Synthesis (Part 2) page 2

Suggested Bibliography

A. Papoulis “Circuits and Systems - A Modern Approach”, Polytechnic Institute of
New York, 1980.

H. Baher: Synthesis of Electrical Networks, New York: J. Wiley, 1984.

W-K. Chen “Passive and Active Filters – Theory and implementations”, Wiley, 1986.

M.E. van Valkenburg “Analog Filter Design”. HRW, 1982. (Good for the active filters
part of the course).

C.S. Williams “Designing Digital Filters”, Prentice Hall, 1986.

T. Kurowski, S. Taranow , “Selected Issues in the Theory of Linear and Non-Linear
Systems", University of Zielona Gora, 2004 (English-Polish). 318493.

additionally, the following books can be useful as reference material:

F.F. Kuo “Network Analysis and Synthesis”. Wiley, 1966. (Useful for Passive network
synthesis).

P. Bowron and F.W. Stephenson. “Active filters for communications and
instrumentation”, McGraw Hill

R.W. Hamming “Digital Filters”. Prentice Hall, 1989. (Useful as introduction to digital
filters).
G.G. Temes and W. LaPatra. “Circuit synthesis and design”, McGraw Hill, 1977.

A. Budak. “Passive and active network analysis and synthesis”, Houghton Mifflin,
1974.

R. Bracewell. “The Fourier transform and its applications”, McGraw Hill, 1978.

Synthesis (Part 1)

INTRODUCTION

What is synthesis?
Let’s start with analysis which seems simpler to understand:
A typical “analysis problem” is represented as:

INPUT CIRCUIT OUTPUT
(given) (given) (required)

• There is always one answer (existence theorem)
• Invariably this answer is unique. (uniqueness theorem)

Now with synthesis the situation is far less clear. In this case we have:

INPUT CIRCUIT OUTPUT
(given) (required) (given)

We now have that:

• Sometimes there is no answer and
• Often the answer is_not unique.

In analysis there are standard tools for a systematic approach:
• circuit theory
• Fourier/Laplace transforms
• convolution.

In synthesis there are far too many design approaches to enumerate here, and
plenty of scope for anyone to invent new ones. It is a less clearly defined problem and
then, there are many ways to approach a solution. There is need of judgement as well
as calculation. It is (arguably at least) more difficult, more open-ended, more
interesting than analysis. Yet analysis remain vital - to test the design ideas before
putting them into practice.

Because so much of synthesis is concerned with filter design, it is easy to suppose
that all design is restricted to the frequency domain. Not so. Often there are clear
requirements in both domains, and usually they conflict. For example, a filter’s
performance may be specified as a particular pass-band in the frequency domain, and


at the same time it may be required to have a restricted transient response. This could
occur in a radar system where bursts of oscillation are used, and the response to one
burst must die away before the next pulse is received.
Usually the requirements are in conflict and a synthesis procedure consists in
finding a compromise. In the last example, the better the discrimination is made in the
frequency domain, the worse is the transient response.

Examples of the need for synthesis

(1) A smoothed DC power supply:
The problem is by no means a trivial one. The response of the circuit to a sine
wave is required to be a constant output.

V V

CIRCUIT
t (required) t

INPUT OUTPUT
(given) (given)

(2) A bandpass filter:
In this case the circuit only lets through frequency components within a certain
frequency range.

V V

CIRCUIT
f (required) f0 f

INPUT SPECTRUM OUTPUT SPECTRUM
(given) (given)

(3) An equalizer (which is a circuit used to compensate for an undesired distortion
or loss, e.g. to compensate for cable attenuation in a transmission system):

Synthesis (Part 1)

V V

...
...
CABLE
(given)
t t
INPUT
(given)

...
... EQUALIZER
(required)
OUTPUT t
(given)

The diagram is trying to convey the idea that a pulse train should ideally be
transmitted through the cable undistorted, but the cable characteristics are such that
different frequency components are transmitted with different velocities so that they
arrive at different times giving a distorted waveform (dispersion). The equalizer is a
circuit whose characteristics are the inverse of the cable characteristics - leading, we
hope, to a faithful pulse train at the receiving end.

(4) Analysis of sampled data:
This is at first sight a less obvious candidate for synthesis procedures. Yet the
concept of a filter is very general and very powerful. When we calculate the average of
a set of readings (measurements), we are “smoothing out the fast variations” (high
frequency components) — we are finding the “DC component” and this process is a
numerical low-pass filter. The analogy with frequency-domain filters becomes closer
when we envisage working out, say, the average of the last five temperature readings in
a series of continuously sampled measurements. Or for instance when the annual
inflation rate is updated every month. The general theme of digital filtering
encompasses any linear operation on the data, and includes integration and
differentiation.

(5) I.F. filter for a TV receiver
In this case, as in most practical cases, the output is not fixed specifically. Instead
we only fix some restrictions to its shape. In all practical cases we are not interested in
obtaining a predetermined fixed function as filter characteristic. Instead, we want an
approximation to an ideal case that satisfy certain restrictions. This is particularly
suitable since as the problem of synthesis not always has an exact solution, this has to
be found as an approximation within a set of “realizable functions”.


i.f.
Tuner i.f. Filter
Amp.

0

?0

?0
Attenuation [dB]

?0

?0

?0

?0

?0

28 30 32 34 36 38 40 42 44 46

Frequency [MHz]

25
Group Delay, [ns]

0

?5

Tolerance masks for the amplitude response and for the group-delay of a television i.f.
filter. The figure also shows a typical response (amplitude and phase) satisfying the
requirements.

Synthesis (Part 1)

The available elements for the synthesis

These are: • Passive circuits,
• active circuits and
• digital filters.

FILTERS
(linear)

ANALOGUE DIGITAL

PASSIVE ACTIVE

Passive circuits:
Components: These are restricted to the use of normal L,C and R electrical components
(lumped or distributed) for the use in filters, equalizers, etc.

Causality: The response of these networks is determined by the signal information
reaching them up to the output instant, i.e., they are limited by causality: the effect
cannot precede the cause.

Limited memory: Another restriction is that of severely limited memory: The energy
dissipation time constants place restrictions on how much data from the past can
influence the output at t = 0, i.e. the present.

Stability: Another restriction (which is also jω
sometimes an advantage of passive circuits) is
stability; i.e. they do not have energy sources
and consequently there cannot be growing
oscillations. This restriction means that poles σ
and zeros of immittances (impedances or
admittances) must be in the left-hand half-
plane.
On the positive side, passive circuits are
simple, reliable, stable, and they can handle


high powers compared with circuits with op-amps.
On the negative side, inductors, which are often necessary, are not compatible
with integrated circuit technology, where circuits are made more or less in a plane.
Inductors are bulky and 3-dimensional. Another fundamental limit arises because, if
you scale an inductor down in size, the Q-factor falls; a 10:1 linear scaling gives a 100:1
reduction in the Q-factor.

Active Circuits
Components: Active circuits include an energy source, such as an amplifier, as well as
resistors and capacitors. Inductors can and usually are avoided.

Active circuits are not limited by stability, – e.g. they can have an exponentially
growing response (within certain limits). This means that poles and zeros of
immittance functions can be located anywhere in
the complex s-plane. jω

They can be made without the need of using
inductors. This is particularly useful since in this
way they are compatible with I.C. technology. σ
Capacitors can be scaled down without affecting
the Q-factor.

For example, a narrow-band filter, which
would need inductors as a passive circuit, can be
made as:
β 

β

where
A0
ω0 ω

V2 A0
This leads to a frequency response: =
V1 1 − βA0
where β is for instance, a twin-T network using only
resistors and capacitors. A0
Q ≈ A0/4 for this circuit.

