Hilbert transform of unit step function

Question

How to calculate Hilbert transform, if it exists, of the signals like $u(t)$, $sgn(t)$.

What properties should a function satisfy for existence of Hilbert transform.

Absolute integrability of a signal $x(t)$ is necessary condition for HT to exist?

Answer 1

Absolute integrability is not a necessary condition for the Hilbert transform to exist. Simple counterexamples are the Hilbert transforms of $\cos(\omega t)$ and $\sin(\omega t)$, given by $\sin(\omega t)$ and $-\cos(\omega t)$, respectively, if $\omega>0$.

Since the functions $u(t)$ and $\textrm{sgn}(t)$ are related by scaling and an additive constant, and since the Hilbert transform of a constant is zero, the Hilbert transforms of the two functions - if they exist - must be scaled versions of each other:

$$\textrm{sgn}(t)=2u(t)-1\\\Rightarrow \mathcal{H}\{\textrm{sgn}(t)\}=2\mathcal{H}\{u(t)\}\tag{1}$$

I believe that the Hilbert transform of these two functions can be computed in the following way. [Edit: this turned out to be true. See below and the comments to RBJ's answer.] We know that the Hilbert transform can be written as a convolution:

$$\mathcal{H}\{x(t)\}=\frac{1}{\pi t}\star x(t)\tag{2}$$

This can be rewritten as

$$\mathcal{H}\{x(t)\}=\frac{1}{\pi}\left(\log(|t|)\right)'\star x(t)=\frac{1}{\pi}\log(|t|)\star x'(t)\tag{3}$$

where $'$ denotes differentiation w.r.t. $t$. The rightmost equality in $(3)$ is true because we generally have

$$(f\star g)'=f'\star g=f\star g'\tag{4}$$

With $x(t)=u(t)$ we get $x'(t)=\delta(t)$, and from $(3)$ we obtain

$$\mathcal{H}\{u(t)\}=\frac{1}{\pi}\log(|t|)\star \delta(t)=\frac{1}{\pi}\log(|t|)\tag{5}$$

From $(1)$ and $(5)$ we get

$$\mathcal{H}\{\textrm{sgn}(t)\}=\frac{2}{\pi}\log(|t|)\tag{6}$$

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The results given by Eqs $(5)$ and $(6)$ can also be derived in the frequency domain. The Fourier transform of the convolution

$$\mathcal{H}\{u(t)\}=\frac{1}{\pi t}\star u(t)$$

is given by

$$\mathcal{F}\left\{\frac{1}{\pi t}\right\}\mathcal{F}\{u(t)\}=-j\;\textrm{sgn}\{\omega\}\left(\pi\delta(\omega)+\frac{1}{j\omega}\right)=-\frac{\textrm{sgn}(\omega)}{\omega}=-\frac{1}{|\omega|}$$

So we have

$$\mathcal{H}\{u(t)\}=\mathcal{F}^{-1}\left\{-\frac{1}{|\omega|}\right\}$$

From this document (p. 27, Ex. 19.4), we know that

$$\mathcal{F}^{-1}\left\{-\frac{1}{|\omega|}\right\}=\frac{1}{\pi}\log(|t|)$$

confirming Eq. $(5)$.

(Note the different scaling in the cited document due to the use of the unitary Fourier transform.)

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EDIT:

The problem can also be solved using the definition of the Hilbert transform of bounded functions:

$$\mathcal{H}\{f(t)\}=\frac{1}{\pi}\left[\mathrm{p.v.}\int_{|\tau|<1}\frac{f(\tau)}{t-\tau}d\tau+\mathrm{p.v.}\int_{|\tau|>1}f(\tau)\left(\frac{1}{t-\tau}+\frac{1}{\tau}\right)d\tau\right]$$

where $\mathrm{p.v.}$ denotes the Cauchy principal value.

For $f(t)=u(t)$ we have

$$\mathcal{H}\{u(t)\}=\frac{1}{\pi}\left[\underbrace{\mathrm{p.v.}\int_{0}^{1}\frac{d\tau}{t-\tau}}_{\displaystyle I_1}+\underbrace{\mathrm{p.v.}\int_{1}^{\infty}\left(\frac{1}{t-\tau}+\frac{1}{\tau}\right)d\tau}_{\displaystyle I_2}\right]$$

If $t$ is outside the interval $(0,1)$, the first integral is simply

$$I_1=\int_{0}^{1}\frac{d\tau}{t-\tau}=\log\left(\frac{t}{t-1}\right)$$

If $t\in (0,1)$ we get

$$I_1=\lim_{\epsilon\to 0^+}\int_0^{t-\epsilon}\frac{d\tau}{t-\tau}+\int_{t+\epsilon}^1\frac{d\tau}{t-\tau}=\log\left(\frac{t}{\epsilon}\right)+\log\left(\frac{-\epsilon}{t-1}\right)=\log\left(\frac{t}{1-t}\right)$$

Since in both cases the argument of the logarithm is positive, both results can be summarized as

$$I_1=\log\left|\frac{t}{1-t}\right|$$

If $t<1$ the second integral is

$$I_2=\lim_{\epsilon\to 0^+}\left[\int_1^{1/\epsilon}\frac{d\tau}{t-\tau}+\int_1^{1/\epsilon}\frac{d\tau}{\tau}\right]=\lim_{\epsilon\to 0^+}\left[\log\left(\frac{t-1}{t-1/\epsilon}\right)+\log\left(\frac{1}{\epsilon}\right)\right]=\\=\lim_{\epsilon\to 0^+}\log\left(\frac{t-1}{\epsilon t-1}\right)=\log\left(1-t\right)$$

