
A:
The
tone
of
the
envelope
detector
is
higher,
indicating
it
is
removing
lower
frequencies.
It
could
be
improved.
2b)
Q:
What
are
these
frequencies?
A:
1.5
MHz
+
500
Hz
=1,500,500
Hz
1.5
MHz
-
500
Hz
=1,499,500
Hz
Q:
Compare
these
tones
with
that
emitting
from
the
radio.
A:
The
tone
from
the
radio
and
the
original
tone
are
the
same.
The
tone
from
the
envelope
detector
is
higher
pitched
and
lower
in
power.
Some
of
the
frequncies
in
the
message
signal
are
eliminated
by
the
envelope
detector.
2c)
Q:
Compare
the
tones
heard
on
the
radio
to
the
original.
A:
They
are
the
same.
Q:
What
is
the
time
domain
complex
envelope
this
wave?
s(t)
=
.5
(2TI10000
+
ß
(—sin2*1000*
+
sin27c3000*
+
—
sin2rr5000
/))
it
3%
5
it
s(t)
=
(.5COS2TT
10000
)
S
2
-
(.5sin2x10000
)
S
Lab
3
Solutions
Page
5
75
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