Results from image theory

Filter theory was initially developed from earlier studies of transmission lines.^[8][9] In this theory, a filter section is specified in terms of its propagation constant and image impedance (or characteristic impedance).

Specifically for the lattice, the propagation function, γ, and characteristic impedance, Z_o, are defined by,^{[4]: 379 [6]}

${\displaystyle \gamma =\ln \left(\ {\frac {\ {\sqrt {{\frac {\ Z_{\mathsf {a}}\ }{Z_{\mathsf {b}}}}+1\ }}\ }{\ {\sqrt {\ {\frac {\ Z_{\mathsf {a}}\ }{Z_{\mathsf {b}}}}-1\ }}\ }}\ \right)=2\ \operatorname {artanh} {\sqrt {{\frac {\ Z_{\mathsf {a}}\ }{Z_{\mathsf {b}}}}\ }}\qquad \;}$ and ${\displaystyle \;\qquad Z_{\mathsf {o}}={\sqrt {\ Z_{\mathsf {a}}\cdot Z_{\mathsf {b}}\ }}}$

Once γ and Z_o have been chosen, solutions can be found for ${\displaystyle \;{\frac {\ Z_{\mathsf {a}}\ }{Z_{\mathsf {b}}}}\;}$ and ${\displaystyle \;Z_{\mathsf {a}}\cdot Z_{\mathsf {b}}\;}$ from which the characteristics of Z_a and Z_b can each be determined. (In practice, the choices for γ and z_o are restricted to those which result in physically realisable impedances for Z_a and Z_b .) Although a filter circuit may have one or more pass-bands and possibly several stop-bands (or attenuation regions), only networks with a single pass-band are considered here.

In the pass-band of the circuit, the product Z_a × Z_b is real (i.e. Z_o is resistive) and may be equated to R_o, the terminating resistance of the filter. So

${\displaystyle Z_{\mathsf {a}}\ Z_{\mathsf {b}}=R_{\mathsf {o}}^{2}\qquad \;}$ or ${\displaystyle \;\qquad {\frac {\ Z_{\mathsf {a}}\ }{R_{\mathsf {o}}}}={\frac {\ R_{\mathsf {o}}\ }{Z_{\mathsf {b}}}}\qquad \;}$ (for frequencies in the passband)

That is, the impedances behave as duals of each other within this frequency range.

In the attenuation range of the filter, the characteristic impedance of the filter is purely imaginary, and

${\displaystyle Z_{\mathsf {a}}=Z_{\mathsf {b}}\qquad \qquad \qquad \;}$ (for frequencies in the attenuation band)

Consequently, in order to achieve a specific characteristic, the reactances within Z_a and Z_b are chosen so that their resonant and anti-resonant frequencies are duals of each other in the passband, and match one another in the stopband. The transition region of the filter, where a change from one set of conditions to another occurs, can be made as narrow as required by increasing the complexity of Z_a and Z_b . The phase response of the filter in the pass-band is governed by the locations (spacings) of the resonant and anti-resonant frequencies of Z_a and Z_b .

For convenience, the normalised parameters  y_o  and  z_o  are defined by

${\displaystyle y_{\mathsf {o}}={\sqrt {{\frac {\ Z_{\mathsf {b}}\ }{Z_{\mathsf {a}}}}\ }}={\sqrt {{\frac {\ z_{\mathsf {b}}\ }{z_{\mathsf {a}}}}\ }}\qquad \;}$ and ${\displaystyle \;\qquad z_{\mathsf {o}}={\sqrt {\ z_{\mathsf {a}}\cdot z_{\mathsf {b}}\ }}={\frac {\ Z_{\mathsf {o}}\ }{R_{\mathsf {o}}}}}$

where normalised values ${\displaystyle \;z_{\mathsf {a}}={\frac {\ Z_{\mathsf {a}}\ }{R_{\mathsf {o}}}}\;}$ and ${\displaystyle \;z_{\mathsf {b}}={\frac {\ Z_{\mathsf {b}}\ }{R_{\mathsf {o}}}}\;}$ have been introduced. The parameter y_o is termed the index function and z_o is the normalised characteristic impedance of the network. The parameter y_o is approximately 1 in the attenuation region; z_o is approximately 1 in the transmission region.^[4]: 383

