Filter Background

Standard filter analysis is for low pass filters of 1 ohm impedance with a high frequency cutoff of 1 radian per second. Such filters may be designed for any desired type of frequency response. Common types are Tschebychev, Butterworth, Eliptic, and Gaussian. Once a low pass filter is designed, it may be readily converted into a high pass, band pass, or band reject filter.

Any filter may be translated from a 1 ohm impedance to any desired impedance without changing its frequency response. To do so, all inductors are multiplied by the desired impedance and all capacitors are divided by it. The cutoff frequency of a 1 radian per second 1 ohm filter may be changed to any desired frequency without changing its impedance. To do so, the inductor and capacitor values are divided by the desired cutoff frequency in radians per second.

The impedance and cutoff frequency may be simultaneously changed to any desired value. To do so, all of the individual inductors and capacitors of the filter are each multiplied by the appropriate scaling value. The value of the scaling inductor is Ls=Z/2πf, and the value of the scaling capacitor is Cs=1/(Z·2πf). In each case, Z is the desired filter impedance in ohms and f is the desired cutoff frequency in Hertz. It is seen that Ls/Cs=Z2 and 1/(LsCs)=(2πf)2. The first equation is the square of the desired impedance and the second equation is the square of the desired cutoff frequency in radians per second.

Any filter must contain at least one resistor to dissipate out of band power. This may take the form of both an input and output resistor which need not necessarily be the same value. Driving impedance may vary from 0 to infinity while the load impedance is 1 ohm. Alternatively, load impedance may vary from 0 to infinity while the driving impedance is 1 ohm.

This paper is a derivation of the values for a Six Pole Butterworth low pass filter. Its amplitude response versus frequency is shown in figure xx. This filter has a 1 radian per second cutoff frequency and its amplitude response there is down by 3 db. It has the characteristic that its response is quite flat and has no ripple in either the pass or reject band. It is the limiting case of the Tschebychev filter for zero ripple. It is designed to be driven from a zero impedance source and terminated with a 1 ohm load.

This type of filter can be driven by a switching amplifier with its output feeding a loudspeaker. It must greatly attenuate the switching frequency and its harmonics but still allow audio frequency components to reach the speaker undisturbed. Such a filter can be designed to have a loss under 1 % thru out most of its passband. The inductors must be operated at no more than half their saturation current to avoid high frequency third harmonic distortion.

Filters designed to be driven by an infinite impedance must be driven by analog amplifier current sources. Their first element must be a capacitor to ground. Filters designed for zero driving impedance may be driven by emitter or source followers as well as zero impedance switches. Their first element must be a series inductor. Filters with a capacitor to ground for their last element can drive high impedance loads such as emitter or source followers. If their last element is a series inductor, they can drive zero impedance loads such as common base or common gate amplifiers.

Input Equation

Assume that the filter shown in figure xx is driven at a level and phase at any given frequency to produce an output voltage of 1 volt at a reference phase of 0°. It is then possible to work backwards toward the input to find out the voltage and phase necessary to produce that output voltage of 1 volt.

EC1=1, IC1=iωC1

IL2=1+iωC1, EL2=iωL22L2C1

EC3=EC1+EL2, EC3=1+iωL22L2C1, IC3=iωC32C3L2-iω3C3L2C1

IL4=IL2+IC3, IL4= (1+iωC1)+(iωC32C3L2-iω3C3L2C1),

IL4=1+iω(C3+C1)-ω2C3L2-iω3C3L2C1

EL4= iωL42L4(C3+C1)-iω3L4C3L24L4C3L2C1

EC5=EC3+EL4, EC5=(1+iωL22L2C1)+(iωL42L4(C3+C1)-iω3L4C3L24L4C3L2C1)

EC5=1+iω[L4+L2]-ω2[L2C1+L4(C3+C1)]-iω3L4C3L24L4C3L2C1

IC5= iωC52C5[L4+L2]-iω3C5[L2C1+L4(C3+C1)]+ω4C5L4C3L2+iω5C5L4C3L2C1

IL6=IL4+IC5

IL6= (1+iω(C3+C1)-ω2C3L2-iω3C3L2C1)+(iωC52C5[L4+L2]-iω3C5[L2C1+L4(C3+C1)]+ω4C5L4C3L2+iω5C5L4C3L2C1)

IL6= 1+iω(C5+C3+C1)-ω2[C3L2+C5(L4+L2)] -iω3(C3L2C1+C5[L2C1 +L4(C3+C1)])+ω4C5L4C3L2+iω5C5L4C3L2C1

EL6= iωL62L6(C5+C3+C1)-iω3L6[C3L2+C5(L4+L2)] +ω4L6(C3L2C1+C5[L2C1+ L4(C3+C1)]) +iω5L6C5L4C3L26L6C5L4C3L2C1

Ein=EC5+EL6,

Ein= (1+iω[L4+L2]-ω2[L2C1+L4(C3+C1)]-iω3L4C3L24L4C3L2C1) +(iωL62L6(C5+C3+C1)-iω3L6[C3L2+C5(L4+L2)]+
ω4L6(C3L2C1+C5[L2C1+ L4(C3+C1)]) +iω5L6C5L4C3L26L6C5L4C3L2C1)

Ein= (1+iω[L6+L4+L2]-ω2[L2C1+L4(C3+C1)+L6(C5+C3+C1)]-iω3(L4C3L2+L6[C3L2+C5(L4+L2)])+
ω4(L4C3L2C1+L6(C3L2C1+C5[L2C1+L4(C3+C1)])+ iω5L6C5L4C3L26L6C5L4C3L2C1)

Filter Coefficients

The above equation for Ein is unwieldy when expressed in terms of all the inductor and capacitor values of the filter. The coefficients of each term involving ω1 thru ω6 will therefore be replaced by a1 thru a6. When expanded, the coefficients yield the following five equations:

L6+L4+L2=a1

L6C5+L4C3+L6C3+L2C1+L4C1+L6C1=a2

L6C5L2+L4C3L2+L6C3L2+L6C5L4=a3

L6C5L2C1+L4C3L2C1+L6C3L2C1+L6C5L4C1+L6C5L4C3=a4

L6C5L4C3L2=a5

L6C5L4C3L2C1=a6

These coefficients can be for any desired 6th degree polynomial, not just the Butterworth polynomial. For example, they can be coefficients of a Tschebychev polynomial. Any type of 6th degree polynomial can be chosen so long as a solution for real component values exists.

The input voltage can now be expressed in a much more manageable form as:
Ein =1+ia1ω-a2ω2-ia3ω3+a4ω4+ia5ω5-a6ω6. Separating real and imaginary parts we get:
Ein=(1-a2ω2+a4ω4-a6ω6)+i(a1ω-a3ω3+a5ω5)

The square of the input amplitude will be the sums of the squares of the real and imaginary parts. This square must be 1+ω12 in order to be a Butterworth response. As the frequency increases from dc, the required input amplitude for a constant output amplitude will smoothly increase. The increase in amplitude is tiny so long as the frequency is in the filter passband.

