Lab 3 - Using the Agilent 54621A Digital Oscilloscope as a Spectrum
Analyzer
Electronics Fundamentals using the Agilent 54621A Oscilloscope
By:
Walter Banzhaf
University of Hartford
Ward College of Technology
USA
Introduction
A spectrum analyzer is an instrument that creates a graph of amplitude versus frequency (contrasted with an
oscilloscope, which produces a graph of amplitude versus time). Spectrum analyzers are used extensively in RF
communications courses to see what frequencies are present in a signal containing information (e.g. an RF carrier
with modulation, such as a signal from an AM or FM broadcast transmitter). Another application of a spectrum
analyzer is to see the harmonics present in a waveform, such as a square wave or a pulse train.
The Agilent 54621A digital oscilloscope can produce displays of amplitude versus frequency by performing a Fast
Fourier Transform (FFT) on the data points in a display of amplitude versus time. While the FFT does not
give all the information and options that a spectrum analyzer does, it is a very useful feature of this instrument.
Equipment Required
•
Agilent 54621A Digital Oscilloscope with two 10X attenuating probes
•
Agilent 33250A or 33120A Function Generator
Procedure A – Measuring a Sine Wave in the Frequency Domain Using FFT:
1. Set the function generator to produce a 1 kHz, 2.83 Vpp sinusoid (this is 1 Vrms), and connect the output of the
generator to the oscilloscope. Use a 50 Ω termination on the generator output, and a 10X probe across the
resistor to connect the oscilloscope.
2. Choose the Default setup of the oscilloscope using the Save/Recall hardkey and then the Default Setup
softkey. Be sure to change the channel 1 Probe Factor (a softkey that says Probe) to 10:1. You can do this by
pressing the oval button labeled “1”, and then turning the control called the “Entry Knob”, located just to the
right of the CRT, below the horizontal section, with an illuminated curved arrow above it.
3. Press the Auto-Scale hardkey; you should see two periods of a sine wave, centered on display, with a sweep
speed of 200 µs per div and vertical sensitivity of 500 mV/div, as shown below:
4. Press the Math hardkey, and then press the FFT softkey. Then, press the Settings softkey to see the
frequency span of the display and the center frequency of the display. As shown below, the display
shows the sine wave in the time domain (upper trace) and in the frequency domain (lower trace). The frequency
domain display is not too useful as shown; there’s something big on the far left edge of the graticule, but it’s
hard to tell any specific information about it.
time/div = 200 µs
FFT sample rate = 1.00 MSa/s
Span = 500 kHz
Center = 250 kHz
Note: the FFT sample rate = 1.00 MSample/second and sweep time = 200 us/div.
The Agilent User’s Guide tells us two key performance specifications of the FFT mode of operation that depend on
the sweep time (time per division):
1) frequency resolution = 0.097656/(time per division), and
2) maximum frequency = 102.4/(time per division)
So, in our display above, the frequency resolution = 0.097656/200 µs = 488.3 Hz, and the maximum frequency
= 102.4/200 µs = 512 kHz. That creates a not very useful display; noise can be seen on the bottom, and there’s
something big on the far left edge of display, and we can’t tell any specific frequency information from this
display.
5. Now we’re going to make the frequency domain display quite a bit more useful, as follows:
Change the sweep speed (using the control in the Horizontal section), pausing to observe the display as each
change is made, from 200 µs to 10.0 ms/div.
Note: the new FFT sample rate = 20.0 kSa/s, freq. res. = 0.097656/10 ms = 9.766 Hz, max. freq. = 102.4/10 ms
= 10240 Hz. This produces a very useful display. Press the Settings softkey to see Span = 10 kHz and Center
(freq.) = 5 kHz (as shown below).
time/div = 10 ms
FFT sample rate = 20 kSa/s
Span = 10.0 kHz
Center = 5.00 kHz
There are two (2) graphs shown here: voltage
vs. time (over a 100 ms interval) and voltage
vs. frequency (in a “window” 10 kHz wide).
Notice that there is a lot of noise (at the bottom of the display), and one big frequency component (vertical line)
located 1 division from the left side of the graticule. Since the center frequency is 5 kHz, and the frequency span
is 10 kHz, the left edge of the display is 0 Hz, the right edge is 10 kHz, each horizontal division is 1 kHz, and the
big voltage component is at 1 kHz (the frequency of the sine wave).
