1. Objectives:
1. Learn how a square wave can be produced from a series of sine waves at different frequencies and
amplitudes.
2. Learn how a triangular can be produced from a series of cosine waves at different frequencies and
amplitudes.
3. Learn about the difference between curve plots in the time domain and the frequency domain.
4. Examine periodic pulses with different duty cycles in the time domain and in the frequency domain.
5. Examine what happens to periodic pulses with different duty cycles when passed through low-pass
filter when the filter cutoff frequency is varied.
3. Data Sheet:
Materials:
One function generator
One oscilloscope
One spectrum analyzer
One LM 741 op-amp
Two 5 nF variable capacitors
Resistors: 5.86 kΩ, 10 kΩ, and 30 kΩ
Theory:
Communications systems are normally studies using sinusoidal voltage waveforms to simplify the analysis.
In the real world, electrical information signal are normally nonsinusoidal voltage waveforms, such as audio
signals, video signals, or computer data. Fourier theory provides a powerful means of analyzing
communications systems by representing a nonsinusoidal signal as series of sinusoidal voltages added
together. Fourier theory states that a complex voltage waveform is essentially a composite of harmonically
related sine or cosine waves at different frequencies and amplitudes determined by the particular signal
waveshape. Any, nonsinusoidal periodic waveform can be broken down into sine or cosine wave equal to
the frequency of the periodic waveform, called the fundamental frequency, and a series of sine or cosine
waves that are integer multiples of the fundamental frequency, called the harmonics. This series of sine or
cosine wave is called a Fourier series.
Most of the signals analyzed in a communications system are expressed in the time domain, meaning that
the voltage, current, or power is plotted as a function of time. The voltage, current, or power is represented
on the vertical axis and time is represented on the horizontal axis. Fourier theory provides a new way of
expressing signals in the frequency domain, meaning that the voltage, current, or power is plotted as a
function of frequency. Complex signals containing many sine or cosine wave components are expressed as
sine or cosine wave amplitudes at different frequencies, with amplitude represented on the vertical axis
and frequency represented on the horizontal axis. The length of each of a series of vertical straight lines
represents the sine or cosine wave amplitudes, and the location of each line along the horizontal axis
represents the sine or cosine wave frequencies. This is called a frequency spectrum. In many cases the
frequency domain is more useful than the time domain because it reveals the bandwidth requirements of
the communications system in order to pass the signal with minimal distortion. Test instruments displaying
signals in both the time domain and the frequency domain are available. The oscilloscope is used to display
signals in the time domain and the spectrum analyzer is used to display the frequency spectrum of signals in
the frequency domain.
In the frequency domain, normally the harmonics decrease in amplitude as their frequency gets higher until
the amplitude becomes negligible. The more harmonics added to make up the composite waveshape, the
more the composite waveshape will look like the original waveshape. Because it is impossible to design a
communications system that will pass an infinite number of frequencies (infinite bandwidth), a perfect
reproduction of an original signal is impossible. In most cases, eliminate of the harmonics does not
significantly alter the original waveform. The more information contained in a signal voltage waveform
4. (after changing voltages), the larger the number of high-frequency harmonics required to reproduce the
original waveform. Therefore, the more complex the signal waveform (the faster the voltage changes), the
wider the bandwidth required to pass it with minimal distortion. A formal relationship between bandwidth
and the amount of information communicated is called Hartley’s law, which states that the amount of
information communicated is proportional to the bandwidth of the communications system and the
transmission time.
Because much of the information communicated today is digital, the accurate transmission of binary pulses
through a communications system is important. Fourier analysis of binary pulses is especially useful in
communications because it provides a way to determine the bandwidth required for the accurate
transmission of digital data. Although theoretically, the communications system must pass all the
harmonics of a pulse waveshape, in reality, relatively few of the harmonics are need to preserve the
waveshape.
The duty cycle of a series of periodic pulses is equal to the ratio of the pulse up time (t O) to the time period
of one cycle (T) expressed as a percentage. Therefore,
In the special case where a series of periodic pulses has a 50% duty cycle, called a square wave, the plot in
the frequency domain will consist of a fundamental and all odd harmonics, with the even harmonics
missing. The fundamental frequency will be equal to the frequency of the square wave. The amplitude of
each odd harmonic will decrease in direct proportion to the odd harmonic frequency. Therefore,
The circuit in Figure 5–1 will generate a square wave voltage by adding a series of sine wave voltages as
specified above. As the number of harmonics is decreased, the square wave that is produced will have
more ripples. An infinite number of harmonics would be required to produce a perfectly flat square wave.
