Mathematical Development of theElliptic Filter

Mark KleehammerQueen’s University

August 26, 2013

The elliptic filter is a very powerful tool for signal processing, however it alsorequires some sophisticated mathematics to properly describe it. Signal process-ing plays a crucial role in a large part in nearly everything electronic today, fromphones, to computers to music. We begin with an abstract introduction to signalprocessing and prove some basic results. Then we will apply the abstract theoryto discuss the Butterworth, Chebyshev and elliptic filters, with a primary focus onthe elliptic filter. However, before we can discuss the elliptic filter, we transitioninto an in depth discussion of elliptic functions.

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Contents

1 Signal Processing 31.1 The Ideal Lowpass Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 The Causal Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 The Transfer Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 The Butterworth Filter 102.1 Butterworth Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2 The Butterworth Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.3 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3 The Chebyshev Filter 153.1 Chebyshev Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.2 Transfer Function H(s) for the Chebyshev Filter . . . . . . . . . . . . . . . . . . . 163.3 The Chebyshev Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.4 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

4 Elliptic Functions 204.1 Elliptic Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204.2 Jacobi’s Elliptic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224.3 The Addition Theorems for the Jacobi Elliptic Functions . . . . . . . . . . . . . . 244.4 Transformations of Jacobi Elliptic Functions . . . . . . . . . . . . . . . . . . . . . 27

4.4.1 The First Degree Transformation . . . . . . . . . . . . . . . . . . . . . . . . 284.4.2 The nth Degree Transformation . . . . . . . . . . . . . . . . . . . . . . . . 29

4.5 The Jacobi Theta Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

5 Elliptic Rational Function 335.1 Statement of the Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335.2 Solution to Problem C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.3 Elliptic Rational Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

5.3.1 Connections Between Texts . . . . . . . . . . . . . . . . . . . . . . . . . . . 425.4 Zeros and Poles of the Elliptic Rational Function . . . . . . . . . . . . . . . . . . 435.5 The Elliptic Rational function for n = 1,2,3 . . . . . . . . . . . . . . . . . . . . . . 45

6 The Elliptic Filter 476.1 The Elliptic Rational Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476.2 The Transfer Function for the Elliptic Filter . . . . . . . . . . . . . . . . . . . . . . 476.3 Transfer Function Poles for Order n . . . . . . . . . . . . . . . . . . . . . . . . . . 486.4 Transfer Function Poles for Orders n = 1,2,3 . . . . . . . . . . . . . . . . . . . . . 496.5 The Elliptic Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 526.6 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

7 Appendix 55

References 71

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1 Signal Processing

A signal is a physical quantity or quality that conveys information [LTE01, p. 1]. For exam-ple a person’s voice is a signal, and it may cause the listener to act in a certain manner. Thisevaluation of the signal is called signal processing. Signals are found in all kinds of places;they could be the voltage produced in an electric circuit, computer graphics, or music. Manyelectronic signals travel through space, and while traveling they tend to pick up other unde-sirable signals such as white noise, or static. The kinds of signals we will focus on will be sinewaves, and we call these signals sinusoidal.

The problem which motivates our discussion is given some signal, we must remove theundesirable frequencies from it. A filter takes a signal as its input, performs some operationson the signal which remove these undesirable frequencies, and outputs a "clean" signal. Al-gebraically we think of a filter as a function that maps signals to signals.

Definition 1.1. A filter is a function that is used to modify or reshape the frequency spectrumof a signal according to some prescribed requirements [LTE01, p. 241]

An everyday example of this problem is used in cell phones all the time. When you makea call (or text, etc...), the phone sends a signal to a satellite in space, and that satellite thenredirects the signal to your friend’s phone. When the signal was traveling through space itpicks up static, so when your friend gets the signal, his/her phone filters out that static. Howdoes the phone filter out the static?

Many electrical engineers are well versed in this problem, and there is a solid amount ofliterature on the subject from an engineering perspective [Cau58] [Dan74] [LTE01], howeverthere is little if any from mathematicians. As such many of the results are explained with littledetail and background, and it’s difficult to see the bigger picture with these details omitted.My goal is to present this material in a straight forward and simple manner, with justificationsgiven for each step. It is important that one can see why we get these results, with less of a"this is how it is" mentality.

Many of the functions we will encounter are functions of the complex frequency s =σ+ iω(where i =p−1) . The variable s comes from the Laplace transform; given some function v(t )(for example a voltage dependent on time) its Laplace transform is defined as [Dan74, p. 2]

V (s) =L [v(t )] :=∫ ∞

0v(t )e−st d t (1.1)

For our purposes we will view the Laplace transformation as taking a function in the timedomain, and translating it into the frequency domain. Typically we will investigate the filterproblem in the frequency domain (since the filter requirements given to an engineer will typ-ically be expressed in the frequency domain), and then use the inverse Laplace transform toexpress our results in the time domain.

Let’s begin by analyzing the following example. Suppose we have a sinusoidal input signalxin(t ) = sinωt , and the engineer only needs this signal for a specific range of frequencies,say [ωa ,ωb]. The engineer will design a filter that preserves those desired frequencies, and

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eliminates everything else. And after the signal has been filtered, the output signal will alsobe a sine wave, but with the amplitude and phase likely altered by the filter, i.e.

xout(t ) =C sin(ωt +ϕ)

for some real constants C ,ϕ [Dan74, p. 3].

Definition 1.2. The transfer function T (s) is defined to be the ratio of the Laplace transformof the output signal (denoted xout(t )), to the Laplace transform of the input signal (denotedxin(t )) [Dan74, p. 3]. Explicitly

T (s) = Xout(s)

Xin(s)= L [xout(t )]

L [xin(t )](1.2)

Often we will consider the function H(s) = 1/T (s), which is referred to as the input/outputtransfer function.

Theorem 1.1. Let xin(t ) = A1 sinωt represent an input signal with amplitude A1 > 0, andxout(t ) = A2 sin(ωt +ϕ) represent the output signal with amplitude A2 > 0. Then

A2

A1= |T (iω)| (1.3)

This means that for a sinusoidal signal, the ratio of the output amplitude to the input am-plitude is equal to the magnitude of the transfer function evaluated at s = iω.

