Analog Phase Samples Approximation from Gain Samples by Discrete Hilbert Transform

Abstract

The Hilbert transform has been recognized as an important method in circuit theory. One of its important applications is related to the gain–phase relationship. Many practical methods have been derived from the Hilbert transform to calculate the analog phase samples from the analog gain samples. However, the use of the discrete Hilbert transform is not so common in phase approximation through gain samples. One reason may be related to the possible presence of aliasing. In this paper, we show that yet another inconvenience can occur when it is desired to obtain some samples of the phase using the discrete Hilbert transform of the gain samples: the minimum-phase sequences of finite length allow only zero total net phase change. Thus, the computation of the analog phase samples using the discrete Hilbert transform is convenient only when its net phase change of the analog circuit is also zero.

Appendices

Appendix A

Let us consider a causal finite-length minimum-phase sequence $\{ x(n) \}$, $n=0,1,\ldots ,N-1$. Its z-transform X(z) can be written as:

$$\begin{aligned} \displaystyle X(z)= & {} x(0) + x(1) z^{-1}+ \cdots + x(N-1)z^{-(N-1)} \nonumber \\ \displaystyle= & {} x (0) \prod _{k=1}^{N-1} (1-z_kz^{-1}) , \end{aligned}$$

(9)

where $z_k=|z_k|e^{j \angle z_k}$ are its zeros inside the unit circle ($0 \le |z_k| <1, \forall k=0,1,\ldots ,N-1$).

Now let us focus on a single term $X_k(z) = 1-z_kz^{-1}$ and compute its net phase change: $\angle X(e^ {j \pi }) - \angle X(e^{- j \pi })$.

First we note that $X_k(e^{j \omega }) = 1-|z_k|e^{j \angle z_k}e^{ -j \omega }= 1 -|z_k| [ \cos ( \omega - \angle z_k) + j \sin (\omega - \angle z_k) ]$ and its real part $1 -|z_k| \cos ( \omega - \angle z_k)$ is always positive. Consequently, its argument is done by

$$\begin{aligned} \angle X_k(e^{j \omega } ) = \arctan \frac{ |z_k| \sin (\omega - \angle z_k) }{1 -|z_k| \cos ( \omega - \angle z_k)}, \end{aligned}$$

it is continuous and its derivative is also continuous. It follows that

$$\begin{aligned} \angle [X(e^ {j \pi }) ] - \angle [ X(e^{- j \pi } ) ]= & {} \arctan \frac{ |z_k| \sin (\pi - \angle z_k) }{1 -|z_k| \cos (\pi - \angle z_k) } \\&- \arctan \frac{ |z_k| \sin (- \pi - \angle z_k) }{1 -|z_k| \cos ( - \pi - \angle z_k) } = 0. \end{aligned}$$

By using (9), we obtain that the overall net phase change of X(z) is zero.

Appendix B

Let us consider a causal system having

$$\begin{aligned} H(s) = K \frac{\displaystyle \prod _{k=1}^M (s-Z_k)}{\displaystyle \prod _{k=1}^N (s-P_k)}, \end{aligned}$$

(10)

where $Z_k=\Re (Z_k) + j \Im (Z_k)$, with $\Re (Z_k) < 0$, and $P_k=\Re (P_k) + j \Im (P_k)$, with $\Re (P_k) < 0$

Let us analyze the net phase change of the term $H_k(s)=s-Z_k$. We have:

$$\begin{aligned} \angle H_k(j \Omega ) = \angle [j \Omega - (\Re (Z_k)+ j \Im (Z_k))] = \arctan \frac{\Omega - \Im (Z_k) }{- \Re (Z_k)} \end{aligned}$$

and it is continuous and its derivative is continuous. We get

$$\begin{aligned} {\displaystyle \angle [ H_k(j \infty ) ] - \angle [ H_k(-j \infty ) ] = \pi . } \end{aligned}$$

The terms that contain poles have the same behavior. Consequently, H(s) has an overall net phase change of zero if and only if $M=N$.