# The Hilbert Transform - Part 2

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

### 3. Interaction with the Fourier Transform

The signal $1/(\pi t)$ has Fourier transform

\begin{eqnarray}
\text{-j sgn}(f) = \begin{cases} \text{-j} & \quad \text{if } f \gt 0\\ 0 & \quad \text{if } f = 0\\ \text{j} & \quad \text{if } f \lt 0\\ \end{cases}
\end{eqnarray}

If $g(t)$ has Fourier transform $G(f)$, then, from the convolution property of the Fourier transform,
it follows that $\hat{g}(t)$ has Fourier transform

\begin{eqnarray}
\hat{G}(f) = -\text{j sgn}(f) G(f)
\end{eqnarray}

Thus, the Hilbert transform is easier to understand in the frequency domain than in the time domain: the Hilbert transform does not change the magnitude of $G(f)$, it changes only the phase. Fourier transform values at positive frequencies are multiplied by −j (corresponding to a phase change of −$\pi$/2) while Fourier transform values at negative frequencies are multiplied by j (corresponding to a phase change of $\pi$/2). Stated yet another way, suppose that $G(f) = a + b$ j for some $f$. Then $\hat{G}(f) = b - a$ j if $f > 0$ and $\hat{G}(f) = - b + a$ j if $f < 0$. Thus the Hilbert transform essentially acts to exchange the real and imaginary parts of $G(f)$ (while changing the sign of one of them).

Energy Spectral Density: Suppose that $g(t)$ is an energy signal. Then, since $|\hat{G}(f)| = |G(f)|$, both $\hat{G}(f)$ and $G(f)$ have exactly the same energy spectral density. Thus, for example, if $G(f)$ is bandlimited to $B$ Hz then so is $\hat{G}(f)$. It also follows that $\hat{g}(t)$ has exactly the same energy as $g(t)$.

Symmetry Properties: If $g(t)$ is real-valued, then $G(f)$ exhibits Hermitian symmetry, i.e., $G(-f) = G^*(f)$. Of course then $\hat{G}(-f) = -\text{j sgn}(-f)G(-f) = [-\text{j sgn}(f)G(f)]^* = \hat{G}(f)^*$, so $\hat{G}(f)$ also exhibits Hermitian symmetry, as expected.
Let $g(t)$ be a real-valued signal. Recall that if $g(t)$ is even (so that $g(-t) = g(t)$) then $G(f)$ is purely real-valued while if $g(t)$ is odd (so that $g(-t) = -g(t)$) then $G(f)$ is purely imaginaryvalued. Now if $G(f)$ is purely real-valued then certainly $\hat{G}(f)$ is purely imaginary-valued (and vice-versa). Thus if $g(t)$ is even, then $\hat{g}(t)$ is odd and if $g(t)$ is odd, then $\hat{g}(t)$ is even.

Orthogonality: If $g(t)$ is a real-valued energy signal, then $g(t)$ and $\hat{g}(t)$ are orthogonal. To see this recall that

\begin{align*}
\left< g(t),\hat{g}(t) \right>&=\int_{-\infty}^{\infty}g(t)\hat{g}^*(t) dt\\
&=\int_{-\infty}^{\infty}G(f)\hat{G}^*(f) df\\
&=\int_{-\infty}^{\infty}G(f)[- \text{j sgn}(f)G(f)]^* df\\
&=\int_{-\infty}^{\infty}\text{j}|G(f)|^2\text{sgn}(f) df\\
&=0,
\end{align*}

where we have used the property that, since $|G(f)|^2$ is an even function of $f$, $|G(f)|^2 \text{sgn}(f)$ is an odd function of $f$ and hence the value of the integral is zero.

Low-pass High-pass Products: Let g(t) be signal whose Fourier transform satisfies $G(f) = 0$ for $|f| > W$ and let $h(t)$ be a signal with $H(f) = 0$ for $|f| < W$. Then \begin{equation*} \mathscr{H}[g(t)h(t)] = g(t)\hat{h}(t), \end{equation*} i.e., to compute the Hilbert transform of the product of a low-pass signal with a high-pass signal, only the high-pass signal needs to be transformed. To see this, let $s(t) = g(t)h(t)$ so that \begin{equation*} S(f) = G(f) * H(f) = G(f) * [H(f) u (f) + H(f) u (−f)], \end{equation*} where $u$ denotes the unit step function. It is easy to see that $G(f) * (H(f) u (f))$ is zero if $f < 0$; similarly $G(f) * (H(f) \text{u}(-f))$ is zero if $f > 0$. Thus

\begin{align*}
\hat{S}(f) &= -\text{j sgn}(f)S(f) \\
&= -\text{j}G(f) * (H(f) \text{u}(f)) + \text{j}G(f) * (H(f) \text{u}(-f)) \\
&= G(f) * [-\text{j}H(f) \text{u}(f) + \text{j}H(f) \text{u}(-f)] \\
&= G(f) * [-\text{j sgn}(f)H(f)] \\
&= G(f) * \hat{H}(f).
\end{align*}

An important special case of this arises in the case of QAM modulation. Assuming that $m_I(t)$ and $m_Q(t)$ are bandlimited to $W$ Hz, then, if $f_c > W$, we have

\mathscr{H}[m_I(t) \cos(2f_ct) + m_Q(t) \sin(2f_ct)] = m_I(t) \sin(2f_ct) − m_Q(t) \cos(2f_ct).

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.