Proof. Compute the Laplace transforms of xin and xout

Xin(s) =L [xin(t )] =∫ ∞

0A1 sin(ωt )e−st d t = A1

ω

s2 +ω2

Xout(s) =L [xout(t )] =∫ ∞

0A2 sin(ωt +ϕ)e−st d t = A2

ωcosϕ+ s sinϕ

s2 +ω2

Therefore

|T (iω)| = A2

A1

∣∣∣∣ωcosϕ+ iωsinϕ

ω

∣∣∣∣= A2

A1

We should take note that, in general the frequency of the input sine wave ω, is not con-stant, and the amplitude A1 depends on the frequency ω. While the theorem above seems toimplicitly assume that these are constant. So to be more precise we will denote the ampli-tude as amp(ω) and reformulate the theorem we just proved (the proof still works the same).Perhaps we should also denote the signals x(t ) as a function of two variables x(ω, t ), howeverthe frequency ω – while non-constant – is not completely independent of time (we cannothave two different frequencies occurring at the same time). So we think of the signal x as afunction of time, the frequency of the signalω is not constant, and we want to filter out somefrequencies of x that may occur.

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Theorem 1.2. Let xin(t ) = ampin(ω)sinωt represent an input signal with amplitudeampin(ω) > 0, and xout(t ) = ampout(ω)sin(ωt +ϕ) represent the output of the filter with am-plitude ampout(ω) > 0. Then

ampout(ω)

ampin(ω)= |T (iω)| (1.4)

We say that a sinusoidal signal has been attenuated if the output signal has a smaller am-plitude than the input [Dan74, p. 3]. With this in mind our problem is to create a filter thatattenuates certain frequencies of the signal, while leaving others invariant.

Regions of low attenuation are called passbands, and we say that the filter passes suchfrequencies. On the other hand, regions of high attenuation are called stopbands, and we saythe filter stops these frequencies. A lowpass filter passes frequencies less than some specificfrequency, and attenuates those higher, and a highpass filter does the opposite [Dan74, p. 4].We will focus our discussion on the lowpass filter.

1.1 The Ideal Lowpass Filter

For the lowpass filter, we pass the frequencies ω with |ω| ≤ ωb , where ωb is some fixed fre-quency [Dan74, p. 8]. That is in the passband ω : |ω| ≤ωb, we want the input amplitude toequal the output amplitude, so their ratio is 1, hence for |ω| ≤ωb we require

|T (iω)| = 1 (1.5)

In the stopband, we want very high attenuation, so the output amplitude should be zero.Thus we also require for |ω| >ωb ,

|T (iω)| = 0 (1.6)

Equivalently, in terms of H(s) = 1/T (s) we desire

|H(iω)| =

1 : |ω| ≤ωb

∞ : |ω| >ωb(1.7)

Such a function is impossible to realize (with the discontinuities), and so we will examinedifferent ways of approximating this ideal. The approximations we deal with will be polyno-mials (or a ratio of polynomials) of degree at most n ∈N, and we say that the filter is of degreen. We think of the degree of the filter as a measure of complexity, and often the approxi-mation problem is to find the lowest degree filter that meets certain requirements [Dan74,Sec. 2.6, 3.7]. A higher degree filter may require more precise (and expensive) parts to realize.

For an approximation, the transition from passband to stopband will not be instantaneous,and so we will have this region of increasing attenuation in-between the passband and stop-band which we call the transition region [Dan74, p. 4].

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Figure 1.1: Plot of the ideal lowpass filter’s transfer function |T (iω)| with a normalized pass-band of [−1,1]

1.2 The Causal Filter

A causal filter depends only on past and present inputs, a filter that depends on future inputsis non-causal [PTVF07]. Only causal filters are realizable (it can operate in real time), since arealizable filter cannot act on future input (it hasn’t happened yet!).

Let xin(t ) and xout(t ) represent the input/output signal at time t respectively, also let Xin(s) =L [xin(t )] and Xout(s) =L [xout(t )]. Furthermore suppose we know the transfer function T (s)for our filter, then we know

Xout(s) = T (s)Xin(s) (1.8)

Then by taking the inverse Laplace transform of both sides we obtain the time domain re-sponse of the filter. That is, we expressed the output signal xout(t ) as a function of the inputxin(t ).

One of the properties of the Laplace transforms is that it converts convolution of functionsin the time domain into multiplication of functions in the frequency domain. More explicitly,let f (t ), g (t ) be real functions, then the convolution of f with g is defined as

( f ∗ g )(t ) =∫ ∞

−∞f (τ)g (t −τ)dτ (1.9)

This property says thatL

[( f ∗ g )(t )

]=L [ f ]L [g ] (1.10)

Denote h(t ) =L −1[T (s)], this function is called the impulse response of the filter [PTVF07].

Definition 1.3. We define a causal filter, to be a filter whose impulse response h(t ) vanishesfor negative t , i.e. h(t ) = 0 for t < 0 [PTVF07].

Then from (1.10) and (1.8) we see the causal filter is

xout(t ) =∫ ∞

0h(τ)xin(t −τ)dτ (1.11)

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In general, computing the inverse Laplace transform can be difficult, however notice thatfor ℜ(a) < 0 (where ℜ(z) denotes the real part of the complex number z)

L[eat ]= ∫ ∞

0e−t (s−a)d t = 1

s −a(1.12)

So conversely the inverse Laplace transform of 1s−a is eat . Therefore, if we can express the

transfer function in partial fractions of this form, then we can find the inverse Laplace trans-form easily. A partial fractions expansion of T (s) (for an nth degree filter) will look like

T (s) =n∑

j=1

α j

s − s jα j = lim

s→s j(s − s j )T (s) (1.13)

where s j are the poles of T (s) (zeros of H). Since the transfer functions we work with will onlyhave simple poles, we can find the constants α j by

α j = lims→s j

s − s j

H(s)−H(s j )= 1

H ′(s j )(1.14)

In the next section we will learn to construct the ratio of the input amplitude to the outputamplitude, that is finding |H(iω)| for a certain type of filter. Since |H(iω)|2 = H(iω)H(−iω),we letω=−i s, so |H(iω)|2 = H(s)H(−s), and then we can find the zeros of H(s)H(−s). Noticethat the function H(s)H(−s) has twice as many roots and poles as H(s).

To find a partial fractions expansion for T , we need the n zeros of H , which we find indi-rectly by finding the 2n zeros of H(s)H(−s). When designing a filter, the zeros of H (polesof T ) must have negative real part in order to apply (1.12), so we associate these zeros ofH(s)H(−s) to H(s) (and the zeros with positive real part are associated to H(−s) ). If we let z j

represent the zeros of H(s)H(−s), then we need

j : ℜ(z j

)< 0= J , and so the partial fractions

decomposition of T is

T (s) = ∑j∈J

α j

s − z j

From here we know the inverse Laplace transform of T is (from (1.12) )

L −1[T (s)]= ∑

j∈Jα j ez j t

Hence the filter is

xout(t ) =∫ ∞

0

(∑j∈Jα j ez jτ

)xin(t −τ)dτ

We phrase these results as a theorem for reference.