Taking the sum of the squares of the real and imaginary parts we get:

1+ω12= (1-a2ω2+a4ω4-a6ω6)2+(a1ω-a3ω3+a5ω5)2=[(1-a2ω2)+(a4ω4-a6ω6)]2+(a1ω-a3ω3+a5ω5)2

Using the rules for the squares of a binomial and trinomial we get:

1+ω12=[(1-a2ω2)2+2(1-a2ω2)(a4ω4-a6ω6)+(a4ω4-a6ω6)2]+
[(a1ω)2+(-a3ω3)2+(a5ω5)2+ 2(a1ω)(-a3ω3)+2(-a3ω3)(a5ω5)+2(a1ω)(a5ω5)]

1+ω12=[(1-2a2ω2+a22ω4)+(2a4ω4-(2a2a4+2a66)+2a2a6ω8)+(a42ω8-2a4a6ω10+a62ω12)]+
[a12ω2+a32ω6+a52ω10 -2a1a3ω4-2a3a5ω8+2a1a5ω6]

1+ω12=1+(a12-2a22+(a22-2a1a 3+2a44+(a32-2a2a4+2a1a5-2a66+(a42-2a3a5+2a2a68 +(a52-2a4a610+a62ω12

Auxiliary Equations

Since a62ω1212, a6 can only be +1. If a6= -1, an odd number of filter components would have to be negative. If the equation for 1+ω12 is to hold, the coefficients of all but the first and last terms on the right side must be zero. The inductor and capacitor values of the filter must be chosen to make this so. Substituting 1 for a6 we have the following five equations in five unknowns for the coefficients of ω2, ω4, ω6, ω8, and ω10.

a12-2a2=0

a22-2a1a3+2a4=0

a32-2a2a4+2a1a5-2=0

a42-2a3a5+2a2=0

a52-2a4=0

From the first equation we get a2=a12/2. Substituting this for a2 in the next three equations yields the following four equations in four unknowns:

a14/4-2a1a3+2a4=0

a32-a12a4+2a1a5-2=0

a42-2a3a5+a12=0

a52-2a4=0

From the last equation a4=a52/2. Substituting this for a4 in the first three equations yields the following three equations in three unknowns:

a14/4-2a1a3+a52=0

a32-a12(a52/2)+2a1a5-2=0

a54/4-2a3a5+a12=0

From the first equation a3=(a14+4a52)/8a1. Substituting this for a3 in the next two equations yields the following two equations in two unknowns:

[(a14+4a52)/8a1]2-a12(a52/2)+2a1a5-2=0

a54/4-[(a14+4a52)a5/4a1]+a12=0

These two equations simplify to:

a18+8a14a52+16a54-32a14a52+128a13a5-128a12=0

a54a1-a14a5-4a53+4a13=0

The second equation can be factored as follows:

a5a1(a53-a13)-4(a53-a13)=(a53-a13)(a5a1-4)=(a5-a1)(a52+a5a1+a12)(a5a1-4)=0

We can set each of the three factors equal to zero which gives these three equations:

a5-a1=0

a52+a5a1+a12=0

a5a1-4=0

From the first equation, a5=a1. The second equation is a quadratic with no real solutions if a1 is real. From the third equation, a5=4/a1 . Substituting this value for a5 in the 8th degree equation gives:

a18+8a1416a1-2+16·256a1-4-32a1416a1-2+128a134a1-1-128a12=a18+128a12+4096a1-4-512a12+512a12-128a12=a18+4096a1-4=0

This reduces to the rather unusual result a112= -4096. Since an even power of a real number must be positive, this equation has no real solution.

This leaves a5=a1 as the only remaining value . Substituting it in the 8th degree equation gives:

a18+8a16+16a14-32a16+128a14-128a12=a18-24a16+144a14-128a12=0

(a12)(a16-24a14+144a12-128)=0

Final Equation

It is impossible for the factor a12 to be zero because L6C5L4C3L2=a5=a1. If a5 is zero at least one of the five components must be zero. Therefore , a12=0 is an invalid root of the eighth degree equation. Dividing both sides of the equation by a12 gives a sixth degree equation which is a cubic in a12. Substituting a12=x in the cubic gives:

x3-24x2+144x-128=0

x=8 is found to be one root of this equation which allows us to factor it as follows:

(x-8)(x2-16x+16)=0

By setting the second factor to zero and solving the resultant quadratic equation, we get x=8±4√3. The three roots are therefore x=a12=8, 8+4√3, and 8-4√3. Taking the square root of each we get a1=2√2=2.8284, √6+√2=3.8637, and √6-√2=1.0352. The negative roots will not be considered since they require negative values for some of the filter components. It will be shown later that the second root is the only valid result. The first and third roots lead to negative values for some of the filter components.

First Root Considered

Substituting x=2√2 for a1 in the first of the five auxilary equations, we get a2=4. Since a1=a5, we can substitute for a5 in the fifth equation to get a4=a2=4. Substituting a1, a2, a4, and a5 in the third equation we get a3=3√2. The six coefficients are therefore:

a1=L6+L4+L2=2√2

a2=L6C5+L4C3+L6C3+L2C1+L4C1+L6C1=4

a3=L6C5L2+L4C3L2+L6C3L2+L6C5L4=3√2

a4=L6C5L2C1+L4C3L2C1+L6C3L2C1+L6C5L4C1+L6C5L4C3=4

a5=L6C5L4C3L2=2√2

a6=L6C5L4C3L2C1=1

From the equations for a6 and a5, L6C5L4C3L2C1=1 and L6C5L4C3L2=a5. We may divide these equations to get
C1=1/a5=1/(2√2). This gives us our first component value C1=√2/4.

Factoring C1 out of the first four terms of the equation for a4 gives
C1(L6C5L2+L4C3L2+L6C3L2+L6C5L4)+L6C5L4C3=a4.

Since L6C5L2+L4C3L2+L6C3L2+L6C5L4=a3, we can substitute a3 in the above equation to get C1a3+L6C5L4C3=a4. Substituting the values for C1, a3, and a4, we get (√2/4)(3√2)+L6C5L4C3=4. This gives L6C5L4C3=5/2. When we divide the above equation for a5 by this equation to get L2=a5/(5/2). This gives us our second component value L2=(4/5)√2.

First Root Third Value

Factoring L2 out of the first three terms of the equation for a3 gives
L2(L6C5+L4C3+L6C3)+L6C5L4=a3.
Factoring C1 out of the equation for a2 gives (L6C5+L4C3+L6C3)+C1(L2+L4+L6)=a2. Substituting 1/a5 for C1 and a1 for L2+L4+L6, we get (L6C5+L4C3+L6C3)+(1/a5)(a1)=4. Since a1=a5, we get L6C5+L4C3+L6C3=3.

Substituting the values for L2, L6C5+L4C3+L6C3 in the above equation for a3 gives [(4/5)√2][3]+L6C5L4=a3=3√2. Simplifying radicals we get L6C5L4=(3/5)√2. Since L6C5L4C3=5/2, we may divide the equations to get C3=[(5/2)]/[(3/5)√2]. Simplifying radicals gives us our third component value C3=(25/12)√2.

First Root Fourth Value

Factoring C3 and C1 out of the equation for a2 we get L6C5+C3(L6+L4)+C1(L6+L4+L2)=a2. From the equation for a1, L6+L4=a1-L2=[2√2]-[4/5)√2]=(6/5)√2 as well as L6+L4+L2=a1=2√2. Substituting these values as well as those for C3 and a2 we get L6C5+[(25/12)√2][(6/5)√2]+[√2/4][2√2]=4. Simplifying radicals gives the disappointing result L6C5= -2. This means that L6 and C5 have opposite signs. One of them has a negative value which makes this an invalid solution. We will complete this solution to illustrate the basic method. Since L6C5L4=(3/5)√2, we may divide this equation by the above equation to get L4=[(3/5)√2]/[-2]. Simplifying radicals we get L4= -(3/10)√2.

First Root Final Values

From the Third Value section we have L6C5+L4C3+L6C3=3. Substituting the values for L6C5, L4, and C3 we get
[-2]+[-(3/10)√2][(25/12)√2]+L6[(25/12)√2]=3,
[-2]-[(75*2/120)]+L6[(25/12)√2]=3,
L6[(25/12)√2]=3+2+5/4=25/4,
L6=[25/4]/[(25/12)√2]=(3/2)√2, and finally
L6=(3/2)√2.