Active filters can be made to behave as if they ω0 ω
contained negative resistance, capacitance and

Synthesis (Part 1)

inductance. A negative resistance can be used to “cancel” losses and in the same way,
negative capacitance can be used to “cancel out” unwanted capacitance, – and this is
better than cancelling by the use of an inductor (tuning for resonance) because the latter
would only work at one frequency. Negative R, L, and C can be simulated with a
circuit:

Active
Z1 = − Z2 Z2
circuit

Another useful device is an active circuit which “looks like an inductor” if the
other port is terminated with a capacitor. It can effectively simulate inductors with
capacitors, – very convenient for integrated circuits.

Active
Z 1 = 1/ Z 2 Z2
circuit

(gyrator)

Digital Filters
Digital filters are algorithms for digital computers or circuits. Digital filtering is
any linear operation performed on data which has been sampled at equally spaced
intervals.
It includes smoothing (averaging), integrating, separating signals (filtering) and
predicting.
Examples are the Fourier transform (or Fast Fourier Transform), important in
signal processing work, the Simpson rule or the trapezoidal rule of integration, the
central difference formula of numerical derivative, etc. All these can be regarded as
digital filters.
We can also find equivalent digital filters to analogue filters. Low-pass and other
analogue filters have their digital counterparts. But digital filters have additionally
some special properties which make them well suited for digital communication
systems, especially when large distances are involved. It is then that analogue systems
are at a particular disadvantage because the attenuation continuously degrades the
signal-to-noise ratio. In digital systems, the signal is completely regenerated at intervals
and then re-transmitted with no loss of information.
There are no impedance-matching problems in the digital domain. Also, two or
more digital filters can have genuinely identical characteristics. These filters are also
programmable so that their characteristics can be changed easily and rapidly - even
almost continuously if needed.


Digital filters can have long memories if required. Initial conditions, far from
dying away, can be stored indefinitely if needed. and the accuracy can be arbitrarily
large, limited only by the word length and rounding/truncation error in the computer.
In some applications they can also “see the future”. For many data-processing
applications, the information is all available on magnetic tape (or any other form of
computer storage) before the calculations begin. In that sense, the filter algorithm can
be written so as to take data from after as well as before the sampling interval for which
the output is being calculated. Thus they are not limited by causality or stability.

Synthesis (Part 1)

SYNTHESIS OF PASSIVE CIRCUITS
Introduction:

The only types of elements allowed are R, L and C (lumped or distributed). The
constraints of stability and causality lead to mathematical conditions on the form of Z(s)
and Y(s).

Immittance is the word used generically for impedances and admittances.
In general, an immittance can be a driving point immittance or a transfer immittance.

I1 I2
U1/I1 : a driving point impedance

U2/I1 : a transfer impedance U1 U2

Basic Ideas on Synthesis
Given the excitation E(s) to a system and the response R(s), the desired system
function H(s) is:
R(s)
H (s ) =
E (s )

which may be a driving point immittance, a transfer immittance or a voltage or current
ratio.
We now have to:
• See whether this function is realizable as a passive network,
• if so, find the network or
• find an approximation.

For example, we might have H(s) = I(s)/U(s) = 3 + 4s in which case the realization
is simple:

Y(s) 4 1/3


Synthesis of 2-ports:
The corresponding synthesis problem for a 2-port is transfer function design (filter
design). A transfer function is a system function for which the variables are defined at
different ports. e.g. if I1(s) is the excitation (input) and U2(s) is the response (output), the
system function to design is the transfer impedance:

Z21(s) = U2(s)/I1(s)

The most important aspect of transfer function synthesis is filter design. For
example, an ideal amplitude response for a low-pass filter is:

H  'ideal' H  approximations

ω ω

As this “ideal” response is not realizable, this leads to the need for finding the
“best” approximation.

Synthesis of Driving Point Immittances: (one-port circuits)

Physical Realizability of a D.P.I.

A network with a finite number of elements (R, L, C) has a driving point immittance of
the form:
a sn + an−1s n−1 + L + a1s + a0
Z(s) (or Y( s)) = n m (1)
bm s + bm−1sm −1 + L + b1 s + b0

where all coefficients ai and bi are real.

Also, due to the constraints imposed by causality and stability, Z(s) (or Y(s)) must
be what is called a positive real (p.r.) function; i.e.:

Re{ Z(s)} > 0 if Re{ s} > 0 (2)

Re{ Z(s)} ≥ 0 if Re{ s} = 0 (3)

These conditions ((1),(2) and (3)) form necessary and sufficient conditions for the
realizability of Z(s) (or Y(s)) with R, L and C components only.

Synthesis (Part 1)

The above set of conditions is very concise and constitutes an elegant way of
defining realizability, but unfortunately it is not very useful from a practical point of
view. To test for a rational function to be p.r., requires to analyse its behaviour for all
values of s ≥ 0, which is not simple.

If we re-write (1) in factorized form:

(s − z1 )(s − z2 )L (s − zn )
Z(s) = K (4)
(s − p1 )(s − p2 ) L (s − pm )

A more convenient set of conditions can be obtained analysing the general
properties of p.r. functions. Some of these are:

I) The constant K must be real and positive.
II) n–m ≤ 1 (Also the lowest powers of numerator and denominator may differ
at most by 1).
III) Poles (and zeros) must be either real or appear in conjugate pairs.
IV) Poles (and zeros) must lie on the left-hand half-plane or on the imaginary axis.
V) Poles on the imaginary axis must be single (but in conjugate pairs) and must have
positive, real residues.
VI) Re{Z(jω)} ≥ 0

Additionally, the sum of two p.r. functions is also p.r. (not necessarily the
difference), and if f(s) is p.r., so is 1/f(s). Perhaps the simplest property to verify is that
if we have a p.r. function written in the form (1), then all coefficients should be real and
positive. Also, if the numerator or denominator are not odd or even polynomials, no
coefficient corresponding to intermediate powers can be missing.

There is some redundancy in the above set but again, conditions I, IV, V and VI
constitute a set of necessary and sufficient conditions and this expanded set is much
easier to verify.

Condition I can be easily verified. The testing of condition IV is a bit more
complicated:
In this case, we have that for a rational function to have no poles (or zeros) in the
right-hand half space, both denominator and numerator must be of a class of
polynomials called ‘Hurwitz’. A ‘Hurwitz’ polynomial is defined by the
conditions:

1) P(s) real when s is real.
2) The roots of P(s) are in the left-hand half-plane or on the imaginary axis.

There is a simple way to test whether a polynomial is Hurwitz or not.


If we write P(s) = m(s) + n(s) where m(s) is the even part of p(s) and n(s) is the odd
part, then the continued fraction expansion of the rational function:

m (s ) 1
R(s) = = q1 (s) +
n( s ) 1
q2 (s ) + has only positive coefficients qi.
K
1
+
qn ( s )

(We will see more about continued fraction expansions later).

(Of course we can get the same result if we factorise the function to form (4) where
the position of each pole and zero are evident).
To check condition V we need a partial fraction decomposition and then we can
check the residues. Finally, condition VI is simply tested examining the form of F(jω).
All this constitutes a rigorous test of realizability. However, the easiest way to
assess realizability of a function is simply trying to realize it.

Examples

as2 + b b
a) Z(s) = = as + clearly realizable as a combination of L and C.
s s
zeros are ±j b a and poles at 0 and ∞ .

b) Z(s) = s + 3 This is OK, (series connection of R and L) one zero at
–3 and a pole at infinity.

c) Z(s) = s + j NO, condition III is not satisfied.

d) Z(s) = s2 + 1 = (s + j)(s – j) NO, condition II is not satisfied (n = 2, m = 0).

s2 + 1
e) Z (s ) = NO, condition V is not satisfied.
s3

Methods of Synthesis of One-Port Networks
We will treat this case first because it is simpler than the synthesis of 2-ports. In
addition, the synthesis of 2-ports can be usually reduced to the synthesis of a d.p.i. In
fact, normally the specification for a 2-port circuit is in the form of a prescribed shape of
Vout/Vin as a function of frequency and this leads to the network requiring a particular
d.p. impedance or admittance in the form of a ratio of polynomials.
There are (at least) two standard approaches to the synthesis of d.p.i.: Foster and
Cauer methods.