Theorem 1.3. Let T (s) = 1/H(s) denote the nth degree transfer function of the filter, and z j

the zeros of H(s)H(−s), and J = j : ℜ(

z j)< 0

. If xin(t ) denotes the input signal at time t , and

xout(t ) the output signal of the filter, then the filtered signal is:

xout(t ) =∫ ∞

0

(∑j∈Jα j ez jτ

)xin(t −τ)dτ (1.15)

with constants α j given by (1.14)

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Much of what we discussed in this section is left out in [LTE01], [Dan74], both of which arestandard texts on signal processing. Most engineers prefer to work in the frequency domain[Dan74, p. 282], and Theorem 1.3 expresses results in the time domain. As such, in these textsfew results are expressed in the time domain, so the results we obtain in the time domain– while not exactly unknown – are often undiscussed. Throughout the course of this paper,we will focus on expressing the casual filters in the time domain, as this way we are dealingdirectly with the signals themselves instead of their frequencies.

Mostly we’ve been concerned with the amplitude |H(iω)|, but in some applications (par-ticularly radar and digital communications) one may be interested in the phase arg H(iω)[Dan74, p. 238]. This phase characteristic affects the time domain response. However for ourdiscussion we will omit this material, and refer the reader to Daniels’ text for more informa-tion [Dan74, Ch. 14].

1.3 The Transfer Function

In our previous discussion, we assumed the transfer function was known, but in general thetransfer function will not be given. The types of filters that we investigate are named after thecorresponding type of polynomial (rational function) used in the approximation.

Consider the normalized passband [−1,1], i.e. ωb = 1 and let ε > 0. In the passband wewant 1−ε< |H(iω)| < 1+ε, and in the stop band we want |H(iω)| to be large. We then choosea polynomial (or ratio of polynomials) P (ω) with |P (ω)| to be close to 0 in the interval [−1,1]and outside this interval we want |P (ω)| to be large. Some typical properties of P (ω) that wedesire are outlined below

1. P (ω) is a polynomial of degree n or a rational function of polynomials of degree at mostn

2. |P (ω)| ≤ 1 in the interval −1 ≤ω≤ 1

3. P (ω) has all of its zeros in the interval −1 <ω< 1

4. |P (ω)| > 1 for |ω| > 1

5. P (1) = 1

Property 1 is chosen so that the approximation can be carried out in a finite sequence ofsteps (a realizable filter cannot contain an infinite process); property 2 is chosen so that theattenuation in a normalized passband of [−1,1] is minimal; it is clear that we do not want anyzeros for ω outside [−1,1] (we want P (ω) to grow large outside this interval), so this is whyproperty 3 is needed; property 4 is chosen so that the attenuation in the stopband is large;and property 5 is just a normalization requirement.

If we want to extend our passband to [−ωb ,ωb], then we apply the transformation ω = ωωb

into the previous equation. Because then by property 2,∣∣∣∣P (ω

ωb

)∣∣∣∣≤ 1 for |ω| ≤ |ωb | (1.16)

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And by applying a linear fractional transformation we can extend the passband to any interval[ωa ,ωb]. Thus once we can construct the normalized lowpass filter, we can simply transformthe results to conform to any passband interval.

To construct a filter with a passband of [ωa ,ωb], an engineer will be given: the maxi-mum attenuation allowed in the passband (denoted Amax), the minimum attenuation in thestopband (denoted Amin), and the frequency ωH where the stopband starts (that is A(ωH ) =Amin ). And so this brings us to the attenuation function A(ω).

Definition 1.4. The attenuation function, which measures the attenuation (in decibels dB)due to the filter at any frequency ω, is [Dan74, p. 3]

A(ω) :=−20log |T (iω)| = 10log |H(iω)|2 (1.17)

Sometimes |H(iω)|2 is called the magnitude or gain of the signal [LTE01, p. 76]. The en-gineer will be given the passband range [ωa ,ωb] and the requirements Amax, Amin in deci-bels, and the frequency ωH where the stopband begins, and then [typically] the problem isto design the filter with the lowest degree n that satisfies these requirements (a higher de-gree will take more time in computation and may require more expensive parts to realize).From these specifications the minimal filter degree n, is determined and the filter can be re-alized [Dan74, Sec. 2.6, 3.7]. However we will not focus on determining the minimal degreefor the filter, and we will assume the filter degree n is known.

Consider a normalized passband of[−1,1]. In the passband we want the attenuation to beminimal, that is A(ω) ≈ 0 for ω ∈ [−1,1], so from (1.17)

|H(iω)|2 ≈ 1

Definition 1.5. Let P (ω) denote the approximating polynomial (ratio of polynomials) usedfor the filter with the properties listed above, and let ε> 0. We set

|H(iω)|2 = 1+ (εP (ω)

)2 (1.18)

Because then for ω ∈ [−1,1] by property 2 we have∣∣∣1+ (εP (ω)

)2∣∣∣≤ 1+ε2 ≈ 1 (1.19)

Also wherever the polynomial has a zero, the attenuation is also zero, i.e. when P (ω) = 0, wehave A(ω) = 0. The constant ε is called the ripple factor (the name will make sense later), andit determines the maximum attenuation in the passband Amax. From (1.19)

Amax = 10log(1+ε2)

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2 The Butterworth Filter

This type of filter is named after the British engineer Stephen Butterworth who designed it[But30]. We shall see that the Butterworth filter has the property that the attenuation in thepassband is maximally flat, this means that the frequencies in the passband are attenuatedsimilarly (the attenuation is as uniform as possible). This is desirable since the frequencies inthe passband are the ones we "like", and the Butterworth filter has very little effect on them.

The Butterworth filter is very simple, so it makes for a good starting point. However due tothe simplicity, this type of approximation tends to be very impractical. Much of our discus-sion here is borrowed from [Dan74, Ch. 2].

2.1 Butterworth Polynomials

The type of approximating polynomial we use for this filter are aptly called Butterworth poly-nomials.

