We talked about splitting up an output into a transient and steady state representations. Now we are going to talk about zero-input and zero-state responses of a system. Consider the following system:

\begin{align*}
y(n) - 0.9y(n-1) = x(n) \quad \lr{\mbox{system is causal}}\\
\end{align*}
 The order of the system is the number of delay elements required for implementation. So we need one initial condition (in this case) to determine the response. What is $y(-1) \neq 0$? In other words, this means that you could pass in the zero signal $x(n) = 0$ and you won't get a zero output; the system is not ZIZO. 

\begin{problem}

What is the zero input response of the system?

\end{problem}


\begin{solution}

\begin{align*}
\mbox{let } x(n) = 0 \quad n \geq 0 \\
\end{align*}

The system's causality implies that we should write $y(n)$ in terms of its past values and solve in the forward direction ( $n=0,1,2,\dots$ ):

\begin{align*}
y(n) = 0.9y(n-1) + x(n) \\
\end{align*}

Remember that we have let $x(n) = 0$ $ \forall $ $n \geq 0$. 

\begin{align*}
y(0) &= (0.9)   y(-1) \\
y(1) &= (0.9)^2 y(-1) \\
y(2) &= (0.9)^3 y(-1) \\
&\vdots  \\
y(n) &= (0.9)^{n+1} y(-1) = y_{ZIR}(n)\\
\end{align*}

This result is called the \emph{zero-input response}, or zero-input solution (homogeneous solution). We have turned off the input but don't get a zero output.

\end{solution}


\begin{problem}

What if $y(-1) = 0$ and $x(n) = u(n)$? In other words, what is the zero-state response of the system?

\end{solution}

\begin{solution}

First we need to find the transform everything into the transform domain, and find the equation for the system. We then pass a unit step through this system.

\begin{align*}
y(n) &= 0.9y(n-1) + x(n) \quad \abs{z} \gt 0.9 \\
\hat{Y}(z) &= z^{-1}0.9\hat{Y}(z) + \hat{X}(z) \\
\hat{Y}(z)(1 - 0.9z^{-1}) &= \hat{X}(z) \\
\hat{H}(z) &= \frac{1}{1 - 0.9z^{-1}} \\
\hat{H}(z) &= \frac{z}{z - 0.9} \\
\end{align*}

We know that the Z-transform of the unit step:

\begin{align*}
\hat{X} &= \frac{1}{1 - z^{-1}} \quad \abs{z} \gt 1 \\ 
\hat{X} &= \frac{z}{z - 1} \quad \abs{z} \gt 1 \\ 
\end{align*}

Convolution in the time domain is multiplication in the Z-domain:

\begin{align*}
\hat{Y}(z) &= \hat{H}(z) \hat{X}(z) \\
\hat{Y}(z) &= \frac{z^2}{(z - 0.9)(z-1)} \quad \abs{z} \gt 1 \\
\end{align*}

Now we break $\hat{Y}(z)$ into its components: 
\begin{align*}
\hat{Y}(z) &= \frac{z^2}{(z - 0.9)(z-1)} \\
\frac{\hat{Y}(z)}{z} &= \frac{A}{(z-0.9)} + \frac{B}{(z-1)} \\
\frac{z}{(z - 0.9)(z-1)} &= \frac{A}{(z-0.9)} + \frac{B}{(z-1)} \\
A &= \left. \frac{z}{z-1} \right|_{z=0.9} = -9\\
B &= \left. \frac{z}{z-0.9} \right|_{z=1} = 10\\
\end{align*}

Now putting it all together:
\begin{align*}
\frac{\hat{Y}(z)}{z} &= \frac{A}{(z-0.9)} + \frac{B}{(z-1)} \\
\frac{\hat{Y}(z)}{z} &= \frac{-9}{(z-0.9)} + \frac{10}{(z-1)} \\
\hat{Y}(z) &= \frac{-9z}{(z-0.9)} + \frac{10z}{(z-1)} \\
y(n) &=-9(0.9)^n u(n) +10 u(n) = y_{ZSR}(n) \\
\end{align*}

This result is the \emph{zero-state response} of the system, (a.k.a. the particular solution).

