1. NATIONAL COLLEGE OF SCIENCE & TECHNOLOGY
Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite
EXPERIMENT 5
Fourier Theory – Frequency Domain and Time Domain
Cauan, Sarah Krystelle P. 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. 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.

4. 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 (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.

8. 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
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.
T = 1.00 ms

9. 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)?
Both the fundamental sine wave and the square wave frequency are 1 kHz. 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)?
The sine wave harmonic frequencies and the sine wave fundamental frequency
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 each harmonic sine wave generators is the quotient of amplitude
of the fundamental sine wave generator to the sine wave harmonic frequencies.
That is why it decreases as the sine wave harmonic frequencies increases.
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.
The waveshape shifted upwards. This was caused 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.

11. 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. The square wave is consist of a fundamental and all odd
harmonics but because of the absence of the sine wave harmonics, the waveshape
became the sine wave fundamental
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.
T = 1.00 ms

12. 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?
The triangular wave frequency and the sine wave fundamental frequency 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 sine wave harmonic frequencies and the sine wave fundamental frequency
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 each harmonic sine wave generators is the quotient of amplitude
of the fundamental sine wave generator to the square of the sine wave harmonic
frequencies. That is why it decreases as the sine wave harmonic frequencies
increases.
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.

13. Question: What happened to the triangular wave curve plot? Explain.
The waveshape shifted upwards. This was caused by the additional 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.
The triangular wave curve plot became sinusoidal. The triangular wave is consist
of a fundamental and all odd harmonics but because the sine wave harmonics are
missing, the waveshape became the sine wave fundamental

14. 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, they both are square wave.
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 calculated duty cycle and the duty cycle setting on the function generator
have difference 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.

15. 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?
For a series of periodic pulse with 50% duty cycle, the frequency domain consists
of a fundamental and odd harmonics while the even harmonics are almost
negligible.
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 each odd harmonic decreases as the fundamental frequency for a
square wave. Also, the plot in the frequency domain consist of a fundamental and
all odd harmonics, with the even harmonics missing
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 11 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.

16. Question: Are the filter input (red) and output (blue) curve plots the same shape, disregarding
any amplitude differences?
No. They do not have the same shape.
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?
Compare with the previous table, the amplitude of the harmonics is lower.
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 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.

17. 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 almost 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.
5.041 kHz
Question: Is this the frequency spectrum you expected for a square wave with duty cycle less
than 50%?
Yes.

18. 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?
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 = 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.
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?

19. 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.

20. 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.