Definition 2.1. The nth order Butterworth polynomial Bn(x) satisfies the following condi-tions [Dan74, p. 9]

1. Bn(x) is an nth degree polynomial

2. Bn(0) = 0

3. Bn(x) is maximally flat at the origin

4. Bn(1) = 1

From property 1, we can write

Bn(x) = c0 + c1x +·· ·+cn xn

Property 2 requires c0 = 0. When we say Bn(x) is maximally flat at the origin, we mean thatwe need as many derivatives as possible of Bn(x) to be 0 at x = 0. So

dBn(x)

d x= c1 +2c2x +·· ·+ncn xn−1

Thus we see c1 = 0. Similarly we see that higher order derivatives are made 0 by making thecorresponding coefficient zero. So we have

Bn(x) = cn xn

Property 4 then requires that cn = 1. Therefore we found the nth order Butterworth polyno-mial to be

Bn(x) = xn

Theorem 2.1. The nth degree Butterworth polynomial is [Dan74, Sec. 2.4]

Bn(x) = xn (2.1)

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Figure 2.1: Plots of the Butterworth polynomial with degree n = 4 and 5 on the interval [−1,1].Notice that near the origin the polynomial is relatively flat, which will cause thefrequencies in the passband to have an almost uniform attenuation. Plots madein R

2.2 The Butterworth Filter

We consider filter that has a normalized passband interval of [−1,1] and ε = 1. We let ε = 1,since the Butterworth filter is maximally flat in the passband, whereas other filters will havea "ripple" effect in the passband and ε will have more meaning. However, we do not mean tosay you cannot have a Butterworth transfer function with a ripple factor ε 6= 1; setting ε = 1merely simplifies the results.

We know the magnitude of the signal is (Definition 1.5)

|H(iω)|2 = 1+ (Bn(ω)

)2 = 1+ω2n

and so by substituting ω=−i s we obtain

H(s)H(−s) = 1+ (−s2)n = 1+ (−1)n s2n (2.2)

Lemma 2.1. The zeros s j of the transfer function H(s)H(−s) with negative real part are

s j = e iθ j

θ j = π2n (2 j −1) if n even

θ j = πn j if n odd

(2.3)

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for j ∈ J with

J =

j :n

2+1 ≤ j ≤ 3n

2

n is even

J =

j :n +1

2+1 ≤ j ≤ 3n −1

2

n is odd

Proof. Notice (2.3) follows immediately from (2.2) for j = 1, ...,2n. So all that needs to be doneis verify the index set J = j : ℜs j < 0. We begin by considering the case where n is even,

ℜs j = cos(θ j ) < 0

⇐⇒ π

2< π

2n(2 j −1) < 3π

2

⇐⇒ n +1

2< j < 3n +1

2

Since j is an integer, we find our bounds on j to be n2 +1 ≤ j ≤ 3n

2 . Similarly when n is oddwe find that n+1

2 ≤ j ≤ 3n−12 .

Now we can easily find the constants α j for the transfer function. Let n be even, −1n = 1and

1

H ′(s)= 1

2ns2n−1

Thus

α j = 1

2n(s j )2n−1 = 1

2ne−iθ j (2n−1) = −1

2ne iθ j

The last equality is due to cosθ j (2n−1) =−cosθ j and sinθ j (2n−1) = sinθ j , which are easilyverified by applying the angle addition identities.

Now consider the case when n is odd. We similarly find

α j = −1

2n(s j )2n−1 = −1

2ne−iθ j (2n−1) = −1

2ne iθ j

This time, since θ j = π j /n, we have cosθ j (2n − 1) = cosθ j and sinθ j (2n − 1) = −sinθ j ,again from the angle addition identities. So the only difference between n being even or oddis the angle θ j and the index set that j runs along. The following theorem summarizes ourresults.

Theorem 2.2 (The Butterworth Filter). Let

Jn =

j :n

2+1 ≤ j ≤ 3n

2

n is even (2.4)

Jn =

j :n +1

2+1 ≤ j ≤ 3n −1

2

n is odd (2.5)

The poles of the transfer function for the Butterworth filter are given by s j = exp(iθ j ) , where θ j

is given in (2.3). We can express T (s) in partial fractions as

T (s) = −1

2n

∑j∈Jn

e iθ j

s − s j(2.6)

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If xin(t ) and xout(t ) represents the input and output signals at time t respectively, then theButterworth filter is given by

xout(t ) =∫ ∞

0

(−1

2n

∑j∈Jn

e iθ j+s jτ

)xin(t −τ)dτ (2.7)

2.3 Conclusions

Lets study the following example problem. We are given the following requirements for alowpass Butterworth filter (with a passband of |ω| ≤ ωb): Amax = 0.1dB, Amin = 30dB andωH /ωb = 1.3, what degree n is necessary [Dan74, p. 12]? The required degree is n = 21, whichis quite when high compared to the other filters we will have at our disposal.

The Butterworth Filter is mathematically simple (compared to the others we will study),but it comes with a cost that it is not very practical. While it keeps the frequencies in thepassband relatively constant, it is not very good at attenuating frequencies in the stopband,and often requires a high degree to meet specific requirements that a filter designer needs.The other filters we discuss will be able to meet the same requirements with a significantlylower degree than the Butterworth filter. So while the Butterworth filter is not very practical,it provides a good introduction to the theory.

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Figure 2.2: Plot of the gain |T (iω)|2 for the Butterworth approximation with n = 5,ε = 0.15to show how well it approximates the ideal lowpass transfer function. While theapproximation isn’t so good, it still may find use in applications where speed isprioritized (since the Butterworth filter is very simple compared to the others wewill discuss).

14

3 The Chebyshev Filter

We want to approximate the ideal lowpass filter function say f (ω), with a function say g (ω)that is accurate in the interval [ωa ,ωb]. With the Butterworth filter, g approximates f wellnear 0, but not so well elsewhere. An approximation g is a Chebyshev approximation of f ifg minimizes max | f (ω)− g (ω)| for ω ∈ [ωa ,ωb] [Dan74, Sec. 3.1]. For this reason sometimes aChebyshev approximation is also called a min-max approximation.

We will find that the transfer function T (s) for a Chebyshev approximation is equiripple inthe passband. Meaning that in the passband, the attenuation oscillates between maximumsand minimums of equal magnitude [Dan74, Ch. 3].

3.1 Chebyshev Polynomials

Definition 3.1. The nth order Chebyshev polynomial is, [Akh70, p. 151]

Tn(x) = cos(n cos−1 x

)(3.1)

While it certainly is not obvious that Tn is an nth degree polynomial, the following lemmaaddresses this.

Lemma 3.1. The nth degree Chebyshev polynomial satisfies the following recursion relation:T0(x) = 1 and T1(x) = x,

Tn+1(x) = 2xTn(x)−Tn−1(x)

Proof. This proof is adapted from [Dan74, p. 29]. First we check the initial conditions of therecursion relation, observe

T0(x) = cos(0cos−1 x

)= 1

T1(x) = cos(1cos−1 x

)= x

Now let y = cos−1 x, then

Tn+1(x) = cos[(n +1)y

]= cos(ny)cos y − sin(ny)sin y

Tn−1(x) = cos[(n −1)y

]= cos(ny)cos y + sin(ny)sin y

⇒ Tn+1 +Tn−1 = 2cos(cos−1 x

)cos

(n cos−1 x

)= 2x cos

(n cos−1 x

)= 2xTn(x)

15

Figure 3.1: Plot of the Chebyshev polynomial for n = 4,5 on the interval [−1,1]. Here we cansee the equiripple property, how the polynomial oscillates between its maximum1 and minimum −1.

