Question

(a) Design a first-order passive high-pass filter with a cutoff frequency of 1000 rad/sec.

Answer 1

Answer:

See the attachment

Verification shown in explanation

Explanation:

wc= 1/(RC)

The value of R and C shown in the attachment can be calculated as follows:

R=100 ohms

1000= 1/(100×C)

C= 10 microfarads

Gain= R/√(R²+Xc²)

at f=500, wc= 3142.59 rad/s

Xc= 1/(2πfC)

Xc=31.83

Gain= 100/√(100²+31.83²)

Gain= 0.953

at fc= 100, wc= 628.32 rad/s

Xc= 159.15

gain= 100/√(100²+ 159.15²)

gain= 0.532

As see form calculations that gain at low frequencies is smaller and gain at very high frequencies is near to 1. This means that at high frequencies Vin≈ Vout

Hence this filter allows high frequencies to pass but blocks low frequencies by lowering the gai