Digital Signal Processing Reference
In-Depth Information
Now we will compute the frequency response for
u ¼ p
/2.
C
0
e
0
C
1
e
p
=
2
C
2
e
p
þ
C
3
e
3
p
=
2
C
4
e
4
p
=
2
H
ð
p
=
2
Þ¼
þ
þ
þ
¼
1
1
þ
3
ð
j
Þþ
5
ð
1
Þþ
3
ð
j
Þþ
1
1
¼
3
So the magnitude of the frequency response has gone from 13
(at
u¼
0) to 3 (at
u ¼ p
/ 2). The phase has gone from 0 degrees
(at
/ 2), although generally we are
not concerned about the phase response of FIR filters. Just from
these two points of the frequency response, we can guess that the
filter is probably some type of low pass filter.
The magnitude is calculated as follows:
Magnitude Z
u ¼
0) to 180 degrees (at
u ¼ p
(X
2
Y
2
)
1/2
j
Our example calculation above turned out to have only real
numbers, but that is because the imaginary components of
¼
X
þ
jY
¼
þ
¼j
Z
p
H
(
/2)
canceled out to zero. The magnitude of
(
p
/2)is:
H
Magnitude
3
A computer program can easily evaluate
j
-3
þ
0j
j¼
p
and plot it for us. Of course, this is almost never done by hand.
Figure 4.3
is a frequency plot of this filter using a FIR Filter
program.
Not the best filter, but it is still a low-pass filter. The frequency
axis is normalized to F
s,
and the magnitude of the amplitude is
(
u
) from
p
to
H
Figure 4.3.
