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

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Aband-pass filter has fc1 = 5khz and fc2 = 88khz.

dayantapia2007

01.01.2022

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

center frequency, f = 20.976 kHz

Given:

Upper cut off frequency, $f_{c1}$ = 88 kHz

Lower cut off frequency, $f_{c1}$ = 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 = $f_{c1}$ - $f_{c1}$

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

Now, to calculate its center frequency, f:

f = $\sqrt{f_{c1}\times f_{c2}}$

f = $\sqrt{88\times 10^{3}\times 5\times 10^{3}}$

f = 20.976 kHz

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