3.2 Transfer Function H(s) for the Chebyshev Filter

Similarly to the Butterworth Filter, we will discuss the normalized Chebyshev filter with apassband of [−1,1] and the ripple factor ε > 0. The magnitude of the input output transferfunction is |H(iω)|2 = 1+ (

εTn(ω))2, and by substituting s = iω,

H(s)H(−s) = 1+[εTn

( s

i

)]2

Lemma 3.2. Let s j =σ j + iω j , then the n roots s j of H(s) (with ℜ(s j ) < 0 ) are

σ j =−sin( π

2n(2 j −1)

)sinh

(1

nsinh−1 1

ε

)(3.2)

ω j = cos( π

2n(2 j −1)

)cosh

(1

nsinh−1 1

ε

)(3.3)

for j = 1,2, ...,n.

Proof. This proof is adapted from [Dan74, Sec. 3.10]. We first find the roots of H(s)H(−s) andthen consider those with σ j < 0. The roots of H(s)H(−s) are given by

Tn

( s j

i

)= cos

(n cos−1 s j

i

)=± i

ε(3.4)

Let, n cos−1 s j

i = x + i y , and so

± i

ε= cos(x + i y) = cos x cos i y − sin x sin i y

= cos x cosh y − i sin x sinh y

16

Then by equating the real and imaginary parts we see,

cos x cosh y = 0 sin x sinh y =∓1

ε(3.5)

Since cosh y > 0, then cos x = 0, hence

x = π

2(2 j −1)

for j = 1,2, ...,2n. So then, y =∓sinh−1 1/ε. Therefore,

n cos−1 s j

i= π

2(2 j −1)∓ i sinh−1 1

ε

Therefore,s j

i=ω j − iσ j = cos

(π

2n(2 j −1)∓ i

nsinh−1 1

ε

)Hence by using the angle addition identity for cosine and equating real and imaginary partswe have,

σ j =±sin( π

2n(2 j −1)

)sinh

(1

nsinh−1 1

ε

)ω j = cos

( π2n

(2 j −1))

cosh

(1

nsinh−1 1

ε

)The roots roots of H(s) are those which σ j < 0, i.e.

σ j =−sin( π

2n(2 j −1)

)sinh

(1

nsinh−1 1

ε

)ω j = cos

( π2n

(2 j −1))

cosh

(1

nsinh−1 1

ε

)for j = 1,2, ...,n.

3.3 The Chebyshev Filter

Now that we have the zeros of H , we are able to express T (s) = 1/H(s) in partial fractions

T (s) =n∑

j=1

α j

s − s jwith α j = 1

dd s H(s j )

(3.6)

We need to find α j , since the derivative of arccos x is −(1−x2

)−1/2, we have

d

d sH(s)H(−s) = 2ε2 cos

(n cos−1 s

i

)sin

(n cos−1 s

i

) −n√1− ( s

i

)2

1

i

= i nε2 cos(n cos−1 s

i

)sin

(2n cos−1 s

i

)p

1+ s2

Therefore

α j =√

1+ s2j

i nε2 cos(n cos−1 s j

i

)sin

(2n cos−1 s j

i

) (3.7)

17

Theorem 3.1. (The Chebyshev Filter): Let s j =σ j + iω j as in Lemma 3.2, and α j as in equa-tion (3.7). If xin(t ) and xout(t ) represents the input and output signals at time t respectively,then the Chebyshev filter is given by

xout(t ) =∫ ∞

0

(n∑

j=1α j e s jτ

)xin(t −τ)dτ (3.8)

3.4 Conclusions

Lets revisit the example filter specifications given at the end of Chapter 2, but now using theChebyshev transfer function. We have the following lowpass Chebyshev filter requirements:Amax = 0.1dB, Amin = 30dB and ωH /ωb = 1.3, what degree n is necessary [Dan74, p. 34]? Forthe Chebyshev filter, the minimal degree is n = 8, which is significantly smaller than the re-quired degree for the Butterworth filter (21).

Now, lets study the following theorem from [Dan74, p. 36].

Theorem 3.2. Suppose H(s) is an nth degree Chebyshev transfer function, and Q(s) is someother nth degree transfer function. If max |Q(ω)| < max |H(ω)| for ω ∈ [ωa ,ωb] (the passbandinterval). Then Q(ω) < H(ω) for ω ∈ (−∞,ωa)∪ (ωb ,∞) (the stopband interval).

Essentially this theorem tells us that if we have some other transfer function Q(s) that hasless attenuation in the passband than H(s), then Q(s) has less attenuation in the stopbandthan H(s). Hence the Chebyshev filter has the greatest stopband performance out of all thefilters for a fixed degree n. However since the Chebyshev filter is equiripple in the passband,some desired frequencies will be attenuated differently than others. Whereas with the But-terworth filter, the attenuation in the passband is maximally flat, so the desired frequenciesare all attenuated similarly.

18

Figure 3.2: Plot of the gain |T (iω)|2 for the Chebyshev approximation with n = 5,ε = 0.15 toshow how well it approximates the ideal lowpass transfer function. We can see theequiripple property of the transfer function in the passband, and the effect of theripple factor ε (the larger ε is, the larger the oscillations will be)

19

4 Elliptic Functions

Before we can discuss the elliptic filter, some background knowledge of elliptic functions isnecessary. In this Chapter we will introduce and gain familiarity with the Jacobi elliptic func-tions. To start we will go over some basic terminology used in the theory of elliptic functions.

We say that a function f (x) is periodic, if there is a nonzero constantΩ such that f (x+Ω) =f (x) [Akh70, p. 1]. The constant Ω is called the period of f ; clearly any integer multiple of Ωis also a period, and so we call Ω a primitive period of f if any other period of f is an integermultiple ofΩ. For example

f (x) = sin x

is a periodic function with primitive periodΩ= 2π.A function of a complex variable is called meromorphic in an open set D if it is differen-

tiable everywhere in D except on a set of isolated points where the function has poles. Forexample

g (z) = 1

z

is a meromorphic function inCwith a pole at z = 0. Meromorphic functions with two distinctprimitive periods are called elliptic functions [Akh70, p. 6].

