# The Hilbert Transform - Part 1

This is part one of a multi-part series that defines The Hilbert Transform and its significance.

### 1. Definition

The Hilbert transform $\mathscr{H}[g(t)]$ of a signal $g(t)$ is defined as

\begin{eqnarray}
\mathscr{H}[g(t)] = g(t) * \frac{1}{\pi t} = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{g(\tau)}{t-\tau} d\tau = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{g(t-\tau)}{\tau} d\tau
\end{eqnarray}

The Hilbert transform of $g(t)$ is the convolution of $g(t)$ with the signal $1/t$. It is the response to $g(t)$ of a linear time-invariant filter (called a Hilbert transformer) having impulse response $1/t$. The Hilbert transform $\mathscr{H}[g(t)]$ is often denoted as $\hat{g}(t)$ or as $[g(t)]^\wedge$.

A technicality arises immediately. The alert reader will already be concerned with the definition (1) as the integral is improper: the integrand has a singularity and the limits of integration are infinite. In fact, the Hilbert transform is properly defined as the Cauchy principal value of the integral in (1), whenever this value exists. The Cauchy principal value is defined—for the first integral in (1)—as

\begin{eqnarray}
\mathscr{H}[g(t)] = \frac{1}{\pi} \lim_{\epsilon \to 0^+} \left(\int_{t-1/\epsilon}^{t-\epsilon} \frac{g(\tau)}{t-\tau} d\tau + \int_{t+\epsilon}^{t+ 1/\epsilon} \frac{g(\tau)}{t-\tau} d\tau \right) \cdot
\end{eqnarray}

We see that the Cauchy principal value is obtained by considering a finite range of integration that is symmetric about the point of singularity, but which excludes a symmetric subinterval, taking the limit of the integral as the length of the interval approaches $\infty$ while, simultaneously, the length of the excluded interval approaches zero. Henceforth, whenever we write an integral as in (1), we will mean the Cauchy principal value of that integral (when it exists).

### 2. Some Basic Properties

Some obvious properties of the Hilbert transform follow directly from the definition. Clearly the Hilbert transform of a time-domain signal $g(t)$ is another time-domain signal $\hat{g}(t)$. If $g(t)$ is real-valued, then so is $\hat{g}(t)$.

Linearity: The Hilbert transform is linear, i.e., if $a_1$ and $a_2$ are arbitrary (complex) scalars, and $g_1(t)$ and $g_2(t)$ are signals, then

\begin{eqnarray}
[a_1g_1(t) + a_2g_2(t)]^\wedge = a_1\hat{g}_1(t) + a_2\hat{g}_2(t).
\end{eqnarray}

(This follows immediately from the fact that the Hilbert transform is the output of a linear system.)

The Hilbert transform of a constant signal: Note that, for any constant $c$, the Hilbert transform of the constant signal $g(t) = c$ is $\hat{g}(t) = \hat{c} = 0$. From linearity it follows that $\mathscr{H}[g(t) + c] = \hat{g}(t) + \hat{c} = \hat{g}(t)$. Thus, like an ideal differentiator, a Hilbert transformer “loses” dc offsets. Later we will define an inverse Hilbert transform which can recover the original signal up to an additive constant (in the same way that integration can undo differentiation only up to an additive constant).

Time-shifting and time-dilation: If $g(t)$ has Hilbert transform $\hat{g}$, then $g(t-t_0)$ has Hilbert transform $\hat{g}(t-t_0)$, and $g(at)$ has Hilbert transform $\text{sgn}(a)\hat{g}(at)$ (assuming a $a\neq0$).

Convolution: The Hilbert transform behaves nicely with respect to convolution, since

$[g_1(t)*g_2(t)]^\wedge = \hat{g}_1(t)*g_2(t) = g_1(t)*\hat{g}_2(t).$

To see this, observe from the associative and commutative properties of convolution that $[g_1(t)*g_2(t)]*\frac{1}{\pi t}$ can be written as $[g_1(t)*\frac{1}{\pi t}]*g_2(t)$ or as $g_1(t)*[g_2(t)*\frac{1}{\pi t}]$.

Time-derivative: The Hilbert transform of the derivative of a signal is the derivative of the Hilbert transform, i.e.,

$\mathscr{H}[\frac{d}{dt}g(t)]=\frac{d}{dt}\mathscr{H}[g(t)]$.

To see this, recall Leibniz’s Integral Rule, which states that

$\frac{d}{dc}\int_{a(c)}^{b(c)}f(x,c) dx = \int_{a(c)}^{b(c)}\frac{\partial}{\partial c}f(x,c)dx+f(b,c)\frac{d}{dc} b(c) - f(a,c) \frac{d}{dc}a(c)$.

In particular, if $a$ and $b$ are definite limits (independent of $c$), we have

$\frac{d}{dc}\int_{a}^{b}f(x,c) dx = \int_{a}^{b}\frac{\partial}{\partial c}f(x,c)dx$.

Now

\begin{align*}
\frac{d}{dt}\mathscr{H}[g(t)]&=\frac{1}{\pi}\frac{d}{dt}\int_{-\infty}^{\infty}\frac{g(t-\tau)}{\tau}d\tau\\
&=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{g'(t-\tau)}{\tau}d\tau\\
&=\mathscr{H}[g'(t)],
\end{align*}
where $g'(t)=\frac{d}{dt}g(t)$.

References:
[1] S. L. Hahn, “Comments on ‘A Tabulation of Hilbert Transforms for Electrical Engineers’,” IEEE Trans. on Commun., vol. 44, p. 768, July 1996.
[2] S. L. Hahn, “Hilbert transforms,” in The Transforms and Applications Handbook (A. Poularakis, Ed.), Boca Raton FL: CRC Press, 2010.