Digital Signal Processing Reference
In-Depth Information
and then sample again at time
1. So from one sample to the
next, our sampled signal will move
¼
m
þ
radians around the unit
circle. If we are sampling at 10 times faster than we are moving
around the unit circle, then it will take 10 samples to get around
the circle, and move 2
u
p
/ 10 radians each sample.
e
j2
p
m/10
x
m
¼
¼
cos(2
p
m / 10)
þ
j sin(2
p
m / 10)
when
/10
To clarify,
Table 4.1
shows x
m
¼
u ¼
2
p
e
j2
p
m/10
evaluated at various
“m” using a complex exponential signal, which is a rotating vector
in the frequency domain. If you want to check using a calculator,
remember that the angles are in units of radians, not degrees.
Table 4.1
e
j0
m
¼
0
x
0
¼
¼
cos (0)
þ
j sin (0)
1
þ
j0
x
1
¼ e
jp/5
m ¼ 1
¼ cos (p /5)þ j sin (p / 5)
0.8090 þ j0.5878
x
2
¼ e
j2p/5
m ¼ 2
¼ cos (2p /5)þ j sin (2p / 5)
0.3090 þ j0.9511
x
3
¼ e
j3p/5
m ¼ 3
¼ cos (3p /5)þ j sin (3p /5)
0.3090 þ j0.9511
x
4
¼ e
j4p/5
m ¼ 4
¼ cos (4p /5)þ j sin (4p /5)
0.8090 þ j0.5878
x
5
¼ e
jp
¼ cos (p) þ j sin (p)
m ¼ 5
1 þ j0
x
6
¼ e
j6p/5
m ¼ 6
¼ cos (6p/5) þ j sin (6p/5)
0.8090 e j0.5878
x
7
¼ e
j7p/5
m ¼ 7
¼ cos (7p /5)þ j sin (7p /5)
0.3090 e j0.9511
x
8
¼ e
j8p/5
m ¼ 8
¼ cos (8p/5) þ j sin (8p/5)
0.3090 e j0.9511
x
9
¼ e
j9p/5
m ¼ 9
¼ cos (9p /5)þ j sin (9p / 5)
0.8090 e j0.5878
x
10
¼ x
0 ¼
e
j2p
¼ cos (2p) þ j sin (2p)
m ¼ 10
1 þ j0
We could next increase x
m
so that we rotate the unit circle
every five samples. This is twice as fast as before.
x
m
¼
e
j2
p
m/5
/5
Now we go back to the filter equation, and substitute the
complex exponential input for x
k-i.
y
k
¼
P
i
¼
N
to
N
C
i
x
k
i
x
m
¼
¼
cos(2
p
m/5)
þ
j sin(2
p
m / 5) when
u ¼
2
p
e
j
u
m
¼
u
þ
u
cos(
m)
j sin(
m)
Insert k-i for m
x
k-i
¼
e
j
u
(k
i)
¼
cos (
u
(k
i))
þ
jsin (
u
(k
i))
Next, replace in x
k-i
in filter equation:
y
k
¼
P
i
¼
N
to
N
C
i
e
j
u
(k
i)
There is a property of exponentials that we frequently need to
use:
e
(a
þ
b)
e
a
e
b
and e
(a-b)
e
a
e
-b
¼
¼