\end{solution}

\begin{problem}

What if $y(-1) = 0$ and $x(n) = u(n)$? 

\end{problem}


\begin{solution}

By superposition, the general solution is the sum or the homogenous and particular solution.
\begin{align*}
y(n) &= y_{\small ZSR}(n) + y_{ZIR}(n) \\
\end{align*}

Is there a way to solve for all of these in one step? Start off by truncating both sides of the equation:
\begin{align*}
y(n) &= 0.9y(n-1) + x(n) \\
y(n)u(n) &= 0.9y(n-1)u(n) + x(n)u(n) \\
\end{align*}

Then Z-transform both sides:

\begin{align*}
\sum \limits_{n=-\infty}^{\infty} y(n)u(n) z^{-n} &= \sum \limits_{n=-\infty}^{\infty}0.9y(n-1)u(n) z^{-n} + \sum \limits_{n=-\infty}^{\infty}x(n)u(n) z^{-n} \\
\sum \limits_{n=0}^{\infty} y(n)z^{-n} &= \sum \limits_{n=0}^{\infty}0.9y(n-1) z^{-n} + \sum \limits_{n=0}^{\infty}x(n) z^{-n} \\
\end{align*}

We need to add a new definition to our toolbox, the \emph{Unilateral Z-transform}:
\begin{nicebox}
\begin{align*}
\mathcal{Y}(z) \bydef \sum \limits_{n=0}^{\infty} y(n) z^{-n}
\end{align*}
\end{nicebox}


This is used in contrast to the \emph{Bilateral Z-transform} which is what we have been using up until this point.

\begin{align*}
\sum \limits_{n=0}^{\infty} y(n)z^{-n} &= 0.9\sum \limits_{n=0}^{\infty}y(n-1) z^{-n} + \sum \limits_{n=0}^{\infty}x(n) z^{-n} \\
\mathcal{Y}(z) &= 0.9\sum \limits_{n=-1}^{\infty}y(m) z^{-(m+1)} + \mathcal{X}(z) \quad \mbox{let } m = n -1\\
\mathcal{Y}(z) &=  z^{-1}0.9\sum \limits_{m=-1}^{\infty}y(m) z^{-m} + \mathcal{X}(z) \\
\mathcal{Y}(z) &= 0.9y(-1) + z^{-1}0.9\sum \limits_{m=0}^{\infty}y(m) z^{-m} + \mathcal{X}(z) \quad \mbox{ (evaluate first term in sum)}  \\
\mathcal{Y}(z) &= 0.9y(-1) + z^{-1}0.9\mathcal{Y}(z) + \mathcal{X}(z)  \\
\end{align*}

We have successfully captured the initial conditions, $y(-1)$ in the formula. Note that for causal systems, $\hat{X}(z) = \mathcal{X}(z)$, the bilateral Z-transform is equal to the unilateral Z-transform.o

\begin{align*}
\mathcal{Y}(z) &= 0.9y(-1) + z^{-1}0.9\mathcal{Y}(z) + \mathcal{X}(z)  \\
\mathcal{Y}(z)(1 - 0.9z^{-1}) &= 0.9y(-1) + \mathcal{X}(z)  \\
\mathcal{Y}(z) &= \frac{0.9y(-1)}{(1 - 0.9z^{-1})} + \frac{1}{(1 - 0.9z^{-1})}\mathcal{X}(z)  \\
\mathcal{Y}(z) &= \frac{0.9y(-1)}{(1 - 0.9z^{-1})} + \hat{H}(z)\mathcal{X}(z)  \\
\mathcal{Y}(z) &= \frac{0.9y(-1)}{(1 - 0.9z^{-1})} + \hat{H}(z)\hat{X}(z)  \\
\mathcal{Y}(z) &= \frac{0.9y(-1)}{(1 - 0.9z^{-1})} + \hat{Y}_{ZSR}(z)  \\
\mathcal{Y}(z) &= \hat{Y}_{ZIR}(n) + \hat{Y}_{ZSR}(z)  \\
\end{align*}