6. Our next task is to measure the amplitude of the 2.83 Vpp 1 kHz signal in the frequency domain. First, turn off
the time-domain display by pressing the oval button labeled “1” in the Vertical section twice. You are now
looking at just the voltage vs. frequency display.
Now press the Quick Meas button, and the oscilloscope automatically tells us that “Max(Math) = -1.5 dB”. All
measurements using the FFT feature will be expressed in dBV (decibels referenced to 1 volt (RMS)). Since our
input sine wave is 2.83 Vpp = 1.0 Vrms, the correct amplitude of our 1 kHz sine wave, expressed in dBV, is
20*LOG(1V) = 0 dBV.
What we are seeing here is a slight error, which brings up another thing to learn about FFT: the type of “window”
that is used to generate the FFT is important. There are three types of windows available (the following is
extracted from the Agilent 54621A User’s Guide):
“Hanning window – window for making accurate frequency measurements or for resolving two
frequencies that are close together.”
“Flat Top window – window for making accurate amplitude measurements of frequency peaks.”
“Rectangular window – good frequency resolution and amplitude accuracy, but use only where there will be
no leakage effects. Use on self-windowing waveforms such as pseudo-random noise, impulses, sine bursts,
and decaying sinusoids.”
To get better amplitude accuracy, change the Window type from Hanning to Flat Top as follows: Press the
Math hardkey, then the Settings softkey, followed by the More FFT softkey. Press the Window softkey and
step the check mark down to Flat Top. Then press Quick Meas hardkey, and notice that the Max(Math) = 0 dB,
as in the display below. It’s really 0 dBV, and the “V” reference is understood.
time/div = 10 ms
FFT sample rate = 20 kSa/s
Span = 10.0 kHz
Center = 5.00 kHz
Window = Flat Top
Amplitude = 0 dBV
The noise at the bottom of the display is about
50 dB below the signal, as each vertical
division = 10 dB.
time/div = 10 ms
FFT sample rate = 20 kSa/s
Span = 10.0 kHz
Center = 5.00 kHz
Window = Flat Top
Amplitude = 0 dBV
The QuickMeas softkey X at Max was used
here to show the frequency = 1 kHz at the
cursor location.
Procedure B – Measuring a Square Wave Using FFT:
1. Assuming you have just completed Procedure A (measuring a sine wave using FFT), leave the oscilloscope set
as it was (10.0 ms/div sweep speed), and FFT turned on. Change the function generator output to produce a 1
kHz, 2.00 Vpp (bipolar) square wave (this is 1 Vrms).
2. The display should look like the one below:
Frequency components can now be seen at 1
kHz, 3 kHz, 5 kHz, 7 kHz and 9 kHz. That is
characteristic of a square wave with a 50% duty
cycle: only odd harmonics (integer multiples of
the fundamental frequency) will be present in its
frequency spectrum.
Note that both X and Y cursors have been
used.
The two X cursors are placed on the 1 kHz and 3
kHz components, and the ∆X = 2.00 kHz is the
difference between 3 kHz and 1 kHz.
The two Y cursors show the amplitudes of the 1
kHz and 3 kHz components, and the ∆Y of –9.69
rd
dB shows that the 3 harmonic amplitude is 9.69
st
dB below the 1 harmonic amplitude.
3. If we want to see the square wave that has the frequency spectrum shown above, all we have to do is press the
oval button labeled “1” in the Vertical section. This turns on the Channel 1 time-domain display. In the left
display below we can see that the square wave is not presented well; it’s hard to see individual cycles. While we
may be tempted to change the sweep speed (time per division) so that individual cycles can be seen in the time
domain, the result of doing this creates other problems in the frequency domain. In the right display below the
sweep speed has been changed to 500 µs /div.
Sweep speed = 10 ms/div
Freq. Resolution = 0.097656/10 ms = 9.8 Hz
Each harmonic can be clearly seen.
Sweep speed = 500 µs /div
Freq. Resolution = 0.097656/500 µs = 195 Hz
Harmonics appear very broadened.
4. Another problem is shown in the display above (on the left): in the time domain the signal appears very noisy.
This contributes to a very noisy display in the frequency domain: the noise “floor” is much higher than it should
be. We can fix this problem using averaging.
