1. NATIONAL COLLEGE OF SCIENCE & TECHNOLOGY
Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite
EXPERIMENT 5
Fourier Theory – Frequency Domain and Time Domain
Boringot, Jeffrey B. September 1, 2011
Signal Spectra and Signal Processing/BSECE 41A1 Score:
Engr. Grace Ramones
Instructor

2. 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. Sample Computation
BW =

4. 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

5. 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 (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
(tO) 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

6. 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.

7. Figure 5 – 1 Square Wave Fourier Series
XSC1
V6
15 R1 1 J1 Ext T rig
+
_ 0
10.0kΩ A B
10 V _ _
+ +
Key = A
V1
R2 J2
9 2
10 Vpk 10.0kΩ 6
1kHz Key = B
0° V2
10 R3 3 J3 R7
100Ω
3.33 Vpk 10.0kΩ
3kHz 0
V3 Key = C
0°
12 R4 4 J4
2 Vpk 10.0kΩ
5kHz Key = D
0° V4
14 R5 5 J5
1.43 Vpk 10.0kΩ
7kHz
0° Key = E
V5 J6
R6 8
0 13
1.11 Vpk 10.0kΩ
9kHz Key = F
0°
Figure 5 – 2 Triangular Wave Fourier Series
XSC1
V6
12 R1 1 J1 Ext T rig
+
_ 0
10.0kΩ A B
10 V _ _
+ +
Key = A
V1
R2 J2
13 2
10 Vpk 10.0kΩ
1kHz Key = B
90° V2
8 R3 3 J3 R7
1.11 Vpk 100Ω
10.0kΩ
3kHz 0
90° V3 Key = C
R4 J4 6
9 4
0.4 Vpk 10.0kΩ
5kHz
90° V4 Key = D
0 11 R5 5 J5
0.2 Vpk 10.0kΩ
7kHz
90° Key = E

8. 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.
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,
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

9. 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.

10. 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 are both 1 kHz. They have the same value.

11. 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 frequency is different with the sine wave
fundamental. The harmonics frequency has lot of ripples.
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.
Question: What happened to the square wave curve plot? Explain why.
The amplitude increased. This is due to the dc voltage applied to the previous circuit.
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.

12. 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.
It became sinusoidal because the harmonic frequency generators had
been eliminated.
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.

13. 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.
T = 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?
They 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)?
Each harmonic frequency is an odd multiple of the fundamental.
What is the relationship between the amplitude of the harmonic generators and the
amplitude of the fundamental generator?

14. 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.
Question: What happened to the triangular wave curve plot? Explain.
The waveshape shifted up. It is because dc voltage is added to a periodic
time varying voltage; the waveshape will be shifted up by the amount of
the dc voltage.
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.

15. 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

16. 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?
They have only diffence of 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

17. 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 wave consists of odd harmonics while the even harmonics are
almost zero.
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.
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, because the more number of harmonics
square wave the more distortion in the input square wave.
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, the input is square wave while the output is a sinusoidal wave.

18. 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 result is lower than the previous table.
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?
There should be less than 5 square wave harmonics to be passed by this
filter. Yes, there have much distortion in the input square wave at the
filter output.
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

19. 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, they have the same shape.
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 = 198.199 µs
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?
The values of both D are the same.
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.

20. Question: Is this the frequency spectrum you expected for a square wave with duty
cycle less than 50%?
Yes, it is what I expected for 50% duty cycle.
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.
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 117 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 = 4.719 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, because BW is inversely proportion to the distortion formed. Then,
the higher the bandwidth, the lesser the distortion formed.
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, they do not have the same shape.

21. 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?
fc is greater than BW required.
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, if the bandwidth is reduced, there will occur much distortion of the
input pulse waveshape at the filter output .
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 inversely proportional. As the number of the harmonics increase, the amplitude
decrease.

22. Conclusion:
Based on the experiment, more harmonics added to the sine wave will result or
generate a more complex waveshape. Square wave is used in the bode plot so harmonics are
easily observed. Triangular wave is used in the spectrum analyzer to see the difference of the
frequencies from each switch. Square wave consists of only fundamental frequency and the
odd harmonics. In direct proportion to the odd harmonic frequency, the amplitude of each
odd harmonic will decrease. For triangular wave, the amplitude of each harmonic will
decrease in direct proportion to the square of the odd harmonic frequency. Moreover,
decreasing the number of harmonic, the more ripple will appear to the generated square wave
curve plot.