMS Value

Problems

ou are at: Basic Concepts - Signals - Signal Features & Measurement eturn to Table of Contents

Why Measure Signals?

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

Power in an audio signal - as you test an audio amplifier's output capabilityFrequency - 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 sinusoidsignals we can note several properties of a sinusoidal signal we might want to measure. Those are tthree 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. Letsay 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 = 2 pf.

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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 thoughtabout. Consider the signal in this figure.

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

V(t) = Vmaxsin(wt + f )

t can't be represented that way because a pure sinusoid has positive and negative extremes of thesame absolute value. One way to characterize this signal is to give the peak-to-peak value of thesignal. The peak-to-peak (or just P-P, and we might represent a voltage as Vpp.) value is the algebra

difference between the largest voltage in the signal and the lowest voltage in the signal. Here wewould 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.

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

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

eak-to-peak Voltage

Peak-to-peak voltage is a pretty simple concept. If you have a signal, the peak-to-peak value othe volage is simply the difference between the largest voltage (usually positive) and the smallestvoltage (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

  Smaller Larger The Same

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ooking at the signal a little closer we might suspect that the signal plotted above has a mathematicdescription something like the following:

V(t) = 2 + 12sin(2 p500t)

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 importan

 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 averagevalue. The average value - denoted by Vavg here - is given by:

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

xample

If we have a signal that is represented by:

V(t) = 2 + 12sin(2 p

500t)This is a periodic signal, and we can compute the average of this signal. Since the average involves antegral over one period of the periodic signal, we are free to choose where the interval starts. Inthis example, the simplest thing is to integrate from t = 0 to t= T, noticing that the signal repeats 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  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:

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(1/T)*(2T) = 2So, 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 longerhistory than most electrical concepts.

When AC and DC power distribution systems were both in existence, there was a need for somstandardization between the two different systems. If you bought a 100 watt light bulb it would oconsume 100 watts if the voltages were the same. But, how do you measure AC voltage in a way thaassures that it will work to produce the same amount of light as the same DC voltage? After all, th

AC voltage is changing value all of the time, so you can't way what voltage it is because it's going tochange. That won't help you with your light bulb. What was needed was a measure of AC voltage tallowed you to use the AC voltage value the same way you used the DC voltage value when youcomputed power.

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

Is this AC voltage 100 v? Well, actually, sometimes it is much bigger than 100v, and it looks likthe positive peak is around +140 v and the negative peak is around -140 v. That means that sometim

t is much larger than 100v. However, there are other times when the voltage is much less than 10and it even passes through 0v every so often! It begins to look like you can't say what this voltagebecause it changes constantly.

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

n the early days of electricity, one of the first products to become widely used was the incandesceight bulb. In the light bulb, the amount of light produced depends upon the temperature of the

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filament, which in turn depends upon the power dissipated as heat. Also, you don't get the perceptthat the light bulb lights up and goes off 120 times a second. (The signal above has a peak - eitherpositive or negative - 120 times a second because it is a 60 Hz. signal.) When the voltage goes throzero and the power goes to zero when the voltage goes to zero. Even though the power goes to zerthe light bulb does not heat up or cool off instantaneously. What really counts in the light bulb isaverage power. Let's compute the average power. We'll do that by assuming that the voltage appeacross 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 comput the average above.

Now, let's look at the evaluation of this integral. We can substitute a general expression for thevoltage 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 integrbecomes:

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

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derivation backwards you will see that the equivalent voltage is the root (the square root) of the m(i.e. the average) of the square (the square of the voltage as a function of time). It's usually calledthe RMS (for Root-Mean-Square) voltage. That's why we gave that title to this section after all. 

xample

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

V(t) = 162.6sin(2 p60t + f )

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

roblems

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

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

nter your answer in the box below, then click the button to submit your answer. You will get a grad

on a 0 (completely wrong) to 100 (perfectly accurate answer) scale.

Your grade is:

2.  What is the RMS voltage for this signal?

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

Check My Answer

Check My Answer

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Your grade is:

How Do You Measure Voltage Signals?

Well, you have to remember, that a voltage signal is a time-varying voltage. It's an acrossvariable, so you treat any instrument's leads just as though they are voltmeter leads, which is whatthey are.

Since it is a time varying voltage, the instrument of choice is often an oscilloscope.Modern oscilloscopes usually can display a waveform and compute values for signal parameters like amplitude, frequency, peak-to-peak voltage and RMS voltage.

roblems

Problem Signal2P01 - Compute signal parameters - Square WaveProblem Signal2P02 - Compute signal parameters - Offset Square WaveProblem Signal2P03 - Compute signal parameters - Offset Sine Wave

Problem Signal2P04 - Compute signal parameters - Repetitive Ramp

Send your comments on these lessons.