Hilbert transform of a function $g(t)$ which is defined in time domain, would result in another function in time domain. Is there any other transformation like Hilbert that the results be in time domain?

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    $\begingroup$ i wish we could discourage or deprecate the use of the symbol "$f(\cdot)$" for functions and leave the symbol "$f$" for ordinary frequency. that is $$f \triangleq \frac{\Omega}{2 \pi}$$ $\endgroup$ Commented May 12, 2020 at 16:26

2 Answers 2


TLDR: if the time variable $t$, and its dual variable ($f$, $\tau$) in the expression of the bivariate kernel have the same homogeneity, (I believe that) you can call it a time-domain transformation.

Here comes the long version. In the continuous domain with primal variable $t\in\mathbb{R}$, a large number of practical or known linear transformations (not called transforms yet) can be expressed as follows. For "many" functions $g(t)$ and a fixed bivariate kernel $K(t,u)$, define the integral of their product over $t$ as:

$$ g_K(u) = \alpha_K\int_{-\infty}^{\infty}g(t) K(t,u)\mathrm{d}t$$

Those specific linear transformations belong to a wider concept of integral operators, that act function spaces. The constant $\alpha_K$ is there for normalization purposes, I will skip it from now on. Similarly, let us not pay attention to the infinite limits.

Such operators are a central piece of analysis, and they exist under several technical conditions (for instance, those with finite energy are Hilbert-Schmidt operators). For us signal/image processing people, there are some examples, with different notations for the second or dual variable:

  • $K(t,f) = e^{-2j\pi tf}$: Fourier transform
  • $K(t,s) = e^{-s t}$: Laplace transform
  • $K(t,\tau) = \frac{1}{t-\tau}$: Hilbert transform (as special case of potential integral operators, with a Cauchy kernel)
  • $K(t,\tau) = e^{-\frac{(\tau-t)^2}{4}}$: Gauss-Weierstrass transform
  • $K(t,\tau) = h(\tau-t)$: convolution with filter defined by $h$ impulse response

Similar expressions (some in higher dimension) can be derived for wavelet transforms, cosine transform, etc.
I don't know of a precise definition about the difference between "transformations" and "transform". In cases I use, I prefer the notion of transform when the representation preserves most of the original information in the function, and especially when it is invertible, or when there exists a reciprocal kernel $\kappa$, such that:

$$ g(t) \sim \alpha_\kappa\int_{-\infty}^{\infty}g_K(u) \kappa(t,u)\mathrm{d}u$$

In the Fourier or Laplace transforms, $(t,f)$ or $(t,s)$ couples appear as products in the kernel, so the primal/dual variables can be seen as inverses of each other. They have reciprocal units (like second and Hertz). For convolution, Gauss and Hilbert transforms, $(t,\tau)$ appears in a difference (it could be a sum as well). So they should have the same physical unit. As a consequence, the output function $g_K(u)$ resides in the same time/temporal domain as $g(t)$.

So I would say: as long as $t$ and $\tau $ appear in the same homogeneous manner in the kernel $K(t,\tau)$, you will get a time-domain transformation.

Honestly, I don't know about generic theorems around that.

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    $\begingroup$ i wish we could discourage or deprecate the use of the symbol "$f(\cdot)$" for functions and leave the symbol "$f$" for ordinary frequency in the continuous-time, continuous-frequency Fourier transform. that is $$f \triangleq \frac{\Omega}{2 \pi}$$ $\endgroup$ Commented May 12, 2020 at 16:27
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    $\begingroup$ if you change "$f(t)$" to "$x(t)$" in your answer, i will change the OP notation. but it should be consistent. $\endgroup$ Commented May 12, 2020 at 16:31
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    $\begingroup$ why did you use "$s$" for goodness sake? $\endgroup$ Commented May 12, 2020 at 16:56
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    $\begingroup$ I like this answer...why aren't there more upvotes? $\endgroup$ Commented Oct 3, 2022 at 2:19
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    $\begingroup$ Your qualitative appreciation is worth a quantitative handful of upvotes $\endgroup$ Commented Oct 3, 2022 at 6:52

This question is hard to answer because it depends on your definition of "transform". The Hilbert transform can be written as a convolution with the kernel

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

and, consequently, it can be represented by a linear time-invariant (LTI) system with an impulse response given by $(1)$. So the application of any LTI system to a signal can be seen as a transform or mapping which doesn't change the domain of the signal. E.g., the application of an ideal band pass filter to an input signal maps that signal to the space of band-limited functions, and in this sense, it is a transform(ation).


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