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Discrete-Time Systems

Module by: Mark A. Davenport. E-mail the author

We begin with the simplest of discrete-time systems, where X=CNX=CN and Y=CMY=CM. In this case a linear operator is just an M×NM×N matrix. We can generalize this concept by letting MM and NN go to , in which case we can think of a linear operator L:2(Z)2(Z)L:2(Z)2(Z) as an infinite matrix.

Example 1

Consider the shift operator Δk:2(Z)2(Z)Δk:2(Z)2(Z) that takes a sequence and shifts it by kk. As an example, Δ1Δ1 can be viewed as the infinite matrix given by

y - 1 y 0 y 1 = 0 0 1 0 0 1 0 0 1 0 0 x - 1 x 0 x 1 y - 1 y 0 y 1 = 0 0 1 0 0 1 0 0 1 0 0 x - 1 x 0 x 1
(1)

Note that Δk2=1Δk2=1 (for any kk and pp) since the delay doesn't change the norm of xx. The delay operator is also an example of a linear shift-invariant (LSI) system.

Definition 1

An operator L:2(Z)2(Z)L:2(Z)2(Z) is called shift-invariant if L(Δk(x))=Δk(L(x))L(Δk(x))=Δk(L(x)) for all x2(Z)x2(Z) and for any kZkZ.

Observe that Δk1(Δk2(x))=Δk1+k2(x)Δk1(Δk2(x))=Δk1+k2(x) so that ΔkΔk 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=Hxy=Hx to denote

y - 1 y 0 y 1 = | | | h - 1 h 0 h 1 | | | x - 1 x 0 x 1 y - 1 y 0 y 1 = | | | h - 1 h 0 h 1 | | | x - 1 x 0 x 1
(2)

Suppose we want to figure out the column of HH corresponding to h0h0. What input xx could help us determine h0h0? Consider the vector

x = 0 1 0 , x = 0 1 0 ,
(3)

i.e., x=δ[n]x=δ[n]. For this input y=Hx=h0y=Hx=h0. What about h1h1? Δ1(x)=δ[n-1]Δ1(x)=δ[n-1] would yield h1h1. In general Δk(x)=δ[n-k]Δk(x)=δ[n-k] tell us the column hkhk. But, if HH is LSI, then

h k = H ( Δ k ( δ [ n ] ) ) = Δ k ( H ( δ [ n ] ) ) = Δ k ( h 0 ) h k = H ( Δ k ( δ [ n ] ) ) = Δ k ( H ( δ [ n ] ) ) = Δ k ( h 0 )
(4)

This means that each column is just a shifted version of h0h0, which is usually called the impulse response.

Now just to keep notation clean, let h=h0h=h0 denote the impulse response. Can we get a simple formula for the output yy in terms of hh and xx? Observe that we can write

y - 1 y 0 y 1 = h 0 h - 1 h - 2 h 1 h 0 h - 1 h 2 h 1 h 0 x - 1 x 0 x 1 y - 1 y 0 y 1 = h 0 h - 1 h - 2 h 1 h 0 h - 1 h 2 h 1 h 0 x - 1 x 0 x 1
(5)

Each column is just shifted down one. (Each successive row is also shifted right one.) Looking at y-1y-1, y0y0 and y1y1, we can rewrite this formula as

y [ - 1 ] y [ 0 ] y [ 1 ] = + x [ - 1 ] h [ 0 ] h [ 1 ] h [ 2 ] + x [ 0 ] h [ - 1 ] h [ 0 ] h [ 1 ] + x [ 1 ] h [ - 2 ] h [ - 1 ] h [ 0 ] + y [ - 1 ] y [ 0 ] y [ 1 ] = + x [ - 1 ] h [ 0 ] h [ 1 ] h [ 2 ] + x [ 0 ] h [ - 1 ] h [ 0 ] h [ 1 ] + x [ 1 ] h [ - 2 ] h [ - 1 ] h [ 0 ] +
(6)

From this we can observe the general pattern

y [ n ] = + x [ - 1 ] h [ n + 1 ] + x [ 0 ] h [ n + 0 ] + x [ 1 ] h [ n - 1 ] + y [ n ] = + x [ - 1 ] h [ n + 1 ] + x [ 0 ] h [ n + 0 ] + x [ 1 ] h [ n - 1 ] +
(7)

or more concisely

y [ n ] = k = - x [ k ] h [ n - k ] . y [ n ] = k = - x [ k ] h [ n - k ] .
(8)

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

y = x * h . y = x * h .
(9)

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