This document describes designing a third-order low-pass Butterworth filter with a cutoff frequency of 3000 rad/s. The resistor value is specified as 2 kΩ. By equating the transfer functions and solving the equations, the component values are calculated as: - Inductor (L) = 1.33 H - Capacitor 1 (C1) = 0.5 F - Capacitor 2 (C2) = 1.5 F

1. (a) (10 pts) The following circuit with the transfer funtion Hcir (s), is to realize a third-order
low-pass filter with 3 frequency w 3000 rad/s. If the resistor is to be 2 ks in your final design, by
using appropriate values Km the scaled values of L, C1 and C2 in the final design, denoted as
LN, CLA of and given that the normalized C respectively. It is third-order Butterworth low pass
filter with 3 dB angular frequency at 1 rad/s has a transfer function 3dBNLP (s) 1 H s3 2s2 2s 1 2
(t) C2 Vout vin(t) Vou (s)- LC C2 Hair (s) C1 C2 in (s) 3 S LCIC2 LCIC2
Solution
Comparing the transfer functions
1/C1 = 2 , C1 = 1/2 = 0.5 F
1/(LC1C2) = 1
Or LC1C2 = 1
PUT THE VALUE OF C1
1/LC2 = 0.5.............................(1)
And (C1+C2)/LC1C2 = 2
Put the value of C1 And LC1C2
(0.5 + C2)/1 = 2
C2 = 1.5 F
So LC1C2 = 1
L = 1/ C1C2
L = 1/(0.5)(1.5)
L = 4/3 = 1.33 H