The shortcomings of image theory

The filter characteristics predicted by image theory require a correctly terminated network. As the necessary terminations are often impossible to achieve, resistors are commonly used as the terminations, resulting in a mismatched filter. Consequently, the predicted amplitude and phase responses of the circuit will no longer be as image theory predicts. In the case of a low-pass filter, for example, where the mismatch is most severe near the cut-off frequency, the transition from pass-band to stop-band is far less sharp than expected.

The figure below illustrates the issue: A lattice filter, equivalent to two sections of constant k low-pass filter, has been derived by image methods. (The network is normalised, with ${\displaystyle \;L=1\;}$ and ${\displaystyle \;C=1\;,}$ so ${\displaystyle \;R_{\mathsf {o}}={\sqrt {\ {\tfrac {\ L\ }{C}}\ }}=1\;}$ and ${\displaystyle \;\omega _{\mathsf {c}}=2\ {\sqrt {\ L\ C\ }}=2~.}$ The left-hand figure gives the lattice circuit and the right-hand figure gives the insertion loss with the network terminated (1) resistively, and (2) in its correct characteristic impedances.

To minimise the mismatch problem, various forms of image filter end terminations were proposed by Zobel and others, but the inevitable compromises led to the method falling out of favour. It was replaced by the more exact methods of network analysis and network synthesis.^{[10][11][12][13]}

Results derived by network analysis

This diagram shows the general circuit for the symmetrical lattice:

Through mesh analysis or nodal analysis of the circuit, its full transfer function can be found. That is,

$${\displaystyle {\frac {v_{\text{out}}}{v_{\text{in}}}}={\frac {Z_{L}(Z_{b}-Z_{a})}{(Z_{a}+Z_{b})(Z_{S}+Z_{L})+2(Z_{a}Z_{b}+Z_{S}Z_{L})}}}$$

The input and output impedances (Z_in and Z_out) of the network are given by

$${\displaystyle {\begin{aligned}Z_{\text{in}}={\frac {2Z_{a}Z_{b}+Z_{L}(Z_{a}+Z_{b})}{Z_{a}+Z_{b}+2Z_{L}}}\\[1ex]Z_{\text{out}}={\frac {2Z_{a}Z_{b}+Z_{S}(Z_{a}+Z_{b})}{Z_{a}+Z_{b}+2Z_{S}}}\end{aligned}}}$$

These equations are exact, for all realisable impedance values, unlike image theory where the propagation function only predicts performance accurately when Z_S and Z_L are the matching characteristic impedances of the network.

The equations can be simplified by making a number of assumptions. Firstly, networks are often sourced and terminated by resistors of the same value R₀ so that Z_S = Z_L = R₀ and the equations become

$${\displaystyle {\begin{aligned}{\frac {v_{\text{out}}}{v_{\text{in}}}}&={\frac {R_{0}(Z_{b}-Z_{a})}{2(Z_{a}+R_{0})(Z_{b}+R_{0})}}\\[1ex]Z_{\text{in}}=Z_{\text{out}}&={\frac {2Z_{a}\,Z_{b}+R_{0}(Z_{a}+Z_{b})}{Z_{a}+Z_{b}+2R_{0}}}\end{aligned}}}$$

Secondly, if the impedances Z_a and Z_b are duals of one another, so that Z_a Z_b = R₀², then further simplification is possible:

$${\displaystyle {\frac {v_{\text{out}}}{v_{\text{in}}}}={\frac {R_{0}-Z_{a}}{2(R_{0}+Z_{a})}}={\frac {Z_{b}-R_{0}}{2(R_{0}+Z_{b})}}\qquad {\text{and}}\qquad Z_{\text{in}}=Z_{\text{out}}=R_{0}}$$

so such networks are constant-resistance networks.

