An AM radio transmitter operating at 3.9 MHz is modulated by frequencies up to 4 kHz. What are the maximum upper and lower side
1 answer:
Answer:
total width bandwidth = 8kHz
Explanation:
given data
transmitter operating = 3.9 MHz
frequencies up to = 4 kHz
solution
we get here upper side frequencies that is
upper side frequencies = 3.9 × + 4 × 10³
upper side frequencies = 3.904 MHz
and
now we get lower side frequencies that is
lower side frequencies = 3.9 × - 4 × 10³
lower side frequencies = 3.896 MHz
and now we get total width bandwidth
total width bandwidth = upper side frequencies - lower side frequencies
total width bandwidth = 8kHz
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Answer:
See attachment
1512 ft
Explanation:
Since the acceleration is either constant or zero, the a−t curve is made of horizontal straight-line segments. The values of t2 and a4 are determined as follows:
Acceleration - Time
0 < t < 6: Change in v = area under a–t curve
V_6 - 0 = (6 s)(4 ft/s2 ) = 24 ft/s
6 < t < t2: Since the velocity increases from 24 to 48 ft/s,
Change in v = area under a–t curve
48 - 24 = (t2 - 6) * 6
t2 = 10 s
t2 < t < 34: Since the velocity is constant, the acceleration is zero.
34 < t < 40: Change in v = area under a–t curve
0 - 42 = 6*a4
a4 = - 8 ft / s^2
The acceleration being negative, the corresponding area is below the t axis; this area represents a decrease in velocity.
Velocity - Time
Since the acceleration is either constant or zero, the v−t curve is made of straight-line segments connecting the points determined above.
Change in x = area under v−t curve
0 < t < 6: x6 - 0 = 0.5*6*24 = 72 ft
6 < t < 10: x10 - x6 = 0.5*4*(24 + 48) = 144 ft
10 < t < 34: x34 - x10 = 48*24 = 1152 ft
34 < t < 40: x40 - x34 = 0.5*6*48 = 144 ft
Adding the changes in x, we obtain the distance from A to B:
d = x40 - 0 = 1512 ft
