EECS 412 Filter Design Project
( Please note: The filter transfer function is the ration
of the coutput voltage to the input voltage )
The D(s) with a cutoff frequency = 1 rad/s in fifth order
is:
(s2 + 1.618s + 1 ) ( s2 + 1.618s +
1 ) ( s + 1 )
Then our transfer function is :
-1/ ( (s2 + 0.618s + 1 ) ( s2
+ 1.618s + 1 ) ( s + 1 ) )
which is equal to :
( -1/ ( s + 1 ) ) ( 1/ (s2 + 1.618s + 1 ) ) (
1/ (s2 + 0.618s + 1 ) )
And, we want the cutoff frequency at 1000 rad/s, so apply
the frequency scaling to each s term:
{ -1/ [ (
s/1000) + 1 ]
}
*
{ 1/ [(s/1000
) 2 + (1.618s/1000) + 1 ) *
{ 1 / [ (
s/1000 ) 2 + ( 0.618s/1000 )
+ 1 ]
}
It finally becomes like this:
{
-303.7737/ ( s + 1000) }
*
{ 1382/ [(s2
+ 1618 s + 1000000 ) *
{ 2382
/ [ s2 + 618s + 1000000 ] }
Use these three equations to apply the Sallen and Key
Filters and the first order Filter, we will have
Our butterworth lowpass filter at cutoff frequency at
1000 rad/s.
( Refrences:
Continuous Time Active Filter Design ( EECS 412 Text Book ),
Introduction to Electric Circuits ( EECS 210 Text Book ),
Microelectronic Circuits ( EECS 340.342 Text Book ).
The main design method of
this project is to refer to the Chapter 16 of Introduction to
Electric Circuits.
)