2.1 Discrete-time systems

We begin with the simplest of discrete-time systems, where $X={\mathbb{C}}^N$ and $Y={\mathbb{C}}^M$ . In this case a linear operator is just an $M\times N$ matrix. We can generalize this concept by letting $M$ and $N$ go to $\infty$ , in which case we can think of a linear operator $L:{\ell }_2\left(\mathbb{Z}\right)\to {\ell }_2\left(\mathbb{Z}\right)$ as an infinite matrix.

Definition 1

Observe that ${\Delta }_{k_1}\left({\Delta }_{k_2}\left(x\right)\right)={\Delta }_{k_1+k_2}\left(x\right)$ so that ${\Delta }_k$ itself is an LSI operator.

Lets take a closer look at the structure of an LSI system by viewing it as an infinite matrix. In this case we write $y=Hx$ to denote

Suppose we want to figure out the column of $H$ corresponding to $h^0$ . What input $x$ could help us determine $h^0$ ? Consider the vector

i.e., $x=\delta \left[n\right]$ . For this input $y=Hx=h^0$ . What about $h^1$ ? ${\Delta }_1\left(x\right)=\delta [n-1]$ would yield $h^1$ . In general ${\Delta }_k\left(x\right)=\delta [n-k]$ tell us the column $h^k$ . But , if $H$ is LSI, then

This means that each column is just a shifted version of $h^0$ , which is usually called the impulse response.

Now just to keep notation clean, let $h=h^0$ denote the impulse response. Can we get a simple formula for the output $y$ in terms of $h$ and $x$ ? Observe that we can write

Each column is just shifted down one. (Each successive row is also shifted right one.) Looking at $y_{-1}$ , $y_0$ and $y_1$ , we can rewrite this formula as

From this we can observe the general pattern

or more concisely

Does this look familiar? It is simply the formula for the discrete-time convolution of $x$ and $h$ , i.e.,