5. Figure 5 – 1 Square Wave Fourier Series
XSC1
Ext T rig
V6 +
R1 J1 _
A B
10.0kΩ + _ + _
10 V
Key = A
V1
R2 J2
10 Vpk 10.0kΩ
1kHz Key = B
0° V2
R3 J3 4
155
0
8
160
14
13
12 R7
109
02
3 100Ω
3.33 Vpk 10.0kΩ
3kHz Key = C
0° V3
R4 J4
2 Vpk 10.0kΩ
5kHz Key = D
0° V4
R5 J5
1.43 Vpk 10.0kΩ
7kHz
0° Key = E
V5 J6
R6
1.11 Vpk 10.0kΩ
9kHz Key = F
0° .
The circuit in Figure 5-2 will generate a triangular voltage by adding a series of cosine wave voltages. In
order to generate a triangular wave, each harmonic frequency must be an odd multiple of the fundamental
with no even harmonics. The fundamental frequency will be equal to the frequency of the triangular wave,
the amplitude of each harmonic will decrease in direct proportion to the square of the odd harmonic
frequency. Therefore,
Whenever a dc voltage is added to a periodic time varying voltage, the waveshape will be shifted up by the
amount of the dc voltage.
6. Figure 5 – 2 Triangular Wave Fourier Series
XSC1
Ext T rig
V6 +
R1 J1 _
A B
10.0kΩ + _ + _
15 V
Key = A
V1
R2 J2
10 Vpk 10.0kΩ
1kHz
90° V2 Key = B
R3 J3 13
12
1
2
3
4
5
8
9
11
0 R7
6
0
1.11 Vpk 100Ω
10.0kΩ
3kHz
90° V3 Key = C
R4 J4
0.4 Vpk 10.0kΩ
5kHz
90° V4 Key = D
R5 J5
0.2 Vpk 10.0kΩ
7kHz
90° Key = E
For a series of periodic pulses with other than a 50% duty cycle, the plot in the frequency domain will
consist of a fundamental and even and odd harmonics. The fundamental frequency will be equal to the
frequency of the periodic pulse train. The amplitude (A) of each harmonic will depend on the value of the
duty cycle. A general frequency domain plot of a periodic pulse train with a duty cycle other than 50% is
shown in the figure on page 57. The outline of peaks if the individual frequency components is called
envelope of the frequency spectrum. The first zero-amplitude frequency crossing point is labelled fo = 1/to,
there to is the up time of the pulse train. The first zero-amplitude frequency crossing point fo) determines
the minimum bandwidth (BW0 required for passing the pulse train with minimal distortion.
Therefore,
7. A
f=1/to 2/to f
Frequency Spectrum of a Pulse Train
Notice than the lower the value of to the wider the bandwidth required to pass the pulse train with minimal
distortion. Also note that the separation of the lines in the frequency spectrum is equal to the inverse of
the time period (1/T) of the pulse train. Therefore a higher frequency pulse train requires a wider
bandwidth (BW) because f = 1/T
The circuit in Figure 5-3 will demonstrate the difference between the time domain and the frequency
domain. It will also determine how filtering out some of the harmonics effects the output waveshape
compared to the original3 input waveshape. The frequency generator (XFG1) will generate a periodic pulse
waveform applied to the input of the filter (5). At the output of the filter (70, the oscilloscope will display
the periodic pulse waveform in the time domain, and the spectrum analyzer will display the frequency
spectrum of the periodic pulse waveform in the frequency domain. The Bode plotter will display the Bode
plot of the filter so that the filter bandwidth can be measured. The filter is a 2-pole low-pass Butterworth
active filter using a 741 op-amp.
Figure 5-3 Time Domain and Frequency Domain
XFG1
XSC1
C1 XSA1
Ext T rig
+
2.5nF 50% _
Key=A A
_
B
_
IN T
+ +
R1 R2 741
30kΩ 30kΩ
42
OPAMP_3T_VIRTUAL
0
6
0
31
R3
C2 R4
5.56kΩ
10kΩ
XBP1
2.5nF 50%
Key=A
R5 IN OUT
10kΩ
8. Procedure:
Step 1 Open circuit file FIG 5-1. Make sure that the following oscilloscope settings are selected: Time base
(Scale = 200 µs/Div, Xpos = 0, Y/t), Ch A (Scale = 5V/Div, Ypos = 0, DC), Ch B (Scale = 50 mV/Div,
Ypos = 0, DC), Trigger (Pos edge, Level = 0, Auto). You will generate a square wave curve plot on the
oscilloscope screen from a series of sine waves called a Fourier series.
