Digital Signal Processing Reference
In-Depth Information
The coefficients
x
0
,x
1
,x
2
,...
and
h
0
,h
1
,h
2
,h
3
,...
are the known samples of
the input
x(n)
and the unit sample response
h(n)
. Either one or both sequences
may be finite or infinite in length. If we multiply the polynomial or the power
series for
X(z)
and
H(z)
, and group all the terms for the coefficients of z
−
n
,in
the polynomial or the power series, we get
x
1
z
−
1
x
2
z
−
2
x
3
z
−
3
X(z)H (z)
=
(x
0
+
+
+
+···
)
h
1
z
−
1
h
2
z
−
2
h
3
z
−
3
×
(h
0
+
+
+
+···
)
x
1
h
0
)z
−
1
x
2
h
0
)z
−
2
=
(x
0
h
0
)
+
(x
0
h
1
+
+
(x
0
h
2
+
x
1
h
1
+
x
3
h
0
)z
−
3
+
(x
0
h
3
+
x
1
h
2
+
x
2
h
1
+
+···
By comparing the coefficients of
z
−
n
in
Y(z)
and those in this expression, we
notice that
y
0
=
(x
0
h
0
)
y
1
=
(x
0
h
1
+
x
1
h
0
)
y
2
=
(x
0
h
2
+
x
1
h
1
+
x
2
h
0
)
y
3
=
(x
0
h
3
+
x
1
h
2
+
x
2
h
1
+
x
3
h
0
)
·
·
(2.51)
·
y
n
=
(x
0
h
n
+
x
1
h
n
−
1
+
x
2
h
n
−
2
+
x
3
h
n
−
3
+···+
x
n
h
0
)
·
·
·
These are the same results as given in (2.7), which are obtained by expanding the
convolution sum
y
n
=
k
=−∞
x(k)h(n
k)
. We can multiply the polynomial
or the power series as
H (z)X(z)
and identify the coefficients of the resulting
polynomial as
y
n
−
=
k
=−∞
h(k)x(n
k)
. [We can also find the coefficients of
H (z)X(z)
by computing the convolution of the coefficients of
H(z)
and
X(z)
.]
Then we would get the following expressions for the coefficients, which are
the same as those given in (2.51):
−
y
0
=
(h
0
x
0
)
y
1
=
(h
0
x
1
+
h
1
x
0
)
y
2
=
(h
0
x
2
+
h
1
x
1
+
h
2
x
0
)
y
3
=
(h
0
x
3
+
h
1
x
2
+
h
2
x
1
+
h
3
x
0
)
·
·
(2.52)
·
Search WWH ::
Custom Search