Since L6C5= -2, we may divide this equation by the above equation to get:

C5=[-2]/[(3/2)√2]

C5= -(2/3)√2

Collecting all six component values we get this solution with two negative component values:

C1=√2/4=0.3535

L2=(4/5)√2=1.1313

C3=(25/12)√2=2.9462

L4= -(3/10)√2= -0.4242

C5= -(2/3)√2)= -0.9428

L6=(3/2)√2=2.1213.

We leave it to the reader to show that the equations for a1 thru a6 are satisfied when the above component values are substituted. It should be noted that if the filter is implemented with active devices replacing some or all components, negative values may have meaning. For example, a negative inductor or capacitor may be implemented with an op amp. Consideration of negative values is beyond the scope of this paper since they can only be used to construct active filters.

Second and Third Roots Considered

We now turn to the two remaining roots of the eighth degree equation in the search for a valid solution. Substituting x=√6±√2 for a1 in the first auxilary equation, we get a2=(8±4√3)/2=4±2√3. Since a1=a5, we can substitute for a5 in the fifth equation to get a4=a2=4±2√3. Substituting a1, a2, a4, and a5 in the third equation we get a3=2√6±3√2. The six coefficients are therefore:

a1=√6±√2

a2=4±2√3

a3=2√6±3√2

a4=4±2√3

a5=√6±√2

a6=1

From the equations for a6 and a5, L6C5L4C3L2C1=1 and L6C5L4C3L2=a5. We may divide these equations to get
C1=1/a5=1/(√6±√2). This gives us our first component value C1=(√6-/+√2)/4.

Factoring C1 out of the first four terms of the equation for a4 gives
C1(L6C5L2+L4C3L2+L6C3L2+L6C5L4)+L6C5L4C3=a4.

Since L6C5L2+L4C3L2+L6C3L2+L6C5L4=a3, we can substitute a3 in the above equation to get C1a3+L6C5L4C3=a4. Substituting the values for C1, a3, and a4, we get
(√6-/+√2)/4)(2√6±3√2)+L6C5L4C3=4±2√3. Simplifying radicals we get L6C5L4C3=(5±3√3)/2=5.0980,-.0980.

Invalid Solutions

Note that for the negative sign on the right side of the previous equation, that L6C5L4C3 will be negative. This is an invalid solution because it requires that some of the component values be negative. All of the values for a1 thru a5 with negative signs are therefore invalid. Hereafter we will use only the valid positive signs for values of a1 thru a5. Actual values for the various components of the filter will be shown truncated to four decimals. This is more than sufficient for any practical filter.

Rejecting values involving negative signs gives the following revised values:

a1=√6+√2

a2=4+2√3

a3=2√6+3√2

a4=4+2√3

a5=√6+√2

The positive sign in the previous value that we obtained for C1 is also invalid. The revised value is:

C1=(√6-√2)/4=0.2588

We have previously shown that the negative sign in the equation for L6C5L4C3 is invalid which gives the revised value
L6C5L4C3=(5+3√3)/2

Since L6C5L4C3L2=a5=√6+√2, we can divide this equation by the one for L6C5L4C3 which gives L2=[√6+√2]/[(5+3√3)/2]. Simplifying radicals gives:

L2=4√2-2√6=0.7578.

Third Value

Factoring L2 out of the first three terms of the equation for a3 gives L2(L6C5+L4C3+L6C3)+L6C5L4=a3.
Factoring C1 out of the equation for a2 gives (L6C5+L4C3+L6C3)+C1(L2+L4+L6)=a2. Substituting 1/a5 for C1, a1 for L2+L4+L6, and the radical value for a5 we get L6C5+L4C3+L6C3)+(1/a5)(a1)=4+2√3. Since a1=a5, we get
L6C5+L4C3+L6C3=3+2√3.

Factoring L2 out of the equation for a3 gives L2(L6C5+L4C3+L6C3)+L6C5L4=a3. Substituting the values for L2, L6C5+L4C3+L6C3, and a3 in the above equation gives (4√2-2√6)(3+2√3)+L6C5L4=2√6+3√2. Simplifying radicals we get L6C5L4=3√2. Since L6C5L4C3=(5+3√3)/2, we may divide the equations to get C3=(5+3√3)/6√2. Simplifying radicals we get

C3=(3√6+5√2)/12=1.2016

Fourth Value

Factoring C3 and C1 out of the equation for a2 we get C5L6+C3(L6+L4)+C1(L6+L4+L2)=a2. From the equation for a1,
L6+L4=a1-L2=(√6+√2)-(4√2-2√6)=3√6-3√2. Also, L6+L4+L2=a1=√6+√2. Substituting these values as well as those for C3 and a2 we get C5L6+[(3√6+5√2)/12][3√6-3√2]+[(√6-√2)/4][√6+√2]=4+2√3. Simplifying radicals gives
L6C5=√3+1.

Since L6C5L4=3√2, we may divide this equation by the above equation to get L4=[3√2]/[√3+1]. Simplifying radicals we get

L4=3(√6-√2)/2=1.5529

Final Values

From the Third Value section we have L6C5+L4C3+L6C3=3+2√3. Substituting the values for L6C5, L4, and C3 we get
[√3+1]+[3(√6-√2)/2][(3√6+5√2)/12]+L6[(3√6+5√2)/12]=3+2√3.

[(18-6√3+10√3-10)/8]+L6[(3√6+5√2)/12]=√3+2

[(2+√3)/2]+L6[(3√6+5√2)/12]=√3+2

L6[(3√6+5√2)/12]=(√3+2)/2

L6=[(√3+2)/2]/[(3√6+5√2)/12]=[6(√3+2)]/[3√6+5√2]=6(√3+2)(3√6-5√2)/(54-50)

L6=6(9√2+6√6-5√6-10√2)/4=3(√6-√2)/2. We finally achieve our result:

L6=3(√6-√2)/2=1.5529

Since L6C5=√3+1, we may divide this equation by the above equation to get:

C5=[√3+1]/[3(√6-√2)/2]=2(√3+1)(√6+√2)/[3(6-2)]=(3√2+√6+√6+√2)/6=(√6+2√2)/3

C5=(√6+2√2)/3=1.7593

Collecting all six component values we get:

C1=(√6-√2)/4=0.2588

L2=4√2-2√6=0.7578

C3=(3√6+5√2)/12=1.2016

L4=3(√6-√2)/2=1.5529

C5=(√6+2√2)/3=1.7593

L6=3(√6-√2)/2=1.5529

Phase Equation

The formula for the input voltage in terms of unity output voltage is Ein=(1-a2ω2+a4ω4-a6ω6)+i(a1ω-a3ω3+a5ω5). If θ is the leading phase angle of the input compared to the output,

tan θ= [(a1ω-a3ω3+a5ω5)/(1-a2ω2+a4ω4-a6ω6)].

If the numerator of this phase equation is zero, the phase angle can be 0°, 180°, or 360°. If the denominator is zero, the phase angle can be 90°, 270°, or 540°. The exact frequencies that these six phase values will be reached can be determined by setting both the numerator and denominator individually equal to zero and solving the resulting equations.

Setting the numerator equal to zero gives an equation of the fifth degree. One of its roots is obviously ω=0. Dividing both sides by ω leaves a5ω4-a3ω2+a1=0 which is a quadratic in ω2. Since a5=a1, we get can divide both sides of the equation by a1 to get ω4-(a3/a12+1=0.

Substituting the numerical value for a3/a1=(3+√3)/2 in this equation and solving it as a quadratic yields
ω2=[3+√3±√(6√3-4)]/4=.5509, 1.8150. The square roots of these radicals cannot be simplified further and can only be expressed numerically. Therefore ω=.7422, 1.3472. This means that the phase will go thru 180° at .7422 radians per second and 360° at 1.3472 radians per second. Note that the roots are reciprocals of each other. This is true any time that the first and last coefficients of a quadratic equation are equal.

