Is a Difference Equation causal, anti causal, or non causal?

Question

A stable discrete-time LTI system is described by the following difference equation:

$$ y[n] - y[n-1] + Cy[n-2] = x[n] $$

where C is a real number. Determine the range of C so that

(a) the system is causal;

(b) the system is anti-causal;

(c) the system is non-causal (i.e., it has a two-sided impulse response).

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It is straight forward to calculate the transfer function:

\begin{align*} Y(z) - z^{-1}Y(z) + Cz^{-2}Y(z) &= X(z) \\ H(Z) = \frac{Y(z)}{X(z)} &= \frac{1}{1 - z^{-1} + Cz^{-2}} \\ \end{align*}

We are given that the system is stable so the ROC must include the unit circle. Therefore there can not be a pole with magnitude \$1\$.

With \$C=0\$, \$H(z)= \frac{1}{1 - z^{-1}}\$, there is a pole at \$1\$ so that is not possible.

With \$C=-2\$, \$H(z)= \frac{1}{1 - z^{-1} - 2z^{-2}} = \frac{\frac{1}{3}}{1+z^{-1}} + \frac{\frac{2}{3}}{1-2z^{-1}}\$, there is a pole at \$-1\$ so that is not possible.

From there, where do I go?

Ultimately this may factor into the form:

$$ \frac{k_1}{1 + a_1 z^{-1}} + \frac{k_2}{1 + a_2 z^{-1}} $$

Is that a causal transfer function or not? The inverse Z transform can yield both a causal and anti-causal impulse response function.

Answer 1

This is a root locus analysis with variable K.

The poles are: $$ p_{1,2}=\frac{1}{2}(1 \pm (1 - 4c)^{1/2}), |p_2|<|p_1| $$

For \$c<=1/4\$, the poles are real. \$p_{1}=1\$ when \$c = 0\$.

For \$c>1/4\$, the poles are complex, and their magnitude is: $$ |p_{1,2}|=\sqrt{\frac{1}{4}(1 + (4c-1))}=\sqrt{c} $$ which is equal to 1 when \$c=1\$.

Hence, for \$c\$, \$p_1\$ and \$p_2\$:

• \$c=0: p_1=1\$ (critical), \$p_2=0\$ (stable)

• \$c=1: p_{1,2}=1/2 \pm i\sqrt(3)/2\$ (underdamped stable,\$<1\$)

• \$c=1/4: p_1=1/2, p_2=1/2\$ (overdamped stable,\$<1\$)

Because the time response is given, with negative delays, the system is casual, independently of \$c\$.

If only \$H(z)\$ were given and no time response, we obtain the explicit division, for inspect each converging term involving infinite sums: $$ H(Z) = \frac{1}{p_1-p_2}(\frac{p_1}{1 - p_1 z^{-1}}-\frac{p_2}{1 - p_2 z^{-1}}) $$

Hence:

For a Causal System (no diverging components on the sum, for any \$c\$):

$$ |p_1 z^{-1}|<1 and |p_2 z^{-1}|<1 \rightarrow max(|p_1|,|p_2|)<|z| $$

if \$c<=1/4\$: \$\frac{1}{2}(1 + (1 - 4c)^{1/2})<|z|\$

if \$c>1/4\$: \$\sqrt{c}<|z|\$

For a Anti-Causal System (both components diverge):

$$ |p_1 z^{-1}|>1 and |p_2 z^{-1}|>1 \rightarrow min(|p_1|,|p_2|)>|z| $$

if \$c<=1/4\$: \$\frac{1}{2}(1 - (1 - 4c)^{1/2})>|z|\$

if \$c>1/4\$: \$\sqrt{c}>|z|\$

For a Non-Causal System (one component diverge and the other converge):

$$ |p_1 z^{-1}|>1 and |p_2 z^{-1}|<1 \rightarrow |p_2|<|z|<|p_1| $$

if \$c<=1/4\$: \$\frac{1}{2}(1 - (1 - 4c)^{1/2})<|z|<\frac{1}{2}(1 + (1 - 4c)^{1/2})\$

if \$c>1/4\$: \$\sqrt{c}=|z|\$

in MATLAB, making:

h = tf([1],[1 -1 0],1); rlocus(h); axis equal;

you obtain the 'variable locus', though the correct interpretation of the c values is artificial here.

Q.E.P.D.