Activity 1 Ac Circuits Ac Voltage and Current - Calculations
Activity 1 Ac Circuits Ac Voltage and Current - Calculations
Activity 1 Ac Circuits Ac Voltage and Current - Calculations
Activity 1
AC CIRCUITS
AC VOLTAGE AND CURRENT – CALCULATIONS
OBJECTIVE
To study sinusoidal voltages and currents in order to understand frequency, period, effective
value, instantaneous power and average power.
DISCUSSION
Alternating Current (AC) is the world standard for driving motors and other electrical equipment.
As the name implies, an alternating current continually and periodically changes direction, going first one
way and then reversing. One may consider an alternating current to be a DC current, which is constantly,
and periodically changing amplitude and direction. The time necessary for the current to undergo one
complete change of amplitude and direction is called a cycle. The number of cycles which occur in one
second is called the frequency, which is measured in Hertz (cycles per second). In North America, AC
system frequency is standardized at 60 Hertz. Much of the remaining world has chosen a 50 Hz standard.
Alternating voltages also reverse polarity in a periodic manner with continually changing
amplitude. The shape of the voltage waveform is dependent upon the manner in which it is produced.
One may construct a device that produces voltage waveforms which are square waves, triangular waves,
etc. However, one type of waveform is most suitable to the use of electric power in transformers and
electric motors, the sine wave. Pure sinusoidal waveforms of a single frequency minimize mechanical
and electrical losses in transformers and motors, thus allowing for the highest efficiency of energy
conversion. If one considers the Fourier Transform of a triangular wave or square wave, which consist of
a large number of sinusoidal waveforms at many frequencies, it becomes clear that there is increased
potential for detrimental effects such as mechanical resonance, eddy currents, sequence currents, etc. A
pure sine wave undergoing electrical transformation (either integral or derivative) remains a pure
sinusoid, minimizing losses. Another advantage of sine wave voltages is that the resulting current
waveforms are also sinusoidal. This is not necessarily true of other waveform shapes. Having consistent
wave shapes reduces the burden of power calculations and the analysis of electric systems.
(None)
PROCEDURE
1) Consider the ideal AC generator pictured below. Assume it produces a sinusoidal voltage output
waveform between terminals A and B.
2) Let the voltage waveform have a peak value of 100 volts, such that the algebraic description is
v(t) = 100 sin (t) volts. Calculate the value of v(t) at 15 degree intervals (i.e. when t = 0,
15, 30, …, 330, 360) and plot them on the graph provided (also, write them down, since you
will need these values again). Connect the plotted points with a smooth curve and label it v(t).
Remember that the waveform will have a negative value for half its period.
3) Read the instantaneous value of v(t) from the graph for each of the following values of t and
record it below:
4) If a load resistance of 2 Ohms is connected across terminals A and B, a current, i(t), will flow.
Knowing the instantaneous value of the voltage from the graph and using Ohm’s Law [ i(t) =
v(t)/R ], calculate and record the instantaneous values of i(t) for the following values of t:
5) Plot the instantaneous current values recorded above on the same graph with v(t) and draw a
smooth curve through the plotted points. Label this curve i(t).
6) Knowing that the instantaneous power, p(t), is the product of the instantaneous voltage and
current, calculate p(t) at every 30 degree interval:
8) Examine the power curve and determine the maximum, or peak, instantaneous power dissipated
by the resistor and the minimum value of instantaneous power. Also make your best estimate as
to the average power dissipated by the resistor.
9) It can be shown mathematically that the average power will equal exactly one-half the peak
instantaneous power. The average power is equivalent to that supplied to the two ohm resistor
from a DC source. If 2500 Watts were dissipated by a 2 Ohm resistor, supplied by a DC source,
what would the DC Voltage of the supply be? (Use the equation, P = V2/R)
10) What is the DC current supplied by the source for those conditions? (Use Ohm’s Law)
Note that the AC values of 100 Volts, peak and 50 Amps, peak deliver the same average power as
DC values of 70.7 Volts, DC and 35.4 Amps, DC. This is a ratio of one over the square root of
two (1/2). Thus, the effective value of an AC signal is its peak value divided by the square root
of two. This is known mathematically as the root-mean-square or RMS value of the waveform,
and leads to the following equations:
QUESTIONS
1) How long does it take, in seconds, for the voltage to change from 0 to maximum (peak) value on
a 60 Hz power system?
2) What is the length of time of the positive portion of one complete cycle of a 60 Hz current
waveform?