Setting the denominator equal to zero gives an equation of the sixth degree, 1-a2ω2+a4ω4-a6ω6=0. Since a6=1 and a4=a2, we have ω6-a2ω4+a2ω2-1=0. This is an equation of the sixth degree but it is of the third degree in ω2. By inspection ω2=1 is a root of this equation. Factoring out (ω2-1) yields
2-1][ω4+(1-a22+1]=0. The second factor is a quadratic in ω2. Substituting a2=4+2√3 and solving for ω2 gives
ω2=[3+2√3±√(17+12√3)]/2=.1585, 6.3055. Therefore ω=.3982 and 2.5110. As in the numerator, these radicals can only be expressed numerically and the roots are reciprocals of each other. Therefore the phase will go thru 90° at .3982 radians per second, 270° at 1 radian per second, and 450° at 2.5110 radians per second.

Phase and Group Delay Graphs

Most phase plotting programs can only display phase of ±180°. The program used here keeps track of quadrants so that 0°/360°, 90°/450°, and 180°/540° can be distinguished from each other. Figure xx shows the phase delay of the filter at various frequencies. It starts at 0° at dc and gradually increases to 540° off the edge of the graph at infinity. At 0.3982 radians per second the delay has become 90°. It then climbs to 180° at 0.7422 radians per second. At cutoff frequency of 1 radian per second, the delay has reached 270°. At 1.3472 radians per second it is now 360°. The delay reaches 450° at 2.5110 radians per second and then slowly climbs to 540° at infinity.

Each frequency at which filter phase delay is a multiple of 90° is significant. It will be shown that the output impedance of the filter alternates between parallel and series resonance as the delay increases from 0° at dc to 540° at infinite frequency. The filter output impedance goes thru parallel resonance at frequencies where the delay is 90°, 270° or 450°. It goes thru series resonance at frequencies where the delay is 180° or 360°.

Filters have different time delays for different frequencies and this can cause problems for wide bandwidth signals such as pulses. Some portions of the signal will pass thru the filter before others and distort pulse shapes. The result is ringing on square waves with fast rise times. High frequency sine waves that have amplitude, phase, or frequency modulation also get out of time with lower frequency sine waves which have similar modulation. When these different frequencies are demodulated, the results will be out of time with respect to each other.

For a fixed time delay of a broadband signal, phase change is proportional to frequency. Higher frequencies will therefore have greater phase delay than lower frequencies. Since phase delay cannot exceed 90° per pole of the filter, the phase delay of the highest frequencies must be limited. For a six pole filter, the maximum phase delay at infinite frequency is 540°. This reduces the time delay of high frequencies which are out of band and otherwise greatly attenuated. For frequencies just below cutoff, phase delay increases at a more rapid rate which increases their time delay.

The low frequency delay of the filter is equal to the time constant of the sum of the inductors with the 1 ohm load. This is 3.8637 seconds. This corresponds to the first linear portion of the phase graph up to about 0.25 radian per second. At this point the phase starts to increase more rapidly with frequency. The group delay increases accordingly until it peaks at roughly 0.95 radians per second. There it has reached about 6.49 seconds. This is about two-thirds more than its low frequency value. It then starts a gradual decrease toward zero at infinite frequency.

Output Impedance

The output impedance may be determined at any frequency by shorting the input and calculating the reactances in a chain as shown in figure xx. Inductive reactances will be denoted by ωL and capacitive reactances by -1/ωC. The reactance of L6 and C5 in parallel is the reciprocal of the sum of their reciprocals. Therefore X5=1/[1/ωL6-ωC5]. Collecting terms and inverting we get
X5=ωL6/(1-ω2L6C5).

To this reactance we add the reactance of L4. This is denoted by
Z4=[ωL6/(1-ω2L6C5)+ωL4]=( ωL6+ωL43L6C5L4)/(1-ω2L6C5). Collecting terms we get
Z4=[ω(L6+L4)-ω3L6C5L4]/(1-ω2L6C5).

Now X3=1/[1/(Z4)-ωC3]= 1/[(1-ω2L6C5)/(ω(L6+L4)-ω3L6C5L4)-ωC3]
X3=1/[(1-ω2L6C5)-ω2C3(L6+L4)+ω4L6C5L4C3]/[ω(L6+L4)-ω3L6C5L4]. Collecting terms and inverting we get
X3=[ω(L6+L4)-ω3L6C5L4]/[1-ω2(C3(L6+L4)+L6C5)+ω4L6C5L4C3].

To this reactance we add the reactance of L2. This is denoted by
Z2=[ω(L6+L4)-ω3L6C5L4]/[1-ω2{C3(L6+L4)+L6C5}+ω4L6C5L4C3]+ωL2=
[ω(L6+L4)-ω3L6C5L4]+ωL23L2{C3(L6+L4)+L6C5}+ω5L6C5L4C3L2]/[1-ω2{C3(L6+L4)+L6C5}+ω4L6C5L4C3].
Collecting terms we get
Z2=[ω(L6+L4+L2)-ω3(L2{C3(L6+L4)+L6C5}+L6C5L4)+ω5L6C5L4C3L2]/[1-ω2{C3(L6+L4)+L6C5}+ω4L6C5L4C3].

Now X1=1/[1/(Z2)-ωC1]=
1/({[1-ω2{C3(L6+L4)+L6C5}+ω4L6C5L4C3]/[ω(L6+L4+L2)-ω3(L2{C3(L6+L4)+L6C5}+L6C5L4)+ω5L6C5L4C3L2]}-ωC1)=
1/({[1-ω2{C3(L6+L4)+L6C5}+ω4L6C5L4C3]-[ω2C1(L6+L4+L2)-ω4C1(L2{C3(L6+L4)+L6C5}+L6C5L4)+ω6L6C5L4C3L2C1]/
[ω(L6+L4+L2)-ω3(L2{C3(L6+L4)+L6C5}+L6C5L4)+ω5L6C5L4C3L2]}). Collecting terms and inverting we get

X1=[ω(L6+L4+L2)-ω3(L2[C3(L6+L4)+C5L6]+L6C5L4)+ω5L6C5L4C3L2]/
[1-ω2[C1(L6+L4+L2)+C3(L6+L4)+C5L6]+ω4[C1(L2{C3(L6+L4)+C5L6}+L6C5L4)+L6C5L4C3]-ω6L6C5L4C3L2C1].
Expanding terms we get:

X1=[ω(L6+L4+L2)-ω3(L2C3L6+L2C3L4+L2C5L6+L6C5L4)+ω5L6C5L4C3L2]/
[1-ω2(C1L6+C1L4+C1L2+C3L6+C3L4+C5L6)+ω4(C1L2C3L6+C1L2C3L4+C1L2C5L6+C1L6C5L4+L6C5L4C3)-ω6L6C5L4C3L2C1].
The coefficients for ω thru ω6 are seen to be identical to the definitions of a1 thru a6 respectively. When we substitute these definitions we get the much more manageable result:

X1=(a1ω-a3ω3+a5ω5)/(1-a2ω2+a4ω46).

Output Impedance Graph

Note that the equation for X1 is identical to the Phase Equation for tan θ. Since a5=a1, a4=a2, and a6=1; this equation reduces to
X1=(a1ω-a3ω3+a1ω5)/(1-a2ω2+a2ω46). In other words, tan θ represents the output impedance of the filter. We plotted tan-1θ in the Phase Delay section. Here we plot tan θ as a function of ω to get output impedance. Changes in filter output impedance need to be considered for things like speaker damping and load variation.