If $t>1$ we have

$$I_2=\lim_{\epsilon\to 0^+}\left[\int_1^{t-\epsilon}\frac{d\tau}{t-\tau}+\int_{t+\epsilon}^{1/\epsilon}\frac{d\tau}{t-\tau}+\int_1^{1/\epsilon}\frac{d\tau}{\tau}\right]=\\=\lim_{\epsilon\to 0^+}\left[\log\left(\frac{t-1}{\epsilon}\right)+\log\left(\frac{-\epsilon}{t-1/\epsilon}\right)+\log\left(\frac{1}{\epsilon}\right)\right]=\\=\lim_{\epsilon\to 0^+}\log\left(\frac{1-t}{\epsilon t-1}\right)=\log(t-1)$$

Again, both results can be summarized as

$$I_2=\log|1-t|$$

Combining these results, we obtain the Hilbert transform of $u(t)$:

$$\mathcal{H}\{u(t)\}=\frac{1}{\pi}(I_1+I_2)=\frac{1}{\pi}\left(\log\left|\frac{t}{1-t}\right|+\log|1-t|\right)=\frac{1}{\pi}\log|t|$$

which agrees with our previous result given by Eq. $(5)$.

Answer 2

To integrate the convolution integral of the Hilbert transform, one must make use of the concept of the Cauchy principal value (p.v.) of an integral with some kinda nasty infinity or singularity in it.

Hilbert Transform:

$$\begin{align} \hat{x}(t) &\triangleq \mathscr{H}\Big\{ x(t) \Big\} \triangleq \frac{1}{\pi t} \ \circledast x(t) \\ \\ &= \mathrm{p.v.} \int\limits_{-\infty}^{\infty} \frac{1}{\pi v} \ x(t-v) \mathrm{d}v \\ \\ &= \lim_{\epsilon \to 0^+} \int\limits_{-1/\epsilon}^{-\epsilon} \frac{1}{\pi v} \ x(t-v) \mathrm{d}v \ + \ \int\limits_{\epsilon}^{1/\epsilon} \frac{1}{\pi v} \ x(t-v) \mathrm{d}v \\ \end{align}$$

That's in general. Now let $x(t)$ be the unit step function (i will leave the symbol "$u$" be used for something else):

$$ x(t) = \tfrac{1}{2}\big(1 + \operatorname{sgn}(t)\big) = \begin{cases} 0 \qquad & t<0 \\ \tfrac12 \qquad & t=0 \\ 1 \qquad & t>0 \\ \end{cases} $$

and the sign function $\operatorname{sgn}(t)$ is

$$ \operatorname{sgn}(t) \triangleq \begin{cases} -1 \qquad & t<0 \\ \ \ 0 \qquad & t=0 \\ \ \ 1 \qquad & t>0 \\ \end{cases} $$

Because of the linearity of convolution and of the fact that the integral of a sum is the sum of integrals, we can think of this as the Hilbert transform of $\tfrac12$ added to the Hilbert transform of the sign function $\operatorname{sgn}(t)$.

The Hilbert transform of $\tfrac12$ or of any constant is zero because the two integrals expressing the principal value above will be exact negatives for any value of $\epsilon$ in the limit. So the problem is just that of the Hilbert transform of $\tfrac12 \operatorname{sgn}(t)$.

In the integral above, let's say $t>0$. The math will be identical with flipped signs for the case $t<0$. For $t=0$ there will simply be the problem of infinity.

$$\begin{align} \mathscr{H}\Big\{ \tfrac12 \operatorname{sgn}(t) \Big\} & = \frac12 \lim_{\epsilon \to 0^+} \left( \int\limits_{-1/\epsilon}^{-\epsilon} \frac{1}{\pi v} \ \operatorname{sgn}(t-v) \mathrm{d}v \ + \ \int\limits_{\epsilon}^{1/\epsilon} \frac{1}{\pi v} \ \operatorname{sgn}(t-v) \mathrm{d}v \right) \\ \\ & = \frac{1}{2\pi} \lim_{\epsilon \to 0^+} \left( \int\limits_{-1/\epsilon}^{-t} \frac{1}{v} (+1) \mathrm{d}v \ + \ \int\limits_{-t}^{-\epsilon} \frac{1}{v} (+1) \mathrm{d}v \ + \ \int\limits_{\epsilon}^{t} \frac{1}{v} (+1) \mathrm{d}v \ + \ \int\limits_{t}^{1/\epsilon} \frac{1}{v} (-1) \mathrm{d}v \right) \\ \end{align}$$

Wow! This is problematic. The two integrals in the interior cancel each other, but the two integrals on the outside team up

$$\begin{align} \mathscr{H}\Big\{ \tfrac12 \operatorname{sgn}(t) \Big\} &= \frac{1}{2\pi} \lim_{\epsilon \to 0^+} \left( \int\limits_{-1/\epsilon}^{-t} \frac{+1}{v} \mathrm{d}v \ + \ \int\limits_{t}^{1/\epsilon} \frac{-1}{v} \mathrm{d}v \right) \\ \\ &= \frac{1}{2\pi} \lim_{\epsilon \to 0^+} 2\int\limits_{t}^{1/\epsilon} \frac{-1}{v} \mathrm{d}v \\ \\ &= \frac{1}{\pi} \lim_{\epsilon \to 0^+} \ -\big(\log(1/\epsilon) - \log(t)\big) \\ \\ &= \frac{1}{\pi} \big( \lim_{\epsilon \to 0^+} \ \log(\epsilon) + \log(t) \big) \\ \end{align}$$

I don't see how that converges to a finite value.

We'll have to get back to this in the future. Meantime, suggestions and corrections from anyone are welcome.