Step 2 Run the simulation. Notice that you have generated a square wave curve plot on the oscilloscope
screen (blue curve) from a series of sine waves. Notice that you have also plotted the fundamental
sine wave (red). Draw the square wave (blue) curve on the plot and the fundamental sine wave
(red) curve plot in the space provided.
Step 3 Use the cursors to measure the time periods for one cycle (T) of the square wave (blue) and the
fundamental sine wave (red) and show the value of T on the curve plot.
T1 = 1.00 ms T2 = 1.00 ms
Step 4 Calculate the frequency (f) of the square wave and the fundamental sine wave from the time
period.
f = 1 kHz
Questions: What is the relationship between the fundamental sine wave and the square wave frequency
(f)?
They have the same frequency.
What is the relationship between the sine wave harmonic frequencies (frequencies of sine wave generators
f3, f5, f7, and f9 in figure 5-1) and the sine wave fundamental frequency (f1)?
The sine wave harmonic frequencies are all odd functions.
What is the relationship between the amplitude of the harmonic sine wave generators and the amplitude
of the fundamental sine wave generator?
The amplitude of the odd harmonics will decrease in direct proportion to odd harmonic frequency.
Step 5 Press the A key to close switch A to add a dc voltage level to the square wave curve plot. (If the
switch does not close, click the mouse arrow in the circuit window before pressing the A key). Run
the simulation again. Change the oscilloscope settings as needed. Draw the new square wave (blue)
curve plot on the space provided.
9. Question: What happened to the square wave curve plot? Explain why.
The square wave shifted upward. It is because of by the additional dc voltage.
Step 6 Press the F and E keys to open the switches F and E to eliminate the ninth and seventh harmonic
sine waves. Run the simulation again. Draw the new curve plot (blue) in the space provided. Note
any change on the graph.
Step 7 Press the D key to open the switch D to eliminate the fifth harmonics sine wave. Run the simulation
again. Draw the new curve plot (blue) in the space provided. Note any change on the graph.
Step 8 Press the C key to open switch C and eliminate the third harmonic sine wave. Run the simulation
again.
Question: What happened to the square wave curve plot? Explain.
Square wave became sinusoidal wave. This is because all the harmonics are missing.
10. Step 9 Open circuit file FIG 5-2. Make sure that the following oscilloscope settings are selected: Time base
(Scale = 200 µs/Div, Xpos = 0, Y/t), Ch A (Scale = 5V/Div, Ypos = 0, DC), Ch B (Scale = 100 mV/Div,
Ypos = 0, DC), Trigger (Pos edge, Level = 0, Auto). You will generate a triangular wave curve plot on
the oscilloscope screen from a series of sine waves called a Fourier series.
Step 10 Run the simulation. Notice that you have generated a triangular wave curve plot on the
oscilloscope screen (blue curve) from the series of cosine waves. Notice that you have also plotted
the fundamental cosine wave (red). Draw the triangular wave (blue) curve plot and the
fundamental cosine wave (red) curve plot in the space provided.
Step 11 Use the cursors to measure the time period for one cycle (T) of the triangular wave (blue) and the
fundamental (red), and show the value of T on the curve plot.
T1 = 1.00 ms T2 = 1.00 ms
Step 12 Calculate the frequency (f) of the triangular wave from the time period (T).
f = 1 kHz
Questions: What is the relationship between the fundamental frequency and the triangular wave
frequency?
Both frequencies are the same.
What is the relationship between the harmonic frequencies (frequencies of generators f3, f5, and f7 in figure
5-2) and the fundamental frequency (f1)?
The frequencies are all odd functions.
What is the relationship between the amplitude of the harmonic generators and the amplitude of the
fundamental generator?
The amplitude of the harmonic generators decreases in direct proportion to the square of the
odd harmonic frequency
Step 13 Press the A key to close switch A to add a dc voltage level to the triangular wave curve plot. Run the
simulation again. Draw the new triangular wave (blue) curve plot on the space provided.
11. Question: What happened to the triangular wave curve plot? Explain.
The wave shifts upward. The reason for this was the additional dc voltage applied to the output.
Step 14 Press the E and D keys to open switches E and D to eliminate the seventh and fifth harmonic sine
waves. Run the simulation again. Draw the new curve plot (blue) in the space provided. Note any
change on the graph.
Step 15 Press the C key to open the switch C to eliminate the third harmonics sine wave. Run the simulation
again.
Question: What happened to the triangular wave curve plot? Explain.
