# Alternative to Hilbert Transform

I have an digital system that needs an analytical signal of length (in time) T. The system is sampling the analog signal with time interval delta t such that T / delta t = 128. So I have 128 samples. I have a FIFO buffer to which, at each delta t, the new sample is stored.

The signal however is not analytical but real-valued. To obtain the analytical signal I do perform an Hilbert transform by taking the FFT of the whole buffer with 128 samples. I set the 2nd half of the FFT outcome to zero and I perform an IFFT on the resulting values.

This Hilbert Transform is using quite a bit resources of my system. I was wondering, would there be an alternative method for obtaining the analytical signal of my last 128 samples?

I was thinking, could I obtain the phase of the signal and using that compute the imaginary component of the signal? Definately I am no expert in signal processing and I'm trying to learn and understand such techniques. Does anyone have any suggestions or maybe explain why my idea wouldn't work (or would) and what other methods could work and why?

• Mart, a practical Hilbert transformer that doesn't need to pass low frequencies can be a very short FIR. (I define "very short" as fewer than 50 taps.) If your signal is not extremely wide-banded (like from nearly DC to nearly Nyquist), a Hilbert transformer that covers the signal bandwidth is not particularly costly. But your DSP is doing complex arithmetic with that analytic function. Ideally, if this is at an Intermediate Frequency (IF) stage, when it's being sampled, the slightly narrowband signal is best around 1/2 of Nyquist. That will allow the Hilbert transformer with fewest taps. Commented Dec 19, 2023 at 18:52

If your signal is relatively narrow band (does not occupy the entire Nyquist bandwidth) then an FIR filter would be more efficient to compute the band-limited Hilbert Transform vs an FFT which provides a reasonable Hilbert Transform over (nearly) the full Nyquist Bandwidth, and specifically if the FIR Hilbert can be implemented with 19 taps or less, which would be based on filter bandwidth required and performance accuracy needed. (See comparison of resources at the end of this post).

If implementing with an FIR filter, choose an odd length so that the compensating delay for the real part of the analytic signal is an integer number of taps.

The remez function in Python scipy.signal (firpm in Matlab and Octave) supports generating Hilbert’s directly with arbitrary bandwidths, so this provides an easy approach for the novice. There are many other alternate FIR based solutions as well (see comments with further suggestions by RBJ).

For example a 51 tap Hilbert over a portion of the Nyquist band is generated as follows:

coeff= sig.remez(19, [0,.1,.2,.3,.4,.5], [0,1,0], type=‘hilbert’)


The compensating delay is 9 samples and the bandwidth is from .2 to .3 of the sampling rate.

This implements a Hilbert Transform with an implementation delay over the specified passband (here we pass .2 to .3 with weight 1 so specifies the passband), results in excellent quadrature phase and +/-0.1 dB of amplitude ripple in the passband.

The analytic signal is a complex waveform with the real part as the original signal and the imaginary part as the Hilbert Transform of the original signal. So to create the analytic signal with the FIR filter above, a 25 sample delayed copy of the input signal represents the delay matched real output of the analytic signal, and the filtered output (as the Hilbert Transform) represents the imaginary output of the analytic signal.

With this solution (or any FIR based Hilbert such as other suggestions recommended by RBJ in the comments) the OP can use an FIR length up to the duration of the buffer but should be one less to be odd length for a significant improvement in processing required. (So up to a 127 length FIR filter). With an odd length filter, the solution will have every other coefficient as 0 including the center tap, but is also an anti-symmetric filter meaning the negative of the first half of the filter will be the same coefficients as the last half of the filter. For this reason the total filter length $$N$$ should be three more than an integer multiple of 4, as otherwise the first and last coefficient will just end up being zero (so use $$N$$ such that $$(N-3)/4$$ is an integer, 127-3 is a multiple of 4, so 127 works). Also with an odd number of coefficients, the delay matched path is an integer which significantly simplifies the delay matching. However, as an FIR filter, a single new output sample is computed based on the FIFO contents (which becomes the filter's memory) after every delta T.

Based on the above points, I will compare resources needed based on an estimate of the number of multiplications and additions compared to the FFT/IFFT approach assuming the OP did maximize the FIR length to 127 coefficients, compared to an FFT of 128 samples. (So $$N=127$$ and we wish to have a result $$N+1=128$$ long).

With the Hilbert FIR, every other coefficient including the center tap is 0 and then due to the assymetry in the coefficients, only half the number of multiplications is required. For $$N$$ coefficients this results in $$(N-3)/2$$ real multiplications and $$N-3$$ real additions, so for 127 coefficients this would be 62 real multiplications, and 124 real additions as shown in the block diagram of the resulting FIR implementation below.

$$x[n]$$ represents every new sample shifted into the FIFO on each clock sample (at each delta t). $$z^{-2}$$ is a 2 sample delay, so every other value in the FIFO is used at any given delta t (clock sample) to compute a new output sample $$y[n]$$. To compare the FFT which does the processing on a complete block, note that it will take 128 samples to fill an output buffer that is 128 samples long. Thus the total processing required to get 128 samples (when $$N=127$$) is $$(N+1)(N-3)/2$$ real multiplications and $$(N+1)(N-3)$$ real additions. Thus approximately $$N^2/2$$ and $$N^2$$ respectively.

For the FFT approach we do an FFT of a real sequence and then an IFFT of the complex result with half the bins zeroed out. This results in $$2(N+1)\log_2(N+1)$$ real multiplications and additions.

Charting this shows that it is difficult to beat the efficiency of computing the Hilbert with an FFT IFFT. The cross-over is at 21 samples, meaning if the Hilbert required could be done with a 19 tap FIR filter, this would be more efficient than using the FFT / IFFT, otherwise the FFT / IFFT would be more efficient.

Zooming in:

• Also, Dan, Mart can design the Hilbert transformer by windowing the ideal impulse response with a really good window, perhaps Kaiser. And, if they have MATLAB, they use use firls() instead of remez(). If you want a maximally flat passband with no ripple anywhere, you can pass a single cycle of $\sin()$ through a smoothstep function and get a Hilbert FIR. Commented Dec 19, 2023 at 19:05
• @robertbristow-johnson yup that works too- the “ideal impulse response” should be the bandlimited one which is the non-causal 1/n with center at n=0 with a value of 0 and every other tap with value =0 (odd length) Commented Dec 19, 2023 at 20:12
• Mart no there is no requirement that Fs> 4x F, you specify the passband of interest to be between DC and Nyquist. As is the case for any Hilbert implementation, it does not converge well near Nyquist or DC. The suggestions by Robert are very good as well but require a little more signal processing knowledge to implement while what I provided is a simple straight forward approach that comes out of the box with firpm (Matlab) or remez (Python). All are fine solutions. Commented Dec 19, 2023 at 20:16
• Note too- the analytic signal is a complex signal with the real part as the signal and the imaginary part as its Hilbert. So for the FIR approach I gave, the real part is the 25 sample delay and the imaginary part is the 51 tap FIR filter output to get the complex I +jQ analytic signal Commented Dec 19, 2023 at 20:19
• @Mart it's an FIR filter so you stream the output on each sample to fill the 128 sample output buffer you desire. See my update comparing resources, it looks like you only save if you can use FIR filter with 19 coefficients or less. Commented Dec 20, 2023 at 14:52