# Impulse response of non-causal system

Suppose I have the following equation, where $x[n]$ is the input and $y[n]$ is the output: $$x[n]-x[n-1]=-10y[n-1]+2y[n]+2y[n+1].$$

The question asks me to compute the impulse response from this equation.

Attempts: $x[n]=1$ when $n=0$; $0$ otherwise. I need to find $h[n]=y[n]$, but, from the equation, it always depends on the future value $y[n+1]$, which is unknown.

One way to solve such a difference equation is to use the $\mathcal{Z}$-transform. Transforming the given difference equation yields

$$X(z)\left(1-z^{-1}\right)=Y(z)\left(-10z^{-1}+2+2z\right)\tag{1}$$

which results in the following transfer function:

$$H(z)=\frac{Y(z)}{X(z)}=\frac{1-z^{-1}}{2z+2-10z^{-1}}=\frac12\frac{z^{-1}-z^{-2}}{1+z^{-1}-5z^{-2}}$$

The poles of $H(z)$, i.e. the zeros of the denominator polynomial in $(1)$, are both real-valued:

$$1+z^{-1}-5z^{-2}=(1-az^{-1})(1-bz^{-1})\tag{2}$$

with $a=\frac{-1+\sqrt{21}}{2}\approx 1.79$ and $b=\frac{-1-\sqrt{21}}{2}\approx -2.79$. Note that $1<|a|<|b|$ is satisfied, so both real-valued poles are outside the unit circle of the complex plane.

With $(2)$, $H(z)$ can be rewritten as

$$H(z)=\frac12\frac{z^{-1}-z^{-2}}{(1-az^{-1})(1-bz^{-1})}=\frac{1}{2(b-a)}\left[\frac{(1-a)z^{-1}}{1-az^{-1}}-\frac{(1-b)z^{-1}}{1-bz^{-1}}\right]\tag{3}$$

This transfer function can have three possible regions of convergence (ROCs), each of them corresponding to a different impulse response:

\begin{align}(1)\quad &|z|>|b|>|a|\\ (2)\quad & |b|>|z|>|a|\\ (3)\quad & |b|>|a|>|z|\end{align}

ROC $(1)$ corresponds to a right-sided impulse response, ROC $(2)$ corresponds to a two-sided impulse response, and ROC $(3)$ corresponds to a left-sided sequence. Only ROC $(3)$ includes the unit circle $|z|=1$, so only the impulse response corresponding to this ROC results in a stable, yet anti-causal, system. The three impulse responses corresponding to the three ROCs given above are

\begin{align}(1)\quad & h_1[n]=\frac{1}{2(b-a)}\left((1-a)a^{n-1}-(1-b)b^{n-1}\right)u[n-1]\\ (2)\quad & h_2[n]=\frac{1}{2(b-a)}\left((1-a)a^{n-1}u[n-1]+(1-b)b^{n-1}u[-n]\right)\\ (3)\quad & h_3[n]=\frac{1}{2(b-a)}\left(-(1-a)a^{n-1}+(1-b)b^{n-1}\right)u[-n]\end{align}

These impulse responses describe three different systems, all of them characterized by the same transfer function given by Eq. $(3)$. $h_1[n]$ corresponds to a causal unstable system, $h_2[n]$ to a non-causal unstable system, and $h_3[n]$ to an anti-causal stable system.

• Actually the question does not state that this system is non-causal, it is my assumption based on the equation. The equation is derived from a 2nd order ODE in continous time, then mapping to DT using Euler approximation. I think your answer might apply in this case – TonyJ Nov 11 '15 at 11:58
• @TonyJ: But the causal impulse response is not stable, so it's not useful in practice. – Matt L. Nov 11 '15 at 12:45