Pi Network Transfer Function RevA

TRANSFER FUNCTION FOR PI FILTER CIRCUIT

Figure 1:

π

Filter Circuit

In the practical application of a π-network, two additional resistances resistances are typically included in the circuit. These two resistors correspond to the source resistance associated with the stage feeding this circuit and along with the input resistance associated with the stage into which it feeds. These resistors have been included in this analysis. analysis. If you choose, you can replace R  S with 0 and R L with very large value if you are interested in seeing what effect that has on the transfer function. The analysis of this passive filter circuit circuit is divided into two major steps. The first step involves using Millman’s Theorem to compute the Thevenin equivalent voltage source and Thevenin impedance section composed of R S and the first first capacitor in the circuit.. The second step involves applying Millman’s Theorem to write the overall expression for the network using the expressions for the Thevenin equivalent voltage and Thevenin impedance to obtain the basic expression. With this expression in hand, it is just a matter of turning the crank to get to the final transfer function for the circuit. This analysis assumes that the two capacitors are of equal value.

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Using Millman’s Theorem, the expression for the Thevenin voltage Vth is:

V i V th

=

 R S 

1  R S 

+

=

1

 X  C V i  X  C 

+  R S 

 X  C 

Equation 1

The Thevenin impedance is the source resistor R S in parallel with the first capacitor C. The expression for this value is:  Z th

=

 X C  RS   X C 

+  RS 

Equation 2

L

Vth

Zth

Vo C

RL

Figure 2

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Redrawing the circuit (see Figure 2) to show the use of the Thevenin values yields a much simpler circuit that lends itself to Millman’s Theorem nicely. Equation 3 below shows the Millman’s derived expression and then the simplification that result from multiplying the numerator and the denominator of equation 2 by the factor x where:  x

= ( Z th +  X  L ) X C RL

V th V O

( Z 

=

t  h

+  X  L )

1

( Z  +  X  ) th

+

 L

1  X C 

+

1

=

 R L  X C V th  X C  R L

+  R L ( Z th +  X  L ) +  X C  ( Z th +  X  L )

 R L Equation 3

 Next, expand the terms in the denominator of equation 3 to get the results shown in equation 4 below. V O

=

 R L X C V th  X C  R L

+  R L Z th +  R L X  L +  X C  Z th +  X C  X  L Equation 4

Equation 5 shows the results of substituting the expressions for Vth (see e quation 1) and Zth (see equation 2) into equation 4.  R L  X C  X C V i V O

 X C 

=  X C  R L

+

 R L  X C  R S   X C 

+  RS 

+  RS 

+  R L X  L +

 X C  X C  R S   X C 

+  RS 

+  X C  X  L

Equation 5

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Equation 6 below is obtained by multiplying the numerator and the denominator of equa tion 5 by the factor x where:  x

V O

=

=  X C  + RS   R L X C  X C V i

(

 X C  R L  X C 

+  RS  ) +  R L RS  X C  +  R L X  L ( X C  +  RS  ) +  X C  X C  RS  +  X C  X  L ( X C  +  RS  ) Equation 6

Equation 7 shows the results of expanding the terms contained in parentheses from equation 6. Both sides of the expression are divided by Vi. V O V i

=

 R L X C  2

 X C   R L

2

+  X C  R L RS  +  R L RS  X C  +  R L X C  X  L +  R L RS  X  L +  X C   RS  +  X C  2

2

 X  L

+  X C  X  L RS 

Equation 7

Equation 8 is obtained by substituting the XC terms and the XL terms with their corresponding values in the s domain. These values are:  X  C 

=

1 C 

and  X  L

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=  sL

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 R L V O ( s ) V i ( s )

=

2

2

 s C   R L 2

2

 s C 

+

 R L RS   sC 

+

 R L RS   sC 

+

 R L L C 

+  R L RS  sL +

 RS  2

2

 s C 

+

 L 2

 sC 

+

 RS  L C 

Equation 8

2

Multiplying the numerator and the denominator in equation 8 by the factor s

V O ( s ) V i ( s )

=

2

C

yields the expression in equation 9.

 R L  R L

+  R L RS  sC  +  R L RS  sC  +  R L Ls

2

C  +  R L RS  s C   L +  RS  3

2

+  sL +  RS  Ls 2 C 

Equation 9

By collecting the terms that share the same exp onent of s in equation 9 the expression in equation 10 is obtained.

V O ( s ) V i ( s )

=

 R L 2

 R L R S  C   Ls

3

+ ( R L LC  +  R S  LC ) s 2 + (2 R L R S C  +  L ) s + ( R L +  R S  ) Equation 10

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By dividing the numerator and the denominator of equation 10 by the coefficient of the s term, equation 11 is obtained.  R L 2

V O ( s )

 R L R S C   L

=  s

3

    1  R +  R S  1   2   2 1  s +  L  s +  +  + + 2  2   R S C   R L C     LC   R L R S C     R L RS C   L Equation 11

At this point it is desirable to have the term in the numerator match the coefficient of the corresponding s term in the denominator. In 0

this example, the coefficient of interest is the coefficient of s the factor:

.

To do this it is necessary to multiply the numerator of equation 11 by

 R L  R L

+  RS  +  RS 

Once this is done, you get the expression in equation 12 below.

    R L +  RS     R L     2   R +  R   R  R C   L V O ( s )  L S   L S       =   V  I  ( s )     1  R L +  RS  1   2   2 1    s 3 +     s +  s + + + 2  2    R C   R C   LC   R  R C   R  R C   L    L   S     L S   L S      Equation 12

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By regrouping the terms in the numerator of equation 12, the expression in equation 13 is obtained. Note that the numerator is now in 0 the form of the product of two terms with one of the terms identical to the c oefficient of the s term in the denominator. This was the

goal, so we are ready to proceed with the final manipulations required to get to the desired results.

     R L +  RS     R L     2    R L RS C   L   R L +  RS    V O ( s )   =  V  I  ( s )     1  R L +  RS   1   2   2 1 3  s +  s +   s +   + + 2  2  R C   R C   LC   R  R C   R  R C   L    L  L S   L S    S        Equation 13

We arrive at the end of this long journey by factoring out that portion of the product in the numerator so that there remains only the term that is equal to the constant term in the denominator and placing it on the outside of the bracketed expression. Note that the term that was factored out is the expression that corresponds to the voltage divider formed by R L and R S. The term in the brackets is the rd

3 -order low-pass transfer function of the circuit.

()  R = V  ( s )  R + R

V O  s  I 

 L

 L

S 

   R L +  RS    2  R  R C   L    L S       1  R L +  RS   1   2   2 1 3  s +  s +   s +   + + 2  2  LC   R  R C   R  R C   L    L S   L S    RS C   R L C       Equation 14

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