1. NATIONAL COLLEGE OF SCIENCE & TECHNOLOGY
Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite
EXPERIMENT 5
Fourier Theory – Frequency Domain and Time Domain
Tagasa, Jerald A. September 06, 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
Frequency
Duty Cycle
First zero crossing point (f)
Bandwidth
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 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

5. 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 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.

6. 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.

7. 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,

8. 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

9. 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 the same.
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)?
They are all odd multiples.
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 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.

10. Question: What happened to the square wave curve plot? Explain why.
It shifted upward. It is because of by the applied 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.
It became sinusoidal wave. This is because all the harmonics are
gone.
Step 9 Open circuit file FIG 5-2. Make sure that the following
oscilloscope settings are selected: Time base (Scale = 200 µs/Div,

11. 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?
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)?
They 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.

12. Question: What happened to the triangular wave curve plot? Explain.
It shifted upward. It is because of by the applied 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.
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 are gone.
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.

13. 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. It is 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.

14. 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 square waves. Yes, there is much 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.
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?
There are less than 5 square wave harmonics. Yes, there is much
distortion of the input square wave at 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
enlargement and run the simulation to one full screen display, then

15. 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
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?

16. The difference is 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.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?
In this frequency plot the amplitude is lower compared with the
frequency plot in Step 29?

17. Conclusion
I conclude that a square wave and a triangular wave can be broken down
into a series of sine function (for square wave) and cosine function (for
triangular wave). Both the square wave and the triangular wave are odd
function and the amplitude of the odd harmonics decrease in direct
proportion to odd harmonic frequency. Add to that, a nonsinusoidal function
is composed of fundamental sine wave and harmonics. The greater the value of
the harmonics, the more it will become complex function. When a dc supply is
added to the circuit, the wave will shift upward but not increase its Vp. If
a circuit has a duty cycle less than 20%, there will have even and odd
harmonics. Lastly, bandwidth is inversely proportional to the up time of the
pulse train. Bandwidth is also inversely proportional to the distortion of
the input.