5th Order Butterworth Low Pass Filter

5th Order Butterworth Low Pass Filter

Fifth Order Butterworth LPF Design a fifth order Butterworth low pass filter with a gain ( ) of 1 and a cutoff frequency 10000 Hz. The normalized Butterworth low pass filter equation is: V2 ( s ) H ( s ) V1 ( s ) R1 R2C1C2 s 2 R1C1 R1C2 R2C2 R1C1 s 1 For a gain ( ) of 1 V2 ( s) 1 H ( s ) V1 ( s) R1 R2C1C2 s 2 R1C1 R1C2 R2C2 (1) * R1C1 s 1

V2 ( s ) 1 H ( s ) V1 ( s ) R1 R2C1C2 s 2 R1C2 R2C2 s 1 The normalized denominator for a fifth order Butterworth low pass filter is: 1 H ( s) ( s 1)(s 2 0.618s 1)(s 2 1.618s 1) First order filter Second order filter Second order filter

Frequency Scaling Replace s with s/ c

First order filter Assume all resistors in the design are 1 k vout 1 / RC vin s 1 / RC 1 k 0.016 F vout 1 / RC vin s / 20000 1 / RC

1 / RC c 20000 1 1000 C 20000 1 1000 C 20000 1 8 C 1.59 10 0.016 F 1000 20000

1 H (s) 2 ( s 0.618s 1) H (s) Defines one of the second order filters 1 s2 s 0.618 1 2 20000 20000 1 H (s) R1 R2C1C2 s 2 R1C2 R2C2 s 1

1 H (s) R1 R2C1C2 s 2 R1C2 R2C2 s 1 1 H (s) 1000 1000 C1C2 s 2 1000C2 1000C2 s 1 1 H (s) 1000 2 C1C2 s 2 1000 C2 C2 s 1

1 1000 2 C1C2 s 2 1000 2C2 s 1 0.618 1000 2C2 20000 1 s2 s 0.618 1 2 20000 20000 0.618 C2 0.0049 F 40000000 1 1000 C1C2 20000 2 2

C1 1 2 6 1.0 10 20000 0.0049 10 6 1 8 C1 5 . 17 10 0.05 F 2 0.0049 20000

1 H (s) 2 ( s 1.618s 1) H ( s) Defines the other second order filters 1 s2 s 1.618 1 2 20000 20000 1 H (s) R1 R2C1C2 s 2 R1C2 R2C2 s 1

1 H (s) R1 R2C1C2 s 2 R1C2 R2C2 s 1 1 H (s) 1000 1000 C1C2 s 2 1000C2 1000C2 s 1 1 H (s) 1000 2 C1C2 s 2 1000 C2 C2 s 1

1 1000 2 C1C2 s 2 1000 2C2 s 1 1.618 1000 2C2 20000 1 s2 s 1.618 1 2 20000 20000 1.618 C2 1.29 10 8 0.013 F 40000000 1 1000 C1C2 2 20000 2

C1 1 2 6 1.0 10 20000 0.013 10 6 1 8 C1 1 . 95 10 0.0195 F 2 0.013 20000