*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 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 has Fourier transform , then, from the convolution property of the Fourier transform,

it follows that 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 , it changes only the *phase*. Fourier transform values at positive frequencies are multiplied by −j (corresponding to a phase change of −/2) while Fourier transform values at negative frequencies are multiplied by j (corresponding to a phase change of /2). Stated yet another way, suppose that j for some . Then j if and j if . Thus the Hilbert transform essentially acts to exchange the real and imaginary parts of (while changing the sign of one of them).

**Energy Spectral Density:** Suppose that is an energy signal. Then, since , both and have exactly the same energy spectral density. Thus, for example, if is bandlimited to Hz then so is . It also follows that has exactly the same energy as .

**Symmetry Properties:** If is real-valued, then exhibits Hermitian symmetry, i.e., . Of course then , so also exhibits Hermitian symmetry, as expected.

Let be a real-valued signal. Recall that if is even (so that ) then is purely real-valued while if is odd (so that ) then is purely imaginaryvalued. Now if is purely real-valued then certainly is purely imaginary-valued (and vice-versa). Thus if is even, then is odd and if is odd, then is even.

**Orthogonality:** If is a real-valued energy signal, then and 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 is an even function of , is an odd function of and hence the value of the integral is zero.

**Low-pass High-pass Products:** Let g(t) be signal whose Fourier transform satisfies for and let be a signal with for . 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 so that
\begin{equation*}
S(f) = G(f) * H(f) = G(f) * [H(f) u (f) + H(f) u (−f)],
\end{equation*}
where denotes the unit step function. It is easy to see that is zero if ; similarly is zero if . 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 and are bandlimited to Hz, then, if , we have

\begin{equation}

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

\end{equation}

**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.

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