Recall ∫ x

0

d tp1− t 2

= sin−1 x (4.1)

We could define sin x by inverting the integral in (4.1), we generalize this idea to a differentkind of integral to define Jacobi’s elliptic functions. Let P (x) be a polynomial. Consider∫

d xpP (x)

(4.2)

If P (x) has degree 2, then (4.2) is called a trigonometric integral and it will be some inversetrig function. If P (x) is of degree 3 or 4, then (4.2) is called an elliptic integral [Akh70, Sec. 17].

4.1 Elliptic Integrals

Historically elliptic integrals arose from the study of the arc length of an ellipse. There arethree kinds of elliptic integrals, but we will only be interested in the first kind; we instead referan interested reader to Akheizer’s text [Akh70, Sec. 17, 29-31] for a more in depth coverage ofthe other kinds.

Definition 4.1. The elliptic integral of the first kind [Akh70, Sec. 24]

u(ϕ,k) =∫ sinϕ

0

d t√(1− t 2)(1−k2t 2)

(4.3)

The parameter k (typically 0 ≤ k < 1 ) is called the modulus of the elliptic integral (sometexts use m = k2 as the modulus), andϕ is called the amplitude. This is the Jacobi form of the

20

elliptic integral, if we let t = sinθ, then d t = cos tdθ =p

1− t 2dθ, and we obtain the Legendreform

u(ϕ,k) =∫ ϕ

0

dθ√1−k2 sin2θ

(4.4)

The final form that we will make use of is the Riemann form, which we get from (4.3) by lettingsin2ϕ= z and t 2 = x, so d x = 2td t = 2

pxd t and

u(z,k) =∫ z

0

d x

2√

x(1−x)(1−k2x)(4.5)

These integrals we defined are sometimes referred to as incomplete elliptic integrals; thecomplete elliptic integral is obtained by setting ϕ=π/2.

Definition 4.2. The complete elliptic integral [Dan74, Sec. 5.5]

K := u(π

2,k

)=

∫ π2

0

dθ√1−k2 sin2θ

=∫ 1

0

d t√(1− t 2)(1−k2t 2)

(4.6)

We conclude this summary of the basics of elliptic integrals by introducing the comple-mentary forms.

Definition 4.3. We define the complementary modulus k ′ so that (k ′)2 +k2 = 1, that is

k ′ =√

1−k2 (4.7)

Similarly we define the complementary complete elliptic integral K ′ as [Dan74, Sec. 5.5]

K ′ := u(π

2,k ′

)=

∫ π2

0

dθ√1− (1−k2)sin2θ

=∫ 1

0

d t√(1− t 2

)(1− (1−k2)t 2

) (4.8)

21

Figure 4.1: Plot of the complete elliptic integral of the first kind K , and the complementaryintegral K ′.

4.2 Jacobi’s Elliptic Functions

By inverting the elliptic integral (4.4), we obtain what are called the Jacobi elliptic functions[Dan74, Sec. 5.6].

Definition 4.4.

elliptic sine sn(u;k) = sinϕ (4.9)

elliptic cosine cn(u;k) = cosϕ (4.10)

And the difference function as

dn(u;k) = dϕ

du(4.11)

Since sin2ϕ+cos2ϕ= 1 we have

sn2(u;k)+cn2(u;k) = 1

Socn(u;k) =

√1− sn2(u;k) (4.12)

Also

dn(u;k) = dϕ

du=

√1−k2 sin2ϕ=

√1−k2 sn2(u;k) (4.13)

So the three basic Jacobi functions can all be expressed in terms of elliptic sine. There are 9other Jacobi elliptic functions, all defined as quotients of these three, however we will onlyneed

cd(u;k) = cn(u;k)

dn(u;k)(4.14)

22

Often we write sn(u) = sn(u;k) when the modulus k is clear from context, and similarly forthe other Jacobi functions. Some useful properties that follow directly from these definitionsare outlined in the following lemma. A more in-depth table of values, and also plots of theJacobi functions, are included in the appendix.

Lemma 4.1. Basic properties of the Jacobi functions

I sn(0;k) = 0 sn(K ;k

)= 1

II cn(0;k) = 1 cn(K ;k

)= 0

III dn(0;k) = 1 dn(K ;k

)= k ′

Proof. We verify the identities for sn(u;k), and the rest follows from equations (4.12) and(4.13). So when u = 0, this means that the elliptic integral in (4.4) is 0, that is∫ ϕ

0

d t√1−k2 sin2 t

= 0

If ϕ = 0 then this integral will be 0, therefore sn(0;k) = sin0 = 0. Similarly if u = K , thenϕ=π/2, and so sn(K ;k) = sinπ/2 = 1.

We like to think of the Jacobi elliptic functions as generalizations of the trigonometric func-tions like sin x. In fact, when k = 0, the Jacobi elliptic functions degenerate into trig functions.Notice in (4.4) that u(ϕ,0) =ϕ, hence

sn(u;0) = sin(u)

cn(u;0) = cos(u)

dn(u;0) = 1

We called the Jacobi functions elliptic, so they must be meromorphic and doubly periodic.It turns out that sn(u) has primitive periods 4K and 2i K ′, and simple poles at u = i K ′ andu = 2K + i K ′ [Akh70, Sec. 25]. We formulate this fact as a theorem for reference.

Theorem 4.1. The Jacobi elliptic function sn(u) has primitive periods 4K and 2i K ′, and simplepoles at u = i K ′ and u = 2K + i K ′.

For now we conclude this section by noting that sn(u) is an odd function and both cn(u)and dn(u) are even. To show sn(u) is odd, we apply the change of variable t =−t to the ellipticintegral

−u =−∫ ϕ

0

d t√1−k2 sin2 t

=∫ −ϕ

0

d t√1−k2 sin2 t

Since sine is odd, sn(−u) = sin(−ϕ) =−sin(ϕ) =−sn(u). Evenness of cn(u) and dn(u) followsimmediately from (4.12) and (4.13), since

cn(−u) =√

1− sn2(−u) = cn(u)

And similarly for dn(u).

23

4.3 The Addition Theorems for the Jacobi Elliptic Functions

Recall the familiar trig identity

sin(α+β) = sinαcosβ+ sinβcosα

This identity allows us to express the trig function of a sum of two arguments α,β in termsof trig functions of the individual arguments. Indeed, such an identity proves to be quiteuseful in the study and applications of trigonometric functions. In this section we will provean analogous identity for elliptic functions, as well as go over a couple interesting examples.

