MAE 2055: Mechetronics I
Mechanical and Aerospace Engineering
Lab Exercise #3
Name
Partner 1
Partner 2
Partner 3
Objectives – Gain an understanding of RMS voltages. Observe the effect of a diode on AC signals.
Pre-lab – complete prior to coming to lab
RMS voltage
Root-mean-square or RMS (or rms) amplitude is one useful way to characterize the magnitude of an AC
(time-varying) electrical signal (voltage or current). The rms amplitude of a signal is a constant, scalar
value that characterizes a signal whose magnitude is a function of time. That is, while the AC signal
varies as a function of time, its rms amplitude is a single constant value (e.g. voltage or current value)
characterizing that signal. The rms value of a signal is a useful quantity, because it can be thought of as
the effective amplitude of a signal, or as a measure of the time-average power that the signal can
provide.
For example, an AC voltage with an rms value of
applied across a resistor will result in timeaverage power dissipation equal to the power dissipation that would result from applying a constant DC
(time-invariant) voltage equal to
across that same resistor. Similarly, an AC current with an rms value
of
applied through a resistor will result in time-average power dissipation equal to the
power dissipation resulting from a
DC current flowing through the same resistor.
To understand why we use the rms amplitude (i.e. why root-mean-square, and why not simply mean or
some other type of statistical averaging?) of a signal, consider a periodic time-varying (AC) voltage
signal,
, applied across a resistor, . The instantaneous power delivered to the resistor (i.e. the
power delivered to the resistor as a function of time) is given by
(1)
Power is the time-rate of energy transfer, and can be expressed as
(2)
where
is the instantaneous energy delivered to the resistor by the signal. The total amount of
energy delivered to the resistor over one period, , can be obtained by equating equations (1) and (2)
and integrating.
(3)
MAE2055 Mechetronics I
Lab Exercise #3
The average power delivered to the resistor over one full period of
total energy delivered during one period by the time of one period.
is then obtained by dividing the
(4)
which can be rewritten as
(5)
Noting the similarity between (1) and (5), motivates the rewriting of (5) as
(6)
where
is the effective amplitude of
and is given by
. By equating (5) and (6) we see that
is, in fact,
,
(7)
Note that the expression for the rms voltage as given in (7) is, in fact, the square root of the mean of a
squared signal, as could be expected for a root-mean-square signal. (The mean of the squared signal,
, is given by the numerator of (5).) The rms value of an AC current is calculated similarly, by
replacing
with
in (7).
The time-average power delivered to a resistive load can then be written in terms of rms signal
amplitudes as
(8)
The important point to note here is that rms current and voltage values allow for the calculation of
power dissipation independent of the nature of the AC signal delivering the power. That is, the signal
may be a sinusoid, a square wave, a triangle wave, white noise, or any other AC signal, but as long as its
rms amplitude is known, the power delivered to a resistive load can be determined. Completely
different AC signals may, of course, have identical rms values.
Now that we have discussed the importance of rms values of AC signals, and how they are calculated,
let’s take a look at the rms values of some common AC signals.
RMS value of sinusoids
A sinusoidal signal, such as the voltage waveform shown in Figure 1, is given by
(9)
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MAE2055 Mechetronics I
Lab Exercise #3
Figure 1. A 2V, 3Hz, sinusoidal voltage signal, v(t).
To calculate
for the signal described by (9), we must first square the signal and make use of a
trigonometric identity for
.
(10)
Next, the mean of (10), call it
(mean-square voltage), is calculated by integrating over a single period
and dividing by the length of the period.
(11)
Finally, the rms value of the sinusoid,
given by (11).
, is obtained by taking the square root of the expression for
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Lab Exercise #3
(12)
Equation (12) is a very important result. It tells us that the rms value of a sinusoid is equal to the peak
value of that sinusoid divided by the square root of two. The 3V sinusoidal voltage signal from Figure 1
is shown again in Figure 2, along with signals representing the intermediate steps in the calculation of
the rms voltage,
. The squared AC signal is shown as a dash-dot curve. Its frequency is twice that of
, and its peak value is
. The dotted line represents the mean value of
, or
. Finally
the rms voltage,
, which is the square root of
, is shown as a dashed line. In this example the rms
value of
is
(13)
Figure 2. Signals representing the intermediate steps involved in the calculation of
for the signal of Figure 1.