Figure x shows the filter output impedance when driven by a zero impedance source. It has three parallel resonances and two series resonances. The inductive reactance starts at zero at dc and gradually increases. At 0.3982 radians per second it undergoes its first parallel resonance. There it assumes a high capacitive reactance which gradually drops to zero at 0.7422 radians per second. This is its first series resonance. It then becomes a progressively higher inductive reactance until cutoff frequency of 1 radian per second is reached. There it undergoes its second parallel resonance and suddenly changes to a high capacitive reactance.

This capacitive reactance then gradually decreases. It becomes zero at 1.3472 radians per second where a second series resonance is reached. The reactance then gradually becomes inductive and increases in value until the third parallel resonance is reached at 2.5110 radians per second. A high capacitive reactance is then suddenly assumed which gradually decreases to zero at infinity. This comes from the reactance of the last capacitor to ground.

It should be noted that the equation for X1 is symmetrical in ω and 1/ω. If we replace ω with 1/ω we get X1=(a1ω-1-a3ω-3+a1ω-5)/(1-a2ω-2+a2ω-4-6). When we multiply both numerator and denominator by ω6 we get X1=(a1ω5-a3ω3+a1ω)/(ω6-a2ω4+a2ω2-1). This is merely the original equation with a negative sign. Since the equation for output impedance is symmetrical in ω and 1/ω, a logarithmic plot of frequency versus impedance is left right symmetrical with opposite sign as shown in figure xx. In other words, if the output impedance at ω is X1, the output impedance at 1/ω is -X1.

Normal Input Impedance

The normal input impedance may be found from Ein/Iin= Ein/IL6. When the values of the components are inserted in each equation and expressed in terms of a1 thru a6 we get Z=[1+ia1ω-a2ω2-ia3ω3+a4ω4+ia5ω5-a6ω6]/[1+i(5/6)a1ω-(2/3)a2ω2-i(a3/2)ω3+(a4/3)ω4+i(a5/6)ω5. Separating real and imaginary parts we get
Z=([1-a2ω2+a4ω4-a6ω6]+i[a1ω-a3ω3+a5ω5])/([1-(2/3)a2ω2+(a4/3)ω4]+i[(5/6)a1ω-(a3/2)ω3+(a5/6)ω5]). Carrying out this complex division gives the impedance as well as its phase.

Figure x shows the plot of the above equation for input impedance and phase angle at various frequencies with a normal load. Under normal conditions the filter input impedance is 1 ohm resistive at dc. It drops slightly to a minimum of .808 ohm with a lagging phase angle of 49.4° at ω=.889. It then gradually increases to infinity with a lagging phase angle approaching 90°. For a drive with flat frequency response, the input current increases slightly at the minimum impedance. However, the phase angle lags enough so that the actual output power is reduced slightly. This results in an output level at that point of about 0.896 when the input is 1. The high frequency attenuation of the filter comes about from the increased input impedance as well as the near 90° phase lag. When the phase lag approaches 90°, slight phase lag increases can result in a greatly reduced output level.

Unloaded Input Impedance

An unterminated six pole filter has three series and two parallel resonances. When a driven filter is unterminated, dangerous voltages or currents can be produced at some frequencies. This may cause destruction of the driving source or some of the filter components. It can also generate high voltage arcs or cause corona. If an unterminated filter is lossless, its input and output impedances will be pure reactances at all frequencies. In practice, inductor Q limitations will cause a small resistive loss.

The input impedance of an unterminated filter may be determined at any frequency by calculating the reactances in a chain as shown in figure xx. Inductive reactances will be denoted by ωL and capacitive reactances by -1/ωC. The reactances to ground will be calculated at five different points which will be denoted by Z2 thru X6. The reactance of L2 and C1 in series is Z2=ωL2-1/ωC1. Combining terms we get
Z2= -(1-ω2L2C1)/ωC1.

The reactance of Z2 in parallel with C3 is the reciprocal of the sum of their reciprocals or
X3=1/{[-ωC1/(1-ω2L2C1)]-ωC3}=1/{[-ωC1-ωC3(1-ω2L2C1)]/(1-ω2L2C1)}. Collecting terms and inverting we get
X3= -(1-ω2L2C1)/[ω(C1+C3)-ω3C3L2C1].

Adding the reactance of ωL4 we get
Z4=X3+ωL4=-(1-ω2L2C1)/[ω(C1+C3)-ω3C3L2C1]+ωL4=-[(1-ω2L2C1)+ωL4{ω(C1+C3)-ω3C3L2C1}]/[ω(C1+C3)-ω3C3L2C1]
Z4= -[1-ω2{L2C1+L4(C1+C3)}+ω4L4C3L2C1]/[ω(C1+C3)-ω3C3L2C1].

The reactance of C5 in parallel with Z4 is the reciprocal of the sum of their reciprocals or
X5=1/{[-ω(C1+C3)+ω3C3L2C1]/[1-ω2[L2C1+L4(C1+C3)]+ω4L4C3L2C1]-ωC5}
X5=1/{[-ω(C1+C3)+ω3C3L2C1]+[-ωC53C5[L2C1+L4(C1+C3)]-ω5C5L4C3L2C1]}/
[1-ω2[L2C1+L4(C1+C3)]+ω4L4C3L2C1].

Collecting terms and inverting we get
X5= -{1-ω2[L2C1+L4(C1+C3)]+ω4L4C3L2C1}/{ω(C1+C3+C5)-ω3[C3L2C1+C5[L2C1+L4(C1+C3)]+ω5C5L4C3L2C1]}.

Adding the reactance of ωL6 we get
Z6=X5+ωL6= -[1-ω2[L2C1+L4(C1+C3)]+ω4L4C3L2C1]+ωL6/ [ω(C1+C3+C5)-ω3{C3L2C1+C5[L2C1+L4(C1+C3)}+ω5C5L4C3L2C1].

Z6= -{1-ω2[L2C1+L4(C1+C3)+L6(C1+C3+C5)]+ω4L4C3L2C1+L6{C3L2C1+C5[L2C1+L4(C1+C3)]}]-ω6L6C5L4C3L2C1]}/
{ω(C1+C3+C5)-ω3[C3L2C1+C5[L2C1+L4(C1+C3)]+ω5C5L4C3L2C1]}.

Expanding terms we get
Z6=-{1-ω2(L2C1+L4C1+L4C3+L6C1+L6C3+L6C5)+
ω4(L4C3L2C1+L6C3L2C1+L6C5L2C1+L6C5L4C1+L6C5L4C3)-ω6(L6C5L4C3L2C1)}/
{ω(C1+C3+C5)-ω3(C3L2C1+C5L2C1+C5L4C1+C5L4C3)+ω5(C5L4C3L2C1)}.

This equation has much in common with the original equation for Ein. Note that the coefficients of all the even powers of ω are the same in both equations. The coefficients of the odd powers of ω are the same as those for the odd powers of ω in the previous equation for IL6. We may substitute the values for the filter components in the present equation to give radicals for the coefficients. Those radicals may in turn be expressed in terms of a1 thru a6 which gives
Z6= -(1-a2ω2+a4ω4-a6ω6)/{(5/6)a1ω-(a3/2)ω3+(a5/6)ω5}. It should be noted that the numerator is the real part of the numerator of the Normal Impedance Equation. The denominator is the imaginary part of the denominator of the Normal Impedance equation.

When the numerator is zero the filter input impedance will be zero. This corresponds to a series resonance point. When the denominator is zero the filter input impedance will be infinite. This corresponds to a parallel resonance point. The numerator of this equation is the same as the denominator of the phase equation which has been solved previously. There we obtained ω=.3982, 1.0000, and 2.5110. These frequencies are the zeros of the unterminated filter where input impedance will be zero.