Synthesis (Part 1)

Cauer synthesis:
The procedure is based on the continued fraction expansion of the quotient of
polynomials and leads to networks of the form:

Foster synthesis:
This is based in a partial fraction expansion and then, in the case of an impedance,
the resultant network is a series connection of elements (or blocks formed with the
parallel connection of two elements) as in:

Similarly, in the case of admittances, the corresponding network is a parallel
connection of single elements (or pairs of elements in series).
In the following we will only be concerned with the Cauer synthesis.

Cauer Synthesis
We start analysing a ladder network where each box represents either R, sL or sC
or their corresponding reciprocals.

Z1 Z3 Z5

Z(s) Y2 Y4 Y6

By inspection, we have:

1
Z(s) = Z1 +
1
Y2 +
1
Z3 +
1
Y4 +
1
Z5 +
Y6

where the quotients are alternately impedances or admittances, moving from left to
right into the ladder network. Since the impedance was specified, the first term must be
an impedance, and the ladder starts with a series element. If the admittance were


required in the form of a continued fraction, the first term would be an admittance and
the ladder network would start with a shunt element.

Obs. Every realizable d.p. immittance can be realized as
a ladder network.

Now, to synthesize a ladder network from a given driving point immittance, we
have to express Z(s) or Y(s) as a continued fraction. Each quotient represents a
physically realizable component.

Example

s4 + 4s2 + 3
Given the (realizable) impedance function: Z (s ) =
s3 + 2s

express it in the form of a continued fraction and so derive the ladder network, giving
all component values.
The first point to note is that there can be more than one continued fraction (and
consequently more than one ladder network) for the required Z(s). Additionally, there
may be some decompositions in the form of continued fraction which may not
correspond to a physically realizable network.

a)We first take numerator and denominator in descending powers of s. We obtain
the continued fraction expansion by repeated division and inversion:

s4 + 4s 2 + 3 2 s2 + 3
Z (s ) = = Z1 + Z' 2 = s + 3 Z1 = s must be an impedance!
s3 + 2s s + 2s

1 s3 + 2s s (1/ 2) s
Now, Y' 2 = = 2 = Y2 + Y' 3 = + 2 Y2 = s/2
Z' 2 2 s + 3 2 2s + 3

1 2 s2 + 3 3
and again, Z' 3 = = = Z3 + Z4 = 4s + Z3 = 4s
Y' 3 (1/ 2) s (1/ 2) s

and finally, Y4 = s/6.

The complete expansion is then:
1
Z (s ) = s + 1
(1/ 2)s +
1
4s +
(1/ 6)s

and the corresponding network:
1 4

Z(s) 1/2 1/6

This process of successively extracting factors proportional to s is called
“removing the poles at infinity”.

b) An alternative synthesis comes from arranging the terms in ascending powers of s,
or “removing the poles at the origin”:

3 + 4s 2 + s4 3 / 2 (5 / 2) s2 + s4
Z (s ) = = + L, etc.
2s + s 3 s 2 s + s3

So the expansion and the corresponding
circuit realization is: 2/3 2/25

3/ 2 1
Z(s) = + 4/5 1 Z(s) 5/4 5
s +
s 25 / 2 1
+
s 1/ 5
s

The Euclid Algorithm
This is not more than a methodical style of book-keeping, that organizes the
continued-fraction algebra.
In general, to find the continued fraction of a d.p.i. function there are always 4
ways of obtaining the expansion, of which 0, 1 or (at most) 2 can be successful:
If we have a d.p.i. in the form of an impedance as in:

Z(s) = F(s), we can as before try in two ways, taking numerator and denominator
in decreasing or increasing order of the powers of s. Also, we can invert the function
taking instead:
Y(s) = 1/F(s) (which of course should correspond to the same d.p. impedance as
before) and try again in the two forms of ordering the powers of s.

6s3 + 4s
If we consider the function Y (s) =
6s2 + 1

the solution with the Euclid algorithm gives:

Y of shunt (a) Z of series
elements elements
s 6s3 + 4s : 6s2 + 1
6s3 + s
3s 6s2 2s
3s 3s 1
0

and the corresponding circuit is:
2

Y(s) 1 3

Now, reordering the powers of s in ascending order:

Y of shunt (b) Z of series
elements elements

4s 4s + 6s3 : 1 + 6s
2

4s + 24s3
3
− 18s

there is no solution since a negative residual is obtained.

If we try now inverting to realize the corresponding Z(s):

Y of shunt (c) Z of series
elements elements

6s2 + 1 : 6s3 + 4s 1/s
6s2
+ 4
− 3
it does not work.

Finally, taking the ascending order in the inverted form:

Y of shunt (d) Z of series
elements elements
1 + 6s2 : 4s + 6s3 1/ (4s)
1 + (6 / 4)s2
8 / (9s) (18 / 4)s2 4s
(18 / 4)s2 6s3 3 / (4s)
0

And the corresponding circuit is:
4 4/3

Z(s) 9/8

Note that we have two possible realizations, both with identical response.

Exercise
6s3 + 8s2 + 4s + 4
Find ladder network(s) corresponding to: Z(s) =
6s 2 + 8s + 1

All of the previous examples correspond to circuits containing only reactive
elements. As we will see later, the realization of circuits containing resistance can be
done following a similar procedure although it is a bit more complicated.

Driving Point Immittance of a Reactive (L, C) Network
If there are only L and C elements in the network, the d.p.i. has some special
properties (or additional restrictions).
This kind of circuits (without R) is important in filter design where a transfer
function without loss is usually desired.
For these circuits, the realizability conditions are more restrictive than for general
RLC networks.

L – C Realizability conditions:

1) Z(s) or Y(s) is the ratio of either an even to an odd polynomial or vice versa.

2) All poles and zeros alternate on the jω axis.

3) All poles and zeros are simple and occur in conjugated pairs.

4) The residues at the poles lim{ (s – pi)Z(s)} must be real and positive.

5) The highest powers of s (and the lowest) in the numerator and denominator must
differ exactly by 1.

6) There must be either a pole or a zero at the origin. (Obviously, the same can be
said for infinity.)

There is visibly a lot of redundancy in the above constraints. A simpler and more
concise statement comes from considering Z(s) (or Y(s)) in a factorised form.

Algebraic Statement of LC Realizability
The d.p. immittance must be of one of the four types:

s(s2 + a1 )(s 2 + a2 ) L (s2 + an )
2
2
2
1) K with 0 < b1 < a1 < … < bn < an
2 2 2
(s 2 + b1 )(s2 + b2 ) L (s2 + bn )
zero at the origin, pole at infinity.

s(s 2 + a1 )(s2 + a2 ) L (s 2 + an )
2 2 2
2) K with 0 < b1 < a1 < … < an < bn+1
2 2 2
(s 2 + b1 )(s2 + b2 ) L (s2 + bn +1 )
zero at the origin, zero at infinity.

(s2 + c1 )(s2 + c2 ) L (s 2 + cn )
2 2 2
3) K with 0 < c1 < b1 < … < cn < bn
2 2 2
s(s2 + b1 )(s 2 + b2 ) L (s2 + bn )
pole at the origin, zero at infinity.

(s 2 + c1 )(s 2 + c2 ) L (s 2 + cn +1 )
2
2
2
4) K with 0 < c1 < b1 < … < bn < cn+1
2 2
s(s 2 + b1 )(s 2 + b2 ) L (s2 + bn )2

pole at the origin, pole at infinity.