5. Return the sweep speed to 10 ms/div, and press the Acquire hardkey, and press the Averaging softkey to
reduce the noise in both time-domain and frequency domain displays.
Notice that the noise has been reduced so much
in the frequency domain display that the noise
floor is now below the bottom of the display.
This will be fixed in the next step.
6. Press Math hardkey, Settings softkey, More FFT softkey, Scale softkey, and change the vertical scale to 20
dB/ (20 dB per vertical division). You can do this by turning the control called the “Entry Knob”, located just to
the right of the CRT, below the horizontal section, with an illuminated curved arrow above it.)
Now we can see the noise floor at the bottom of
the display. And, some new frequency
components are now visible: the even harmonics
at 2 kHz, 4 kHz, 6 kHz and 8 kHz.
While the even harmonics are now visible, they
are very small (about 60 dB below the odd
harmonic amplitudes).
Procedure C – What’s All This “Harmonic” Stuff Anyhow?
Nearly 200 years ago, a French fellow named Jean Baptiste Joseph Fourier was doing some very important
theoretical work in science and mathematics. Among his quotes is “Mathematics compares the most diverse
phenomena and discovers the secret analogies that unite them.” His name is among the 72 famous French
scientists and mathematicians immortalized on a plaque at the base of the Eiffel tower in Paris. He feared having
his head removed during the French revolution; fortunately for us, he was spared the guillotine.
Fourier determined that any periodic, non-sinusoidal waveform can be reproduced by adding up an infinite number
of harmonics of the waveform’s fundamental frequency. The fundamental frequency is 1/T, where T is the period of
the waveform. For our purposes, one can say that a square wave voltage contains an infinite number of harmonics,
which can be seen on a spectrum analyzer. While a more complete coverage of Fourier series can be found in
many sources, here the basics of Fourier series for a square wave will be presented.
If a square wave is bipolar (positive level and negative level are equal in magnitude and opposite in polarity), we
can find out the amplitude of each harmonic (an) using the following formula:
an = 4A/(nπ), where n = number of the odd harmonic (for n = even numbers, an = 0), and A is the peak
amplitude of the square wave (A = 1 2 of the peak-peak value).
So, using our 1 kHz, 2 Vpp square wave as an example, we will calculate the amplitudes of its first seven
harmonics. A 2 Vpp square wave has a peak amplitude of 1 V, so A = 1V.
First Harmonic Amplitude: 4(1V)/(1π) = 1.273 Vpeak, frequency = 1(1 kHz) = 1 kHz
Second Harmonic Amplitude: 0 (because it’s an even harmonic)
Third Harmonic Amplitude: 4(1V)/(3π) = 0.424 Vpeak, frequency = 3(1 kHz) = 3 kHz
Fourth Harmonic Amplitude: 0 (because it’s an even harmonic)
Fifth Harmonic Amplitude: 4(1V)/(5π) = 0.255 Vpeak, frequency = 5(1 kHz) = 5 kHz
Sixth Harmonic Amplitude: 0 (because it’s an even harmonic)
Seventh Harmonic Amplitude: 4(1V)/(7π) = 0.182 Vpeak, frequency = 7(1 kHz) = 7 kHz
etc.
However, our oscilloscope, and spectrum analzyers, always display amplitudes of frequency components based on
their RMS values, and often in dBV. The table below shows us the values of the first seven harmonics, expressed in
Vpeak, Vrms, and dBV:
Harmonic
Number
Frequency
Amplitude in
volts peak
Amplitude in
volts RMS
Amplitude in
dBV
1
1 kHz
1.273 Vp
0.9003 V
-0.91 dBV
2
2 kHz
0 Vp
0V
*
3
3 kHz
0.424 Vp
0.3001 V
-10.45 dBV
4
4 kHz
0 Vp
0V
*
5
5 kHz
0.255 Vp
0.1801 V
-14.9 dBV
6
6 kHz
0 Vp
0V
*
7
7 kHz
0.182 Vp
0.1286 V
-17.8 dBV
*Theoretically even harmonic amplitudes are 0 V. Expressed in dBV, 0 V = 20 LOG(0V) = - ∞ dBV. Since noise
amplitude is larger than 0 V, or - ∞ dBV, we see noise, or very small even harmonics, instead
of - ∞ dBV.