When we set the denominator equal to zero we get (5/6)a1ω-(a3/2)ω3+(a5/6)ω5=0. Since a5=a1, we get
(5/6)a1ω-(a3/2)ω3+(a1/6)ω5=0. Dividing both sides by a1/6 gives ω5-(3a3/a13+5ω=0. Since ω=0 is a root of this equation, we divide both sides by ω to get ω4-(3a3/a12+5=0. Since (3a3/a1)=(9+3√3)/2, we can substitute this value and solve the equation as a quadratic to get ω2=[(9+3√3±√(28+54√3)]/4=0.7930, 6.3050. These radicals cannot be simplified further. Taking the square root of each of these values gives ω=0.8905, 2.5109. These frequencies are the poles of the unterminated filter where input impedance will be infinite.

Unloaded Impedance Graph

Figure x shows the reactive input impedance of the unterminated filter at various frequencies, ignoring resistive losses. It has three series and two parallel resonances. Near dc, input impedance is a very large capacitive reactance which decreases with frequency. This comes from the three shunt capacitors in the filter. At 0.3982 radians per second, the reactance goes thru zero and becomes inductive. This is its first series resonance point. Here the output impedance goes thru a parallel resonance. The inductive reactance gradually increases until the frequency 0.8905 radians per second is reached. There it undergoes parallel resonance and suddenly becomes a large capacitive reactance. There is no corresponding point on the output impedance. The capacitive reactance rapidly decreases now and becomes zero at the cutoff frequency of 1 radian per second. There it undergoes its second series resonance. This point corresponds to a parallel resonance in the output impedance. The inductive reactance now proceeds to climb again until it reaches about +3.6i.

At 2.5109 radians per second, the filter undergoes a second parallel resonance. There is no corresponding point on the output impedance. Input impedance again rapidly changes from a large inductive reactance to a large capacitive reactance. That capacitive reactance then rapidly decreases to zero with a sudden change to its third series resonance at 2.5110 radians per second. This corresponds to a parallel resonance point on the output impedance. The parallel and series resonance points are separated by only one part in 28,000 and are so closely spaced that the graph cannot display them. The reactance now suddenly becomes inductive and rapidly returns to about +3.6i. There it continues a gradual climb to infinity. This comes from the reactance of the first inductor driving the first capacitor which is acting as a short.

When the input impedance of the unterminated filter goes thru zero, high currents on the input and high voltages on the output can be produced. This occurs at 0.3982, 1.0, and 2.5110 radians per second. Ordinarily the last frequency will not be a problem, so long as the switching frequency is not equal to it. The normalized switching frequency will usually be at 4 radians per second or higher. Measures must be taken to prevent damage at the first two lower frequencies. Note that the three parallel resonance points on the previous filter output impedance graph correspond to the dangerous series resonance points on the unterminated filter input impedance. The two series resonance points on filter output impedance do not correspond to any points on the unterminated filter input impedance.

Load Variation Effects

Figure x shows the frequency response of the filter for various loads. When the load is not 1 ohm, resonance effects may change the output level. Load changes have no effect on the output at series resonance points. Parallel resonance points cause the output to rise for loads above 1 ohm, and fall for loads below 1 ohm. The effects change gradually between these points. The five graphs in figure xx are for loads of 2.000, 1.414, 1.000, 0.707, and 0.500 ohms. The output level increases to a maximum at ω=0.3982 for loads over 1 ohm and decreases to a minimum there for loads under 1 ohm. This is the first parallel resonance point. The output level returns to 0.9863 at ω=0.7422 for all loads. This is the first series resonance point. The slight level reduction here is from normal filter attenuation as cutoff is approached.

At cutoff frequency of ω=1, output level is above or below the normal level of 0.707 when the load is above or below the normal 1 ohm. This comes from the presence of the second parallel resonance point. Above ω=1 the response of the filter drops for all loads until ω=1.3472 is reached. This is the second series resonance point. The level there is equal to the normal filter attenuation 0.1649 (1/6.0629) and is independent of load. At ω=2.5110, the third and final parallel resonance point occurs. Normal level there is 0.003988 (1/250.7) which cannot be seen on the graph. The level there rises or falls with loads above or below normal just like the two previous parallel resonance points.

Shorted Load Input Impedance

New resonances are produced when the output of the filter is shorted. They must be considered when designing current limiting for a switching amplifier. An easy way to calculate the input impedance with a shorted load is to substitute an infinite value for C1 and use the equation for input impedance of an unterminated filter. Any terms not containing C1 can be eliminated because they are of negligible value.

The input impedance of the unterminated filter is
Z6= -[1-ω2(L2C1+L4C1+L4C3+L6C1+L6C3+L6C5)+
ω4(L4C3L2C1+L6C3L2C1+L6C5L2C1+L6C5L4C1+L6C5L4C3)-ω6(L6C5L4C3L2C1)]/
[ω(C1+C3+C5)-ω3(C3L2C1+C5L2C1+C5L4C1+C5L4C3)+ω5(C5L4C3L2C1)].

Eliminating all terms not containing C1 gives
Z= -[-ω2(L2C1+L4C1+L6C1)+ ω4(L4C3L2C1+L6C3L2C1+L6C5L2C1+L6C5L4C1)-ω6(L6C5L4C3L2C1)]/
[ω(C1)-ω3(C3L2C1+C5L2C1+C5L4C1)+ω5(C5L4C3L2C1)].

We can now divide both numerator and denominator by ωC1 which gives
Z=[ω(L2+L4+L6)-ω3(L4C3L2+L6C3L2+L6C5L2+L6C5L4)+ω5(L6C5L4C3L2)]/[1-ω2(C3L2+C5L2+C5L4)+ω4(C5L4C3L2)].

The coefficients for the numerator are the definitions of a1, a3, and a5. We may substitute the values for the filter components in the denominator to give radicals for the coefficients. Those radicals may in turn be expressed in terms of a2 and a4 which result in
Z=[a1ω-a3ω3+a5ω5[/[1-(2a2/3)ω2+(a4/3)ω4].

The numerator is identical to the numerator of the Phase Equation which has been previously solved when set equal to zero. This yielded ω=0 and ω2=[3+√3±√(6√3-4)]/4=.5509, 1.8150. The square roots of these radicals cannot be simplified further and can only be expressed numerically. Therefore ω=.7422, 1.3472. This means that the shorted filter will go thru series resonance at .7422 and 1.3472 radians per second as well as having zero impedance at dc. These are the same series resonance points on the filter output impedance.

We can set the denominator equal to zero to find the poles of the shorted filter. Since a2=a4, this gives 1-(2a2/3)ω2+(a2/3)ω4=0. Dividing both sides of the equation by a2/3 gives ω4-2ω2+3/a2=0. Substituting the radical value 4+2√3 for a2 and solving this equation as a quadratic gives ω2=(2±√[4-4(3/a2)])/2=1±√[1-(3/a2)]=(2±√[6√3-8])/2. The square root of this radical cannot be simplified further and can only be expressed numerically. Therefore ω=0.4760, 1.3316. This means that the shorted filter will go thru parallel resonance at .4760 and 1.3316 radians per second. These points do not correspond to any points on the filter output impedance graph.

Shorted Load Graph

Figure x shows the reactive input impedance of the shorted filter at various frequencies, ignoring resistive losses. It has two parallel and two series resonances. Near dc, input impedance is a small inductive reactance which gradually increases until the frequency of 0.4760 radians per second is reached. Here the filter undergoes its first parallel resonance and suddenly becomes a large capacitive reactance. There is no corresponding point on the normal filter output impedance graph. This capacitive reactance gradually decreases and becomes zero at the 0.7422 radians per second. It now undergoes its first series resonance. This corresponds to an identical series resonance on the normal filter output impedance graph. The reactance now becomes inductive and proceeds to climb again until it reaches about +1.8i. There it proceeds to climb more and more rapidly.