Examples

(s 2 + 1) s(s2 + 1)
3 is OK (type 3) 7 is not OK
s(s2 + 4) (s2 + 4)

The simplest way to display LC realizability is to say that, in the pole/zero and K
representation of equation 4 in page 13; K is real and positive, and the pole/zero
distribution along the jω axis is one of the following 4 possibilities (same numbering as
above):

1 2 3 4

Pole at ∞ Zero at ∞ Zero at ∞ Pole at ∞

Networks containing Resistance
It is also possible to synthesize networks containing all three circuit elements
using the Cauer method, i.e. obtaining a continued-fraction expansion leading to a
ladder network, as for example in the following exercise.

Exercise
Use the Cauer method to derive the
2/3 1/3
following circuit from:

4 + 5s + s2 Z(s) 5/18 2/19 5/2
Z(s) =
6 + 5s + s2

Important Note: In the synthesis of these cases, it is often necessary to rearrange the
terms of the polynomials during the process in order to keep the quotients positive (i.e.
realizable).
In the case of circuits containing resistances, another elementary synthesis
procedure (apart from the removal of poles at the origin and at infinity) is the removal
of min Re{ Z(jω)} .
If we have a function Z(s) which has a minimum on the jω axis at ω = ω1:

We can decompose Z(s) as:
Re{Z(j ω) } Z(s) = Z1(s) + Z2(s)

Z1
Z2
K1

ω1 ω

We have already seen the case where Z1 corresponds to a pole at the origin (a/s)
or at infinity (as). We can also extract a constant of value K provided that K ≤ K1 since
in that case the remainder will still satisfy the condition: Re{ Z2(jω)} ≥ 0.

Example
jω

7s + 2
Y (s) = is a p.r. function:
2s + 4 ? ?/7 σ

Re{Y(j ω) } We can see that:
7/2

4 + 7ω 2
Re{Y ( jω )} =
8 + 2ω 2
1/2
and the minimum occurs at ω = 0.
ω
We can then remove Y1 = 1/2 and the remainder is still p.r. (e.g. realizable)

Y(s)= 1/2 + Y2 ; Y2(s)= 3s / (s + 2)
Y2(s) can be treated in the same way or, considering that Y2(0) = 0, the pole of
Z2(s) = 1/Y2(s) at the origin can be removed. But, in this case it is simpler to recognise
that:
Z2(s) = 1/3 + 2/3s,
1/3

2 3/2
and then the complete network is:

Problem 1
Determine which of the following functions are realizable as driving point
impedances. Find a ladder circuit realization whenever possible.

s2 + s + 4 s 5 + 20s3 + 64s s2 + 1
a) b) c)
s+5 s 4 + 10s2 + 9 (s + 1)2

s 4 + 2s3 + 3s 2 + 2 s + 1 s 2 + 7s + 12
d) e)
s 4 + 2s 3 + 2 s2 s 2 + 3s + 2

s+4 s 6 + 6s4 + 11s 2 + 6
f) 2 g)
s + s + 15 6s 5 + 24s3 + 22s

SYNTHESIS OF PASSIVE TWO-PORTS NETWORKS (FILTER DESIGN)

Transfer Function Synthesis
The synthesis of two-port networks can be easily reduced to the synthesis of a
d.p.i. and then solved using the methods described earlier. However, one of the most
important problems in filter design is that very rarely the specifications are in the form
of a defined function of s as we have seen in the previous chapter. Instead, the
specifications are usually given in terms of restrictions over the behaviour of the
“frequency response” i.e. the shape of the curve (amplitude)2 versus frequency or phase
versus frequency. That is, some characteristics are given of a function of ω and to
synthesize a network we will first need to find a precise function of s which behaves as
prescribed on the jω axis. Then, one of the important aspects of this process is the
approximation.

Filter design strategy:

1.- From the specification to the transfer function
Often the specifications are given (or are transformed) in terms of a normalized
low-pass prototype. This makes the synthesis procedure much easier and the resultant
filter can be converted back to any other form of response and its element values re-
scaled. In this terms, the specifications are given in the form of values to:

Transmission
– cut-off frequency fc
100%

– maximum attenuation in the passband.

– rate of “fall-off” in the stop band.
fc f

Scaling of component values

Impedance Scaling
If we start with a ‘normalized’ filter designed with easy numbers:

Impedance level = 1 Ohm
low-pass
Cut-off frequency: filter 1
ωc = 1 rad/sec

To change “impedance level” from 1 Ohm to say, R0 Ohm, just multiply all individual
impedances by R0. Then the impedance scaling factor is kL = R0:

Rnew = kL Rn where Rn, Ln and Cn are the values in the
Lnew = kL Ln normalized filter.
Cnew = Cn /kL

Frequency Scaling:
If we now want to change the frequency ωc to ω0 instead of 1 rad/sec:
We want each component to have the same behaviour at ωc = ω0 as they had at 1
rad/sec, that is, we need the impedance values to be the same at the new frequency as
they were for 1 rad/sec: So, the frequency scaling factor is kf = ω0, and

R remains unchanged (since the impedance of resistors is independent of freq.)
Lnew = Lnorm/kf (to have: ω0Lnew = 1 Lnorm)
Cnew = Cnorm/kf (to have: ω0Cnew = 1Cnorm)

Another form of specifying a filter is using the so called “Insertion Loss” function:

network load

The insertion loss (in dB) as a function of frequency is defined as:

Power in load when network is absent
Insertion loss = 10 log10
Power in load when network is present

Insertion loss
[dB]

3

ωc ω

The cut-off frequency is the frequency at which some specified insertion loss is
obtained e.g. 3 dB. The 3-dB point is the most common choice but it is not universal.
Again in this case the practical specifications are given in terms of restrictions to
the behaviour of this function.

In order to synthesize a filter we have to use an approximation which often is a
polynomial to fit the design curve. There are multiple choices where the most
commonly used are the Butterworth, Chebyshev, or Bessel responses, each of them
corresponds to a different choice of approximation polynomials. The ultimate choice
should also consider other aspects of the filter performance as it is for example the
transient response.

Choice of filter characteristics
Let’s consider a low-pass filter with variable parameters:

V1 V2

The frequency response will change with the element values:

How can we adjust the element values to
V2 / V1 get the “best” characteristic?

What is the “best” response?

f

There are different common types of response or families of filters:

1) Maximally flat response (Butterworth filters)
∂ nV
For a specific value of V2 at a given ωc, we ask for: =0
∂ω n ω = 0
for n as large as possible.
The more derivatives equal to 0 at ω = 0, the “flatter” the response results at the origin.

2) Equal ripple response (Chebyshev filters)
This can be characterized by the conditions or specifications:
For a given f0, V(f0) and fc: minimize the value of
τ (the maximum ripple in the passband). 1
τ
or equivalently:
s
Given τ, f0, fc: minimize V(f0) (maximize the rate
of fall-off).
fc f0

The Chebyshev approximation will give the maximum rate of fall-off in the stop
band for a maximum allowed ripple in the pass-band.

3) Elliptic filter
1
For given cutoff frequency fc and ripple levels τ1 τ1

and τ1, minimize f0.
τ
or equivalently, (maximize the rate of fall-off 2

between fc and f0.
fc f0 f
In this case we allow ripple in the stop-band as well as in the pass-band.

4) Optimum or “L” filter

The response is monotonic with ω

∂V
At a specified fc, maximize:
∂ω fc f

5) Linear phase response (Bessel filters)
In this case we want an approximation to a linear phase response. But now, why
linear phase?
An ideal delay line (only introduces a time delay T) has a system function:

H(s) = K e–sT

Then, the frequency response is: H(jω) = K e–jωT ; that is:
Amplitude response = K, a constant and
Phase response = φ(ω) = –ωT ∴ varies linearly with ω.