3) In a 50 Hz system, what is the time duration of one complete cycle? If standard residential wiring
is rated at 120 Volts, rms, what is its peak value?
4) An incandescent lamp rated at 100 Watts, emits a certain amount of illumination when energized
at 120 Vrms. Would the amount of illumination increase, decrease or remain the same when the
lamp is energized at 120 Vdc? Explain.
5) Explain in your own words what is meant by the terms effective voltage and effective current:
CONCLUSION
Write your observation/conclusion about this activity.
ACTIVITY 2
PHASE ANGLE, AC POWER
OBJECTIVE
To explore the concept of phase angle and study the relationship between real and apparent
power.
DISCUSSION
Most AC circuits, however, contain other elements in addition to pure resistances. Under such
conditions, the current waveform will not be in phase with the voltage. In an AC circuit where the peak
value of the current waveform occurs later in time than the peak of the voltage waveform, the current is
said to lag the voltage. Such a condition is shown below in figure B.
Figure C above depicts a current whose peak value occurs earlier in time than the peak value of the
voltage. This current in said to lead the voltage.
It is possible that the current could lead or lag the voltage by as much as 90 electrical degrees (/2
radians). In figure D below, the current lags the voltage by 90 degrees. It is interesting to note that the
current maximum occurs when the voltage is ZERO. Likewise, the voltage maximum occurs at the
current zero. Although this seems highly improbable, it occurs because the AC circuit consists of an
element which is capable of storing and releasing energy, such as an inductor or capacitor. These
elements absorb energy during part of the AC cycle and return it to the circuit later in the cycle. The
storing and releasing of energy accounts for the phase angle difference between voltage and current.
Loads that are purely inductive will have a current that lags voltage by 90 degrees. Loads that are purely
capacitive will have a current that leads voltage by 90 degrees, such as figure E. In either case, a
capacitive or inductive load will absorb power for one quarter cycle and then release it all back to the
circuit the next quarter cycle. Capacitors and inductors are called reactive elements and as such, dissipate
almost no real power.
PROCEDURE
1) Construct the circuit below using Multisim. Using multimeter, measure the load’s
resistance/impedance, voltage and current. Calculate the apparent power S, supplied to the load
(S = VI) and complete the table below.
XMM1
500Ω
V1
120 Vrms R1 R2
60 Hz 1.0kΩ 1.0kΩ
0°
2) The voltage and current waveforms for this circuit are shown in figure F below. Use the
oscilloscope to obtain the waveform of the voltage given the parameters: V/div = 100V/div, T/div
= 10ms/Div then plot the graph of current and power in the same graph.
Take a screenshot.
F. Instantaneous Current, Voltage and Power for a Resistive Load
3) The instantaneous power, p(t), is also plotted in figure F. Notice that the power waveform is
sinusoidal but at double the frequency (two cycles of power per one cycle of voltage or current).
Recall the trig identity for SIN2A and answer the following questions:
a. Does the power curve ever have a negative value for a resistive load? _________
b. Is the power real? _______________
c. Write the trigonometric identity which proves that the average power is actually ½ the peak
power.
4) Construct the circuit below using Multisim. Using multimeter, measure the load’s
resistance/impedance, voltage and current. Calculate the apparent power, S, supplied to the load (S = VI)
and complete the table below.
XMM1
4) The voltage and current waveforms for this circuit are shown in figure G below. Use the
oscilloscope to obtain the same figure given the parameters: V/div = 100V/div, T/div = 10ms/Div
then plot the graph of current and power in the same graph.
Notice that the current maximum occurs at a voltage zero and vice-versa. Take a screenshot
5) The following table contains v(t) and i(t) values at 45 degree intervals. Complete the table by
calculating the instantaneous power, p(t) = v(t)i(t) VA at each interval.
6) Plot the calculated values of p(t) on the graph in figure G and draw a smooth curve connecting the
points. Remember that the power curve will be sinusoidal at twice the frequency of the voltage
and current.
Are all the peaks of the power curve of the same magnitude? ___________
Does the area enclosed under the positive portions of the curve equal the area enclosed under the
negative portions of the curve? _______________
The average or real power for one complete cycle (360 degrees) must be _________ W.
9) Construct the circuit below using Multisim. Using multimeter, measure the load’s resistance,
voltage and current. Calculate the apparent power, S, supplied to the load. (S = VI) and
complete the table below.
