The Hilbert transform of a function is defined as :

$$\mathscr{H}\big\{f(t)\big\} = \lim_{T\rightarrow \infty}\frac{1}{\pi}\int\limits_{-T/2}^{T/2}\frac{f(\tau)}{t-\tau}\, \mathrm{d}\tau$$

Okay but what does it do exactly? What information do we take from it?

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    $\begingroup$ That's one of the definitions. If you really want to be anal about it, you gotta do this "principal value" (p,v.) thing. $\endgroup$ Jun 20, 2023 at 2:23
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    $\begingroup$ Relevant $\endgroup$ Jun 20, 2023 at 20:08

3 Answers 3


If the OP is interested in what is the practical purpose of the Hilbert Transform, the rest of this post applies. Similar to the Fourier, Laplace, and Z transforms, the Hilbert Transform can be used to simplify signal processing operations and implementations. I concur with Fat32's conclusion in that I cannot think of a case where the Hilbert Transform has a direct "analysis" utility as we do with the Fourier Transform.

I gave the presentation Demystifying the Hilbert Transform at the DSP Online Conference last year (2022) that goes into much further detail in answering this very question with much more intuition as well as the math. It's a broad question (the presentation is 85 pages!) but I will attempt to provide a high level summary to this good question below.

In my opinion, the Hilbert Transform's interesting properties are best explained in the frequency domain. Below shows the Hilbert Transform as a time domain process, where a signal $x(t)$ is convolved with the function $1/(\pi t)$. Below that is the same process in the frequency domain (convolution in time is multiplication in frequency) where the frequency spectrum of $x(t)$ (it's Fourier Transform), has all positive frequencies multiplied by $-j$ and all negative frequencies multiplied by $+j$. Multiplying by $j$ is a 90° rotation in phase.

Hilbert Transform

Where the use of the Hilbert Transform gets interesting is in its use in creating the "Analytic Signal" which is given as:

$$x_a(t) = x(t) + j\hat{x}(t)$$

$x_a(t)$ is complex and is the analytic signal for the real signal given as $x(t)$. $\hat{x}(t)$ is the Hilbert Transform of the real signal $x(t)$. The Analytic Signal, $x_a(t)$, has the convenient property of not having any negative frequency components!

Analytic Signal

If the concept of "negative frequency" is not clear, this post and the graphic below may help further understand this. A big hint to some is knowing that the equation $Ke^{j\phi}$ is simply a phasor with magnitude $K$ and angle $\phi$. If the phasor rotates, the result is a single frequency tone, either positive or negative as $Ke^{j\omega t}$ or $Ke^{-j\omega t}$ (I know this is a big "Ah-ha!" to many getting introduced to signal processing):

Negative Frequencies

Thus we can, in implementation, create a complex signal (two outputs: one as the real component and the other as the imaginary component) with the result treated as a single complex waveform that has no negative frequency components. Common implementations in analog or digital form are shown below:

Actual Hilberts

We can use the Hillbert and its ability of eliminating all negative frequencies for the following applications:

  • Fine and Course Frequency Translation
  • IQ Modulation and Demodualation
  • Filter Simplification and Elimination
  • Envelope Extraction
  • Measurement of Instantaneous Frequency
  • Deriving the Minimum Phase Response from a Magnitude Response

The first example, Frequency Translation, is demonstrated with the block diagram below.

Frequency Translation

As an attempt to explain this briefly, note that the input spectrum is "one-sided" in that we have removed all negative frequencies (using the Hilbert Transform!), with the result input centered at a tone given by $e^{j\omega_c t}$. The Local Oscillator is also complex (as the complex conjugate - rotates the other way- of the Analytic Signal of a real Local Oscillator, producing the signal $e^{-j\omega_c\Delta t}$. Multiplying the two is just a matter of adding the exponents! Prior to using the Hilbert, the input would have been a real signal given as $\cos(\omega_c t)$.

Here's some additional background that may help with further understanding the diagram above:

Euler's Formula relates a real sinusoid to it's positive and negative frequency components:

$$2\cos(\omega_c t) = e^{j\omega_c t} + e^{-j\omega_c t}$$

The Analytic Signal for $\cos(\omega_c t)$ is the positive frequency component $e^{j\omega_c t}$:

The Hilbert Transform of $\cos(\omega_c t)$ is $\sin(\omega_c t)$

Thus the Analytic Signal for $x(t) = \cos(\omega_c t)$ is:

$$x_a(t) = x(t) + j\hat{x}(t) = \cos(\omega_c t) + j \sin(\omega_c t) = e^{j\omega_c t}$$

The output is the product of the input with the Local Oscillator:

$$e^{j(\omega_c - \omega_\Delta) t } = e^{j\omega_c t}e^{-j\omega_\Delta t}$$

The resulting frequency translated output is given as $e^{j(\omega_c-\omega_\Delta)t}$. If we write this in terms of its real and imaginary components we have:

$$e^{j(\omega_c-\omega_\Delta)t} =\cos((\omega_c-\omega_\Delta)t) + j\sin ((\omega_c-\omega_\Delta)t)$$

