Why Measure Signals?You are at: Basic Concepts - Signals - Signal Features & MeasurementWhat Can You Measure In A Signal?
Peak-To-Peak Value
Average Value
RMS ValueProblems
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.
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.
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) = V_{max}sin(wt + f)
There are three parameters here.
V(t) = V_{max}sin(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 V_{pp}.) value is the algebraic difference between the largest voltage in the signal and the lowest voltage in the signal. Here we would have:
V_{pp} = 14 - (-10) = 24v
Here are some other things you might want to measure for a sinusoidal signal.
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 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.
V_{pp} = 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.
Let's examine another aspect of a signal that is often important. Signals can have an average value. The average value - denoted by V_{avg} here - is given by:
Let's look at an example - the signal we saw earlier.
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:
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 V^{2}/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.
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 V^{2}/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 V_{eq}), 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.
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.
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?
P2. What is the RMS voltage for this signal?Well, you have to remember, that a voltage signal is a time-varying voltage. It's an across variable, so you treat any instrument's leads just as though they are voltmeter leads, which is what they are.