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

the answer is a)

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electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

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