# Causality and stability of digital systems

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As we have seen, the transfer function H(Z) is the Z-transform of the impulse-response signal $h_{k}$ . A causal system is one for which $h_{k}$ is one sided; that is, $h_{k}=0$ for $k<0$ . The reason is that a causal system cannot produce a response to the spike input before the spike occurs. Because the spike occurs at time zero, the response $h_{k}$ thus must be zero before time zero. Certainly, all physical systems operating in real time must be causal.

Suppose now that we take a causal system and watch its impulse-response signal $h_{k}$ as the time index k becomes larger and larger. If the infinite sum

 {\begin{aligned}\sum _{k=0}^{\infty }{|}h_{k}|\end{aligned}} (48)

is finite, then the impulse response damps out quite rapidly, and we say that the system is stable. However, if this infinite sum diverges, we say the system is unstable. The problem of modifying the elements of an unstable system to make it stable is one of the major problems of control theory.

Let us look at two examples, which we call Example A and Example B. Let a be a complex constant. In Example A, the prototype causal digital signal is the causal signal given by the geometric signal

 {\begin{aligned}h_{k}=0\mathrm {\;} \;\;\;\;{\rm {for}}\;\;\;\;k=-{\rm {l,}}-{2,}\ldots \\h_{k}=a^{k}\mathrm {\;} \;\;\;{\rm {for}}\;\;\;\;k=0,1,2,\ldots .\end{aligned}} (49)

Its Z-transform is the power series

 {\begin{aligned}H\left(Z\right)=\sum _{k=0}^{\infty }{a^{k}}Z^{k}=\left[1+aZ+{\left(aZ\right)}^{2}+{\left(aZ\right)}^{3}+\dots \right]={\frac {1}{1-aZ}}.\end{aligned}} (50)

This power series converges for $|aZ|<1$ , which is $|Z|<|a{|}^{-1}$ . That is, the region of convergence is the interior of the circle of radius $|a{|}^{-1}.$ .

Two cases are important. Case A1 is for $|a|<1$ (Figure 6a). In such a case, the region of convergence includes the unit circle, and the causal signal $h_{k}$ is stable. Case A2 is for $|a{\rm {|>}}1$ (Figure 6b). In that case, the region of convergence does not include the unit circle, and the causal signal $h_{k}$ is unstable. Figure 6.  (a) For a stable causal digital signal, the region of convergence is the interior of a circle, where the interior includes the unit circle. (b) For an unstable causal digital signal, the region of convergence is the interior of a circle, where the interior does not include the unit circle.

In Example B, the prototype anticausal digital signal is the anticausal signal given by the geometric signal

 {\begin{aligned}h_{k}=-a^{k}\mathrm {\;\;\;for\;\;\;} k=-{\rm {l,}}-{2,\ldots }\\h_{k}=0\mathrm {\;\;for\;\;\;} k=0,1,2,\ldots .\end{aligned}} (51)

Its Z-transform is the Laurent series that involves only negative powers of Z given by

 {\begin{aligned}H\left(z\right)=-\sum _{k=-\infty }^{-1}{a^{k}}Z^{k}=-{\left(aZ\right)}^{-1}-{\left(aZ\right)}^{-2}-{\left(aZ\right)}^{-3}-\ldots \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=-(aZ)^{_{-1}}[1+aZ^{-1}+aZ^{-2}-...]={\frac {-(aZ)^{-1}}{1-(aZ^{-1})}}={\frac {1}{1-aZ}}.\end{aligned}} (52)

This Laurent series converges for $|aZ{\rm {|>}}1$ , which is $|Z|>|a{|}^{-1}$ . That is, the region of convergence is the exterior of the circle of radius $|a{|}^{-1}$ .

Two cases are important. Case B1 is for $|a|<1$ (Figure 7a). In such a case, the region of convergence does not include the unit circle, and the anticausal signal $h_{k}$ is unstable. Case B2 is for $|a{\rm {|>}}1$ (Figure 7b). In that case, the region of convergence includes the unit circle, and the anticausal signal $h_{k}$ is stable. Figure 7.  (a) For an unstable anticausal digital signal, the region of convergence is the exterior of a circle, where the exterior does not include the unit circle. (b) For a stable anticausal digital signal, the region of convergence is the exterior of a circle, where the exterior includes the unit circle.

The above two examples show that the causal signal given by equation 50 and the anticausal signal given by equation 52 have the same expression, ${\rm {1/}}\left(1-aZ\right)$ , for their Z-transforms. However, the regions of convergence are different. The region of convergence for the causal signal is inside the circle of radius $\left({\rm {1/|}}a|\right)$ , and the region of convergence for the anticausal signal is outside the circle of radius $\left({\rm {1/|}}a|\right)$ . Thus, the complete specification of the Z-transform requires giving the region of convergence as well as the algebraic expression.

Let us summarize. Given the Z-transform $\left(1{\rm {\ }}-aZ\right)$ , where $|a|<1$ , we have two choices. If we choose the power-series representation (equation 50), we obtain the stable causal signal

 {\begin{aligned}{\rm {\ldots ,0,\ 0,\ 1,\ }}a^{1},\ a^{2},\ a^{3},\ldots ,\end{aligned}} (53)

where the entry 1 occurs at time index $k=0$ . Because $|a|<1$ , this signal damps geometrically in the positive time direction. On the other hand, if we choose the Laurent-series representation (equation 52), we obtain the unstable anticausal signal

 {\begin{aligned}...,a^{-3},\ a^{-2},\ a^{-1}{\rm {,\ 0,0,0,0,}}\dots ,\end{aligned}} (54)

where the first 0 occurs at time index $k=0$ . Because $|a|<1$ , the entries in this signal grow geometrically in the negative time direction. Thus, the choice of Laurent series leads not only to an anticausal digital signal but also to an unstable one. Clearly, such a choice is to be avoided.