XMM1
9) The voltage and current waveforms for this circuit are shown in figure H below. Use the
oscilloscope to obtain the same figure given the parameters: V/div = T/div =
Notice that the current maximum occurs at a voltage zero and vice-versa. . Take a screenshot
10) The following table contains v(t) and i(t) values at 45 degree intervals. Complete the table by
calculating the instantaneous power, p(t) = v(t)i(t) VA at each interval.
11) Plot the calculated values of p(t) on the graph in figure H and draw a smooth curve connecting the
points. Remember that the power curve will be sinusoidal at twice the frequency of the voltage
and current.
Are all the peaks of the power curve of the same magnitude? ___________
Does the area enclosed under the positive portions of the curve equal the area enclosed under the
negative portions of the curve? _______________
13) The average or real power for one complete cycle (360 degrees) must be _________ W
QUESTIONS
1) If in one complete cycle, the instantaneous power, p(t), is always positive, then the load must be a
pure __________________.
2) In the space below, make a rough sketch of a current waveform lagging a voltage waveform by 60
degrees.
3) A wattmeter reads zero watts when the current lags or leads the voltage by 90 degrees. Explain this
phenomenon.
4) In a 60 Hz system, if the current leads the voltage by 45 degrees, what is the time difference, in
milliseconds, between the occurrence of each respective peak?
CONCLUSION
Activity 3
MESH ANALYSIS
OBJECTIVE
To solve the unknown parameters in an AC circuit using mesh analysis, both manually and
through the use of MATLAB commands.
DISCUSSION
A mesh is defined as a set of branches that forms a loop but that encloses no other branch of the
circuit. It is a simplified procedure to arrive at similar results more expeditiously. On each mesh we
identify a circulating mesh current. The mesh currents are identical with the link current defining the
fundamental loops. Then every branch current equals either a mesh current or the difference between two
meshes current; just like the relationship between the branch voltages in a node voltage analysis.
To fully understand mesh analysis, let us consider solving the circuit below. Take note that it has
a ‘supermesh’—a current source contained between two essential meshes.
As shown in the figure below, meshes 3 and 4 form a supermesh due to the current source between the
meshes. For mesh 1, KVL gives
For mesh 2, I2 = - 3
We can use MATLAB to solve these equations. We first cast the equations as
Or AI = B
EQUIPMENT
PROCEDURE
Using manual computation, solve for the needed parameters in the circuits below. After that,
solve the same circuit using MATLAB and compare the results you got from the two methods. Take a
screenshot/copy of all the commands you used.
SCHEMATICS
1) Calculate Io.
2) Calculate Io.
3) Calculate Io.
CONCLUSION
Activity 4
AC THEVENIN’S THEOREM
OBJECTIVE
Thevenin’s Theorem will be examined for the AC case. The Thevenin source voltage and
Thevenin impedance will be determined experimentally and compared to theory. Loads will be examined
when driven by both an arbitrary circuit and that circuit’s Thevenin equivalent to determine if the
resulting load potentials are indeed identical. Both resistive and complex loads will be examined as well
as well source impedances that are inductive or capacitive.
DISCUSSION
Thevenin’s Theorem states that any linear two port network can be replaced by a single voltage
source with series impedance. While the theorem is applicable to any number of voltage and current
sources, this exercise will only examine single source circuits for the sake of simplicity. The Thevenin
voltage is the open circuit output voltage. This may be determined experimentally by isolating the portion
to be Thevenized and simply placing an oscilloscope at its output terminals.
The Thevenin impedance is found by replacing all sources with their internal impedance and then
applying appropriate series-parallel impedance simplification rules. If an impedance meter is available, an
easy method of doing this in the lab is to replace the sources with appropriate impedance values and apply
the impedance meter to the output terminals of the circuit portion under investigation.
EQUIPMENT
(1) AC Function Generator
(1) Oscilloscope
Components
(1) 0.1 μF actual:__________________ (1) 1.0 k 𝞨 actual:__________________
(1) 0.47 μF actual:__________________ (1) 1.5 k 𝞨 actual:__________________
(1) 10 mH actual: ________________ (1) 2.2 k 𝞨 actual:__________________
(1) 50𝞨 actual:__________________
SCHEMATICS
Figure 4.1
Figure 4.2
Figure 4.3
PROCEDURE
1. For the circuit of figure 4.1, calculate the voltage across the 1 k𝞨 load using R1=1.5 k𝞨, R2=2.2 k𝞨,
and C=0.47 μF, with a 2 V p-p 1 kHz source. Record this value in Table 4.1. Also calculate the expected
Thevenin voltage and Thevenin impedance. Record these values in Table 4.2.