1) Return to the display of the 2 Vpp, 1 kHz square wave in step 5 of Procedure B, on page 5.
2) Use the Cursor function, and measure the amplitude of each of the odd harmonics, and record them in the
table below:
Harmonic
Number
Frequency
Amplitude in
dBV
(calculated)
1
1 kHz
-0.91 dBV
Amplitude in dBV
(measured)
3
3 kHz
-10.45 dBV
5
5 kHz
-14.9 dBV
7
7 kHz
-17.8 dBV
Procedure D – Exploring Amplitude Modulation in the Frequency Domain
Amplitude Modulation (commonly called AM) is a way to put “intelligence” (although if you listen to many AM radio
stations the term “intelligence” may seem inappropriate) on a radio-frequency (RF) carrier by varying the amplitude
of the RF carrier.
First, let’s take a look at an RF carrier in the time domain, with and without modulation. The two displays below
have no amplitude modulation (called 0% modulation):
50 kHz RF Carrier, 0% Modulation
10 µs/division
Individual cycles of carrier easily visible
50 kHz RF Carrier, 0% Modulation
100 µs/division
Individual cycles of carrier hard to see
The two displays below both show amplitude modulation in the time domain:
50 kHz RF Carrier, 2 kHz 50% Modulation
Carrier peak-peak amplitude increased (and
decreased) by 50% (2.25 Vpp at modulation
peaks), with a period of 500 µs.
Modulating freq. = 1/T = 1/(500 µs) = 2 kHz
50 kHz RF Carrier, 2 kHz 100% Modulation
Carrier peak-peak amplitude increased (and
decreased) by 100% (3.00 Vpp at modulation
peaks), with a period of 500 µs.
Modulating freq. = 1/T = 1/(500 µs) = 2 kHz
On the next page, we will see the RF carrier with amplitude modulation, shown in both the time domain and the
frequency domain. AM creates side frequencies, which will be seen in the frequency domain. As we saw
earlier in this introduction to using the oscilloscope as a spectrum analyzer, sometimes we have take control and
change the Time/Div, Frequency Span, Center Frequency, turn on Averaging, etc. to make the display show what
we want to see. You should be able to duplicate the displays below, to become familiar with frequency domain
displays of AM signals.
50 kHz RF Carrier, 0%
Modulation
Time Domain:
Sweep speed = 500
µs/division
Frequency Domain:
Span Freq. = 100 kHz,
Center Freq. = 50 kHz
20 dB/div
50 kHz RF Carrier, 50% Modulation
Modulating freq. 5 kHz
Sweep speed = 500 µs/division
Span Freq. = 100 kHz, Center Freq. = 50 kHz
20 dB/div
50 kHz RF Carrier, 100% Modulation
Modulating freq. 5 kHz
Sweep speed = 500 µs/division
Span Freq. = 100 kHz, Center Freq. = 50 kHz
20 dB/div
This expanded view of the display above
shows the two side frequencies that were
created when a 50 kHz carrier is amplitude
modulated by a 5 kHz sine wave.
50 kHz RF Carrier
50% Modulation
Modulating freq. 5 kHz
Sweep speed = 500 µs/div
1000 kHz RF Carrier
100% Modulation
Modulating freq. 10 kHz
Sweep speed = 50 µs/division
Span Freq. = 50 kHz
Center Freq. = 1.00 MHz
20 dB/div
You can see one significant problem with the display above: the Frequency Resolution is too large, resulting in the
individual frequency components (lower side frequency, carrier and upper side frequency) looking much broader
than they really are.
Frequency Resolution = .097656/(time
per division) = .097656/(50 µs) = 1.95 kHz.
While we might be tempted to increase the
time per division (perhaps from 50 µs to
100 µs) in order to make the frequency
resolution smaller, there’s another FFT
parameter that is affected by time per
division: the Maximum Frequency.
Maximum Frequency = 102.4/(time per
division), so at 50 µs/div the max. freq. =
102.4/50 µs = 2.05 MHz. This is fine for a 1
MHz carrier frequency. But, if we change
the time per division to 100 µs, the
maximum frequency is just above 1.0 MHz,
which prevents the display from showing
the upper side frequency. The display above shows what happens when the time/division is changed to 100 µs: no
frequencies above 1 MHz are displayed.