If we excite with e(t) (or E(s)), the response is:

R(s) = K E(s) e–st and then,
r(t) = K e(t–T) u(t–T),

which is the same function as in the input, but delayed by a time T.
Then a linear phase response will modify the amplitude but not the relative phase
of each frequency component.

Frequency ω
0.1 0.2 0.3 0.4 0.6 0.8 1 2 3
1
0
-1
-2
-3
-4
-5
Magnitude, dB

-6
-7 Amplitude response of
-8 n=3 Chebyshev filter
-9 with 1 dB ripple
-10 Butterworth response
-11 with n=3
-12
-13
-14
Amplitude responses of third order filters (n = 3).

Frequency ω
0.1 0.2 0.3 0.4 0.6 0.8 1 2 3
2
0
-2
-4
-6
-8
Magnitude, dB

-10
0.4 0.6 0.8 1
-12 0.0

-14 -0.5

-16 -1.0

-18 -1.5

-20 -2.0

-22 -2.5

-24 -3.0

-26
Amplitude responses of Butterworth filter of order 3 (solid line) and Optimum
filter of order 3 (dashed line).

Transient Response of Low-Pass Filters
In this section we will compare the transient response of the filters discussed
earlier. In particular, we will compare the step response of the filters according to a set
of figures of merit. In relation to the general response to the step excitation, we can
define the following parameters or figures of merit:

1. Rise time The rise time tR of the step response is defined here as the time required
for the step response to rise from 10% to 90% of the final value as shown
in the figure.
2. Ringing Ringing is an oscillatory transient occurring in the response of the filter
to a sudden change in input (such as a step). A measure of the ringing in a
step response is given by its settling time.
3. Settling time This is the time ts beyond which the step response does not differ
from the final value in more than, say, ± 2%, as shown in the figure.
4. Delay time, tD Delay time is the time which the step response requires to reach
50% of its final value.
5. Overshoot. The overshoot in the step response is the difference between the peak
value and the final value, expressed as a percentage of the final value.

ts

1.0
Amplitude (normalized)

0.9

Overshoot
tD
0.5

0.1 tR
0.0

0 2 4 6 8 10 12 14 16 18 20
Time t, sec
Figures of merit for step response.

Most of the filter characteristics mentioned earlier are related to the frequency
response, in particular, to the bandwidth and to the phase linearity. The rise time and
the delay time are closely related to each other but have little connection with
overshoot.
We will now examine qualitatively these relationships.
Rise time and bandwidth have an inverse relationship in a filter. The wider the
bandwidth, the smaller is the rise time and vice versa. This could be understood by

noting that a limited performance of a filter at high frequencies slows down the increase
of the output in response to an abrupt step in the input, causing a long rise time.
It has been found experimentally that the relationship is very closely inversely
proportional so that we can write the following expression:

TR × (Bandwidth) = Constant

This particular parameter, the rise time, is an important criterion in pulse
transmission.

Overshoot is normally caused by an “excess” gain at high frequencies. By this we
mean a low-pass frequency response with a peak value in the high frequency part of the
pass-band.
magnitude,[dB]

frequency(Log)
Comparison of a peaked response and a simple response from an RC filter.

The step responses of Butterworth filters of orders n = 3, 7 and 10 are shown in the
next figure. Note that as n increases, the overshoot increases. This is because the higher
order Butterworth filters have flatter magnitude characteristics (i.e. there is more gain at
higher frequencies than in the lower order filters), although the response is never of the
“peaked” type.
Ringing is due to sharp cutoff in the filter magnitude response and we can see that
also increases when the order increases.

1.2
1.1
1.0

0.9
0.8
0.7
0.6 n=3
n=7
0.5
n = 10
0.4
0.3
0.2
0.1
0.0

0 2 4 6 8 10 12 14 16 18 20
Time, sec

Step response of normalized Butterworth low-pass filters

1.2
1.1
1.0

0.9
0.8
0.7
0.6
0.5 Step Responses
0.4
Butterworth n = 3
0.3
Chebyshev n = 3
0.2
Bessel n = 3
0.1
0.0

0 2 4 6 8 10 12 14 16 18 20
Time, sec
Comparison of filter transient response

The last figure compares the step response of a Bessel (linear phase) filter of order
n = 3 to the response of an n = 3 Chebyshev filter with 1-dB ripple. Rise times cannot be
compared because the bandwidths have not been adjusted to be equal. However, we
can compare their ringing and settling times. The Chebyshev filter has a sharper cutoff,
and therefore, more ringing and longer settling time than the Bessel filter. Note also
that the overshoot of the Bessel filter is negligible. This is characteristic of this class of
filters.
The decision of which filter is best depends strongly upon the particular situation.
In certain applications, such as in the transmission of music, phase is not important. In

these cases, the sharpness of the cutoff may be the dominant factor so that the
Chebyshev or the Optimum filter are preferred to the others.
If we are dealing with pulse transmission instead, the requirement of the system is
usually that the output sequence has approximately the same shape as the input
sequence, except for a time delay of T = T2 – T1, as shown in the next figure. It is clear
that a filter with a long rise time is not suitable, because the pulses would “smear” over
each other as shown in the figure. The same can be said for long settling times. Since a
pulse transmission system must have linear phase to ensure undistorted harmonic
reconstruction at the receiver, the best filter for these systems is one with linear phase
and small rise and settling times.

System
with short
rise and
settling
times
Input pulse train (a) Output pulse train

System
with long
rise and
settling
times

Input pulse train (b) Output pulse train

Smearing of pulses in systems with long rise and settling times.

Prototype filter networks
Typical frequency response and simple LC realizations

Low-pass

pass stop

f

High-pass

stop pass

f

Band-pass

stop stop
pass

f

Band-stop

stop
pass pass

f

Two-port reactive circuit synthesis Filter synthesis
From the transfer function to the circuit

There are (at least) two ways of starting with some desired response and then,
trying to realize it systematically.
1) The “Darlington procedure” starts with a desired insertion loss function, as a
function of the frequency and then continues to find a reactive two-port to give this
function.
2) The procedure we will follow is similar, but instead of the insertion loss
function, it takes a transfer function, effectively, the ratio of output to input voltage or
current.

The impedance matrix Z is defined
from: I1 I2
U=ZI in matrix form
U2
or in full: U1

U1= Z11 I1 + Z12 I2
U2=Z21 I1 + Z22 I2 from where:

U1 U1
Z11 = Z12 =
I1 I =0
I2 I
2 1= 0

U2 U2
Z21 = Z22 =
I1 I I2 I
2 =0 1= 0

Z11 and Z22 are driving point immittances, as they are concerned with a 1-port
(with the other pair of terminals “open-circuited”).
Z12 and Z21 are transfer impedances, being concerned with voltages and currents
in different ports.
If one was concerned with filter design where the load was effectively an infinite
impedance (open circuit), the above should be the starting equations; but two-ports are
usually designed to give a specified response with a real, resistive load.

