Digital Signal Processing Reference
In-Depth Information
If you remember your scientific notation, this makes sense.
For example:
10
2
10
3
10
5
10
(2
þ
3)
¼
100
1000
¼
100,000
¼
¼
Now back to the filter equation:
y
k
¼
P
i
¼
N
to
N
C
i
e
j
u
(k
i)
¼
P
i
¼
N
to
N
C
i
e
j
u
k
e
j
u
i
Let's do a little algebra trick. Notice the term e
j
u
k
does not
contain the term i used in the summation. So we can pull this
term out in front of the summation.
y
k
¼
P
i
¼
N
to
N
C
i
e
j
u
i
Notice the term e
j
u
k
is just the complex exponential we used as
an input.
y
k
¼
e
j
u
k
x
k
P
i
¼
N
to
N
C
i
e
j
u
i
Voila! The expression
P
i
¼
N
to
N
C
i
e
j
u
i
gives us the value of
the frequency response of the filter at frequency
u
. It is solely
a function of
and the filter coefficients.
This expression applies a gain factor to the input, x
k
,to
produce the filter output. Where this expression is large, we are in
the passband of the filter. If this expression is close to zero, we are
in the stopband of the filter.
Let's give this expression a less cumbersome representation.
Again, it is a function of
u
, which we expect, because the charac-
teristics of the filter vary with frequency. It is also a function of the
coefficients, C
i
, but these are assumed to be fixed for a given filter.
Frequency response
u
[
P
i
[
L
N
to
N
C
i
e
j
u
i
Now in reality, it is not as bad as it looks. This is the generic
version of the equation, where we must allow for an infinite
number of coefficients (or taps). But suppose we are determining
the frequency response of our 5 tap example filter.
H
(
)
[H
u
¼
P
i
¼
0to4
C
i
e
j
u
i
and {C
0
,C
1
,C
2
,C
3
,C
4
}
{1, 3, 5, 3, 1}
Let's find the response of the filter at a couple of different
frequencies. First, let
(
u
)
¼
we
are putting a constant level signal into the filter. This would be
x
k
¼
1
for all values k.
H
u ¼
0. This corresponds to DC input
e
(0)
¼
C
0
þ
C
1
þ
C
2
þ
C
3
þ
C
4
¼
1
þ
3
þ
5
þ
3
þ
1
¼
13
This one was simple, since e
0
1. The DC or zero frequency
response of the filter is called the gain of the filter. Often, it may be
convenient to force the gain
¼
1, which would involve dividing all
the individual filter coefficients by
¼
(0). The passband and
stopband characteristics are not altered by this process, since all
the coefficients are scaled equally. It just normalizes the
frequency response so the passband has a gain equal to one.
H