At 1.3316 radians per second, the shorted filter undergoes a second parallel resonance. This does not correspond to any point on the normal filter output impedance graph. Input impedance again rapidly changes from a large inductive reactance to a large capacitive reactance. That capacitive reactance then rapidly decreases to zero with a sudden change to series resonance at 1.3472 radians per second. This corresponds to an indentical series resonance point on the normal filter output impedance graph. The parallel and series resonance points are separated by only one part in 85 and are so closely spaced that the graph can barely display them. The reactance now becomes inductive and rapidly returns to about +1.8i. It then continues a gradual climb to infinity. This comes from the reactance of the first inductor driving the first capacitor which is acting as a short.

When the input impedance of the shorted filter goes thru zero, high currents on the input can be produced even with low drive voltages. This occurs at dc, 0.7422, and 1.3472 radians per second. At the later two frequencies, high voltages on the two unshorted capacitors can be produced. Measures must be taken to prevent damage at these three frequencies. Serious phase changes in the current will also occur for loads less than the nominal termination impedance of 1 ohm. These phase changes complicate the design of current limiting circuitry.

Output Impedance Model

Near dc, the shunt capacitors of the filter have little effect and output impedance of the filter is inductive. It may be modeled there by the sum of L2, L4, and L6. At very high frequencies, the series inductors have little effect and the output impedance of the filter is capacitive. It may be modeled there by C1. A crude model of the filter would be the circuit of figure xx where L=L2+L4+L6 and C=C1. This model is correct near dc and at high frequencies. We have previously shown that (L2+L4+L6)C1=1. Therefore the circuit exhibits the correct parallel resonance at 1 radian per second but its impedance near resonance is incorrect. It further fails to exhibit the two other parallel resonances at 0.3982 and 2.5110 radians per second. The two series resonances at 0.7422 and 1.3472 radians per second are missing as well.

The exact values for the resonances at 0.3982, 0.7422, 1.0000, 1.3472, and 2.5110 radians per second were previously derived in the Phase Equation section. They will be denoted here as 1/b, 1/a, 1, a, and b respectively. A potentially accurate model of filter output impedance is the circuit of figure xx. Three parallel resonant circuits at 1/b, 1, and b radians per second are in series. At those three frequencies, the output impedance goes thru parallel resonance. If the impedance of the parallel resonant circuit for 1/b is the same as the one for b, their reactances at 1 radian per second will be equal and opposite. They will cancel and leave the reactance of the parallel resonant circuit of 1 radian per second undisturbed.

Between 1/b and 1 radian per second, the bottom parallel resonant circuit is capacitive and the middle and top circuits are inductive. This equivalent capacitor and two inductors will therefore form a series resonant circuit. If the components are all chosen correctly, this series resonance will occur at 1/a radians per second. Between 1 and b radians per second, the bottom and middle parallel resonant circuits are capacitive and the top circuit is inductive. These two equivalent capacitors and inductor will therefore form another series resonant circuit. If the components are again chosen correctly, this series resonance will occur at a radians per second. The model will then precisely describe the output impedance of the filter at all frequencies.

Output Impedance Model Solution

The series combination of L7, L8, and L9 must be equal to L2+L4+L6. Also, the series combination of C7, C8, and C9 must be equal to C1. The impedance of the top resonant circuit at b radians per second is √(L9/C9). The impedance of the middle resonant circuit at 1 radian per second is √(L8/C8) and will be denoted by y. The impedance of the bottom resonant circuit at 1/b radians per second is √(L7/C7). The impedances of the top and bottom resonant circuits must be equal and will be denoted by x.

Assume a circuit is series resonant at a frequency fr with the inductor having a reactance of x and the capacitor having a reactance of -x. At a frequency f, its total impedance will be x[(f/fr)-(fr/f)]. If f>fr this is inductive and positive, if f<fr it is capacitive and negative. Assume a circuit is parallel resonant at a frequency fr with the inductor having a reactance of x and the capacitor having a reactance of -x. At a frequency f, its total impedance will be x/[(fr/f)-(f/fr)]. If (f>fr) this is capacitive and negative, if fr<f it is inductive and positive. Therefore the series and parallel impedances are:

Zs=x[(f/fr)-(fr/f)]
Zp=x/[(fr/f)-(f/fr)]

The L7C7 circuit resonates at frequency 1/b radians per second with an impedance of x ohms. At 1/a radians per second its impedance is Z7=x/([(1/b)/(1/a)]-[(1/a)/(1/b)])=x/[(a/b)-(b/a)]. Since (a/b)< 1 , Z7 is negative and therefore capacitive. The L8C8 circuit resonates at frequency 1 radian per second with an impedance of y ohms. At 1/a radians per second its impedance is Z8=y/[1/(1/a)]-[(1/a)/1]=y/[a-(1/a)]. Since a>1 , Z8 is positive and therefore inductive. The L9C9 circuit resonates at frequency b radians per second and also has an impedance of x ohms. At 1/a radians per second its impedance is Z9=x/([b/(1/a)]-[(1/a)/b])=x/[ab-(1/ab)]. Since ab>1 , Z9 is also positive and therefore also inductive. This capacitor and two inductors will form a series resonant circuit at 1/a radians per second if component values are chosen correctly.

It is left to the reader to show that at a frequency of a radians per second, that Z7 has the same capacitive reactance as the inductive reactance that Z9 had at 1/a radians per second. The reader will further see that Z8 will display a capacitive reactance at a radians per second that is equal to its inductive reactance at 1/a radians per second. Lastly the reader can show that Z9 will exhibit a capacitive reactance at a radians per second equal to the inductive reactance shown by Z7 at 1/a radians per second. This proves that if the six components of the impedance model are series resonant at 1/a radians per second, they will also be series resonant at a radians per second.

Impedance Ratios

If the model is to be series resonant at 1/a radians per second, then Z7+Z8+Z9=0. This means that
x/[(a/b)-(b/a)]+y/[a-(1/a)]+x/[ab-(1/ab)]=0 . Separating unknowns we get y/[a-(1/a)]=x(1/[(b/a)-(a/b)]-1/[ab-(1/ab)]) or
y/x=[a-(1/a)]{1/[(b/a)-(a/b)]-1/[ab-(1/ab)]}. We will denote this ratio by k and its numeric value is k=0.2598 . Therefore y/x=k. We now know the ratio of the impedance of the middle parallel circuit to the impedances of the top and bottom parallel circuits. Our task now is to use this ratio to compute the values of the six components in the model circuit.

The impedance of L8C8 circuit is y ohms and it resonates at 1 radian per second. The inductive reactance there is y ohms and therefore L8=y. The impedance of L7C7 circuit is x ohms and it resonates at (1/b) radians per second. Since y/x=k, x=y/k. Since the inductive reactance is y/k ohms, therefore L7=(y/k)/(1/b)=yb/k. The impedance of L9C9 circuit is also x ohms and it resonates at b radians per second. Its inductive reactance at b radians per second is x=y/k ohms. Therefore L9=(y/k)/b=y/kb.

We have shown that L7+L8+L9=L2+L4+L6. Since L2+L4+L6=a1, Substituting the above values we get yb/k+y+y/kb=a1. Collecting terms we get y[(b/k)+1+(1/kb)]=a1. Substituting numeric values for b, k, and a1 we get y=0.3167. Substituting numeric values for y, k, and b in the equations for the three inductors we get:

L7=3.0613
L8=0.3167
L9=0.4855 .

It is seen that the series combination of L7, L8, and L9 equals L2+L4+L6=a1=3.8637 .

C7, C8, and C9 can now be chosen to resonate with L7, L8, and L9 at 1/b, 1, and b radians per second respectively. This gives:

C7=2.0596
C8=3.1565
C9=0.3266 .

