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a. Determine R for a series RC high-pass filter with a cutoff frequency (fc) of 8 kHz. Use a 100 nF capacitor. b. Draw the schem

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a) 199.04 ohms

b) attached in image

c) -0.696dB

Explanation:

We are given:

Fc = 8Khz = 8000hz

$C = 100nF = 100*10^-^9F$

a)Using the formula:

$F_c = \frac{1}{2pie*Rc}$

$8000= \frac{1}{2*3.14*R*100*10^-^9}$

$R =\frac{1}{2*3.14*100*10^-^9*8000}$

R = 199.04 ohms

b) diagram is attached

c) $H(w) = \frac{V_out(w)}{Vin(w)} = \frac{1}{1-j\frac{wc}{w}}$

$H(F) = \frac{1}{1-j\frac{fc}{f}}$

At F = 20KHz and Fc= 8KHz we have:

$H(F)= \frac{1}{1-j\frac{8}{20}} = \frac{1}{1-j(0.4)}$

$|H(F)|= \frac{1}{\sqrt{1^2+(0.4)^2}}$

=0.923

|H(F)| in dB = 20log |H(F)|

=20log0.923

= -0.696dB

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