If we connect a resistor of value R (a load) to the output terminals, the second of
the impedance matrix equations now becomes:

U2 = Z21 I1 + Z22 I2 = –R I2

I2 − Z21
so that: CT (s) =
=
I1 R + Z22
Starting now with the admittance matrix I = Y U, would similarly give:

U2 − Y21
UT(s) = = where G = 1/R
U1 G + Y22

Synthesis of I2/I1 or U2/U1 with a 1 Ohm load
We will now consider the synthesis of a reactive ladder network with a 1 Ohm
termination to meet a specified voltage-transfer or current-transfer function; either

− Y21 − Z21
UT(s) = or CT (s) =
1 + Y22 1 + Z22

It can be demonstrated that as the numerators are transfer functions of reactive
two-ports, they must be odd functions of the complex frequency s (although we will
skip the proof), that is: Z21(s) = –Z21(–s).
Secondly, we can remember (from page 20) that Z22 and Y22, being d.p.i’s are
ratios of either odd to even or even to odd polynomials in s.
Suppose that one of the transfer functions above, say CT(s), is given as the ratio of
two polynomials in s, like:
P(s)
CT (s) =
E(s) + O(s)

where the polynomial in the denominator has been split up into its even and odd parts,
and P(s) is either even or odd. The problem now is how to rearrange this equation into
the form of the previous equation above.
The answer is to divide both numerator and denominator by either E(s) or O(s),
the choice being such that the resultant numerator is an odd function of s, – which is
necessary. In other words, we have to choose the part with the opposite parity to P(s).
The denominator now becomes either: 1 + O(s)/E(s) or 1 + E(s)/O(s) which
corresponds to the form of the denominator of the equation at the beginning. From this
last expression we can easily separate the form of Z22, which is the d.p.i. for a reactive
circuit (the open-circuit output impedance on page 33) and so can be routinely realized.

Example
Suppose we want to realize the following current-transfer function:

2 − Z21
CT (s) = 3 2 =
s + 3s + 4s + 2 1 + Z22

Comparing this with the previous equations, we see that P(s) is even in s and so
must be divided by the odd part of the denominator: (s3 + 4s).
Back to the first equation, we have:

2 3s 2 + 2
Z21 = 3 and Z22 =
s + 4s s3 + 4s

There is still one last problem to solve in order to realize this Z22. Using the Euclid
algorithm, there are various possibilities and then we may find more than one solution
to this stage. (Note that since we are synthesizing Z22, the first elements to emerge are
those closest to the output terminals.

Z Y
3s2
+ 2 : s3
+ 4s s/ 3
9/10
s3 + 2s / 3 C
9 s/ 10 3s2 10s / 3
5/3 1/3
2 10s / 3 5s / 3
0 D
and
Z Y
1/ (2s) 3s + 2 2
: s 3
+ 4s 2/5 2
s / 2 + 2
2 C
5 s2 / 2 4s 8 / (5s)
5/8
5 / (2s) 5 s2 / 2 3
s
0 D

Two possibilities seem to emerge, but these are the open-circuit impedances
looking into the output terminals of the network of page 33, which should therefore
look like:
9/10 2/5 2
C B
C
B

5/3 1/3 1 5/8 1

A A D
D
Here we have attached the 1 Ohm resistive load to the C and D terminals of the
circuits, so that we have indeed realized Z22. Points A and B appear to be the only
terminals left above to become the port 1 of the circuit in page 33.
The right hand circuit above is, in fact, a band-pass filter and the left-hand is low-pass.
The desired current-transfer function CT(s) satisfies:

CT(s) → 1 as s → 0, and CT(s) → 0 as s → ∞

which is consistent only with the left-hand circuit above and not with the other.
So, in the end only one circuit emerges successfully from the two possibilities.

We now need to consider how to find sensible transfer functions from the filter
specifications.

Obtaining a transfer function from a desired frequency response
A filter specification in the frequency domain usually consists of a specification of
the power as a function of the frequency or equivalently, the amplitude response:
V2/V1(ω) or I2/I1(ω). This specifies a real function of ω, so the problem is, how to
convert this into a complex function of s, namely a transfer function VT(s) or CT(s).
Suppose that a desired response for a low-pass filter can be specified by:

2
I2 K 2 2
= 2 = CT ( j ω ) = h( −ω ) (1)
I1 1 + Pn (ω )

Pn(ω2) is a polynomial of the form: a1ω2 + a2ω4 + … + anω2n, and K is the d.c.
“gain”. This clearly corresponds to a low-pass since all “zeros of transmission” are at
s = ∞.
Now we want to find CT(s). We first note that:

CT(jω)2 = CT(jω)· CT*(jω) = CT(jω) · CT(–jω)

since CT(s) is the ratio of polynomials with real coefficients.
We can now define an extension of this function into the complex plane as:

2 2
h(s ) = h(−ω )
−ω 2 = s 2

and, substituting –ω2 by s2 in (1) we have:

CT(s) · CT(–s) = h(s2)

We observe that the roots of CT(–s) are the negatives of those of CT(s). Then, from
h(s2)we can separate CT(s) from CT(–s) simply selecting the roots that lie in the left-
hand half-plane. This can be seen clearly with an example:

Butterworth Filters:
If we consider a Butterworth or maximally flat response, as many derivatives of
h(–ω2) must vanish at ω = 0. This is equivalent to asking for as many derivatives as
possible of P(ω2) w.r.t. ω2 to vanish at ω = 0, and consequently, all (n–1) first coefficients
of P must be zero and the polynomial has the form:

P(ω2) = an ω2n Exercise: Prove this.
The cut-off frequency (3-dB point) is then given by:

2 1
CT( jω ) = 1/ 2 = 2
1 + anω c n

And a value of an = 1 will give a normalized cut-off frequency of ωc= 1.

Finally, then: CT(jω)2 = 1/(1 + ω2n) = CT(jω) · CT(–jω)

In particular, if we consider a second order filter with n = 2, so that:
CT(jω) · CT(–jω) = 1/(1 + ω4), and using –ω2 = s2:

CT(s)· CT(–s) = 1/(1 + s4) ,

the roots are all on the unitary circle and are symmetrically distributed with respect to
the jω axis.
We can separate then:

±1 s2 s1
CT (s) =
( s − s2 )( s − s3 )

since these are the roots on the left-hand half-plane.

We now have:
±1 s3 s4
CT ( s ) = 2
s + 2s + 1

On pages 34 to 36 we saw how to synthesize a CT function. For this example this
will give:
1/ 2 2

and
2 1/ 2

s2 + 1
as the two possible realizations of: Z22 =
2s

The first is the network for a low-pass filter with response:
1
CT(s)· CT(–s) = , so our second order Butterworth filter is as in the figure:
1 + s4
1/ 2

2 1

Similarly, for a third order Butterworth filter:

1 1
CT ( jω ) ⋅ CT( − j ω ) = 6 =
1+ ω 1 − (− ω 2 )3

1 1
CT (s) ⋅ CT (− s) = 6 =
1− s (1 + 2s + 2 s + s )(1 − 2s + 2s2 − s3 )
2 3

In general this factorised form can be written as:

1 1
CT (s) ⋅ CT (− s ) = and CT (s) =
Bn (s) ⋅ Bn (− s) Bn ( s)

where Bn(s) is called Butterworth polynomial of order n.

The synthesis approach used above can be applied to Butterworth filters of any
order, and the adequate factorisation (the corresponding Bn(s)) is given in the following
table:

n a1 a2 a3 a4 a5 a6 a7 a8
1 1
2 1 n
2
Bn ( s ) = ∑ ai s i with a0 = 1
3 2 2 1
i =0
4 2.613 3.414 2.613 1
5 3.236 5.236 5.236 3.236 1
6 3.864 7.464 9.141 7.464 3.864 1
7 4.494 10.103 14.606 14.606 10.103 4.494 1
8 5.126 13.138 21.848 25.691 21.848 13.138 5.126 1

The same polynomials can be written in factorised form to give directly the
location of the roots:

n Butterworth Polynomials (Factorised form)
1 s+1
2 s2 + 2 s + 1
3 (s2 + s + 1)(s + 1)
4 (s2 + 0.7653s + 1)(s2 + 1.84776s + 1)
5 (s + 1)(s2 + 0.6180s + 1)(s2 + 1.6180s + 1)
6 (s2 + 0.5176s + 1)(s2 + 2 s + 1)(s2 + 1.9318s + 1)
7 (s + 1)(s2 + 0.4450s + 1)(s2 + 1.2456s + 1)(s2 + 1.8022s + 1)
8 (s2 + 0.3986s + 1)(s2 + 1.1110s + 1)(s2 + 1.6630s + 1)(s2 + 1.9622s + 1)

The figure shows the Frequency, ω
amplitude response of some of the 0.1 0.2 0.3 0.4 0.6 0.8 1 2 3

Butterworth polynomials. Note
0
that the response gradually
-2
approaches the ideal as the order
-4
increases.
-6
The amplitude response

Magnitude, dB
-8
curve shows how, with the 3-dB
-10
point at unit frequency, the curve
-12 n=3
approaches the rectangular
-14
“ideal”, with the response falling n=5
-16
off like 1/ωn as ω increases (20n -18
n=7
dB per decade or 6n dB per -20
octave). -22
But the response to the step -24
in page 34 show the price of this -26
increasing “sharpness” in the
frequency response. We see rise time, delay time, settling time, overshooting and
ringing all getting worse.