Finally, for normalised networks, with R₀ = 1,

$${\displaystyle {\frac {v_{\text{out}}}{v_{\text{in}}}}={\frac {(1-z_{a})}{2(1+z_{a})}}={\frac {(z_{b}-1)}{2(1+z_{b})}}\qquad \qquad {\text{and}}\qquad z_{\text{in}}=z_{\text{out}}=1}$$

If the impedances Z_a and Z_b (or the normalised impedances z_a and z_b) are pure reactances, then the networks become all-pass, constant-resistance, with a flat frequency response but a variable phase response. This makes them ideal as delay networks and phase equalisers.

When resistors are present within Z_a and Z_b then, provided the duality condition still applies, a circuit will be constant-resistance but have a variable amplitude response. One application for such circuits is as amplitude equalisers.

T to lattice

Pi to lattice

Common series element

Common parallel element

Combining two lattices into one

Lattice to T (see also the next section)

Shunt

This lattice-to-T conversion only gives a realisable circuit when the evaluation of (Z_b − Z_a) / 2 gives positive valued components. For other situations, the bridged-T may provide a solution, as discussed in the next section.

Synthesis with z parameters

z-parameters, or Impedance parameters, are one set from the family of parameters that define a two-port network, with input and output values defined by I₁, I₂, V₁ and V₂,^{[12]: 254 [25]: 29} as shown in the figure.

Two-port Network

Equations defining network behaviour in terms of z-parameters are

$${\displaystyle {\begin{aligned}E_{1}=z_{11}I_{1}+z_{12}I_{2}\\E_{2}=z_{21}I_{1}+z_{22}I_{2}\end{aligned}}}$$

where the z-parameters are defined under open circuit conditions (see Impedance parameters) so they are sometimes referred to as "open-circuit parameters".^[26] They are defined thus^[4]: 136

$${\displaystyle z_{11}=\left[{\frac {V_{1}}{I_{1}}}\right]{\text{ with }}I_{2}=0\qquad \qquad z_{12}=\left[{\frac {V_{1}}{I_{2}}}\right]{\text{ with }}I_{1}=0}$$

$${\displaystyle z_{21}=\left[{\frac {V_{2}}{I_{1}}}\right]{\text{ with }}I_{2}=0\qquad \qquad z_{22}=\left[{\frac {V_{2}}{I_{2}}}\right]{\text{ with }}I_{1}=0}$$

For the symmetrical lattice, the relationships between z-parameters and the lattice impedances are easily found, and they are

$${\displaystyle z_{11}=z_{22}={\frac {Z_{a}+Z_{b}}{2}}\qquad z_{12}=z_{21}={\frac {Z_{b}-Z_{a}}{2}}}$$

$${\displaystyle Z_{I}={\sqrt {Z_{a}Z_{b}}}}$$

So,

$${\displaystyle Z_{a}=z_{11}-z_{12}\,,\quad Z_{b}=z_{11}+z_{12}}$$

Sometimes synthesis of a lattice can be achieved by simply apportioning parts of an expression in z₁₂, or in z₁₁ and z₁₂, directly to the impedances Z_a and Z_b, as in the following example.

Example 1

Consider z₁₂ to be given by^[21]: 229

$${\displaystyle z_{12}={\frac {p^{5}}{(p^{2}+1)(p^{2}+2)}}}$$

This can be expanded into partial fractions, to give

$${\displaystyle z_{12}={\frac {p}{p^{2}+1}}-{\frac {4p}{p^{2}+2}}+p}$$

Allocate terms to Z_a and Z_b, accordingly, so giving

$${\displaystyle Z_{a}={\frac {8p}{p^{2}+2}}}$$

and ${\displaystyle Z_{b}={\frac {2p}{p^{2}+1}}+2p={\frac {2p^{3}+4p}{p^{2}+1}}}$

The lattice network which has these solutions for Z_a and Z_b is shown in the left-hand circuit, below. It can be converted to an unbalanced form by, firstly, extracting the common parallel inductors and, secondly, by then extracting series common capacitors. This gives the ladder network shown in the right-hand circuit.