2. Build the circuit of figure 4.1 using R1=1.5 k𝞨, R2=2.2 k𝞨, Rload=1 k𝞨 and C=0.47 μF. Set the
generator to a 1 kHz sine wave at 2 V p-p. Make sure that the Bandwidth Limit of the oscilloscope is
engaged. This will reduce the signal noise and make for more accurate readings. Measure the load voltage
and record in Table 4.1 as VLoad Original.
3. Remove the load and measure the unloaded output voltage. This is the experimental Thevenin voltage.
Record it in Table 4.2.
4. Replace the voltage source with a 50 𝞨 resistor to represent its internal impedance. Set the impedance
meter to 1 kHz and measure the resulting impedance at the open load terminals. This is the experimental
Thevenin impedance. Record these values in Table 4.2 and compare with the theoretical values.
5. Using the decade resistance box and capacitor, build the Thevenin equivalent circuit of figure 4.2 and
apply the 1 k𝞨 load resistor. Measure the load voltage and record in Table 4.1. Compare with the values
of the original (non-Thevenized) circuit and determine the deviation between the original and Thevenized
circuits.
6. To verify that Thevenin’s Theorem also works with an inductive source and a complex load, repeat
steps 1 through 5 in like manner but using figure 4.3 with R1=1.5 k𝞨, R2=2.2 k𝞨, L=10 mH, Rload=1
k𝞨 with Cload=0.1 μF. Set the generator to a 10 kHz sine wave at 2 V p-p. Record results in Tables 4.3
and 4.4.
DATA TABLES
Table 4.1
Table 4.2
Table 4.3
Table 4.4
QUESTIONS
1. How does the AC version of Thevenin’s Theorem compare with the DC version?
2. Would the Thevenin equivalent circuits be altered if the source frequency was changed? If so, why?
3. Based on the results of this exercise, would you expect Norton’s Theorem for AC to behave similarly
to its DC case?
CONCLUSION
Activity 5
SERIES RESONANCE
OBJECTIVE
This exercise investigates the voltage relationships in a series resonant circuit. Of primary
importance are the establishment of the resonant frequency and the quality factor, or Q, of the circuit with
relation to the values of the R, L, and C components.
DISCUSSION
A series resonant circuit consists of a resistor, a capacitor, and an inductor in a simple loop. At
some frequency the capacitive and inductive reactances will be of the same magnitude, and as they are
180 degrees in opposition, they effectively nullify each other. This leaves the circuit purely resistive, the
source “seeing” only the resistive element. Consequently, the current will be at a maximum at the
resonant frequency. At any higher or lower frequency, a net reactance (the difference between XL and
XC) must be added to the resistor value, producing a higher impedance and thus, a lower current. As this
is a simple series loop, the resistor’s voltage will be proportional to the current. Consequently, the resistor
voltage should be a maximum at the resonant frequency and decrease as the frequency is either increased
or decreased. At resonance, the resistor value sets the maximal current and consequently has a major
effect on the voltages developed across the capacitor and inductor as well as the “tightness” of the voltage
versus frequency curve: The smaller the resistance, the tighter the curve and the higher the voltage seen
across the capacitor and inductor. The Q of the circuit can be defined as the ratio of the resonant reactance
to the circuit resistance, Q=X/R, which also corresponds to the ratio of the resonant frequency to the
circuit bandwidth, Q=F0/BW.
EQUIPMENT
(1) AC Function Generator
(1) Oscilloscope
Components
(1) 10 nf actual:__________________
(1) 10 mH actual:__________________
(1) 47 𝞨 actual:__________________
(1) 470 𝞨 actual:__________________
Schematics
Figure 5.1
PROCEDURE
Low Q Circuit
1. Using Figure 5.1 with R=470𝞨, L= 10 mH, and C=10 nF, determine the theoretical resonance
frequency and Q, and record the results in Table 5.1. Based on these values determine the upper and
lower frequencies defining the bandwidth, f1 and f2, and record them in Table 5.1.
2. Build the circuit of Figure 5.1 using R=470𝞨, L=10 mH and C=10 nF. Place a probe across the
resistor. Set the output of the generator to a 1 V p-p sine wave. Set the frequency to the theoretical
resonance frequency of Table 5.1. Make sure that the Bandwidth Limit of the oscilloscope is engaged for
both channels. This will reduce the signal noise and make for more accurate readings.