Now we can transform these responses back to the time domain:
\begin{align*}
\hat{Y}_{ZIR}(z) &= \frac{0.9y(-1)}{(1 - 0.9z^{-1})}  \\
y_{ZIR}(n) &= 0.9y(-1)(0.9)^n u(n) \\
y_{ZIR}(n) &= y(-1)(0.9)^{n+1} u(n) \\
\end{align*}

\begin{align*}
\hat{Y}_{ZSR}(z) &= \hat{H}(z)\hat{X}(z) \\
y_{ZSR}(n) &= -9(0.9)^n u(n) + 10 u(n) \\
\end{align*}

Now for the final result:

$$
y(n) = y(-1)(0.9)^{n+1} u(n) - 9(0.9)^n u(n) + 10 u(n) 
$$

Here we find all of the various responses:

\begin{align*}
y_{ZIR}(n) &= y(-1)(0.9)^{n+1} u(n) \\
y_{ZSR}(n) &= -9(0.9)^n u(n) + 10 u(n) \\
y_{tr}(n) &= y(-1)(0.9)^{n+1} u(n) -9(0.9)^n u(n)\\
y_{ss}(n) &= 10 u(n) \\
\end{align*}

Note that the steady-state response is equal to the transfer function evaluated at $z = 1$:

\begin{align*}
y_{ss}(n) &= \left. \hat{H}(z) \right|_{z=1} = 10 u(n)
\end{align*}

The natural response of the system, the contribution from the pole (which is inside of the unit circle), dies out. What remains is the response from the input $x(n) = 1$ $\forall n$. In other words, $H$ is BIBO, so the natural response or response due to the poles of the system decays, so for $n$ sufficiently large, the system can't distinguish between $x(n) = 1$ $\forall n$ and $x(n) = u(n)$.

\end{solution}


\begin{example}


Given the following system, find zero-input response.
\begin{align*}
y(n) - \alpha y(n-1) + \beta y(n-2) &= x(n) \\
\end{align*}

Take the unilateral Z-transform:

\begin{align*}
\mathcal{Y}(z) - \alpha \sum \limits_{n=0}^{\infty} y(n-1)z^{-n} + \beta \sum \limits_{n=0}^{\infty} y(n-2)z^{-n} = \mathcal{X}(z) \\
\mathcal{Y}(z) - \alpha z^{-1} \sum \limits_{m=-1}^{\infty} y(m)z^{-m} + \beta z^{-2} \sum \limits_{\ell=-2}^{\infty} y(\ell)z^{-\ell} = \mathcal{X}(z) \\
\mathcal{Y}(z) - \alpha y(-1) - \alpha z^{-1} \sum \limits_{m=0}^{\infty} y(m)z^{-m} + \beta y(-2) + \beta y(-1) z^{-1} + \beta z^{-2} \sum \limits_{\ell=0}^{\infty} y(\ell)z^{-\ell} = \mathcal{X}(z) \\
\mathcal{Y}(z) - \alpha y(-1) - \alpha z^{-1} \mathcal{Y}(z) + \beta y(-2) + \beta y(-1) z^{-1} + \beta z^{-2} \mathcal{Y}(z) = \mathcal{X}(z) \\
\end{align*}

Since we are finding the zero-input response (homogeneous solution), we set $x(n) = 0 \ztp 0$:
\begin{align*}
\mathcal{Y}(z) - \alpha y(-1) - \alpha z^{-1} \mathcal{Y}(z) + \beta y(-2) + \beta y(-1) z^{-1} + \beta z^{-2} \mathcal{Y}(z) &= 0 \\
\mathcal{Y}(z)\lr{ 1 - \alpha z^{-1} + \beta z^{-2}  }  &= \alpha y(-1) - \beta y(-2) - \beta y(-1) z^{-1} \\
\mathcal{Y}(z)  &= \frac{\alpha y(-1) - \beta y(-2) - \beta y(-1) z^{-1}}{\lr{ 1 - \alpha z^{-1} + \beta z^{-2}  }} \\
y_{ZIR}(n)  &= \iztrans{\frac{\alpha y(-1) - \beta y(-2) - \beta y(-1) z^{-1}}{\lr{ 1 - \alpha z^{-1} + \beta z^{-2}  }}} \\
\end{align*}

\end{example}


Zero-State and Zero-Input Responses of LTI Systems