It became sine wave, because the harmonic sine waves had already been eliminated.
Step 16 Open circuit FIG 5-3. Make sure that following function generator settings are selected: Square
wave, Freq = 1 kHz, Duty cycle = 50%, Ampl – 2.5 V, Offset = 2.5 V. Make sure that the following
oscilloscope settings are selected: Time base (Scale = 500 µs/Div, Xpos = 0, Y/T), Ch A (Scale = 5
V/Div, Ypos = 0, DC), Ch B (Scale = 5 V/Div, Ypos = 0, DC), Trigger (pos edge, Level = 0, Auto).
You will plot a square wave in the time domain at the input and output of a two-pole low-pass
Butterworth filter.
Step 17 Bring down the oscilloscope enlargement and run the simulation to one full screen display,
then pause the simulation. Notice that you are displaying square wave curve plot in the time
domain (voltage as a function of time). The red curve plot is the filter input (5) and the blue
curve plot is the filter output (7)
Question: Are the filter input (red) and the output (blue) plots the same shape disregarding any amplitude
differences?
Yes.
12. Step 18 Use the cursor to measure the time period (T) and the time (fo) of the input curve plot (red) and
record the values.
T= 1 ms to = 500.477µs
Step 19 Calculate the pulse duty cycle (D) from the to and T
D = 50.07%.
Question: How did your calculated duty cycle compare with the duty cycle setting on the function
generator?
The difference is 0.07%.
Step 20 Bring down the Bode plotter enlargement to display the Bode plot of the filter. Make sure that
the following Bode plotter settings are selected; Magnitude, Vertical (Log, F = 10 dB, I = -40 dB),
Horizontal (Log, F = 200 kHz, I = 100 Hz). Run the simulation to completion. Use the cursor to
measure the cutoff frequency (fC) of the low-pass filter and record the value.
fC = 21.197
Step 21 Bring down the analyzer enlargement. Make sure that the following spectrum analyzer settings
are selected: Freq (Start = 0 kHz, Center = 5 kHz, End = 10 kHz), Ampl (Lin, Range = 1 V/Div), Res
= 50 Hz. Run the simulation until the Resolution frequencies match, then pause the simulation.
Notice that you have displayed the filter output square wave frequency spectrum in the
frequency domain, use the cursor to measure the amplitude of the fundamental and each
harmonic to the ninth and record your answers in table 5-1.
Table 5-1
Frequency (kHz) Amplitude
f1 1 5.048 V
f2 2 11.717 µV
f3 3 1.683 V
f4 4 15.533 µV
f5 5 1.008 V
f6 6 20.326 µV
f7 7 713.390 mV
f8 8 25.452 µV
f9 9 552.582 mV
Questions: What conclusion can you draw about the difference between the even and odd harmonics for a
square wave with the duty cycle (D) calculated in Step 19?
The even harmonics is much lower compared with the odd harmonics, proving that the
wave is an odd function.
What conclusions can you draw about the amplitude of each odd harmonic compared to the fundamental
for a square wave with the duty cycle (D) calculated in Step 19?
The amplitude of odd harmonics decreases in direct proportion with the odd harmonic
frequency. Also, the plot in the frequency domain consist of a fundamental and all odd
harmonics, with the even harmonics missing
13. Was this frequency spectrum what you expected for a square wave with the duty cycle (D) calculated in
Step 19?
Yes.
Based on the filter cutoff frequency (fC) measured in Step 20, how many of the square wave harmonics
would you expect to be passed by this filter? Based on this answer, would you expect much distortion of
the input square wave at the filter? Did your answer in Step 17 verify this conclusion?
There are 21 square waves. Yes, I expect much distortion of the input square wave at the
filter,.
Step 22 Adjust both filter capacitors (C) to 50% (2.5 nF) each. (If the capacitors won’t change, click the
mouse arrow in the circuit window). Bring down the oscilloscope enlargement and run the
simulation to one full screen display, then pause the simulation. The red curve plot is the filter
input and the blue curve plot is the filter output.
Question: Are the filter input (red) and output (blue) curve plots the same shape, disregarding any
amplitude differences?
No.
Step 23 Bring down the Bode plotter enlargement to display the Bode plot of the filter. Use the cursor
to measure the cutoff frequency (Fc of the low-pass filter and record the value.
fc = 2.12 kHz
Step 24 Bring down the spectrum analyzer enlargement to display the filter output frequency spectrum
in the frequency domain, Run the simulation until the Resolution Frequencies match, then
pause the simulation. Use cursor to measure the amplitude of the fundamental and each
harmonic to the ninth and record your answers in Table 5-2.
