In part 2 of this series, we looked at a swept-tuned spectrum analyzer and how it could sweep a frequency span of interest from 1 MHz to 1.1 MHz with a sweep time of 50 msec. As **Figure 1** shows, the analyzer readily identifies a signal at 1.03 MHz but misses the intermittent signals shown in red.

**How can we make sure to capture intermittent signals?**

**Figure 2** shows one possibility. We eliminated the analyzer’s sweep generator and replaced its single swept-tuned filter with a bank of parallel filters, each with a successively higher center frequency. For reasonable frequency resolution, you may need a lot of filters, leading to a complex, expensive instrument.

**Is there an alternative?**

Yes. Digital technology contributes to significant improvements in signal analysis. **Figure 3** shows an analyzer architecture that digitizes an input signal, enabling the instrument to apply a fast Fourier transform (FFT) to derive frequency and phase information, as discussed in an earlier series. In addition, it can display signals in a modulation domain.

**Would this be considered a real-time signal analyzer?**

Be careful when talking about “real-time.” Any electronic circuit imposes some latency, and the vector signal analyzer is no exception. Unlike the analog analyzer, which displays a signal as soon as it detects it (after accounting for filtering and other delays), the vector signal analyzer must acquire a complete time-domain record before executing the FFT. So, the minimum latency will equal the acquisition time, which depends on the signal and your RBW requirements, plus the FFT execution time, which depends on the analyzer’s DSP engine.

**Can you provide an example?**

Consider the example in Figure 1. We’ll need a sampling frequency greater than the Nyquist frequency, so for a back-of-the-envelope calculation, let’s choose *f _{S}* = 2.5 MHz. Some of the signals of interest are only 10 kHz apart, so let’s choose a resolution bandwidth of

*f*= 5 kHz. From our earlier series on FFT, we know that the number of FFT points N =

_{RBW}*f*/

_{S}*f*= 2.5 MHz/5 kHz = 500. If we’re using an FFT algorithm that requires N to be a power of 2, we can choose N = 512. For

_{RBW}*f*= 2.5 MHz, our sampling interval is 1/2.5 MHz = 400 nsec, so our time record length is 512*400 ns = 204.8 µsec.

_{S}**Figure 4**at the top shows time record 1 in Figure 1.

**Q: How does that time record get processed?**

Figure 4 shows four possible scenarios. In scenario a, when the record 1 acquisition is completed, execution of the FFT on record 1 begins immediately, as does the acquisition of record 2. The FFT execution time is less than the acquisition time, so the FFT engine is ready to start processing record 2 as soon as record 2’s acquisition is completed, and the process continues. This is as close as we can come to real-time.

**Q: What about the second scenario?**

Scenario b is untenable. Execution time is longer than acquisition time, and we will soon run out of memory. Scenario c represents a possible solution to this problem—delay each acquisition to let the FFT engine catch up. With this approach, however, some data (represented by the gray areas) is never acquired, and we return to the problem of Figure 1, where some intermittent signals are never detected. Scenario d represents a solution to this problem. If the FFT execution time is more than the acquisition time but less than twice the acquisition time, we can add a second FFT engine. In the figure, FFT1 executes the odd-numbered records, and FFT2 executes the even-numbered records. This approach can be expanded with more FFT engines in parallel.

**The title of this series mentions vector network analyzers, but we haven’t discussed them yet.**

The analog and digital analyzers we’ve examined so far represent an evolutionary approach to addressing similar sets of measurements. The vector network analyzer has separate use cases that set it apart. We will take a closer look at the final part of this series.

Continue to Part 4.

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