Example:
Find an expression to calculate the minimum order n of a normalized Butterworth
low-pass filter, to have an attenuation of at least αs dB at a frequency ωs outside the
passband.
ω ω ω
c s
We can define the attenuation α as
3
α = –20 log T dB.

(It is simply the negative of the amplitude in
dB).
For a normalized Butterworth:
αs
2 1 1
T = 2n = Atten.
1 +ω 1 + (ω / ω c ) 2 n α dB

then α(ω) = 10 log (1 + ω2n)

∴ α/10 = log (1 + ω2n)

α 10 α 10 α 10
10 = 1+ ω 2n ⇒ ω 2n = 10 − 1 ⇒ 2nlog ω = log(10 − 1)

log(10αs /10 − 1)
and finally, n=
2 log ω s

If we select for example, αs = 20 dB, and ωs = 2 (ωs = 2ωc)
n results:
n = (log 99)/(2 log 2) = 3.3146 = 4

Problem 2
a) Consider the general form of a Butterworth frequency response:

2 1
T(ω ) =
1 + (ω / ω c )
2n

Find an expression to calculate the minimum value of n (order of the filter) necessary to
satisfy the following specifications:
ω p ωs ω
where α = −20 log10 T dB
α max
α max : Maximum attenuation
allowed in passband.

α min α min : Minimum attenuation
required for ω > ω s

(Note that the 3-dB cutoff frequency
is not given)
α dB

b) Given the values αmax = 0.5 dB, ωp =105 [rad/sec], αmin = 20 dB, ωs = 2×105
[rad/sec], find the minimum value of n (approximated to an integer) to satisfy this
specification. Choose the values given for ωs and αmin to determine the value of the 3-
dB cutoff frequency ωc using the calculated value of the order of the filter.

Problem 3

Synthesize a normalized Butterworth or maximally flat filter of order 4 giving the
value of all components.
Using this normalized prototype, design a low-pass filter with cutoff frequency of
ωc = 106 rad/sec and terminated with a resistive load of 600Ω.
.

Chebyshev filters (equal ripple)

Similarly to the case of Butterworth filters, the Chebyshev response comes from:

1
CT (s) ⋅ CT (− s) =
1+ ε 2 Cn (ω )
2

where the Cn(ω) are the Chebyshev polynomials and ε is a parameter used to control
the amount of ripple in the passband.

Let's first examine some of the properties of the Chebyshev polynomials:

≤ 1 ω ≤ 1
• Cn (ω )  Cn C7 C
> 1 ω > 1 3
C2

• All zeros are in the interval [–1,
C1
1]. 1
C0

• Maxima and minima are all in
-1 1
the same interval and have the
same magnitude.
-1
• For given n, the slope of the
polynomial at ω > 1 is the
highest that can be obtained with
any polynomial of the same
order.

Some of the first polynomials are:

C0(ω) = 1
C1(ω) = ω
C2(ω) = 2ω2 – 1 In general we can write:
C3(ω) = 4ω3 – 3ω
C4(ω) = 8ω4 – 8ω2 + 1 Cn(ω) = cos (n cos–1ω)
C5(ω) = 16ω5 – 20ω3 + 5ω
C6(ω) = 32ω6 – 48ω4 + 18ω2 – 1
C7(ω) = 64ω7 – 112ω5 + 56ω3 – 7ω
C8(ω) = 128ω8 – 256ω6 + 160ω4 – 32ω2 + 1
C9(ω) = 256ω9 – 576ω7 + 432ω5 – 120ω3 + 9ω
C10(ω) = 512ω10 – 1280ω8 + 1120ω6 – 400ω4+ 50ω2 – 1

How a Chebyshev polynomial can be used to form a filter response?

C3

1
Consider for example the
third order polynomial C3:
-1 1

-1

2
ε 2 C3

ε2
2 2
Now, we make: ε C3

2
1+ ε C3
2

If we now form: 1+ ε
2

2 2
1 + ε C3
1

2
CT

1
And finally we get:

2 1 1
CT = 2 2
1 + ε C3 1+ ε
2

We saw in page 37 that the poles in the Butterworth response were all located on
the unitary circle. In the case of Chebyshev filters, the poles are on an ellipse as shown
in the figure. There we can compare the position of the roots for Butterworth and
Chebyshev responses of order 6.

The actual shape of the ellipse depends
on the amount of ripple (ε).
The smaller the ripple, the nearest is the
ellipse to the circle.

Problem 4

Find a passive ladder circuit realization of a low-pass Chebyshev filter of order
five with 0.5 dB ripple in the passband. Scale component values to give a cutoff
frequency ωc of 106 [rad/sec] (fc = 1/(2π) MHz) and terminated with a resistive
impedance load of 600 Ohm.

Often, in the design process it is convenient to write the denominator polynomial
of the transfer function CT(s) in a factorised form.
The corresponding factorisation of the denominator D(s) of CT(s) for Chebyshev
filters is given in the following table for two different amounts of ripple.

 n /2 2
 ∏ ( s + b1i s + b0 i )
 i =1
n = 2, 4,6,L
1
CT (s) = D( s) =  ( n −1) / 2
D (s )  (s + b )
1 ∏ (s + b i s + b0 i )
2
n = 3,5,L


1
i =1

Note: The symbol Π means multiple product.

Order Factorised form: coefficients of D(s)
Approximation n b1 b11 b01 b12 b02 b13 b03
2 1.414214 1
Maximally flat 3 1 1 1
(Butterworth) 4 0.765367 1 1.847759 1
5 1 0.618034 1 1.618034
6 0.517638 1 1.414214 1 1.931852 1

2 1.425624 1.516203
0.5 dB 3 0.626457 0.626456 1.142448
equal-ripple 4 0.350706 1.063519 0.846680 0.356412
(Chebyshev) 5 0.362320 0.223926 1.035784 0.586245 0.476767
6 0.155300 1.023023 0.424288 0.590136 0.579588 0.156997

2 1.097734 1.102510
1 dB 3 0.494171 0.494171 0.994205
equal-ripple 4 0.279072 0.986505 0.673739 0.279398
(Chebyshev) 5 0.289493 0.178917 0.988315 0.468410 0.429298
6 0.124362 0.990733 0.339763 0.557720 0.464125 0.124707

Frequency transformations
As mentioned before, a low-pass prototype can be converted to any other type of
frequency response using an appropriate frequency transformation.
We already introduced frequency scaling by which a frequency ω is transformed
to a new frequency ω’ by:
ω’ = kf ω

and then: ω ′L′ = ω L = (k f ω )( L k f ) ∴ L′ = L k f
ω ′C′ = ω C = ( k f ω )(C k f ) ∴ C′ = C k f (1)
R’ = R

For example, the impedance of the inductor L’ at the new frequency ω’ is the same as
that of the inductor L at the old frequency ω.
If we write in general the old frequency as a transformation of the new one:
ω = T(ω’) or s = T(s’), for what functions T(s’) can the corresponding relations still
apply?
The old impedances are:
sL = T(s’) L
s C = T(s’) C (2)
R = R’

We can see that the only condition is that T(s’)L and T(s’)C, or indeed simply T(s’),
must be realizable as a d.p.i.

