Download Experiment 1-7

PHYSC 3322
Experiment 1.7
28 June, 2017
Frequency Analysis of Electronic Signals
Purpose
This experiment will introduce you to the basic principles of frequency domain analysis
of electronic signals. In particular, you will study the Fast Fourier Transform (FFT),
which is a convenient and powerful tool for performing frequency domain analysis on
a variety of signals.
Equipment
HP 54603B oscilloscope, HP 54659B measurement/storage module, and universal
function generators.
Background
Normally, when a signal is measured with an oscilloscope, it is viewed in the time
domain. That is, the vertical axis is voltage and the horizontal axis is time. For many
signals, this is the most logical and intuitive way to view them. However, when the
frequency content of the signal is of interest, it makes sense to view the signal in the
frequency domain. In this case, the horizontal axis becomes frequency. Using the
Fourier theory, one can mathematically relates the time domain with the frequency
domain. The Fourier transform is given by:

V(f) 
 v(t) e
-i2ft
dt

where v(t) is a voltage signal represented in the time domain and V(f) is its Fourier
transform in the frequency domain. Some typical signals represented in the time
domain and the frequency domain are shown in the figure below.
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28 June, 2017
The discrete version of the Fourier transform shown in the above equation is called the
Discrete Fourier Transform (DFT). The Fast Fourier Transform (FFT) is a
computationally efficient algorithm for computing the DFT. It takes digitized time
domain data and computes the frequency domain representation. The FFT function in
the HP 54603 scope (equipped with the HP 54659B Measurement/Storage Module)
uses 1024 time-domain data points and produces a frequency-domain display with 512
points. This display extends in frequency from fmin to feff/2, where feff is the effective
sampling rate of the time record. The effective sampling rate feff is the reciprocal of the
time between two adjacent samples and depends on the Time/Div setting of the scope.
For the HP 54603 scope, feff is given by: feff=1024/(10 *Time/Div). The minimal
frequency fmin is given by: fmin= feff/1024, which is equal to the reciprocal of the total
duration of the time record.
Procedure
The following four laboratory exercises are designed to illustrate the relationship
between frequency resolution, effective signal sampling rate, spectral leakage,
windowing, and aliasing in the FFT. Prior to the experiment, you need to check and
make sure that the Measurement/Storage Module (HP 54659B) has been installed on
the back of the scope.
Exercise 1. This exercise illustrates the relationship between the effective sampling rate
and the resulting frequency resolution for spectral analysis using the FFT. The spectral
leakage properties of the rectangular and Hanning windows are also demonstrated.
Connect a 3.5V, 1kHz sinusoidal input signal to channel 1 of the scope. Display the
time-domain waveform and measure the exact frequency of the signal using the Time
Measurements functions of the scope. Next, go to the Function menu (by pressing the
+- key), turn on Function 2, and adjust the FFT Menu settings, such as Units/Div, Cent.
Freq., Freq. Span, and Time/Div, so that you can get a nice peak in the FFT display.
(See the attached Measurement/Storage Module User Guide, p.28, for assistance in
adjusting the FFT Menu settings.) Use the Curses and Find Peak features to measure
the fundamental frequency of the peak. Use the Time/Div control to set the sampling
rate to 50 kSa/s. Adjust the FFT settings and find the main lobe width for the Hanning
window and the rectangular window, respectively. Note the difference in the spectral
leakage for the two windows. Print out the FFT spectra under the two different
windows. (You may need to connect your scope to a computer and use the computer to
download the data and print them out). Repeat the above measurements using an
effective sampling rate of 10 kSa/s.
Answer the following questions:
1.
Should the effective sampling rate be increased or decreased in order to improve
the frequency resolution of the FFT?
2.
Is there a limit on the spectral resolution capabilities of a fixed, 1024 point FFT?
3.
Does the Hanning window exhibit more or less spectral leakage when compared
with the rectangular window?
Exercise 2. The goal of this exercise is to study the relationship between the effective
sampling rate feff and aliasing. The upper frequency feff/2 for the FFT spectrum is also
called the folding frequency. Frequencies that would normally appear above f eff/2 (and
hence outside the useful range of the FFT) are folded back into the frequency domain of
the display. These unwanted frequency components are called aliases, since they
erroneously appear under the alias of another frequency. Aliasing is avoided if the
effective sampling rate is greater than twice the bandwidth of the signal being
measured. Connect a 3.5V, 10 kHz sinusoidal input signal to channel 1 of the scope.
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Experiment 1.7