Synthesis from the open-circuit transfer function

The open-circuit voltage-ratio transfer function T can be obtained in terms of z₁₁ and z₁₂,^[22]: 43 since with I₂ = 0

$${\displaystyle T={\frac {V_{2}}{V_{1}}}={\frac {z_{12}}{z_{11}}}}$$

so from an expression for T, which gives the ratio of z_12, and z₁₁, it may be possible to obtain circuits for Z_a and Z_b.

In practice, T may be expressed in the form

$${\displaystyle T(p)=K\cdot {\frac {N(p)}{D{p}}}}$$

where N(p) and D(p) are polynomials in p, the complex frequency variable, and K is a constant factor less or equal to unity.

For a given expression for T, it is often possible to find expressions (and hence circuits for Z_a and Z_b), provided the value chosen for K is small enough.

Now, for the lattice, ${\displaystyle T={\frac {z_{11}}{z_{12}}}={\frac {Z_{b}-Z_{a}}{Z_{b}+Z_{a}}}={\frac {1-{\frac {Z_{a}}{Z_{b}}}}{1+{\frac {Z_{a}}{Z_{b}}}}}}$

Rearranging ${\displaystyle {\frac {Z_{a}}{Z_{b}}}={\frac {1-T}{1+T}}={\frac {D-KN}{D+KN}}}$

The procedure^[24] evaluates the numerator and denominator of the expression as polynomials in p and then apportions factors to Z_a and Z_b. A loss term K, with K < 1, may be needed to aid realization.

Example 2

Derive a lattice network with voltage-ratio transfer function T₂ given by^[22]: 345

$${\displaystyle T_{2}=K\cdot {\frac {p^{2}+1}{p^{2}+5p+4}}}$$

with ${\displaystyle K=1}$ and

$${\displaystyle {\begin{aligned}D-KN&=p^{2}+5p+4-p^{2}-1=5p+3\\D+KN&=p^{2}+5p+4+p^{2}+1=2p^{2}+5p+5\end{aligned}}}$$

$${\displaystyle {\frac {Z_{a}}{Z_{b}}}={\frac {5p+3}{2p^{2}+5p+5}}={\frac {1+3/5p}{1+(2p^{2}+5)/5p}}}$$

Choose ${\displaystyle Z_{a}=1+{\frac {3}{5p}}}$ and ${\displaystyle Z_{b}=1+{\frac {2p}{5}}+{\frac {1}{p}}}$

The lattice realization of T₂ is shown below, on the left. The unbalanced network, on the right, is obtained by first extracting the common series resistors and then extracting capacitance.

Example 3

An L-C circuit has a transfer function T₃ given by

$${\displaystyle T_{3}=K\cdot {\frac {(p^{4}+1)}{(p^{2}+1)(p^{2}+2)}}}$$

This is realizable with K = 0.05,^[24] so

$${\displaystyle {\frac {D-KN}{D+KN}}={\frac {0.95(p^{4}+3.1579p^{2}+2.0526}{1.05(p^{4}+2.8571p^{2}+1.9524}}={\frac {Z_{a}}{Z_{b}}}}$$

Factorizing top and bottom gives

$${\displaystyle {\frac {Z_{a}}{Z_{b}}}={\frac {0.95(p^{2}+0.9153)(p^{2}+2.2426)}{11.05(p^{2}+1.1312)(p^{2}+1.7259)}}}$$