3. Adjust the frequency in small amounts, up and down, until the maximum voltage is found. This is the
experimental resonance frequency. Record it in Table 5.1. Note the amplitude (it should be approximately
equal to the source voltage of 1 V p-p). Sweep the frequency above and below the resonance frequency
until the experimental f1 and f2 are found. These will occur at a voltage amplitude of approximately .707
times the resonant voltage (i.e., the half-power points). Record these frequencies in Table 5.1. Also,
determine and record the experimental Q based on the experimental f0, f1, and f2.
4. Transcribe the experimental frequencies of Table 5.1 to the top three entries of Table 5.2. For all of the
frequencies in Table 5.2, measure and record the voltage across the resistor. Also measure and record the
inductor and capacitor voltages. Note that the inductor and capacitor will have to be swapped with the
resistor position in order to maintain proper ground reference with the oscilloscope.
5. Based on the data from Table 5.2, plot VR, VC, and VL as a function of frequency.
6. Change R to 47𝞨 and repeat steps 1 through 5 but using Tables 5.3 and 5.4 for high Q.
Multisim
7. Build the circuit of Figure 5.1 in Multisim. Using AC Analysis, plot the voltage across the resistor from
1 kHz to 100 kHz for both the high and low Q cases and compare them to the plots derived from Tables
5.2 and 5.4. Be sure to include the 50 Ω source resistance and coil resistance in the simulation.
Data Tables
Low Q Circuit
Table 5.1
Table 5.2
High Q Circuit
Table 5.3
Table 5.4
QUESTIONS
2. Are the VC and VL curves the same as the VR curves? If not, why?
3. In practical terms, what sets the limit on how high Q may be?
CONCLUSION
Activity 6
PARALLEL RESONANCE
OBJECTIVE
This exercise investigates the voltage relationships in a parallel resonant circuit. Of primary
importance are the establishment of the resonant frequency and the quality factor, or Q, of the circuit with
relation to the values of the R, L, and C components.
DISCUSSION
A parallel resonant circuit consists of a resistor, a capacitor, and an inductor in parallel, typically
driven by a current source. At some frequency the capacitive and inductive reactances will be of the same
magnitude, and as they are 180 degrees in opposition, they effectively nullify each other. This leaves the
circuit purely resistive, the source “seeing” only the resistive element. At any lower or higher frequency
the inductive or capacitive reactance will shunt the resistance. The result is a maximum impedance
magnitude at resonance, and thus, a maximum voltage. Any resistance value in series (such as the
inductor’s coil resistance) should be transformed into a parallel resistance in order to gauge its effect on
the system voltage. The combined parallel resistance sets the Q of the circuit and can be defined as the
ratio of the combined resistance to the resonant reactance, Q=R/X, which also corresponds to the ratio of
the resonant frequency to the circuit bandwidth, Q=f0/BW.
Equipment
(1) AC Function Generator
(1) Oscilloscope
Components
(1) 10 nF actual:__________________
(1) 10 mH actual:__________________
(1) 2.2 k𝞨 actual:__________________
(1) 10 k𝞨 actual:__________________
(1) 100 k𝞨 actual:__________________
SCHEMATICS
Figure 6.1
PROCEDURE
1. Using Figure 6.1 with Rs=100 k𝞨, Ra=2.2 k𝞨, L=10 mH, Rcoil=7 𝞨 and C=10nF, determine the
theoretical resonance frequency and Q, and record the results in Table 6.1. Based on these values
determine the upper and lower frequencies defining the bandwidth, f1 and f2, and record them in Table
6.1 also.
2. Build the circuit of Figure 6.1 using Rs=100 k𝞨, Ra=2.2 k𝞨, L=10 mH and C=10nF. Set the output of
the generator to a 10 V p-p sine wave at the theoretical resonant frequency. The large value of Rs
associated with the voltage source will make it appear as a current source equal to approximately .1 mA
p-p, assuming the parallel branch impedance is much less than Rs. Place a probe across the parallel
branch. Set the frequency to the theoretical resonance frequency of Table 6.1. Make sure that the
Bandwidth Limit of the oscilloscope is engaged for both channels. This will reduce the signal noise and
make for more accurate readings.
3. Adjust the frequency in small amounts, up and down, until the maximum voltage is found. This is the
experimental resonant frequency. Record it in Table 6.1. Note the amplitude. Sweep the frequency above
and below the resonance frequency until the experimental f1 and f2 are found. These will occur at a
voltage amplitude of approximately .707 times the resonant voltage (i.e., the half-power points). Record
these frequencies in Table 6.1. Also, determine and record the experimental Q based on the experimental
f0, f1, and f2.
