Digital Signal Processing Reference
In-Depth Information
Our x
k
sequence will be defined as {x
0
,x
1
,x
2
,x
3
,x
4
,x
5
}
¼
{
1, 1,
2, 1, 4,
1} and x
k
¼
0 for k
<
0 and for k
>
6 (everywhere else).
¼
P
i
¼
0toN
1
C
i
x
Let's start by computing y
i
We can see that the subscript on x will be negative for all i
1
1
¼
0
to 4. In this example y
k
¼
0. That is, until there is a non-
zero input x
k
, the output y
k
will also be zero. Things start to
happen at k
0 for k
<
¼
0, because x
0
is the first non-zero input.
¼
P
i
¼
0toN
y
1
C
i
x
¼
(1)(0)
þ
(3)(0)
þ
(5)(0)
þ
(3)(0)
1
1
i
0
y
0
¼
P
i
¼
0oN
þ
(1)(0)
¼
1
C
i
x
0
i
¼
þ
þ
þ
(1)(-1)
(3)(0)
(5)(0)
(3)(0)
1
y
1
¼
P
i
¼
0toN
þ
(1)(0)
¼
1
C
i
x
1
i
¼
(1)(1)
þ
(3)(
1)
þ
(5)(0)
þ
(3)(0)
2
y
2
¼
P
i
¼
0oN
þ
(1)(0)
¼
1
C
i
x
2
i
¼
(1)(2)
þ
(3)(1)
þ
(5)(-1)
þ
(3)(0)
0
y
3
¼
P
i
¼
0toN
þ
(1)(0)
¼
1
C
i
x
3
i
¼
(1)(1)
þ
(3)(2)
þ
(5)(1)
þ
(3)(
1)
9
y
4
¼
P
i
¼
0oN
þ
(1)(0)
¼
C
i
x
4
i
¼
(1)(4)
þ
(3)(1)
þ
(5)(2)
þ
(3)(1)
1
19
y
5
¼
P
i
¼
0toN
þ
(1)(
1)
¼
1
C
i
x
5
i
¼
þ
þ
þ
(1)(
1)
(3)(4)
(5)(1)
(3)(2)
23
y
6
¼
P
i
¼
0toN
þ
(1)(1)
¼
1
C
i
x
6
i
¼
(1)(0)
þ
(3)(
1)
þ
(5)(4)
þ
(3)(1)
22
y
7
¼
P
i
¼
0toN
þ
(1)(2)
¼
1
C
i
x
7
i
¼
(1)(0)
þ
(3)(0)
þ
(5)(
1)
þ
(3)(4)
8
y
8
¼
P
i
¼
0toN
þ
(1)(1)
¼
1
C
i
x
8
i
¼
(1)(0)
þ
(3)(0)
þ
(5)(0)
þ
(3)(
1)
1
y
9
¼
P
i
¼
0oN
þ
(1)(4)
¼
1
C
i
x
9
i
¼
(1)(0)
þ
(3)(0)
þ
(5)(0)
þ
(3)(0)
þ
(1)(
1)
¼
1
y
10
¼
P
i
¼
0toN
1
C
i
x
10
i
¼
(1)(0)
þ
(3)(0)
þ
(5)(0)
þ
(3)(0)
0
y
11
¼
P
i
¼
0toN
þ
(1)(0)
¼
1
C
i
x
11
i
¼
(1)(0)
þ
(3)(0)
þ
(5)(0)
þ
(3)(0)
0
y
12
¼
P
i
¼
0toN
þ
(1)(0)
¼
1
C
i
x
12
i
¼
(1)(0)
þ
(3)(0)
þ
(5)(0)
þ
(3)(0)
0
This is definitely tedious. There are a couple of things to
notice. Follow the input x
4
¼
þ
(1)(0)
¼
4 (in bold) in our example. See how it
moves across, from one multiplier to the next. Each input sample
x
k
will be multiplied by each tap in turn. Once it passes through
the filter, that input sample is discarded and has no further
influence upon the output. In our example, x
4
is discarded after
computing y
8
.
