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  

$$-i \hat \omega \eq 2 {1 - Z \over 1 + Z}$$ (42)

is easily solved for Z as  

$$Z \eq {1 + \hat \omega /2 \over 1 - \hat \omega /2}$$ (43)

These approximations are often useful. They are truncations of the exact power series expansions  

$$-i\omega \eq - \, \ln \, e^{i\omega} \eq -\ln \, Z \eq 2 \le... ...er 3} {(1 - Z)^3 \over (1 + Z)^3} + {1 \over 5} \cdots \right]$$ (44)

and  

$$Z \eq e^{i\omega} \eq {e^{i\omega /2} \over e^{-i\omega /2}}... ...! + \cdots \over 1 - i\omega /2 + (i\omega /2)^2 /2! + \cdots}$$ (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.

 

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

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 Z^1/2

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

 

$$\omega \, \Delta t \quad\approx\quad 2 \, \tan \, \omega \Delta t /2$$ (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.

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

 

$$P \eq {1 \over -i\omega + \varepsilon}$$ (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
Figure 23 Properties of the integration operator.

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 f_t 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.