28 June, 2017
Use Autoscale to check the time-domain waveform. Turn on the FFT Function and
adjust the FFT Menu settings, so that you can get a nice peak in the central region of the
FFT display. Use the Time/Div control to set the sampling rate to 50 kSa/s. Now you
can slowly increase the frequency of the sinusoid to roughly 24 kHz, allowing the FFT
display to stabilize at several points along the way. You should see the FFT peak move
to the right as the sinusoidal frequency is increased. Print out a FFT spectrum. Aliasing
occurs as the frequency of the sinusoid exceeds 25 kHz. As the frequency is swept from
25 kHz to 50 kHz, the main peak moves to left on the display. As the frequency is
further increased from 50 kHz to 75 kHz, the main peak moves to the right. Set the
sinusoid frequency to 40 kHz. Use the Cursors and Find Peaks functions to measure
the peak frequency displayed by the FFT. Because of aliasing, the FFT should
erroneously indicate that the peak occurs at about 10 kHz. Print out this FFT spectrum.
Repeat the above measurement using an effective sampling rate of 100 kSa/s.
Answer the following questions:
4.
If a 120 kHz sinusoid is sampled at 50 kSa/s, at what frequency will the aliased 100
kHz component appear in the FFT display?
5.
Is aliasing affected by the choice of window functions?
6.
Since the spectral resolution of the FFT is limited by the effective sampling rate,
and since we must sample above the signal frequency to avoid aliasing, is it
possible to improve spectral resolution by zero padding?
Exercise 3. This exercise demonstrates the use of the FFT for analyzing the spectral
content of a square wave and a triangle wave. Unlike the sinusoidal wave, which has
only one frequency component, the square and triangle waves have many frequency
components. As a result, their FFT spectra will show many peaks. The amplitude of
these peaks can be calculated exactly and the table below shows the Fourier Series
coefficients (magnitude only) of a square wave and a triangle wave.
Square Wave
Triangle Wave
Harmonic
Magnitude (dB)
Harmonic
Magnitude (dB)
1
-9.943
1
-4.842
3
-19.485
3
-26.924
5
-23.922
5
-35.788
7
-26.845
7
-41.618
9
-29.028
9
-45.963
11
-30.771
11
-49.423
13
-33.222
13
-52.295
Connect a 2V, 500kHz square wave to channel 1 of the scope. Use Autoscale to check
the time-domain waveform. Turn on the FFT Function and adjust the FFT Menu
settings, so that you can see many peaks in the FFT display. Use the Time/Div control
to set the sampling rate to 100 MSa/s. Find the relative amplitude difference between
the peaks and compare your results with the theoretical values shown in the table.
Notice that the flat top window provides the most accurate measurement of the peak
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amplitude. Next, go to the sampling rate of 5 MSa/s and show how the higher
harmonics are aliased and appear in the low frequency range of the FFT display. Make
sure to print out your FFT spectra. Repeat the above measurements for a 1V, 500kHz
triangle wave.
Answer the following questions:
7.
Is it necessary to have a stable time-domain display in order to analyze the
frequency content of a signal?
8.
Is prior knowledge of the signal’s bandwidth required?
9.
Is it still possible to obtain useful information from the FFT display when
components of the signal are aliased?
Exercise 4. This last exercise illustrates how the FFT can be used to analyze the spectral
content of a signal, which consists of a sum of two sinusoids. A similar analysis using
time-domain techniques can be difficult to perform on an oscilloscope. Connect two
sinusoidal sources in parallel to a 1k resistor. You may set the first source V1 at 3.5V,
1kHz and the second source V2 at 3.5V, 2kHz. Use a scope to obtain a time-domain
display of the voltage across the resistor. Notice that you may need to use the Run and
Stop controls to take “snapshots” of the resulting oscilloscope traces, because the timedomain display of the two sinusoidal waves is unstable. Turn on the FFT Function and
adjust the FFT Menu settings, so that you can see two peaks on the FFT display. Find
the frequency location of the two peaks and get a hard copy of your FFT display.
Slowly reduce the frequency of the source V2 and find the minimum frequency
difference, at which the two peaks can still be resolved. (You may need to adjust the
FFT Menu settings to “zoom-in” on the two peaks.) Adjust the Time/Div control to see
how the frequency resolution is affected by the sampling rate. Now, increase the
frequency of the source V2 up to 20kHz and watch how the FFT display changes with
the frequency increase. Notice that when the sampling rate is reduced to 20 kSa/s, the
high frequency term V2 aliases to a lower frequency near the 1kHz term.
Answer the following questions:
10. How does the presence of a high frequency component affect the spectral
resolution for analyzing narrow band components of the signal?
11. Based on the results of this exercise, why would it be difficult to use the FFT to
analyze the bandwidth of a “typical” amplitude modulated signal?
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