pi network transfer function reva

1 of 7 Hgmjr APR-5-2006

TRANSFER FUNCTION FOR PI FILTER CIRCUIT

Figure 1: π Filter Circuit

In the practical application of a π-network, two additional 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. If you choose, you can replace RS with 0 and RL 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 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 RS and the 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.


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

SC

iC

CS

S

i

th RXVX

XR

RV

V+

=+

= 11

Equation 1

The Thevenin impedance is the source resistor RS in parallel with the first capacitor C. The expression for this value is:

SC

SCth RX

RXZ+

=

Equation 2

Figure 2

Vth Vo

RL

L

C

Zth


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:

( ) LCLth RXXZx +=

( )

( )( ) ( )LthCLthLLC

thCL

LCLth

Lht

th

O XZXXZRRXVXR

RXXZ

XZV

V++++

=++

+

+= 111

Equation 3

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

LCthCLLthLLC

thCLO XXZXXRZRRX

VXRV++++

=

Equation 4

Equation 5 shows the results of substituting the expressions for Vth (see equation 1) and Zth (see equation 2) into equation 4.

LCSC

SCCLL

SC

SCLLC

SC

iCCL

O

XXRXRXXXR

RXRXRRX

RXVXXR

V+

+++

++

+=

Equation 5


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

SC RXx +=

( ) ( ) ( )SCLCSCCSCLLCSLSCLC

iCCLO RXXXRXXRXXRXRRRXRX

VXXRV+++++++

=

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.

SLCLCSCLSLLCLCSLSLCLC

CL

i

O

RXXXXRXXRRXXRXRRRRXRXXR

VV

+++++++= 222

2

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:

sCX C

1= and sLX L =


CLR

sCL

CsRsLRR

CLR

sCRR

sCRR

CsR

CsR

sVsV

SSSL

LSLSLL

L

i

O

+++++++=

22222

22

)()(

Equation 8

Multiplying the numerator and the denominator in equation 8 by the factor s2C2 yields the expression in equation 9.

CLsRsLRLCsRRCLsRsCRRsCRRRR

sVsV

SSSLLSLSLL

L

i

O2232)(

)(+++++++

=

Equation 9

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

( ) ( ) ( )SLSLSLSL

L

i

O

RRsLCRRsLCRLCRLsCRRR

sVsV

++++++=

2)()(

232

Equation 10


By dividing the numerator and the denominator of equation 10 by the coefficient of the s3 term, equation 11 is obtained.

LCRRRR

sCRRLC

sCRCR

s

LCRRR

sV

SL

SL

SLLS

SL

L

O

2223

2

1211)(

++

++

++

=

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 this example, the coefficient of interest is the coefficient of s0. To do this it is necessary to multiply the numerator of equation 11 by the factor:

SL

SL

RRRR

++

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

( )( )

++

++

++

++

=

LCRRRRs

CRRLCs

CRCRs

RRRR

LCRRR

sVsV

SL

SL

SLLS

SL

SL

SL

L

I

O

2223

2

1211

Equation 12


By regrouping the terms in the numerator of equation 12, the expression in equation 13 is obtained. Note that the numerator is now in

the form of the product of two terms with one of the terms identical to the coefficient of the s0 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.

( )( )

++

++

++

+

+

=

LCRRRRs

CRRLCs

CRCRs

RRR

LCRRRR

sVsV

SL

SL

SLLS

SL

L

SL

SL

I

O

2223

2

1211

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 RL and RS. The term in the brackets is the

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

( )( )

++

++

++

+

+=

LCRRRRs

CRRLCs

CRCRs

LCRRRR

RRR

sVsV

SL

SL

SLLS

SL

SL

SL

L

I

O

2223

2

1211

Equation 14

pi network transfer function reva

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