If we proceeded to take the real only portion of this complex output, we would eliminate two of the otherwise four multipliers required in actual implementation, and the result would be $\cos((\omega_c-\omega_\Delta)t)$ which conveniently reduced to what is also called an "Image Reject Mixer" as given in the diagram below:

Image Reject Mixer

I summarized the main points above, but I do recommend accessing this presentation for the complete answer well beyond what I am able to include in the short Q&A format here. Alternatively this is an important segment covered in my "DSP for Wireless Communications" course which also includes additional requisite background. This course is running soon (July 2023) through the IEEE and routinely offered at both links below:



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    $\begingroup$ Hi!So is it a tool of convenience?Because we understand better phase shifts than negative frequencies , it help us replace any negative frequencies in the frequency domain with phase shifts? $\endgroup$
    – Volpina
    Jun 20, 2023 at 1:40
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    $\begingroup$ Yes it is definitely a tool of convenience, as an analytical way to describe physical implementations we can actually do within practical limits. But I don't come to that conclusion about phase shifts and negative frequencies. For DSP I think it is important to really understand "phase shifts" and negative frequencies. There are serious misconceptions on what a "phase shift" is, often confused with time delay. (although those are related, they are distinctly different). $\endgroup$ Jun 20, 2023 at 1:48
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    $\begingroup$ Ok I will watch the presentation $\endgroup$
    – Volpina
    Jun 20, 2023 at 1:51
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    $\begingroup$ Also there are a lot of great posts here that detail additional features and considerations with the Hilbert Transform, use the search to discover more already here! $\endgroup$ Jun 20, 2023 at 1:52
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    $\begingroup$ @Volpina the pedant in me wants to say: that's what you should expect. A transform doesn't give us any information that we didn't already have; it helps us look at the information we do have from a different angle. $\endgroup$
    – hobbs
    Jun 20, 2023 at 16:21

To my knowledge, and I'm not as deeply interested in it as Dan seems to be, the continuous-time Hilbert transform integral itself does not return a meaningfully interpretable information related with its input. In other words, it does not analyse its input, unlike a Fourier or Laplace transform would do.

In this respect, an analyser, according to linear vector-spaces point of view upon which signal processing theory is based on, performs a decomposition of its input with respect to a set of base vectors that span a vector-space of interest. The transform integral merely quantifies the similarity or overlap (i.e., correlation) between its input and given basis. From this point of view, therefore, a Hilbert transformer is not analysing (aka decomposing) its input; I cannot see the base vectors at all.

Furthermore, the very name of the Hilbert transformer is also 90 degrees phase shifter. And I think it's already sufficient to deduce from its name only (without even considering its mathematical definition) that an HT is merely shifting its input by 90 degrees in phase; turning sines into cosines and vice versa. Its action is not to correlate but to shift the phase of its input: a very useful operation that finds many interesting applications in theory and practice.

The transform, very possibly, was introduced in backwards: from a frequency-domain definition to a time-domain integral.

First, the theoretical concept of a complex-valued analytic signal emerges. It has the property that its frequency spectrum is null for negative frequencies. Such a signal has many practical utilizations in communication technology as its bandwidth is confined to a single side band etc.

Then, the frequency-domain characterization of a transform is defined such that it would allow generation of an analytic signal from a given input by an LTI system.

Then the time-domain impulse response of the LTI system whose frequency response was specified is obtained, turns out to be:

$$ h(t) = \frac{1}{\pi t} $$

The since this is an LTI system, it's output for a given input is given by a convolution sum:

$$ y(t) = h(t) \star x(t) $$

$$ y(t) = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{x(\tau)}{t-\tau} d\tau $$

And that's the Hilbert transform integral. So, the integral is merely a reflection of the frequency-domain design of the transform.

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    $\begingroup$ Ha! Great first sentence, I feel like I get that a lot about many of the things that pique my interest! $\endgroup$ Jun 20, 2023 at 1:57
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    $\begingroup$ @DanBoschen your graphics are fascinating ! as always. $\endgroup$
    – Fat32
    Jun 20, 2023 at 1:58
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    $\begingroup$ @DanBoschen I used it in the sense that given $\hat{x}(7) = 3$ means nothing(?), unlike if it were a Fourier transform telling about frequency content of the signal being transformed. Note that it's a time to time transform. And it does not analyse the input from any aspect. $\endgroup$
    – Fat32
    Jun 20, 2023 at 2:05
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    $\begingroup$ Yes agreed, it really isn't an "analysis tool" in contrast to the Fourier Transform. Good point....although deriving the minimum phase from the magnitude response perhaps covers this. $\endgroup$ Jun 20, 2023 at 2:06
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    $\begingroup$ @DanBoschen thanks, although that being said, an envleope detector, can be an example of it indirect utilization, where it returns some information about the signal. That's an indirect interpretation though. And I cannot go further but leave it up to you :-)) $\endgroup$
    – Fat32
    Jun 20, 2023 at 2:11

In many ways the Hilbert transform is different from other common transforms used in signal processing, such as the Fourier transform, the Laplace transform, or the $\mathcal{Z}$-transform. These other transforms perform functional decomposition, which is useful for analyzing and processing signals. The Hilbert transform describes the relation between the real and imaginary parts of signals with specific properties. In signal processing, Hilbert transform relations are important in the following three cases:

  1. analytic signals: the real and imaginary parts of analytic signals are related via the Hilbert transform.
  2. causal systems: the real and imaginary parts of the frequency response of causal systems are related via the Hilbert transform.
  3. minimum-phase systems: the (logarithm) of the magnitude of the frequency response of minimum-phase systems is related to the phase response via the Hilbert transform.