Theorem 4.2. The addition theorem for elliptic sine.

sn(u + v) = snu cn v dn v + sn v cnu dnu

1−k2 sn2 u sn2 v(4.15)

Proof. The following method of proof is due to Darboux and Akhiezer [Akh70, Sec. 28]. Con-sider the equation

d x√(1−x2

)(1−k2x2

) + d y√(1− y2

)(1−k2 y2

) = 0 (4.16)

It has a transcendental integral∫ x

0

d x√(1−x2

)(1−k2x2

) +∫ y

0

d y√(1− y2

)(1−k2 y2

) = A (4.17)

where A is an arbitrary constant. Our strategy is to find an algebraic integral for (4.16) andcompare it to the transcendental integral in (4.17). Proceeding we let

u =∫ x

0

d x√(1−x2

)(1−k2x2

) (4.18)

v =∫ y

0

d y√(1− y2

)(1−k2 y2

) (4.19)

Notice that by inverting these integrals we have x = sn(u) and y = sn(v). Also we have by(4.17)

u + v = A

We consider an equivalent system of equations to (4.16), by letting the first term equal d t andthe second equal −d t , so then

d x/d t =√(

1−x2)(

1−k2x2)

d y/d t =−√(

1− y2)(

1−k2 y2) (4.20)

Square both sides,

(d x/d t )2 = (1−x2

)(1−k2x2

)(d y/d t )2 = (

1− y2)(

1−k2 y2) (4.21)

24

Now we differentiate (d x/d t )2

2d x

d t

d 2x

d t 2 =−2xd x

d t

(1−k2x2)−2k2x

d x

d t

(1−x2)

⇒ d 2x

d t 2 = x(2k2x2 −1−k2)

⇒ d 2 y

d t 2 = y(2k2 y2 −1−k2)

From here we have

yd 2x

d t 2 −xd 2 y

d t 2 = 2k2x y(x2 − y2)

We recognize the left hand side as a derivative, which gets us to

d

d t

(y

d x

d t−x

d y

d t

)= 2k2x y

(x2 − y2) (4.22)

Using equations (4.21), we obtain after some labor(y

d x

d t+x

d y

d t

)(y

d x

d t−x

d y

d t

)= y2

(d x

d t

)2

−x2(

d y

d t

)2

= y2 −x2 +k2x4 y2 −k2x2 y4

= (y2 −x2)(1−k2x2 y2) (4.23)

So by dividing (4.22) by (4.23) we have

dd t

(y d x

d t −x d yd t

)y d x

d t −x d yd t

=2k2x y

(y d x

d t +x d yd t

)k2x2 y2 −1

We recognize that both sides are logarithmically differentiated, that is

d

d tln

(y

d x

d t−x

d y

d t

)= d

d tln

(k2x2 y2 −1

)Integrating gives us

yd x

d t−x

d y

d t=C

(1−k2x2 y2)

for some constant C . With (4.20) in mind we obtain the algebraic form of the integral in (4.16)

y√

(1−x2)(1−k2x2)+x√

(1− y2)(1−k2 y2)

1−k2x2 y2 =C (4.24)

From here we can now establish the addition formula for elliptic sine by recalling that in-tegrals in (4.18) and (4.19) give us x = snu and y = sn v respectively. Therefore from (4.24)

sn v cnu dnu + snu cn v dn v

1−k2 sn2 u sn2 v=C (4.25)

25

We compare the algebraic integral obtained in (4.24) to the transcendental integral (4.17).Now since (4.25) is a consequence of (4.17) , (4.18) , (4.19), this constant C must be somefunction of the other constant A, that is C = f (A). Since A = u + v , we have

snu cn v dn v + sn v cnu dnu

1−k2 sn2 u sn2 v= f (u + v)

Now we just need to figure out this function f . Let v = 0 then sn v = 0, cn v = dn v = 1, hencewe conclude that f (u) = snu and the result follows.

The addition formulas for the other functions can be derived from Theorem 4.2 by usingthe simple relations we already discussed, however we will not go through the lengthy com-putations.

Corollary 4.1. The addition theorems for cn(u + v), dn(u + v), and cd(u + v)

cn(u + v) = cnu cn v − snu dnu sn v dn v

1−k2 sn2 u sn2 v

dn(u + v) = dnu dn v −k2 snu cnu sn v cn v

1−k2 sn2 u sn2 v

cd(u + v) = cdu cd v − snu sn v

1−k2 snu sn v cdu cd v

Recall the familiar identity

sin(x + π

2

)= cos(x)

By adding the quarter period π/2 to the argument of trigonometric sine, we obtain anothertrigonometric function (cosine) of the same argument. We like to think of the elliptic func-tions as generalizations of trig functions, and so we investigate what happens when we addthe quarter period K to the argument of elliptic sine. With the addition theorem we see

sn(u +K ) = snu cnK dnK + snK cnu dnu

1−k2 sn2 u sn2 K

= cnu dnu

1−k2 sn2 u

= cnu

dnu= cdu (4.26)

Because of this identity sometimes people prefer to call the function cd(u) elliptic cosineinstead of cn(u). Also from the addition theorem we can see sn(u +2K ) =−sn(u), similar tosin(x +π) =−sin(x).

26

Figure 4.2: Plot cd(x;k) and sn(x;k) showing that cd(x) = sn(x +K ).

4.4 Transformations of Jacobi Elliptic Functions

Our goal for this section is to develop the relations that transform an elliptic function of theform y = sn2(u/M ;λ) with some constant M , to x = sn2(u;k). More specifically we investigatewhat happens when the periods of y are linearly related to the periods of x. Here we borrowsome of the discussion from [Akh70, Sec. 35].

First considerd x√

4x(1−x)(1−k2x)= M

d y√4y(1− y)(1−λ2 y)

(4.27)

where M is a constant. We are trying to find an algebraic dependence between x and y , thattransforms the first elliptic differential into the second. To simplify the problem, we confineourselves to determining the integral of (4.27) that assigns x = 0 to y = 0. That is we want todetermine the algebraic dependence between x and y that follows from the relation∫ x

0

d t√4t (1− t )(1−k2t )

= M∫ y

0

d t√4t (1− t )(1−λ2t )

(4.28)

Now we set ∫ x

0

d t√4t (1− t )(1−k2t )

= u

We now replace (4.28) with the parametric equations

x = sn2(u;k)y = sn2(u/M ;λ)

(4.29)

since with the Riemann form of the elliptic integral (4.5), the upper limit x = sin2ϕ= sn2(u;k).

27

The problem now is to determine the conditions where the elliptic functions x and y areconnected by an algebraic relation. We will not go too deep with the details for this problem,but for the interested reader they can be readily found in Akhiezer’s text [Akh70, Ch. 6].

