Answer:
a) 397.89 ohm
b) attached below
c) 0.707 as a ratio
Gain in dB = 20 log 0.707 = -3 dB
d) 0.8087
Gain in dB = 20 log [tex]|\frac{Vout}{Vin}|[/tex] = -1.844 dB
Explanation:
A) Find the appropriate resistor
c = 0.01 uf
fc = 40 kHz
cut-off frequency ; fo = [tex]\frac{1}{2\pi RC }[/tex]
from the above equation R = [tex]\frac{1}{2\pi foC}[/tex] = 397.89 ohm
B) sketch of the circuit is attached
C) The gain of the filter at the cutoff frequency
fc = 40 kHz,
C = 0.01 uF ⇒ [tex]\frac{-j}{2\pi foC }[/tex] = -j 397.89
Vout = Vin * ( R / R- C )
Vout = Vin * ( 397.89 / (397.89 - j 397.89))
Vout = [tex]\frac{1}{\sqrt{2} }[/tex] Vin ∠45⁰
therefore gain = |[tex]\frac{Vout}{Vin }[/tex]| = [tex]\frac{1}{\sqrt{2} }[/tex] = 0.707 as a ratio
Gain in dB = 20 log 0.707 = -3 dB
D) Gain of filter at 55 kHz
c = 0.01 uF = [tex]\frac{-J}{2\pi foC }[/tex] = -j 289.373 ohms
Vout = Vin * [tex]\frac{R}{R-C}[/tex]
= Vin * ( 397.89 / ( 397.89 - j 289.373))
Gain in ratio [tex]|\frac{Vout}{Vin}|[/tex] = 0.8087 ∠ 36.03⁰
therefore gain in ratio = 0.8087
Gain in dB = 20 log [tex]|\frac{Vout}{Vin}|[/tex] = -1.844 dB