4. Transcribe the experimental frequencies of Table 6.1 to the top three entries of Table 6.2. For all of the
frequencies in Table 6.2, measure and record the voltage across the parallel branch.
5. Based on the data from Table 6.2, plot the parallel branch voltage as a function of frequency.
6. For high Q, change Ra to 10 k𝞨 and repeat steps 1 through 5 but using Tables 6.3 and 6.4.
DATA TABLES
Low Q Circuit
Table 6.1
Table 6.2
High Q Circuit
Table 6.3
Table 6.4
QUESTIONS
3. In practical terms, what sets the limit on how high Q may be?
CONCLUSION
Activity 7
FREQUENCY RESPONSE
OBJECTIVE
In this lab you will be able to understand the frequency response of a circuit using the bode
plotter in Multisim and bode function in the MATLAB.
DISCUSSION
The transfer function H(ɷ) (also called the network function) is a useful analytical tool
for finding the frequency response of a circuit. In fact, the frequency response of a circuit is the
plot of the circuit’s transfer function H(ɷ) versus ɷ, with ɷ varying from ɷ = 0 to ɷ = ∞.
PROCEDURE
Bode plot of Dynamic System: bode(sys) creates a Bode plot of the frequency response of a
dynamic system model sys. The plot displays the magnitude (in dB) and phase (in degrees) of the
system response as a function of frequency. bode automatically determines frequencies to plot based
on system dynamics.
If sys is a multi-input, multi-output (MIMO) model, then bode produces an array of Bode plots, each
plot showing the frequency response of one I/O pair.
Bode plot over a specified frequency range: bode(___,ɷ) plots system responses for frequencies
specified by ɷ.
If ɷ is a cell array of the form {ɷmin, ɷmax}, then bode plots the response at frequencies ranging
between ɷmin and ɷmax. If ɷ is a vector of frequencies, then bode plots the response at each
specified frequency. You can use w with any of the input-argument combinations in previous
syntaxes.
1) Create a Bode plot over a specified frequency range. Use this approach when you want to
focus on the dynamics in a particular range of frequencies.
2) In the MATLAB, type the following:
7) Wait until the bode plot generates. Save this and include in your lab report.
The graph is only referred to as a Bode plot when using a logarithmic scale.
Lin—Sets the horizontal axis to a linear scale.
F—The final frequency
I—The initial frequencyofofthe
thesweep.
sweep.Click
Clickininthe
theunit
unitfield
fieldto
toselect
selectthe
thedesired
desiredunits
unitsfor
forthe
the
frequency.
When measuring voltage gain, the vertical axis shows the ratio of the circuit’s output voltage
to its input voltage. For a logarithmic base, the units are decibels. When measuring phase,
the vertical axis always shows the phase angle in degrees. Regardless of the units, you can
set initial (I) and final (F) values for the axis using the Bode plotter’s controls.
Log—Sets the vertical or y-axis to a logarithmic scale.
Lin—Sets the vertical or y-axis to a linear scale. This is the only option when Phase is
Vertical
I—The initial value for the vertical or y-axis. Click in the unit field to select the desired units
for this axis. If Log is selected, the unit is dB (decibels) and cannot be changed. If Phase is
selected, the unit is Deg (degrees) and cannot be changed.
F—The final value for the vertical or y-axis . Click in the unit field to select the desired units
for this axis. If Log is selected, the unit is dB (decibels) and cannot be changed. If Phase is
selected, the unit is Deg (degrees) and cannot be changed.
Reverse—Reverses the graphical display’s background color; toggles between white and
Save—Saves results as a bode data (.bod) or binary (.tdm) file. If you choose to save the
results as a.tdm file, the Data resampling settingsdialog box displays. Along with the
Controls
plotter’s Save button, you can also save simulation results in the Grapher.
Set—Click to display the Settings dialog box. Enter the desired number of Resolution
points and click Accept to return to the Bode plotter.
1.0kΩ 1.0kΩ
V2
V1
C1 12 Vrms L1
12 Vrms 1µF 60 Hz 1mH
60 Hz 0°
0°
6) Move the cursor until it reaches approximately -3dB. Record this cut-off frequency in the table
above.
7) Observe what happened to the bode plot.
APPLICATION
Given the circuit below, obtain the transfer function Io(s)/Ii(s) and bode plot using both MATLAB and
MULTISIM.
CONCLUSION