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THE BILINEAR TRANSFORM

Z transforms and Fourier transforms are related by the relations $Z = e^{i\omega}$ and $i\omega = \ln Z$.A problem with these relations is that simple ratios of polynomials in Z do not translate to ratios of polynomials in $\omega$ and vice versa. The approximation  
 \begin{displaymath}
-i \hat \omega \eq 2 {1 - Z \over 1 + Z}\end{displaymath} (42)
is easily solved for Z as  
 \begin{displaymath}
Z \eq {1 + \hat \omega /2 \over 1 - \hat \omega /2}\end{displaymath} (43)
These approximations are often useful. They are truncations of the exact power series expansions  
 \begin{displaymath}
-i\omega
\eq - \, \ln \, e^{i\omega}
\eq -\ln \, Z
\eq 2 \le...
 ...er 3}
 {(1 - Z)^3 \over (1 + Z)^3}
+ {1 \over 5} \cdots \right]\end{displaymath} (44)
and  
 \begin{displaymath}
Z
\eq e^{i\omega}
\eq {e^{i\omega /2} \over e^{-i\omega /2}}...
 ...! + \cdots \over
 1 - i\omega /2 + (i\omega /2)^2 /2! + \cdots}\end{displaymath} (45)

For a Z transform B(Z) to be minimum phase, any root Z0 of 0 = B(Z0) should be outside the unit circle. Since $Z_0 = \exp \{i[\mbox{Re} (\omega_0) + i\, \mbox{Im} (\omega_0)]\} $and $\mid Z_0 \, \mid = e^{-\mbox{Im}(\omega_0)}$, it means that for a minimum phase $\mbox{Im} \, (\omega_0)$ should be negative. (In other words, $\omega_0$ is in the lower half-plane.) Thus it may be said that $Z = e^{i\omega}$ maps the exterior of the unit circle to the lower half-plane. By inspection of Figures 20 and 21, it is found that the bilinear approximation (42) or (43) also maps the exterior of the unit circle into the lower half-plane.

 
2-20
Figure 20
Some typical points in the Z-plane, the $\omega$-plane, and the $\hat \omega$-plane.

2-20
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2-21
2-21
Figure 21
The points of Figure 20 displayed in the Z plane, the $\omega$ plane, and the $\hat \omega$-plane.


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Thus, although the bilinear approximation is an approximation, it turns out to exactly preserve the minimum-phase property. This is very fortunate because if a stable differential equation is converted to a difference equation via (42), the resulting difference equation will be stable. (Many cases may be found where the approximation of a time derivative by multiplication with 1 - Z would convert a stable differential equation into an unstable difference equation.)

A handy way to remember (42) is that $-i\omega$ corresponds to time differentiation of a Fourier transform and (1 - Z) is the first differencing operator. The (1 + Z) in the denominator gets things ``centered" at Z1/2

To see that the bilinear approximation is a low-frequency approximation, multiply top and bottom of (42) by Z-1/2

   \begin{eqnarray}
-i\hat \omega
&= & 2\ {Z^{-1/2} - Z^{1/2} \over
 Z^{-1/2} + Z^{...
 ..., \omega /2} \nonumber \\  \hat \omega
&= & 2 \, \tan \, \omega /2\end{eqnarray}
(46)
Equation (46) implicitly refers to a sampling rate of one sample per second. Taking an arbitrary sampling rate $\Delta t$, the approximation (46) becomes

 
 \begin{displaymath}
\omega \, \Delta t \quad\approx\quad 2 \, \tan \, \omega \Delta t /2\end{displaymath} (47)
This approximation is plotted in Figure 22. Clearly, the error can be made as small as one wishes merely by sampling often enough; that is, taking $\Delta t$ small enough.

 
2-22
Figure 22
The accuracy of the bilinear transformation approximation.

2-22
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From Figure 22 we see that the error will be only a few percent if we choose $\Delta t$ small enough so that $\omega_{\max} \Delta t \leq 1$. Readers familiar with the folding theorem will recall that it gives the less severe restraint $\omega_{\max} \Delta t < \pi$. Clearly, the folding theorem is too generous for applications involving the bilinear transform.

Now, by way of example, let us take up the case of a pole $1/ -i\omega$ at zero frequency. This is integration. For reasons which will presently be clear, we will consider the slightly different pole

 
 \begin{displaymath}
P \eq {1 \over -i\omega + \varepsilon}\end{displaymath} (48)
where $\varepsilon$ is small. Inserting the bilinear transform, we get

   \begin{eqnarray}
P 
&= & {1 \over 2 [(1 - Z)/(1 + Z)] + \varepsilon}
 \eq {0.5(1...
 ...& {0.5(1 + Z) \over (1 + \varepsilon /2)
 - Z(1 - \varepsilon /2)}\end{eqnarray}
(49)
By inspection of (49) we see that the time-domain function is real, and as $\varepsilon$ goes to zero it takes the form (.5, 1, 1, 1, ...). (Taking $\varepsilon$ positive forces the step to go out into positive time, whereas $\varepsilon$ negative would cause the step to rise at negative time.) The properties of this function are summarized in Figure 23.

 
2-23
2-23
Figure 23
Properties of the integration operator.


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EXERCISES:

  1. In the solution to diffusion problems, the factor $F(\omega) = 1/(-i\omega)^{1/2}$ often arises as a multiplier. To see the equivalent convolution operation, find a causal, sampled-time representation ft of $F(\omega)$ by identification of powers of Z in
    \begin{eqnarray}
(f_0 + f_1 Z + f_2 Z^2 + \cdots)^2 \eq
1/(-i\omega) \quad\simeq\quad {1 \over 2}(1 + Z)/(1 - Z) \nonumber \\ \end{eqnarray}
    Solve numerically for f0 through f7.


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Next: About this document ... Up: One-sided functions Previous: NOTCH FILTER AND POLE
Stanford Exploration Project
10/30/1997