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Derivation of the FIR transfer function for the frequency response of the Hilbert transform

The ideal frequency response of the Hilbert transform is expressed as

$\displaystyle H_{IHT}(\omega)=-i\,{\rm {sgn}}\,\omega= -i\frac{\omega}{\left\ve...
... \textrm{$-\pi<\omega<0$}\\ -i, & \textrm{$0<\omega<\pi$} \end{array} \right. .$ (15)

From equations 4 and A-5, we obtain the difference as $ –1/\left\vert\omega\right\vert$ . For

$\displaystyle {\rm {sgn}}\,x=\frac{x}{\sqrt{x^2}}=xf(x^2),x\neq0$ (16)

and $ f(u)=\displaystyle\frac{1}{\sqrt{u}},u>0$ , the Taylor series of $ f(u)$ at center c is expressed

$\displaystyle f(u)=\frac{1}{\sqrt{c}}\left[1+\sum_{m=1}^{\infty} \frac{(2{\rm {m}}-1)!!}{(2{\rm {m}})!!}\left(1-\frac{u}{c}\right)^m\right],$ (17)

where $ (2{\rm {m}}-1)!!=1 \cdot 3 \cdot 5 \dots (2 {\rm {m}}- 1)$ , $ (2{\rm {m}})!!=1\cdot 3\cdot 5\dots(2{\rm {m}})$ . Consequently, the signum function sgn$ x$ is expressed

$\displaystyle {\rm {sgn}}\,x=\frac{x}{\sqrt{c}}\left[1+\sum_{m=1}^{\infty} \frac{(2{\rm {m}}-1)!!}{(2{\rm {m}})!!}\left(1-\frac{x^2}{c}\right)^m\right].$ (18)

We substitute sin$ \omega$ for $ x$ , based on sgn$ \omega$ =sgn(sin$ \omega$ ) for $ –\pi<\omega<\pi$ , truncate the series at the first M terms, and obtain the sinusoidal power series of the signum function as

$\displaystyle \rm {sgn}\,\omega=\frac{\rm {sin}\omega}{\sqrt{c}} \left[1+ \sum_...
...{sin}^2 \omega}{c}\right)^m+\circ((1-\frac{\rm {sin^2\omega}}{c})^{M+1})\right]$ (19)

The series in A-9 converges for $ \displaystyle
-1<1-\frac{{\rm {sin}}\omega}{c}<1$ ; that is, $ c$ has to be larger than 1/2. On the other hand, the expansion center $ c$ in the $ x$ -domain is associated to the frequency center in the $ \omega$ -domain via the relation $ c={\rm {sin}}^2\omega_c$ . Therefore, $ c={\rm {sin}}^2\omega_c$ must be less than or equal to 1. Accordingly, $ c$ is constrained by $ 1/2<c\;{\leq}\;1$ and the corresponding $ \omega_c$ is within the range $ [\pi/4,\pi/2]$ . Clearly, the ideal frequency response is well approximated within the middle frequency band. Multiplying A-9 by $ - i$ and substituting $ \displaystyle\frac{z-z^{-1}}{2i}$ for sin$ \omega$ , the transfer function for the zero phase FIR of the Hilbert transform is expressed as

$\displaystyle H_{HT}\rm {(z,c)}\approx-\frac{z-z^{-1}}{2\sqrt{c}} \left\{1+\sum...
...2m)!!}\left[1+\frac{1}{c} \left( \frac{z-z^{-1}}{2} \right)^2\right]^m \right\}$ (20)

To obtain the causal transfer function, $ H_{HT}(z,c)$ is multiplied by $ z^{-2M-1}$ and the resultant transfer function of the FIR Hilbert transform of the (2$ M$ +2)th-order is

$\displaystyle \hat{H}_{HT}\rm {(z,c)}\approx-\frac{1-z^{-2}}{2\sqrt{c}} \left\{...
... \left[z^{-2}+\frac{1}{c} \left( \frac{1-z^{-2}}{2} \right)^2\right]^m \right\}$ (21)

For $ M$ =0, the transfer functions of equations A-4 and A-11 are approximated as

$\displaystyle \hat{H}_{HT}\rm {(z,c)}\approx-\frac{1-z^{-2}}{2\sqrt{c}}$ (22)

$\displaystyle \hat{F}_{DD}\rm {(z)}\approx-\frac{1-z^{-2}}{2}$ (23)

We compare equations A-12 and A-13, and we conclude that these two transfer functions in middle frequency band of the frequency domain differ by the constant coefficient $ \displaystyle\frac{1}{\sqrt{c}}$ .


next up previous [pdf]

Next: Bibliography Up: Appendix A: Hilbert transform Previous: Derivation of the FIR

2015-05-07