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Example 1: Resolving Closely Spaced Signals (with Resolution Bandwidth)
Stepping Through a Measurement of Two Signals of
Unequal Amplitude
This example resolves a third-order intermodulation distortion product
with a frequency separation of 700 kHz and an amplitude separation of
about 60 dB.
1. Connect two signal sources to the spectrum analyzer INPUT 50 Ω.
Set the frequency of one source to 10 MHz and the other source to
10.7 MHz. Set both sources to an amplitude of about −10 dBm.
2. Press
on the spectrum analyzer to start the procedure from
PRESET
a preset state and set the spectrum analyzer center frequency to
10.35 MHz.
3. Set the span to 5 MHz.
4. Set the resolution bandwidth to 100 kHz and the video bandwidth to
1 kHz. See Figure 2-4 on page 55.
To resolve two signals of unequal amplitude, the resolution bandwidth
must be less than or equal to the frequency separation of the two
signals (the same as resolving two equal amplitude signals). However,
in this case the largest resolution bandwidth that will resolve the two
unequal signals is determined primarily by the shape factor of the IF
filter, rather than by the 3 dB bandwidth. Shape factor is defined as the
ratio of the 60 dB bandwidth to the 3 dB bandwidth of the IF filter, as in
Figure 2-3.
The wider IF filters in this spectrum analyzer have shape factors of
15:1 or better. The IF filters less than or equal to 100 Hz have a better
shape factor of 5:1 or better. Therefore, to resolve two signals of unequal
amplitude, the half-bandwidth of a filter at the point equal to the
amplitude separation of the two signals must be less than the frequency
separation of the two signals.
Chapter 2
Making Measurements
53
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