Digital Signal Processing Reference
In-Depth Information
The magnitude response is:
G
m
1
þð
x
o
Þ
2
j
H
ð
x
Þj¼
q
where x
o
¼
1
RC
p
:
Now we have:
H
ð
0
Þ¼
G
m
;
H
ð1Þ¼
0;
j
H
ð
x
o
Þj¼
G
m
=
2
p
:
RC
Hence, the system is a LPF with cutoff frequency x
o
¼
1
:
p
Tutorial 38
Q: Determine and plot the spectrum at all points shown in the signal processing
system below. Consider only the frequency range 0-40 Hz.
Anti−aliasing
LPF
Sampler
& ADC
Reconstruction
LPF
DSP
r
(
t
)
p
(n)
v
(n)
H
a
(
s
)
H
r
(
f
)
x
(
t
) =
sin(10
π
t
)
H
(
z
)
y
(
t
)
Differentiator
(1−
z
−1
)
f
s
= 40 Hz
(1/
f
s
)sinc(
f
/
f
s
)
100/(
s
+ 100)
Solution:
The signal frequency is 2pf
o
¼
10p
!
f
o
¼
5 Hz.
X
ð
f
Þ¼
1
2j
d
ð
f
f
o
Þ
1
2j
d
ð
f
þ
f
o
Þð
Tables
Þ)
X
ð
f
j j¼
1
2
d
ð
f
f
o
Þþ
1
2
d
ð
f
þ
f
o
Þ:
¼
0
:
47
For
|R(f)|
we
have:
A
¼
0
:
5 H
a
ð
j2pf
o
Þ
j
j¼
0
:
5
100
j10p
þ
100
(see
the
figure
below).
For
¼
18
:
81
e
j2p5
=
40
¼
14
:
4,
B
1
¼
18
:
8 H
ð
e
j2pf
0
T
s
Þ
|V(f)|
we
have:
and
¼
18
:
8 H
ð
e
j
ð
2p
2pf
o
T
s
Þ
Þ
¼
18
:
8 H
ð
e
j2pf
o
T
s
Þ
¼
B
1
:
B
2
¼
18
:
8 H
ð
e
j2p
ð
f
s
f
o
Þ
T
s
Þ
For |Y(f)| we have:
1
40
5
40
C
¼
B
1
H
r
ð
f
0
Þ
j
j¼
14
:
4
sinc
¼
0
:
35
and
D
¼
B
2
H
r
ð
35
Þ
j
j¼
0
:
05
:
Note: no need here to divide (1 - z
-1
) by T
s
for accurate differentiation, since the
sampler performs this function.
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