Should I use a spectrum, signal, or vector network analyzer?

An analog spectrum analyzer’s sweep time can hide intermittent unwanted signals.

In part 1 of this series, we examined a simplified block diagram of a traditional analog spectrum analyzer, which includes an RF frontend, mixer, voltage-controlled oscillator (VCO), intermediate-frequency (IF) stage, sweep generator, envelope detector, and display. We also looked at resolution bandwidth (RBW), the bandwidth of a bandpass filter in the IF stage.

What’s the relationship between the input-signal frequencies, IF, RBW, VCO frequency, and sweep time?
Figure 1 shows two input signals with frequencies 1,030 kHz and 1,060 kHz. Let’s assume an IF of 400 kHz with an RBW of 2 kHz.

Recall from an earlier series on intermodulation that when we combine two frequencies in a mixer, we get the sums and differences. In Figure 1, we get sums and differences of each input frequency and the VCO frequency. Because the input frequencies are much higher than the IF, we can ignore the sums and focus on the differences. The table in the figure shows some examples. When the sweep generator output is 4.1 V, the VCO produces a 615-kHz signal, and the difference frequencies are 415 Hz and 445 Hz, both of which the IF filter blocks. When the sweep-generator voltage rises to 4.2 V, however, the VCO frequency becomes 630 kHz, and mixer output f₁–f_VCO equals exactly 400 kHz. This signal appears at the output of the IF stage, where it will be detected and displayed. Note that the display uses the sweep-generator output to position the pulse representing the 1,030-kHz input properly. Similarly, the analyzer detects and displays f₂ when f_VCO equals 660 kHz.

Could you elaborate on how sweep time relates to RBW?
Sweep time depends on our frequency span of interest f_SPAN, the RBW, and a constant k that equals the filter rise time multiplied by the RBW:

The constant k is often 2 or 3; for our example, we’ll choose 2. In Figure 2, f_SPAN equals 100 kHz, so we can calculate the sweep time:

We tend to think of a spectrum analyzer scanning from left to right along the frequency domain, but it also scans from top to bottom in the time domain. The process is continuous, but Figure 2 shows snapshots for every 10 kHz left to right and every 5 ms top to bottom.

Note that the analyzer sees nothing through the first 10 ms. But at 15 ms, it detects our 1,030-kHz signal from Figure 1 and writes it to the display. It again sees nothing until 30 ms, when it sees our 1,060-kHz signal. Then, it continues to the end of our span of interest, finding two other signals along the way.

What’s the orange signal at 1.08 MHz?
In part 1, I arbitrarily used Boston AM radio station frequencies as examples and continue that here. While there are stations at 1.03, 1.06, and 1.09 MHz, there isn’t one at 1.08 MHz. The orange signal could be a rogue transmitter that the FCC might be interested in.

What about the red pulses at 1.02 kHz?
Those represent an intermittent signal that the analyzer misses. Note the red signal appears sometime after 20 msec, whereas the analyzer conducted its scan for 1.02-MHz signals at 10 msec. In addition, the signal disappears after 40 msec, and you can’t count on capturing it during subsequent sweeps. Indeed, it may reappear at another frequency.

What’s the significance?
Such signals may be random interference, or if you are in the signals-intelligence (SIGINT) business, they may come from an adversary using frequency-hopping techniques to avoid detection. If we repeat the sweep many times, we might capture one of these signals, but it’s a matter of chance, not certainty.

So, how can we capture those signals?
We’ll look at the details in part 3. Meanwhile, you might want to review our recent series on the fast Fourier transform.