Question

(a) Design a high-pass filter with a cutoff frequency of 40 kHz. Use0.01 uF capacitor and an appropriate resistor. B) Sketch and label the circuit. C) What is the gain of the filter at the cutoff frequency? Give your answer both as ratio (Vout/Vin) and in dB. D) What is the gain of the filter at a frequency of 55 kHz? Give your answer both as a ratio (Vout/Vin) and in dB.

Answer 1

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
`|(Vout)/(Vin)|` = -1.844 dB

Step-by-step explanation:

A) Find the appropriate resistor

c = 0.01 uf

fc = 40 kHz

cut-off frequency ; fo =
`(1)/(2\pi RC )`

from the above equation R =
`(1)/(2\pi foC)` = 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 ⇒
`(-j)/(2\pi foC )` = -j 397.89

Vout = Vin * ( R / R- C )

Vout = Vin * ( 397.89 / (397.89 - j 397.89))

Vout =
`(1)/(√(2) )` Vin ∠45⁰

therefore gain = |
`(Vout)/(Vin )`| =
`(1)/(√(2) )` = 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 =
`(-J)/(2\pi foC )` = -j 289.373 ohms

Vout = Vin *
`(R)/(R-C)`

= Vin * ( 397.89 / ( 397.89 - j 289.373))

Gain in ratio
`|(Vout)/(Vin)|` = 0.8087 ∠ 36.03⁰

therefore gain in ratio = 0.8087

Gain in dB = 20 log
`|(Vout)/(Vin)|` = -1.844 dB