Aband-pass filter has fc1 = 5khz and fc2 = 88khz. calculate bandwidth (bw) and center frequency (fo) for this filter.

Bandwidth of the Band Pass Filter, BW = 88 - 5 = 83 kHz

center frequency, f = 20.976 kHz

Given:

Upper cut off frequency, = 88 kHz

Lower cut off frequency, = 5 kHz

Solution:

Bandwidth of the Band Pass Filter is the difference between its upper cut-off frequency and its lower cut-off frequency and thus is given by:

Bandwidth, BW = -

Bandwidth of the Band Pass Filter, BW = 88 - 5 = 83 kHz

Now, to calculate its center frequency, f:

f =

f =  

f = 20.976 kHz

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.