Digital Signal Processing Reference
In-Depth Information
1
Diode
ac
voltage
source
R
L
Load
v
1
(
t
)
Figure 5.30
Half-wave rectifier.
v
s
(
t
)
B
cos
1
(
t
)
B
t
B
(a)
t
nT
1
T
f
(
t
)
=
rect
n
T
2
T
2
0
T
1
2
T
1
t
T
1
2
1
2
T
1
= ,T =
(b)
v
1
(
t
)
T
1
T
1
2
T
1
t
0
(c)
Figure 5.31
Waveforms for Example 5.18.
After some relatively simple algebraic manipulation, we find
V
1
(v)
by using the fre-
quency shift and linearity properties of the Fourier transform:
v
1
(t) = f(t)B cos(v
1
t) = f(t)
B
2
[e
jv
1
t
+ e
-jv
1
t
]
B
2
f(t)e
jv
1
t
+
B
2
f(t)e
-jv
1
t
.
=
Using Table 5.1, we see that
B
2
F(v - v
1
) +
B
2
F(v + v
1
).
V
1
(v) =
