From Fourier transformation and Nyquist principle, assuming that the signal bandwidth is Bw and the number of sampling points is N, so in the time domain, the time interval Δt should be 1/Bw , while the time range is supposed to be N*1/Bw.

However, when we used the Network Analyzer with setting the stop frequency fs, end frequency fe and points number N, we got the frequency domain response (S11).

Here, fs = 390MHz, fe=440MHz, N=801. SoΔt should be 2E-8 s.

Then we transfered to time domain as was show in other figure, and set the start time ts = 0, end time te = 7E-6s. Strangely, we found there were 801 points in the arrange [ts, te] andΔt was 7E-6/801=0.87E-8 s. It violated the Fourier principle. Why?

However, when we used the Network Analyzer with setting the stop frequency fs, end frequency fe and points number N, we got the frequency domain response (S11).

Here, fs = 390MHz, fe=440MHz, N=801. SoΔt should be 2E-8 s.

Then we transfered to time domain as was show in other figure, and set the start time ts = 0, end time te = 7E-6s. Strangely, we found there were 801 points in the arrange [ts, te] andΔt was 7E-6/801=0.87E-8 s. It violated the Fourier principle. Why?

Given a frequency response F(w), the inverse fourier transform is defined for all time and is in fact a continous function. There is no lower limit on the time resolution. There is no violation of any sampling theory.

It is only the "short-cuts" in the IFFT that create the appearance of a discrete limit on the time repsonse.

In our VNA, we use an Inverse Fourier Transform, and NOT an IFFT.