USE VCVS TOPOLOGY - VOLTAGE CONTROLLED VOLTAGE SOURCE 

CHOOSE A = 1 ... GOOD FOR OP-AMPS AND EMITTER FOLLOWERS


IN A HIGH PASS FILTER:  
               -----------R1-----
               I                I
               I                I
               I                I
V(IN)----C1--------C2------ A=1 ---- V(OUT)
                       I
                       I
                       R2
                       I
                       I
                      ---

BIAS CAN BE SUPPLIED THROUGH R2 ...


IN A LOW PASS FILTER:
               -----------C2-----
               I                I
               I                I
               I                I
V(IN)----R1--------R2------ A=1 ---- V(OUT)
                       I
                       I
                       C1
                       I
                       I
                      ---

BIAS MUST BE SUPPLIED THROUGH SOURCE ...

IN BOTH CASES, THERE IS NO INPUT BUFFER AMP, SO THE SOURCE 
IMPEDANCE MUST BE  RELATIVELY LOW AND/OR CONSTANT.  

THERE IS ALSO NO INVERSION OF THE INPUT SIGNAL.  

FOR THE LPF, THE EQUATIONS REDUCE TO 

B0 = 1 / (R1*R2*C1*C2)   AND    B1 = (1/C2)*(1/R1 + 1/R2) 
               OR
B0 = G1*G2 / (C1*C2)   AND   B1 = (G1 + G2) / C2  

WHERE              H(S) = B0 / (S^2 + B1*S + B0) 
IS THE SECOND-ORDER LP FILTER RESPONSE TRANSFER FUNCTION.

NOTE: HIGHER ORDER FILTERS CAN BE MADE FROM A CASCADE OF FIRST 
AND SECOND ORDER SECTIONS.  
IF WE CHOOSE R1 = R2 = R = 1, THEN WE ONLY HAVE TO FIND THE 
CAPACITOR VALUES TO GET A FILTER AND RE-NORMALIZE FOR THE RŐS WE 
REALLY WANT AND THE FREQUENCIES WE REALLY WANT.   

THEN     C2 = 2 / B1     AND      C1 = B1 / (2*B0)  


EXAMPLE 1:   2-POLE BUTTERWORTH LP FILTER (RIPPLE = 0)  (W3 = 1)

              B1 = 1.414    B0 = 1

SO         C2 = 2 / 1.414 = 1.414    AND     C1 = 1.414 / 2*1) = 0.707 


EXAMPLE 2:  2-POLE CHEBYCHEV LP FILTER (RIPPLE = 0.5 DB) (WR = 1) 

               B1 = 1.425    B0 = 1.516 

SO          C2 = 2 / 1.425 = 1.4    AND     C1 = 1.425 / (2*1.516) = 0.47  

IF R = 10K,  THEN   C2 = 1.4 E -4   AND   C1 = 0.47 E -4  

IF F(RIPPLE) = 2.23 KHZ,  WR = 14.0 E3,   AND  C1 = 10 E-9   AND   C2 = 3.36 E-9  



WE CANNOT JUST CHOOSE THE CAPACITORS, LIKE    C1 = C2 = 1 .

A SOLUTION FOR G1:     

G1 = B1 / 2 ± (B1^2 - 4*B0)^0.5 / 2.    
G1 IS ONLY REAL IF B0  R1 = 20K, R2 = 5K

J. E. MATZ - 28 FEB 97