0
\$\begingroup\$

I am reading about how spectrum analyzers function and I am somewhat confused about how the Fourier transform is computed.

It seems to me the argument made in the Wikipedia per signal reconstruction assumes the signal is bandlimited - if the function of the spectrum analyzer assumes bandlimitedness of the input, doesn't its output become circular(IE it only produces a correct approximation when the signals frequency range is already known)?

The other way to compute this is through direct numerical integration of the Fourier transform which will always produce a correct approximation but that doesn't appear to be the system used.

\$\endgroup\$
  • 1
    \$\begingroup\$ Real ADCs don't have infinite analog bandwidth, so the signal they actually sample is always (at least approximately) bandlimited. \$\endgroup\$ – The Photon Nov 6 '17 at 21:00
  • \$\begingroup\$ Second, it's not common at all to use the output from a spectrum analyzer to reconstruct a signal. \$\endgroup\$ – The Photon Nov 6 '17 at 21:05
  • \$\begingroup\$ " its output become circular(IE it only produces a correct approximation when the signals frequency range is already known)" ... that is fundamentally the definition of sampled data systems. Not all spectrum analyzers use sampled data and Fourier transforms, but all that do have this limitation. \$\endgroup\$ – Brian Drummond Nov 6 '17 at 21:31
  • \$\begingroup\$ Hmm then it seems all measurements are essentially circular. It works if you're attempting to compute the fourier transform of a known signal but if you have known input you get something which is essentially meaningless(IE the output could in fact have horrible aliasing - I think numerical integration provides a better bound for this computation then arbitrarily applying the sampling theorem?) How does the FCC do testing at top end frequencies to make sure people aren't leaking outside of their bandrange? \$\endgroup\$ – FourierFlux Nov 6 '17 at 22:35
  • 1
    \$\begingroup\$ You will always have some window effect when calculating a DFT on a time-limited array of samples. This effect can be made (if you are careful) no worse than the various errors caused by the limitations of the analog circuits (mixers, filters, power sensors) in an equivalent analog spectrum analyzer. \$\endgroup\$ – The Photon Nov 6 '17 at 23:41
2
\$\begingroup\$

I think you are conflating two issues: aliasing and windowing.

Aliasing depends on the sampling frequency. Aliasing happens when input signals are present at the sampler with frequencies higher than the Nyquist rate, which is half the sampling frequency. Aliasing can be minimized by use of an anti-aliasing filter at the sampler input and with oversampling. When sampling an IF, the anti-aliasing filter might be a band-pass filter rather than low-pass.

Windowing depends on the total duration of the sampling window. Windowing reduces the spectral resolution of the DFT, when considered as an estimate of the spectrum of the input signal.

You can reduce the windowing effect by taking a longer sampling window, but of course that makes the measurement slower.

Analog spectrum analyzers also suffered from spectral leakage because IF bandpass filters cannot be made perfectly narrow. And similarly were able to obtain higher spectral resolution by reducing the "resolution bandwidth (RBW)" with the similar trade-off of slowing the measurement.

Practically, digital spectrum analyzers aren't designed to be ideal. They're designed to be better than the analog spectral analyzer they replaced, or to provide similar performance at a lower price. It's well known that the spectral resolution is limited, and its not expected to be able to perfectly reconstruct the input signal from the spectrum analyzer measurement.

\$\endgroup\$
0
\$\begingroup\$

if the function of the spectrum analyzer assumes bandlimitedness of the input, doesn't its output become circular(IE it only produces a correct approximation when the signals frequency range is already known)?

No, because just knowing/assuming bandlimitedness doesn't mean you know the exact frequency spectrum; that's much more than just the fact that it becomes small enough above some limit. So there is no circularity.

The other way to compute this is through direct numerical integration of the Fourier transform which will always produce a correct approximation but that doesn't appear to be the system used.

No, because in order to do a numerical Fourier transform you have to sample the signal in time domain which also requires the signal to be band limited.

\$\endgroup\$
  • \$\begingroup\$ No, this is incorrect. You can numerically integrate the fourier transform of a time limited signal to produce its actual fourier transformex, when you actually integrate a time limited signal symbolically you get an exact value for the fourier transform, it's not bandlimited but that's irrelevant - no time limited signals are. \$\endgroup\$ – FourierFlux Nov 6 '17 at 20:38
  • 1
    \$\begingroup\$ ...also your comment is unclear. What do you mean by "numerically integrate the fourier transform"? And what does symbolical integration have to do with the whole thing? \$\endgroup\$ – Curd Nov 6 '17 at 20:51
  • 1
    \$\begingroup\$ I haven't mentioned time limitedness at all! My point is that you need a minimum sampling frequency for a given bandwidth. \$\endgroup\$ – Curd Nov 6 '17 at 20:53
  • 1
    \$\begingroup\$ Please answer my question: what do you mean by "computing the FT directly"? How would you do that numerically (without sampling the signal in TD)? \$\endgroup\$ – Curd Nov 6 '17 at 21:05
  • 1
    \$\begingroup\$ @FourierFlux The fourier transform doesn't operate on discrete time (i.e. sampled) signals. That's the discrete Fourier transform \$\endgroup\$ – τεκ Nov 6 '17 at 23:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.