While the derivation of why we use rms values, equations (1) through (8), and the derivation of the rms
value of a sinusoidal signal, equations (9) through (12), are important to understand, in practice it is very
rare to analytically calculate the rms value for an AC signal in this way. Instead, because the majority of
signals for which we are interested in knowing the rms value are sinusoids, conversion back and forth
between peak and rms values is accomplished through the application of equation (12).
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MAE2055 Mechetronics I
Lab Exercise #3
1) Most of us know that that voltage supplied at a standard wall outlet is 120V, 60Hz, but what many
people don’t realize is that the specified 120V is the rms value of the power supply voltage.
Calculate the peak value of the voltage supplied at an electrical wall outlet, and write a
mathematical expression for this power line voltage.
RMS value of square waves
Along with sinusoid, another very common type of signal you will encounter when dealing with electrical
circuits is a square wave. Square waves, such as the signal shown in Figure 3, switch back and forth
between two DC voltage levels,
. For the example square wave shown in Figure 3,
.
Figure 3. A 4Vpeak, 3Hz, square wave.
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Lab Exercise #3
2) Derive an expression for the rms value of a square wave as a function of the peak voltage, . Your
expression should be similar to equation (12). While you could go through a rigorous mathematical
derivation similar to what was done for sinusoidal signals, that is not necessary. Instead, think about
what happens when you square the time-varying square wave signal – what sort of signal results?
What is the mean value of that squared signal? And, what is the square root of that mean value?
Alternatively, think about how much power would be dissipated in a resistor across which a square
wave voltage is applied. What is the power dissipation when
? How about when
? What is the DC voltage value that would result in equivalent power dissipation?
Write your expression for
arrived at that expression.
for a square wave below, and provide an explanation for how you
3) What is the rms value for the square wave shown in Figure 3?
------------------------------------------ End of the pre-lab ------------------------------------------
Have your instructor initial here to verify completion of the pre-lab.
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Lab Exercise #3
---------------------------------- To be completed in the lab ---------------------------------330Ω
VLED
Vs
LED1
4) In this lab you’ll be building the above circuit on your breadboard. The voltage source shown in the
schematic represents the function generator set for a 10Vpp, 10Hz sinusoid. Write the expression for
this sinusoid as a function of time, and calculate the rms voltage of the signal,
.
5) Build the circuit on your breadboard.
6) What is the LED doing? Explain why you think it’s behaving this way. Remember an LED is a lightemitting diode, and diodes act like check valves for electrical current – they allow current flow in
one direction only.
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Lab Exercise #3
7) Probe Vs with the scope. Use the automated measurements to measure the peak-to-peak and rms
values of the signal (
and
). (Note: the accuracy of the rms measurement will increase if you
increase the number of periods of the signal that are visible on-screen.)
Does the relationship
hold for the measured values? (It should. If it doesn’t, figure out
why and fix the problem.)
8) Now set the function generator to output a 10Vpp square wave. What are the measured peak-topeak and rms values of the square wave? Ensure that this agrees reasonably well with the value
predicted by the expression you derived in question 3 in the pre-lab. (Note: If your square wave
signal has any overshoot, you may find that the
measurement provides a more relevant
measurement than
.)
9) Observe the LED when you switch the function generator between a sinusoidal and a square wave
output (Look carefully – the change may be subtle). What do you see? Why do you think this is the
case?
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Lab Exercise #3
10) Set the output of the function generator back to a sinusoid. Probe VLED with the scope. Sketch the
signal measured at VLED on the axes below. Label time (sec/DIV) and voltage (V/DIV) axis scales as
they appear on the display.
11) Provide a detailed explanation of why the waveform looks the way it does. That is, explain what is
happening in the circuit during the different portions of the waveform. (e.g. Why does the voltage
across the LED behave the way it does? What current is flowing, and when? When is
?
Why? When do they differ? Why?)
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