Analytic Signal Block in Simulink Gives Reduced Amplitude for the Imaginary Part

Question

In Simulink, I have connected a sine wave to zero-order hold and zero-order hold to analytic signal block, as in the figure below. My purpose is to transform the real signal to the corresponding analytic signal by adding complex part, in order to be able to do some phase shifting operations. Zero-order hold block is to discretize the signal, as the analytic signal accepts only discrete signals. Now, the problem is this: when I discretize the signal "more finely" by setting the sample time of zero-order hold to a smaller number (eg. $0.00003$), imaginary part of the analytic signal output is reduced significantly, while the real part staying normally, as seen in the figure below. On the contrary, when the sample time is set to a greater number (eg. $0.0003$) amplitudes of the real and the imaginary parts of the analytic signal output are the same, as expected.

According to the Matlab's documentation, analytic signal block's output is like this: $y=u+j H \{\ u \}\ $, where $H \{\ \}\ $ denotes the Hilbert transform and $j$ = $ \sqrt{-1}$. So, in the setup mentioned above it seems like when the continuous signal is sampled wıth a higher frequency, Hilbert transform reduces the amplitude.

Why would Hilbert transform reduce the amplitude, or is there a problem with my setup in Simulink?

Answer 1

If the filter has a fixed order then increasing the sampling rate will make its response scale with frequency, thus the transition width increases relative to the sampling frequency. For example: if $f_s=1\;\mathrm{kHz}$ and $\omega_{tw}=10\;\mathrm{Hz}$ for a given order then keeping the order and increasing $f_s$ tenfold will make $\omega_{tw}$ increase tenfold.

I don't have Matlab but according to the documentation you linked the block perfoms a Hilbert transform using a FIR (which makes sense). Since I also don't have your other details, I'm testing this with a Kaiser window and the Kaiser order formula, using $A_s=-40\;\mathrm{dB}$ and $\omega_{tw}=1\;\mathrm{Hz}$. When $f_s=10\;\mathrm{Hz}$ the order is $23$ (length $24$). Using a $1\;\mathrm{Hz}$ signal at the input results in an analytic signal at the output equal in amplitude with the input. Here is the result in LTspice (note: the input V(x) is not delayed, so the two signals are not in quadrature):

Keeping the order and increasing $f_s=100\;\mathrm{Hz}$ results in $\omega_{tw}=10\;\mathrm{Hz}$, and for the same $1\;\mathrm{Hz}$ input the output is reduced:

In order to avoid this the order needs to be increased. In this case it ends up being $224$ (length $225$):

If you compare the 1st and the 3rd pictures you'll see the similarities in the delays, unlike the 2nd picture, and that makes sense since the phase is fixed so the outcome needs to end up delayed by the same amount if you change the sampling frequency. I haven't posted any frequency domain responses, but you should be able to see for yourself, easily.