Choose, say,

$${\displaystyle Z_{a}={\frac {0.9048(p^{2}+2.2426)}{(p^{+}1.1312)}}\qquad {\text{and}}\qquad Z_{b}={\frac {p(p+1.7259)}{p^{2}+0.9153}}}$$

Z_a and Z_b can be realized as LC ladder networks, with Z_a having a shunt inductor as first element and Z_b having a series inductor as first element, as shown in the left-hand figure. This lattice can be converted to unbalanced form, by the methods given earlier, to give the component values of the right-hand figure,

Darlington synthesis

The Darlington method forms the basis for synthesis of lossless two terminal-pair networks with resistive termination for prescribed transfer characteristics.^[27][10]

Network for Darlington's Method

The figure shows the basic network configuration. The associated transfer impedance is

$${\displaystyle T={\frac {E_{2}}{I_{1}}}}$$

The first step is to express the input impedance Z_I of a terminated network in terms of its z-parameters. This is ^[21]

$${\displaystyle Z_{I}={\frac {(z_{11}z_{22}-z_{12}^{2})+z_{11}R}{z_{22}+R}}}$$

in which z₁₁, z₂₂ and z₁₂ are z-parameters of the network, as defined earlier. For a normalized network, put R = 1, and rearrange the expression thus:

$${\displaystyle Z_{I}=z_{11}\cdot {\frac {(z_{11}z_{22}-z_{12}^{2})/z_{11}+1}{z_{22}+1}}}$$

In practice, Z_I consists of a ratio of two polynomials in p:

$${\displaystyle Z_{I}={\frac {N(p)}{D(p)}}={\frac {m_{1}+n_{1}}{m_{2}+n_{2}}}}$$

where m₁ and n₁ are the even and odd parts of the numerator polynomial, respectively and m₂ and n₂ are the even and odd parts of the denominator polynomial, respectively.

Rearranging ${\displaystyle Z_{I}={\frac {m_{1}}{n_{2}}}\cdot {\frac {(n_{1}/m_{1})+1}{(m_{2}/n_{2})+1}}}$

By comparing the two expressions for Z_I, the following relationships are suggested

$${\displaystyle z_{11}={\frac {m_{1}}{n_{2}}}\qquad \qquad z_{22}={\frac {m_{2}}{n_{2}}}\qquad \qquad z_{12}={\frac {\sqrt {m_{1}m_{2}-n_{1}n_{2}}}{n_{2}}}}$$

Example 4

Consider a network with Z_I given by

$${\displaystyle Z_{I}={\frac {2.1351p^{3}+1.915p^{2}+2.4969p+1}{1.327p^{2}+1.18p+1}}}$$

So,

$${\displaystyle {\begin{aligned}m_{1}&=1.915p^{2}+1&n_{1}&=2.1351p^{3}+1\\m_{2}&=1.327p^{2}+1&n_{2}&=1.18p\end{aligned}}}$$

So solutions for z₁₁, z₂₂ and z₁₂ are

$${\displaystyle z_{11}={\frac {1.915p^{2}+1}{1.18p}}=1.6229p+{\frac {1}{1.18p}}}$$

i.e. z₁₁ is an inductor of 1.6229H in series with a capacitor of 1.18F.

$${\displaystyle z_{22}={\frac {1.327p^{2}+1}{1.18p}}=1.1246p+{\frac {1}{1.18}}}$$

i.e. z₂₂ is an inductor of 1.1246H in series with a capacitor of 1.18F

$${\displaystyle {\begin{aligned}z_{12}&={\frac {\sqrt {(1.915p^{2}+1)(1.327p^{2}+1)-(2.1351p^{3}+2.4969p)}}{1.18p}}\\&={\frac {\sqrt {0.0218p^{4}+0.2957p^{2}+1}}{1.18p}}\quad ={\frac {0.1479p^{2}+1}{1.18p}}\quad =0.1253p+{\frac {1}{1.18p}}\end{aligned}}}$$

By extracting a series inductance of 0.4983p = (1.6229p – 1.1246p) from z₁₁, the remaining network becomes symmetrical with

$${\displaystyle {\begin{aligned}z_{11}^{\text{(new)}}=z_{22}=1.1246p+{\frac {1}{1.18p}}\\z_{12}=0.1253p+{\frac {1}{1.18p}}\end{aligned}}}$$

The components of a symmetrical lattice can be calculated from Z_a = z₁₁ − z₁₂ and Z_b = z₁₁ + z₁₂.

