0

Let an IAR system be defined by the following difference equation:

$$y[n]-\frac{1}{4} y[n-2]=x[n]+3x[n-1]$$

and an input signal $x[n]=(-0.5)^n$.

The transfer function is defined as $H^z(z)=\frac{1+3z^{-1}}{1-0.25 z^{-2}}$. The value $z=-0.5$ is a pole of the system, hence, it is not in the ROC. We cannot use $y[n]=H^z(-0.5)x[n]$ to solve, as we would for any other eigen-signal.

Is there any other way to solve this? What can we say about the output? Is it infinity? is it undefined?

Matt L.
  • 78,282
  • 5
  • 66
  • 149
havakok
  • 603
  • 1
  • 4
  • 17

1 Answers1

0

If the input is $x[n]=\left(-\frac12\right)^n$ for all $n$, then the output is infinite, because the convolution sum doesn't converge. Note that also the input grows without bounds for negative $n$.

If, on the other hand, the input starts at $n=0$, i.e., $x[n]=\left(-\frac12\right)^nu[n]$ then you can use the $\mathcal{Z}$-transform to compute the output:

$$Y(z)=X(z)H(z)\tag{1}$$

With $$X(z)=\frac{1}{1+\frac12 z^{-1}}\tag{2}$$

you obtain

$$\begin{align}Y(z)&=\frac{1+3z^{-1}}{\left(1+\frac12 z^{-1}\right)^2\left(1-\frac12 z^{-1}\right)}\\&=\frac{a}{\left(1+\frac12 z^{-1}\right)^2}+\frac{b }{1+\frac12 z^{-1}}+\frac{c}{1-\frac12 z^{-1}}\tag{3}\end{align}$$

with some real-valued constants $a$, $b$, and $c$.

Inverse $\mathcal{Z}$-transform of $(3)$ gives

$$y[n]=-2a\;n\left(-\frac12\right)^n+b\left(-\frac12\right)^n+c\left(\frac12\right)^n\tag{4}$$

EDIT: First, to clarify, I use a basic example that shows that a double pole in the $\mathcal{Z}$-transform is no problem. Assume a system with impulse response $h[n]=(1/2)^nu[n]$ and an input signal $x[n]=(1/2)^nu[n]$. Computing the output in the time domain is straightforward:

$$\begin{align}y[n]&=u[n]\sum_{k=0}^n\left(\frac12\right)^k\left(\frac12\right)^{n-k}\\&=u[n]\left(\frac12\right)^n\sum_{k=0}^n1\\&=(n+1)\left(\frac12\right)^nu[n]\tag{5}\end{align}$$

The $\mathcal{Z}$-transform of the output sequence is simply given by the multiplication of the $\mathcal{Z}$-transforms $H(z)$ and $X(z)$:

$$Y(z)=\frac{1}{\left(1-\frac12z^{-1}\right)^2}\tag{6}$$

The inverse $\mathcal{Z}$-transform of $(6)$ equals the result $(5)$, which is very easy to confirm.

The probable cause of the OP's confusion could be the fact that for an input signal $x[n]=z_0^n$, $-\infty<n<\infty$, the output is given by

$$y[n]=H(z_0)z_0^n$$

if and only if $z_0$ is inside the ROC of $H(z)$, because otherwise the convolution sum doesn't converge and the expression $H(z_0)$ is meaningless. However, in the computation of the outputs $y[n]$ in the examples above we never evaluate $H(z)$ for any $z$ outside the ROC, so the situation is entirely different.

Matt L.
  • 78,282
  • 5
  • 66
  • 149
  • I think your solution for $x[n]=(-0.5)^nu[n]$ is incorrect. How are you using $H^z(z)$ for $z\notin ROC$? – havakok Jan 07 '21 at 12:53
  • You did not evaluate it. You say "of course you can use it to multiply it with the Z-transform of the input". I disagree. Can you support your claim? – havakok Jan 07 '21 at 13:56
  • Yes, that fact is well known in the general case and I want you to support that it can also be done in the specific case of an eigen-signal which is a pole of the transfer function. A book or a paper that supports this delicate point or mathematical reasoning. – havakok Jan 07 '21 at 14:05