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Lab 3 - Using the Agilent 54621A Digital as a 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 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 , 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 , such as a 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

Procedure A – Measuring a in the 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 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 s 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 s 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 s = 488.3 Hz, and the maximum frequency = 102.4/200 s = 512 kHz. That creates a not very useful display; 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 s 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 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 of the 1 kHz and 3 kHz components, and the Y of –9.69 dB shows that the 3 rd harmonic amplitude is 9.69 dB below the 1 st 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 s /div.

Sweep speed = 10 ms/div Sweep speed = 500 s /div Freq. Resolution = 0.097656/10 ms = 9.8 Hz Freq. Resolution = 0.097656/500 s = 195 Hz Each harmonic can be clearly seen. 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 (a n) using the following formula: an = 4A/(n π), where n = number of the odd harmonic (for n = even numbers, a n = 0), and A is the peak 1 amplitude of the square wave (A = 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)/(1 π) = 1.273 Vpeak, frequency = 1(1 kHz) = 1 kHz Second Harmonic Amplitude: 0 (because it’s an even harmonic) Third Harmonic Amplitude: 4(1V)/(3 π) = 0.424 Vpeak, frequency = 3(1 kHz) = 3 kHz Fourth Harmonic Amplitude: 0 (because it’s an even harmonic) Fifth Harmonic Amplitude: 4(1V)/(5 π) = 0.255 Vpeak, frequency = 5(1 kHz) = 5 kHz Sixth Harmonic Amplitude: 0 (because it’s an even harmonic) Seventh Harmonic Amplitude: 4(1V)/(7 π) = 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 Frequency Amplitude in Amplitude in Amplitude in Number volts peak volts RMS dBV 1 1 kHz 1.273 Vp 0.9003 V -0.91 dBV 2 2 kHz 0 Vp 0 V * 3 3 kHz 0.424 Vp 0.3001 V -10.45 dBV 4 4 kHz 0 Vp 0 V * 5 5 kHz 0.255 Vp 0.1801 V -14.9 dBV 6 6 kHz 0 Vp 0 V * 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 Frequency Amplitude in Amplitude in dBV Number dBV (measured) (calculated) 1 1 kHz -0.91 dBV

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 50 kHz RF Carrier, 0% Modulation 10 s/division 100 s/division Individual cycles of carrier easily visible 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 50 kHz RF Carrier, 2 kHz 100% Modulation Carrier peak-peak amplitude increased (and Carrier peak-peak amplitude increased (and decreased) by 50% (2.25 Vpp at modulation decreased) by 100% (3.00 Vpp at modulation peaks), with a period of 500 s. peaks), with a period of 500 s. Modulating freq. = 1/T = 1/(500 s) = 2 kHz Modulating freq. = 1/T = 1/(500 s) = 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 s/division

Frequency Domain: Span Freq. = 100 kHz, Center Freq. = 50 kHz 20 dB/div

50 kHz RF Carrier, 50% Modulation 50 kHz RF Carrier, 100% Modulation Modulating freq. 5 kHz Modulating freq. 5 kHz Sweep speed = 500 s/division Sweep speed = 500 s/division Span Freq. = 100 kHz, Center Freq. = 50 kHz Span Freq. = 100 kHz, Center Freq. = 50 kHz 20 dB/div 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 s/div

1000 kHz RF Carrier 100% Modulation Modulating freq. 10 kHz Sweep speed = 50 s/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 s) = 1.95 kHz. While we might be tempted to increase the time per division (perhaps from 50 s to 100 s) 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 s/div the max. freq. = 102.4/50 s = 2.05 MHz. This is fine for a 1 MHz carrier frequency. But, if we change the time per division to 100 s, 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 s: no frequencies above 1 MHz are displayed.