So

$${\displaystyle Z_{a}=1.1246p+{\frac {1}{1.18p}}-0.1253p-{\frac {1}{1.18p}}=0.9993p,}$$

i.e. an inductor of 0.9993H. and

$${\displaystyle Z_{b}=1.1246p+{\frac {1}{1.18p}}+0.1253p+{\frac {1}{1.18p}}=1.2499p+{\frac {2}{1.18p}},}$$

i.e. an inductor of 1.2499H in series with a capacitor of 0.59F

The circuit is shown in the left hand figure below. It can be easily converted into the unbalanced form shown in the right hand figure. It is a low-pass filter with pass-band ripple of 1.25 dB, with −3 dB at 0.169 Hz, a null in the stop band at 0.414 Hz, and stop-band attenuation beyond the null frequency below −40 dB.

Synthesis of constant-resistance lattice networks

If the impedances Z_a and Z_b are duals, and normalised, so that

$${\displaystyle Z_{a}Z_{b}=1}$$

then the image impedance Z_I becomes a pure resistance. A symmetrical lattice fulfilling this condition is a "constant resistance lattice".

Such a lattice, terminated in 1 ohm, is shown below.

Terminated Constant-R Lattice

This has the transfer function

$${\displaystyle {\frac {E_{2}}{E_{1}}}={\frac {E_{2}}{I_{1}}}=T={\frac {1-Z_{a}}{1+Z_{a}}}}$$

in which T is the transfer impedance with a 1-ohm load in contrast to the open-circuit transfer impedance z₂₁. Rearranging this, gives

$${\displaystyle Z_{a}={\frac {1-T}{1+T}}}$$

The constant resistance lattice is thus seen to offer a possible approach to the synthesis of transfer functions.

It is the case that a constant resistance lattice is no less general than any other lattice, which means that any realizable transfer impedance can be realized in the form of a constant resistance lattice,.^{[20]: 233 [21]: 480} Such networks are very convenient, because there is no mismatch between sections or with resistive terminations. Consequently, the overall insertion loss of a cascade of constant resistance sections is simply to sum total of the individual sections. Conversely, a given complicated transfer impedance may be decomposed into multiplicative factors, whose individual lattice realizations, when connected in cascade, represent a synthesis of that transfer impedance. So, although it is possible to synthesize a single lattice with complicated impedances Z_a and Z_b, it is practically easier to construct and align a cascade of simpler circuits.

Example 5

The all-pass response of the first order is

$${\displaystyle T_{5}={\frac {c-p}{c+p}}}$$

This has a zero located at +c and a pole at –c in the complex frequency plane. It has a response where the phase varies with frequency, but the magnitude of T₅ is unity at all frequencies.

Using the expression for Z_a as a function of T, from earlier, gives

$${\displaystyle Z_{a}={\frac {1-T_{5}}{1+T_{5}}}={\frac {p}{c}}}$$

So Z_a is an inductance with value 1/c and, consequently, Z_b is a capacitor of value 1/c. The network, normalised to 1 ohm, is shown in the left-hand figure below.

Example 6

The all-pass response of the second order is

$${\displaystyle T_{6}={\frac {p^{2}-ap+b}{p^{2}+ap+b}}}$$

This has two zeroes located at ${\displaystyle x\pm y}$ and two poles at ${\displaystyle -x\pm y}$ where a = 2x and b = x² + y². For such a response, the phase varies with frequency, but the magnitude of T₆ is unity at all frequencies.