It is not immediately apparent how the three cases above are related. Of course, there must be a relation, because in all three cases the Hilbert transform relates the real and imaginary parts of functions or sequences with given properties. The relation between the three cases is the following:

If there is a "causality" constraint in one domain, i.e., if functions or sequences vanish for negative argument, then the real and imaginary parts of their (inverse) Fourier transform are related via the Hilbert transform.

Let's take a look at the three cases above. Analytic signals have no components at negative frequencies. Hence, the "causality" constraint exists in the frequency domain ($H(\omega)=0$ for $\omega<0$). Consequently, there's a Hilbert transform relationship between the real and imaginary parts of the inverse Fourier transform of $H(\omega)$. Hence, the real and imaginary parts of the complex-valued time domain signal are Hilbert transforms of each other.

Causal systems (or signals) are dual to analytic signals. Here, the causality condition is in the time domain: $h(t)=0$ for $t<0$. Consequently, the Fourier transform of $h(t)$, i.e., the frequency response, satisfies a Hilbert transform relation between its real and imaginary parts.

For minimum-phase systems we take a look at the logarithm of the frequency response:

$$\ln H(\omega)=\ln |H(\omega)| + j\phi(\omega)$$

The inverse Fourier transform of $\ln H(\omega)$ is called the complex cepstrum and it can be shown that for minimum-phase systems, the complex cepstrum is causal. Consequently, the real and imaginary parts of $\ln H(\omega)$, i.e., the log-magnitude and the phase, are related via the Hilbert transform.

In sum, in all three cases we have a "causality" condition, i.e., a function that vanishes for negative argument (the spectrum for analytic signal, for causal signals the signal itself, and the cepstrum for minimum-phase systems). This has as a consequence that the complex functions obtained via the (inverse) Fourier transform are related by the Hilbert transform. For analytic signals, this relation is satisfied in the time domain, for causal signals it is satisfied in the frequency domain, and for minimum-phase systems, it is satisfied in the logarithmic frequency domain, i.e., between log-magnitude and phase.

Even though the Hilbert transform is probably less commonly used for analyzing or processing signals than the Fourier transform, we can make use of the relation it describes between real and imaginary parts of functions with certain properties (see above). E.g., given a magnitude response, we can compute the corresponding phase response such that the resulting complex frequency response is a minimum-phase response. Similarly, given a real-valued signal, we can compute an imaginary part such that the resulting complex-valued signal is analytic. Finally, knowing that a signal is causal, we can determine the imaginary part of the frequency response if just its real part is known (for instance through measurement). Hence, systems computing the Hilbert transform are used regularly. These systems are just specific allpass filters (phase shifters). In discrete time, Hilbert transformers can be implemented either as finite impulse response (FIR) or infinite impulse response (IIR) filters, or they can be implemented in the frequency domain using the FFT (as in the Matlab function hilbert.m).

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    $\begingroup$ //"... the Hilbert transform is not a tool for analysis or processing..."// terms like "tool" or "analysis" or "processing" are pretty broad to make a blanket assertion like that. I have used the Hilbert transform both as a tool for analysis and as a tool for processing audio signals. $\endgroup$ Jun 20, 2023 at 17:41
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    $\begingroup$ @robertbristow-johnson: I wrote the last paragraph to clarify that assertion, but I agree that the wording is a bit broad. Will change that. However, I would still claim that the Hilbert transform is not a common tool for analyzing signals, in the sense that the Fourier transform or similar transforms are. What can you say about a signal after taking its Hilbert transform? $\endgroup$
    – Matt L.
    Jun 20, 2023 at 17:53
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    $\begingroup$ The Hilbert transform is a tool, but like other tools, may be a component of a larger operation. The Hilbert transform, along with the appropriately delayed original and a little more arithmetic, can tell me what the instantaneous amplitude or instantaneous frequency of a time-varying sinusoid are. $\endgroup$ Jun 20, 2023 at 18:10
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    $\begingroup$ @robertbristow-johnson: Yes, that's the computation of the analytic signal. I mention that in the last paragraph. $\endgroup$
    – Matt L.
    Jun 20, 2023 at 19:13
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    $\begingroup$ Not that we can also say, similarly that the Fourier Transform is not necessarily just a tool for analysis and decomposition but, like the Hilbert, a tool for simplifying processing (using the FT for fast convolution for example). I see the two as quite similar in that regard, they are both equally mathematical "tools". $\endgroup$ Jun 21, 2023 at 1:51

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