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:ℓ2(Z)→ℓ2(Z)L:ℓ2(Z)→ℓ2(Z) as an
infinite matrix.
Consider the shift operator Δk:ℓ2(Z)→ℓ2(Z)Δk:ℓ2(Z)→ℓ2(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 Δkℓ2=1Δkℓ2=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.
An operator L:ℓ2(Z)→ℓ2(Z)L:ℓ2(Z)→ℓ2(Z) is called shift-invariant
if L(Δk(x))=Δk(L(x))L(Δk(x))=Δk(L(x)) for all x∈ℓ2(Z)x∈ℓ2(Z) and
for any k∈Zk∈Z.
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)