For this characteristic, Z_a is found from

$${\displaystyle Z_{a}={\frac {1-T_{6}}{1+T_{6}}}={\frac {ap}{p^{2}+b}}}$$

So Z_a is a parallel combination of a capacitance 1/a and an inductance with value a/b. Similarly Z_b is an inductor 1/a in series with a capacitor of value a/b and the network is shown at the right hand side below.

The lattice networks can be converted to unbalanced circuits by using the properties of lattices with common elements in both Z_a and Z_b, shown earlier, and Bartlett’s Bisection theorem.^[16]: 28

First order all-pass

In the case of the second order network, when a2 > b (i.e. L1 > L2 or C2 > C1 or y > √3x), it is necessary to use the circuit containing mutually coupled coils for the second order all-pass network.

A cascade of second order networks with, maybe, a single first order network, can be used to give a high order response. For example, the article Lattice delay network gives pole-zero locations for many all-pass transfer functions which approximate to a linear phase characteristic. That article also includes some examples.

Synthesis of amplitude equalizers

A typical transmission path has increasing loss with frequency and this can be corrected by cascading the system with an equalizing network that has a rising response with frequency. In this regard, one circuit configuration that is commonly used to provide the necessary equalization is shown in the figure labelled 'Lattice - basic equaliser circuit', given earlier (in the section on 'Unbalanced Equivalents'). As stated there, the insertion loss of the normalized circuit is given by ${\displaystyle T={\frac {1}{1+Z_{1}(p)}}}$, so Z₁ can be found from ${\displaystyle Z_{1}(p)={\frac {1}{T(p)}}-1}$

If some residual ripple on the response is permitted, then a simple correcting network may suffice for Z₁ and Z₂, but this ripple may be reduced as much as desired by adopting more complicated correcting networks. Choosing locations for the poles and zeros for Z₁ and Z₂ may be aided by the straight line asymptotic method.^[28]

Example 7

A transfer function which has a rising response over a limited frequency range is

$${\displaystyle T_{7}(p)={\frac {3.3333+10.3333p+p^{2}}{30.003+31.003p+p^{2}}}={\frac {(0.3333+p)(10+p)}{(1+p)(30.003+p)}}}$$

Note that the response approaches unity at high frequencies. It can be realized as a bridged-T or lattice in which Z₁ is an R-C network.

Z₁ can be found from ${\displaystyle Z_{1}(p)={\frac {1}{T(p)}}-1}$. So ${\displaystyle Z_{1}(p)={\frac {26.6697+20.6697p}{3.3333+10.3333p+p^{2}}}}$

The admittance Y₁, where Y₁ = 1/Z₁ can be expressed as a continued fraction containing four terms, thus

$${\displaystyle Y_{1}(p)={\frac {p^{2}+10.333p+3.3333}{20.6697p+26.6697}}=0.04838+{\cfrac {1}{2.2857+{\cfrac {1}{0.4747p+{\cfrac {1}{5.7153}}}}}}}$$

So Z₁ can be realized as an R-C ladder network, in the Cauer manner,^[21] and is shown as part of the bridged-T circuit below. Z₂ is the dual of Z₁, and so is an R-L circuit, as shown. The equivalent lattice circuit is shown on the right–hand side.

Constant resistance low-pass filters

High order low-pass filters can be obtained by cascading an appropriate number of simpler constant resistance low-pass sections.^[21]: 484

The first of these low-pass sections, with just a single pole, has the response

$${\displaystyle T_{1}={\frac {k_{1}}{p+a}}\qquad so\qquad Z_{a1}={\frac {1-T_{1}}{1+T_{1}}}={\frac {p+(a-k_{1})}{p+(a+k_{1})}}}$$

Provided ${\displaystyle k_{1}\leq a}$ this is realizable impedance, where Z_a1 is a combination of two resistors and an inductor, as shown in the left-hand circuit below, and Z_b1 is the dual of Z_a1. This is easily transformed into an unbalanced form, as shown on the right.