It is seen that the series combination of C7, C8, and C9 equals C1=0.2588 . The output impedance model is now complete.

Unloaded Impedance Model

Near dc, the series inductors have very little effect on input impedance of the unloaded filter which is capacitive. It may be modeled there by the sum of C1, C3, and C5. At high frequencies C5 acts like a short and the input impedance of the filter is inductive. It may be modeled there by L6. A crude model of the filter would be the circuit of figure xx where L=L6 and C=C1+C3+C5. This model is correct near dc and at high frequencies. However, it exhibits an incorrect series resonance at 0.4472 radians per second. It further fails to exhibit the three series resonances at 0.3982, 1, and 2.5110 radians per second. The two parallel resonances at 0.8905 and 2.5109 radians per second are missing as well.

The exact values for the two parallel resonances at 0.8905 and 2.5109 radians per second were derived in a previous section. It was shown that their product is √5. They will be denoted as √5/c and c, respectively. The resonant frequencies at 0.3982, 0.8905, 1.0000, 2.5109, and 2.5110 radians per second will be denoted as 1/b, √5/c, 1, c, and b respectively. Note that c is only slightly less than b. A potentially accurate model of filter output impedance is the circuit of figure xx. Three series resonant circuits at 1/b, 1, and b radians per second are in parallel. At those three frequencies, the input impedance goes thru series resonance.

Between 1/b and 1 radian per second, the left series resonant circuit is inductive and the center and right circuits are capacitive. This equivalent inductor and two equivalent capacitors will therefore form a parallel resonant circuit. If the components are all chosen correctly, this parallel resonance will occur at √5/c radians per second. Between 1 and b radians per second, the left and center parallel resonant circuits are inductive and the right circuit is capacitive. These two equivalent inductors and equivalent capacitor will therefore form another parallel resonant circuit. If the components are again chosen correctly, this parallel resonance will occur at c radians per second. The model will then precisely describe the input impedance of the unterminated filter at all frequencies.

Unloaded Impedance Model Solution

The parallel combination of L7, L8, and L9 must be equal to L6. Also, the sum of C7, C8, and C9 must be equal to the sum of C1, C3, and C5. The impedance of the left resonant circuit at 1/b radians per second is √(L7/C7)and will be denoted by x. The impedance of the middle resonant circuit at 1 radian per second is √(L8/C8) and will be denoted by y. The impedance of the right resonant circuit at b radians per second is √(L9/C9) and will be denoted by z.

The L7C7 circuit resonates at frequency 1/b radians per second with an impedance of x ohms. At √5/c radians per second its impedance is Z7=x/([(√5/c)/(1/b)]-[(1/b)/(√5/c)])=x/[b√5/c-c/(b√5)]. Since b√5/c>1 , Z7 is positive and therefore inductive. The L8C8 circuit resonates at frequency 1 radian per second with an impedance of y ohms. At √5/c radians per second its impedance is Z8=y/[(√5/c)/1]-[1/(√5/c)]=y/[√5/c-(c/√5]. Since √5/c< 1 , Z8 is negative and therefore capacitive. The L9C9 circuit resonates at frequency b radians per second and has an impedance of z ohms. At √5/c radians per second its impedance is Z9=z/([(√5/c)/b)]-[b/(√5/c)])=z/[√5/(bc)-(bc/√5)]. Since √5/(bc)< 1 , Z9 is also negative and therefore also capacitive. This inductor and two capacitors will form a parallel resonant circuit at √5/c radians per second if component values are chosen correctly.

The impedance of the parallel resonant circuit at √5/c radians per second is the reciprocal of the sum of the reciprocals of the reactances of its three branches. Since that value is infinite, it is more practical to deal with the reciprocal of that impedance which is zero. Therefore 1/Z7+1/Z8+1/Z9=0. Substituting the individual values of each branch gives [b√5/c-c/(b√5)]/x+[√5/c-(c/√5]/y+[√5/(bc)-(bc/√5)]/z=0. It will be clear shortly that is easier to solve for 1/x, 1/y, and 1/z than for x, y, and z.

The L7C7 circuit resonates at frequency 1/b radians per second with an impedance of x ohms. At c radians per second its impedance is Z7=x/([c/(1/b)]-[(1/b)/(c)])=x/[bc-1/(bc)]. Since bc>1 , Z7 is positive and therefore inductive. The L8C8 circuit resonates at frequency 1 radian per second with an impedance of y ohms. At c radians per second its impedance is Z8=y/[(c/1)-(1/c)]=y/[c-1/c]. Since c> 1 , Z8 is positive and therefore also inductive. The L9C9 circuit resonates at frequency b radians per second and has an impedance of z ohms. At c radians per second its impedance is Z9=z/([c/b)]-[(b/c)]). Since c/b< 1 , Z9 is negative and therefore capacitive. The two equivalent inductors and this equivalent capacitor will form a parallel resonant circuit at c radians per second if component values are chosen correctly.

The reciprocal of the impedance of the parallel resonant circuit formed at c radians per second is also zero. This is the sum of the reciprocals of the reactances of its three branches or 1/Z7+1/Z8+1/Z9=0. Substituting the individual values of each branch gives [bc-1/(bc)]/x+[c-1/c]/y+[(c/b)-(b/c)]/z=0. We now have two equations in three unknowns where the unknowns are 1/x, 1/y, and 1/z. We must derive a third equation in these same unknowns to be able to solve the system.

At 1/b radians per second the capacitive reactance of C7 is x=1/[(1/b)C7] or C7=b/x. In like manner at 1 radian per second, the capacitive reactance of C8 is y or C8=1/y. Finally, at b radians per second, the capacitive reactance of C9 is z or C9=1/(bz). Adding these three we get C7+C8+C9=b(1/x)+1/y+(1/b)(1/z). We have shown that C7+C8+C9=C1+C3+C5. Substituting the values of C1+C3+C5 we find that their sum is (5/6)a1. This gives us our third equation b(1/x)+1/y+(1/b)(1/z)=(5/6)a1.

Overview

The six pole filter has the unexpected result L1=L3. This gives it an important cost advantage since only two different inductor types are required. The four and eight pole filters do not have this feature. Also note that C1=sin(90°/6)=sin 15° and L6 is 6 sin 15°. This leads to the author's conjecture for which he would be grateful for a proof. It holds for zero input impedance filters and is known to hold for filters of two thru ten poles.

  1. The last element in an n pole filter is sin(90°/n).
  2. The first element is n·sin(90°/n). As n becomes large, this value therefore approaches π/2=1.5707.
  3. The sum of the inductors/capacitors (even/odd) is 1/[sin(90°/n)].
  4. The sum of the capacitors/inductors (even/odd) is [(n-1)/n][sin(90°/n)].
  5. The product of all the values is 1.

Knowing these rules make it possible to derive the values for the two, four, and six pole filters without laborious mathematical calculations. To derive the six pole values by this method, you must make use of the fact that the first two inductors are equal. Since sin(90°/7) and sin(90°/9) cannot be expressed exactly in radicals, the values for seven and nine pole filters can only be a decimal approximation. This is true for most of the higher order filters.

As a good rule of thumb, inductors for filters cost ten times as much as capacitors and are several times larger in size. Good filter design will minimize the number of inductors used. Filters designed to be driven by a zero input impedance with an odd number of poles require one more inductor than capacitor. If the extra inductor is used, it makes sense to add another capacitor on the output. This will turn it into an even pole filter of substantially better performance with only a small increase in cost.

If a filter is driven by a current source, its first element must be a capacitor. An odd number of poles should be chosen for this filter since it uses one more capacitor than inductor. Much better performance will again result with only a small increase in cost.

10:00 AM 3/25/2010 πθ°Ωω±√·