Given the Z-transform $\left({\rm {1\ }}-aZ\right)$ , where $|a{\rm {|>}}1$ , again we have two choices. If we choose the power-series representation (equation 50), we obtain the unstable causal signal

 {\begin{aligned}{\rm {...,\ 0,\ 0,\ 1,\ }}a^{1},\ a^{2},\ a^{3},\ldots ,\end{aligned}} (55)

where the entry 1 occurs at time index $k=0$ . Because $|a{\rm {|>}}1$ , this signal grows geometrically in the positive time direction. On the other hand, if we choose the Laurent-series representation (equation 52), we obtain the stable anticausal signal

 {\begin{aligned}...,\ a^{-3},{\rm {\ }}a^{-2},\ a^{-1}{\rm {,0,\ 0,\ 0,\ 0,}}\ldots ,\end{aligned}} (56)

where the first 0 occurs at time index $k=0$ . Because $|a{\rm {|>}}1$ , this signal damps geometrically in the negative time direction. Thus we have no clear-cut choice, as we did when $|a|<1$ . We cannot have both causality and stability; we must choose one or the other. The usual choice is equation 56, which gives us stability at the expense of causality.

We now wish to present the method of partial fractions to find the impulse response of a digital ARMA$\left(p,\ q\right)$ system from its transfer function

 {\begin{aligned}H\left(Z\right)={\frac {{\beta }_{0}+{\beta }_{1}Z+{\beta }_{2}Z^{2}+\ldots +{\beta }_{q}Z^{q}}{1+{\alpha }_{1}Z^{1}+{\alpha }_{2}Z^{2}{\rm {+\ldots +}}{\alpha }_{p}Z^{P}}}={\frac {{\beta }_{0}\left(1-b_{1}Z\right)\left(1-b_{2}Z\right)\ldots \left(1-b_{q}Z\right)}{\left(1-a_{1}Z\right)\left(1-a_{2}Z\right)\ldots \left(1-a_{p}Z\right)}}.\end{aligned}} (57)

The right-hand expression is the factored form of the transfer function. The $a_{i}^{-1}$ are its poles and the $b_{i}^{-1}$ are its zeros. For simplicity of presentation, we will assume that $p>q$ , so that $H\left(Z\right)$ is a proper fraction in Z. In addition, we will suppose that all the poles $a_{1}^{-1}$ , $a_{2}^{-1}$ ,..., $a_{p}^{-1}$ are simple ones. We now write $H\left(Z\right)$ as the sum of partial fractions

 {\begin{aligned}H\left(Z\right)=\sum _{i=1}^{p}{\frac {{\rm {A}}_{i}}{1-a_{i}Z}}.\end{aligned}} (58)

The constants $A_{i}$ are determined by means of the equation

 {\begin{aligned}A_{i}=\left(1-a_{i}Z\right)H\left(Z\right){|}_{Z=a_{i}^{-1}}.\end{aligned}} (59)

We now divide the poles into two sets: those with magnitude greater than one, which we designate as $a_{1}^{-1}$ , $a_{2}^{-1}$ , $\ldots$ , $a_{r}^{-1}$ and those with magnitude less than one, which we designate as $a_{r+{1}^{-1}}$ , $a_{r+{2}^{-1}}$ , $\ldots$ ,$a_{p^{-1}}$ . For the first set, we pick the regions $|Z|<|a_{i}^{-1}|$ as the regions of convergence. For the second set, we pick the regions $|Z|>|a_{i}^{-1}|$ as the regions of convergence. Thus, all the regions include the unit circle. It follows that the stable impulse response is the two-sided signal given by

 {\begin{aligned}h_{k}=A_{1}a_{1}^{k}+A_{2}a_{2}^{k}+\ldots +A_{r}a_{r}^{k}\mathrm {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;for\;\;\;\;\;\;} k=-1,-2,...\;\;\;\;\\h_{k}=-A_{r+1}a_{r+1}^{k}-A_{r+2}a_{r+1}^{k}-\ldots -A_{p}a_{p}^{k}\mathrm {\;\;\;\;\;\;for\;\;\;\;\;} k=0,1,2,....\end{aligned}} (60)

Unless otherwise stated, it is assumed that an ARMA$\left(p,\ q\right)$ system is both stable and causal. The above expression for the impulse response shows that the system is causal if and only if the set of poles of magnitude less than one is empty. Thus we have proved the following theorem: All the poles $a_{1}^{-1}$ , $a_{2}^{-1}$ ,..., $a_{p}^{-1}$ of a stable causal digital ARMA$\left({\rm {p,\ }}q\right)$ system $H\left(z\right)$ have magnitude greater than one. In the case in which $p>q$ and in which the poles are distinct, the stable causal impulse response has the form

 {\begin{aligned}h_{k}=A_{1}a_{1}^{k}+A_{2}a_{2}^{k}+\ldots -+A_{p}a_{p}^{k}\mathrm {\;\;\;for\;\;\;} k=0,1,2,\ldots ,\end{aligned}} (61)

where the coefficients are given by equation 59.