The second of the filter sections, with two poles, has the response

$${\displaystyle T_{2}=k_{2}{\frac {(p+a)}{(p^{2}+b_{1}p+b_{0})}}}$$

So the lattice impedance Z_a2 is given by:

$${\displaystyle Z_{a2}={\frac {p^{2}+(b_{1}-k_{2})p+(b_{0}-a.k_{2})}{p^{2}+(b_{1}+k_{2})p+(b_{0}+a.k_{2})}}}$$

Certain conditions have to be met to ensure this is a realizable network,^[21]: 486 which are

$${\displaystyle a\geq 0;\qquad k_{2}\leq b_{1};\qquad k_{2}\leq {\frac {b_{0}}{a}}.}$$

Also

$${\displaystyle b_{1}-k_{2}\iff k_{2}\leq b_{1}-a.}$$

The conditions set limits on the value of the constant multiplier k₂ in the expression for T₂.

The circuit for the lattice elements Z_a2 is shown on the left, below, and that for the dual elements Z_b is shown on the right.

Impedance Z_a

Impedance Zb

Component values for Z_a are,

$${\displaystyle {\begin{aligned}R_{1}&={\frac {b_{1}-k_{2}-a}{2k_{2}}}&L&={\frac {1}{2k_{2}}}\\C&={\frac {2k_{2}}{b_{0}-ab_{1}+a^{2}}}{2ak_{2}}&R_{2}&={\frac {b_{0}-ab_{1}+a^{2}}{2ak-2}}\end{aligned}}}$$

and those for the impedances Z_b2 are:

$${\displaystyle {\begin{aligned}R_{3}&={\frac {1}{R_{1}}}\\R_{4}&={\frac {1}{R_{2}}}\\C_{2}&=L_{1}\quad {\text{and}}\quad L_{2}+C_{1}\end{aligned}}}$$

The unbalanced version of this lattice is as shown below:

By cascading a number of the first and second order circuits, of the type just developed, it is possible to derive higher order low-pass networks of the type:

$${\displaystyle T_{n}(p)={\frac {k}{p^{n}+b_{n-1}p^{n-1}+\cdots +b_{2}p^{2}+b_{1}p+b_{0}}}}$$

The lattice networks so obtained can be converted to an unbalanced form, provided the value of k is sufficiently small.

Example 8

A maximally flat third-order normalized low pass filter has the transfer function

$${\displaystyle T_{8}={\frac {k}{p^{3}+2p^{2}+2p+1}}}$$

This can be expanded as

$${\displaystyle T_{8}={\frac {k_{1}}{p+a}}\cdot {\frac {k_{2}(p+a)}{p^{2}+p+1}}\cdot {\frac {1}{p+1}}}$$

So a cascade of three lattices will give the required result.

If an unbalanced circuit is required, we have to accept some overall loss. By choosing k₁ = k₂ = a = 0.5, then the network shown below is obtained. This circuit has an overall loss of four times, whereas the conventional L-C ladder network^[1]: 605 has no loss (but is not a constant resistance network).

Computer-aided design methods

The development of mainframe and then personal computers, in the final quarter of the twentieth century, permitted the rapid development of numerical processing techniques. Initially, computers were used as an aid to network analysis^[29] then to optimization methods such as the minimax method,^[30] in the design of phase equalizers^[31] and filters^[32]), before being applied to network synthesis directly. Overviews of the software developments in the field of synthesis have been given in Taylor & Huang^[33] and Kuo.^[12]: 438

Only a few of the early synthesis programs have dealt with lattice networks, but S-Filsyn (a powerful synthesis and analysis program^[34] ) provides some coverage of lattice and bridged-T circuits.