Transformation from low-pass to high-pass

Consider the transformation s’ = 1/s or s = 1/s’ = T(s’)
From equations (2), the impedance of the old elements should be now:

sL = (1/s’) L = 1/s’C1
sC = (1/s’) C = 1/s’L1
R = R’

Then, an inductor of value L is transformed to a capacitor of value C1 = 1/L, and
similarly: L1 = 1/C and all resistors remain unchanged.
For example the low-pass filter calculated in page 35 transforms to the high-pass
version:

9/10 10/9

5/3 1/3 1 → 3/5 3 1

The circuit on the left (the low-pass filter from page 35) has identical insertion loss,
and I2/I1 at frequency ω as the circuit on the right at frequency 1/ω.
Note that one can follow the s’ = 1/s transformation by a frequency scaling, or do
it in one step by the transformation: s’= K/s. (or s = T(s’) = K/s’)

Low-pass to band-pass transformation
A transformation Τ(s’): s’ → s is possible providing that T(s’) represents a
realizable immittance.
We know that the frequency scaling, s’ = as (T(s’) = s’/a) is realizable as it is also
the low-pass to high-pass s’ = b/s (or s = T(s’) = b/s’). Clearly, a combination of both
is also realizable, so we can examine:

s' b
s = T( s' ) = + and
a s'

 s' b 
s = T( s' ) = 1  + 
 a s' 

We can re-write the first transformation as:
ω  s' ω 0 
s = T( s' ) = 0  +  where ∆ω = ω2 – ω1
∆ω  ω 0 s' 

and ω2/ω0 = ω0/ω1, giving: ω02 = ω1ω2.

Using this transformation, a low-pass frequency response will transform as:

0 1 ω 0 ω ω ω ω
1 0 2

dB dB
→ s'
s

The individual elements will be transformed as:

2 ∆ω L
L s' L ω 0 L
sL = T (s' ) L = + ω2 L
0 ∆ω
∆ ω s' ∆ω
and
∆ω
ω 2C
0
2
C s' C ω 0C
sC = T( s' )C = +
∆ ω s' ∆ω
C
∆ω

In this way, the prototype low-pass filter of Chebyshev response (for 1 dB ripple):
(Exercise: Find the corresponding element values)

1.333

1.509 1.012 1

would transform to:
∆ω 1.333
1.333ω 0 ∆ω
2

∆ω 1.509 ∆ω 1.012 1
1.509ω 2
0 ∆ω 1.012ω 2
0 ∆ω

This is a Chebyshev filter with band-pass response. If we consider ω = ω0 all LC
pairs are at resonance giving perfect transmission, – corresponding to d.c. in the low-
pass filter.

Low-pass to band-stop transformation
Finally, from the second expression in last page we get a low-pass to band-stop
transformation. (The formula is the reciprocal of the previous one).

∆ω
s = T(s' ) = where again ∆ω = ω2 – ω1 and ω02 = ω1ω2.
 s' ω 0 
ω0  +
ω 0 s' 


The low-pass and the corresponding transformed response are:

0 1 ω 0 ω ω ω ω
1 0 2

dB dB
→ s'
s

The individual elements will be transformed as:

L∆ω
2
ω0
L s' ∆ω
sL = T (s' )L = L
(s' 2 +ω 2 )
0
1
L∆ω
and
C∆ω
C s' ∆ω 1
sC = T(s' )C = 2 C ω2 C∆ω
(s' + ω 2 )
0
0

Richard’s transformation
Let's now try the transformation: s = T(s’) = tanh(s’)
or using a normalizing factor ω0: s = tanh(s’/ω0)
The question now is: what circuit component has a d.p.i. of the form:

s = tanh(s’/ω0)
= tanh(jω’/ω0) = jtan(ω’/ω0)

The impedance of a piece of transmission line of electric length θ with
characteristic impedance Zc and terminated with a load Zl is:

Zl + jZc tanθ
Zc
Zc + jZl tanθ
Then, a short-circuited piece with Zc = 1 and length: θ = βl= ω’/ω0 has the
impedance:
jtan(ω’/ω0)

Now, what component has an admittance of jtanθ ?
It is easy to see that this corresponds to an open-circuited piece of transmission line of
the same length l and characteristic impedance Zc.
So, with this transformation, the circuit components will transform:

sL = jLtan(ω’/ω0) (short circuit, Zc = L)
sC = jCtan(ω’/ω0) (open circuit, Yc = C)

L Zc = L
→

C
→ Yc = C

In both cases the electric length is θ = ω’/ω0
A network with a number of components L1, L2, … and C1, C2, … transforms to
short-circuited lines with Zc = L1, L2, … and open-circuited lines with Yc = C1, C2, …
simultaneously provided that ω0 is the same for all lines.
This transformation results in a periodic mapping, since tan(θ) is periodic with
period π. For non-dispersive lines, where electric length is proportional to frequency
(as assumed in the equations above), this means the transformation is periodic in
frequency, the period being the frequency at which the line length is λ/2.
Using this transformation, the simple second order Butterworth filter of page 37 is
transformed to:
Zc = 1 / 2
1/ 2

→ 1
2 1

Yc = 2

Both pieces of transmission lines or “stubs” have the same Zc in this case, – of
0.707 Ohm or 70.7 Ohm if we want a load of 100 Ohm.
The curves showing the corresponding transformation of the frequency response
are:
ω ω
0 1 ω 0 π/ 2 π 3π / 2 2π 0

dB dB
→ s'
s

The resulting filter looks dual-purpose, in containing both band-pass and band-
stop regions; but the maximally flat characteristic at d.c. of the low-pass prototype
strictly transforms to ω’/ω0 = π, 2π, 3π, etc.
This transformation is widely used at microwave frequencies, –say, 0.5 to 30 GHz.
The circuit above is not very convenient as it is and in practice some other
transformations are used (Kurodo transformations and “unit elements”) so that only
shunt elements are needed (because that is more convenient in planar microwave
circuits).
This transformation allows us to add transmission lines to the set of possible
elements for realizing passive filters, –that is, to L, C, and R.
These filters are very easy to make in microstrip, – a sort of printed-circuit board
for use at microwave frequencies.

conducting strip

dielectric

ground plane
For instance, a mask necessary for a 3-element filter would look like:

input output

The 3 elements are shunt open-circuit stubs.

The 3 free variables allow us to choose the desired third order response
(Chebyshev, Butterworth, or any other).

Problem 5
Starting from the Butterworth approximation for CT(jω)2, find the function
CT(s) corresponding to a third order low-pass filter and realize it as a ladder circuit.
Apply the low-pass to high-pass transformation to find a normalized third order
high-pass filter.
These could be used as part of a 2-loudspeaker design as in the figure.
Verify that the impedance looking into the two circuits in series is 1 Ω, for all
frequencies.
Finally, scale the two filters to provide: loudspeakers of 8 Ω and a cross-over
frequency of 5000/π Hz. The latter, taken as cutoff frequencies, would provide 3 dB
attenuation to each speaker.

low-pass 'woofer'

high-pass 'tweeter'

Problem 6

A low-pass system function employing a third order Bessel polynomial (not
normalized to 3dB cutoff at ωc = 1) is:

15
H(s) = 3 2
s + 6s + 15s + 15

Find the corresponding frequency scaling to give a 3dB cutoff frequency of 1
rad/sec.
For this low-pass prototype, find a ladder realization and apply the necessary
transformations to get a band-pass filter with ω1 = 2×104 rad/sec, ω2 = 8×104 rad/sec
and a resistive load of 600 Ω.
Note: The 3-dB cutoff frequency of the transfer function above is ωc = 1.75568

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