Table 5-2
Frequency (kHz) Amplitude
f1 1 4.4928 V
f2 2 4.44397µV
f3 3 792.585 mV
f4 4 323.075 µV
f5 5 178.663mV
f6 6 224.681 µV
f7 7 65.766 mV
f8 8 172.430 µV
f9 9 30.959 mV
Questions: How did the amplitude of each harmonic in Table 5-2 compare with the values in Table 5-1?
The amplitude of the harmonics is lower compare with the previous result.
Based on the filter cutoff frequency (fc), how many of the square wave harmonics should be passed by this
filter? Based on this answer, would you expect much distortion of the input square wave at the filter
output? Did your answer in Step 22 verify this conclusion?
Based on the fc, there are less than 5 square wave harmonics. Yes, there is much distortion
of the input square wave at output.
14. Step 25 Change the both capacitor (C) back to 5% (0.25 nF). Change the duty cycle to 20% on the
function generator. Bring down the oscilloscope enlargement and run the simulation to one full
screen display, then pause the simulation. Notice that you have displayed a pulse curve plot on
the oscilloscope in the time domain (voltage as a function of time). The red curve plot is the
filter input and the blue curve plot is the filter output.
Question: Are the filter input (red) and the output (blue) curve plots the same shape, disregarding any
amplitude differences?
Yes.
Step 26 Use the cursors to measure the time period (T) and the up time (to) of the input curve plot (red) and
record the values.
T= 1 ms to =
Step 27 Calculate the pulse duty cycle (D) from the to and T.
D = 19.82%
Question: How did your calculated duty cycle compare with the duty cycle setting on the function
generator?
Their difference is 0.18%
Step 28 Bring down the Bode plotter enlargement to display the Bode plot of the filter. Use the cursor
to measure the cutoff frequency (fC) of the low-pass filter and record the value.
fC = 21.197 kHz
Step 29 Bring down the spectrum analyzer enlargement to display the filter output frequency spectrum
in the frequency domain. Run the simulation until the Resolution Frequencies match, then
pause the simulation. Draw the frequency plot in the space provided. Also draw the envelope
of the frequency spectrum.
5.041 kHz
Question: Is this the frequency spectrum you expected for a square wave with duty cycle less than 50%?
Yes.
Step 30 Use the cursor to measure the frequency of the first zero crossing point (fo) of the spectrum
envelope and record your answer on the graph.
fo = 5.041 kHz
15. Step 31 Based on the value of the to measured in Step 26, calculate the expected first zero crossing
point (fo) of the spectrum envelope.
fo = 5.045 kHz
Question: How did your calculated value of fo compare the measured value on the curve plot?
They have a difference of 0.004 Hz
Step 32 Based on the value of fo, calculate the minimum bandwidth (BW) required for the filter to pass
the input pulse waveshape with minimal distortion.
BW = 5.045 kHz
Question: Based on this answer and the cutoff frequency (fc) of the low-pass filter measure in Step 28,
would you expect much distortion of the input square wave at the filter output? Did your answer in Step 25
verify this conclusion?
No, there is less distortion of the input square wave at the filter output. The higher the
bandwidth, the lesser the distortion.
Step 33 Adjust the filter capacitors (C) to 50% (2.5 nF) each. Bring down the oscilloscope enlargement
and run the simulation to one full screen display, then pause the simulation. The red curve plot
is the filter input and the blue curve plot is the filter output.
Question: Are the filter input (red) and the output (blue) curve plots the same shape, disregarding any
amplitude differences?
No..
Step 34 Bring down the Bode plotter enlargement to display the Bode plot of the filter. Use the cursor to
measure the cutoff frequency (fc) of the low-pass filter and record the value.
fc = 4.239 kHz
Questions: Was the cutoff frequency (fc) less than or greater than the minimum bandwidth (BW) required
to pass the input waveshape with minimal distortion as determined in Step 32?
The fc is greater than the BW required to pass the input waveshape with minimal distortion
Based on this answer, would you expect much distortion of the input pulse waveshape at the filter output?
Did your answer in Step 33 verify this conclusion?
No, there will have much distortion in the input waveshape at the output if the bandwidth is
reduced.
Step 35 Bring down the spectrum analyzer enlargement to display the filter output frequency spectrum
in the frequency domain. Run the simulation until the Resolution Frequencies match, then
pause the simulation.
Question: What is the difference between this frequency plot and the frequency plot in Step 29?
It is less than the amplitude in the frequency plot in Step 29.
