Digital Signal Processing Reference
In-Depth Information
F
(
w
)
2.0
1.5
2
w
1.0
0.5
−
20
−
10
10
20
Frequency,
ω
Figure 8-5
The spectrum of a square wave fits within the envelope of 2/
ω
at high
frequencies.
rise and fall times are realized. The easiest way to apply this filtering is to use
a simple low-pass one-pole filtering response, such as an
RC
network. The step
response of a simple one-pole filter is
v
out
v
input
−
e
−
t/τ
=
1
(8-6)
where
v
input
is the input voltage to the filter,
v
out
the output voltage, and
τ
the time
constant. If the rise times are defined with the 10% and 90% voltage magnitude
points, the time constant required to degrade a step to a specific
t
10
-
90%
can be
calculated. The rise time of a unit step after it passes though a one-pole filter
with a time constant of
τ
is calculated as
t
10
-
90%
=
t
90%
−
t
10%
=
2
.
3
τ
−
0
.
105
τ
=
2
.
195
τ
(8-7)
−
e
−
t
10%
/τ
Note that
t
10%
and t
90%
are calculated from
0
.
1
−
e
−
t
90%
/τ
. An example is shown in Figure 8-6, where the time constant was cal-
culated assuming an
RC
network with
R
=
=
1
and 0
.
9
=
1
50
and
C
=
5 pf.
The 3-dB bandwidth of a one-pole filter is
1
2
πτ
=
f
3dB
1
2
πf
3dB
→
τ
=
Solving for
τ
and substituting into (8-7) produces the well-known relationship
between the spectral content of an edge and the rise time:
2
.
195
2
πf
3dB
≈
0
.
35
f
3dB
t
10
-
90%
=
2
.
195
τ
=
(8-8)
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