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
\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.

 

About the author

Frank KschischangFrank R. Kschischang received the B.A.Sc. degree (with honors) from the University of British Columbia, Vancouver, BC, Canada, in 1985 and the M.A.Sc. and Ph.D. degrees from the University of Toronto, Toronto, ON, Canada, in 1988 and 1991, respectively, all in electrical engineering. He is a Professor of electrical and computer engineering at the University of Toronto, where he has been a faculty member since 1991. During 1997–98, he was a visiting scientist at MIT, Cambridge, MA and in 2005 he was a visiting professor at the ETH, Zurich.
His research interests are focused primarily on the area of channel coding techniques, applied to wireline, wireless and optical communication systems and networks. In 1999, he was a recipient of the Ontario Premier’s Excellence Research Award and in 2001 (renewed in 2008) he was awarded the Tier I Canada Research Chair in Communication Algorithms at the University of Toronto. In 2010, he was awarded the Killam Research Fellowship by the Canada Council for the Arts. Jointly with Ralf Koetter, he received the 2010 Communications Society and Information Theory Society Joint Paper Award. He is a Fellow of IEEE and of the Engineering Institute of Canada. During 1997–2000, he served as an Associate Editor for Coding Theory for the IEEE TRANSACTIONS ON INFORMATION THEORY. He also served as technical program co-chair for the 2004 IEEE International Symposium on Information Theory (ISIT), Chicago, and as general co-chair for ISIT 2008, Toronto. He served as the 2010 President of the IEEE Information Theory Society. He is currently serving as Editor-in-Chief of the IEEE Transactions on Information Theory.

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