4.4.1 The First Degree Transformation

There are actually two first degree transformations, however we will only discuss one, since itis far more useful to our purposes than the other. This analysis has been adapted from [Akh70,Sec. 37].

We know the periods of x = sn2(u;k) are 2K and 2i K ′, similarly we denote by 2L and 2i L′

the periods of sn2(v ;λ). Then y = sn2(u/M ;λ) has periods 2ML and 2i ML′. This transforma-tion results from setting

ML = i K ′ i ML′ =−K

So going from x to y we essentially interchanged the roles of K and K ′.Consider the ratio y/x, which is a rational function of sn2(u;k). At u = i K ′, we have y =

sn2(L;λ) = 1, and x = sn2(i K ′;k) has a second order pole, hence at u = i K ′, y/x has a secondorder zero. Similarly when u = K , y/x has a second order pole. Thus

y

x= A

sn2(u;k)−1

for some constant A (because both sides have matching poles, zeros and periods). If we let utend to zero, on the right we have −A, and on the left after applying L’Hospital’s rule twice weobtain 1/M 2 (see the appendix for differentiation formulas). Explicitly we found that 1/M 2 =−A. Continuing

y = A sn2(u;k)

sn2(u;k)−1

Let u = i K ′, and we see that 1 = A, hence M =±i . Now set u =−K + i K ′, and we have

sn2(L+ i L′;λ) = sn2(−K + i K ′;k)

sn(−K + i K ′;k)−1= sn2(K + i K ′;k)

sn(K + i K ′;k)−1

by periodicity. Now since sn(K + i K ′;k) = 1/k we obtain

1

λ2 = 1/k2

1/k2 −1

Solving for λ yields λ= k ′. Let M =−i so we have

sn2(i u;k ′) =− sn2(u;k)

1− sn2(u;k)

And we have obtained the following theorem.

Theorem 4.3. The imaginary transformation of elliptic sinus

sn(i u;k ′) = isn(u;k)

cn(u;k)(4.30)

28

The formulas for cn(i u),dn(i u) and cd(i u) follow from (4.30), and they are

cn(i u;k ′) = 1

cn(u;k)

dn(i u;k ′) = dn(u;k)

cn(u;k)

cd(i u;k ′) = 1

dn(u;k)

These transformations show us how to deal with pure imaginary arguments of the ellipticfunctions in terms of real variables. So this transformation combined with the addition the-orems allows us to express the Jacobi elliptic functions of any complex variable, in terms ofreal variables.

The other first degree transformation results from letting i ML′ = K + i K ′ and ML = K , werefer the interested reader to [Akh70, Sec. 36] for the details of this transformation.

4.4.2 The nth Degree Transformation

Here we will skip most of the details, and instead provide the framework for the second prin-cipal transformation so that the results have some meaning. The second principal nth degreetransformation entails division of one of the periods by an integer n. We consider the follow-ing relations

x = sn(u;k) y = sn( u

M ;λ)

L = KM L′ = K ′

nM

Again we consider the ratio y/x, which is an even function of u. Also we find that the periodsof y/x are 2K and 2i K ′, since

sn(u +2K +2i K ′;k

)=−sn(u;k) =−x

sn

(u +2K +2i K ′

M;λ

)=−sn

( u

M;λ

)=−y

Then since y/x has the same periods as sn2(u;k), it must be a rational function of sn2(u;k).From here we will skip ahead to the results, and refer the more interested reader to Akhiezer’stext [Akh70, Sec. 40].

Theorem 4.4 (Second Principal nth degree transformation). By subjecting the periods to thefollowing transformations

L = K

ML′ = K ′

nM

We have

sn( u

M;λ

)= 1

Msn(u;k)

b n2 c∏

r=1

1+ sn2(u;k)c2r

1+ sn2(u;k)c2r−1

(4.31)

29

with

M =b n

2 c∏r=1

sn2(2r−1

n K ′;k ′)sn2

(2rn K ′;k ′) (4.32)

λ=n∏

r=1

θ24

(r K ′nK ;k ′

)θ2

4

((2r−1)K ′

2nK ;k ′) (4.33)

cr =sn2

( rn K ′;k ′)

cn2( r

n K ′;k ′) (4.34)

Where bxc denotes the floor function, and θ4 is one of the Jacobi theta functions, whichwe will discuss in the next section. The first principal nth degree transformation theorem isincluded below, as it will be useful later.

Theorem 4.5 (First Principal nth degree transformation). By subjecting the periods to thefollowing relations

L = K

nML′ = K ′

M

We have

λ= knbn/2c∏r=1

c22r−1 (4.35)

M =bn/2c∏r=1

c2r−1

c2r(4.36)

cr = sn2( r

nK ;k

)(4.37)

if n is odd

sn( u

M;λ

)= 1

Msn(u;k)

n−12∏

r=1

1− sn2(u;k)c2r

1−k2c2r sn2(u;k)(4.38)

if n is even

sn( u

M+L;λ

)=

n2∏

r=1

1− sn2(u;k)c2r−1

1−k2c2r−1 sn2(u;k)(4.39)

30

4.5 The Jacobi Theta Functions

The Jacobi theta functions are periodic, entire functions that can be defined as Fourier seriesthat rapidly converge (about 4 terms should suffice for most calculations). They depend on aparameter q with |q | < 1, we define [BE55, Sec 13.19]

Definition 4.5.

θ1(v, q

)= 2q1/4∞∑

j=0(−1) j q j ( j+1) sin

(2 j +1

)πv (4.40)

θ2(v, q

)= 2q1/4∞∑

j=0q j ( j+1) cos

(2 j +1

)πv (4.41)

θ3(v, q

)= 1+2∞∑

j=1q j 2

cos(2 jπv

)(4.42)

θ4(v, q

)= 1+2∞∑

j=1(−1) j q j 2

cos(2 jπv

)(4.43)

Sometimes the notation θ0 = θ4 is used.Consider the function sn(u;k), with periods 4K and 2i K ′, the parameter q is related to

these by [BE55, Sec 13.19]

q = e−πK ′K (4.44)

Alternatively, a more efficient computation for q for a given k can be found in [Akh70,Sec. 45], under the assumption that ∣∣∣∣∣1−p

k ′

1+pk ′

∣∣∣∣∣< 1

which holds when 0 < k < 1. On this note let

2l = 1−pk ′

1+pk ′ (4.45)

Now q can be computed from the following rapidly converging series

q = l +2l 5 +15l 9 +150l 13 +·· · (4.46)

For the nth degree transformation, we used the following notation in the computation ofλ, θ4(w ;k) (a semicolon instead of a comma). This was deliberate to emphasize that we aregiven k and we must compute q from (4.45) and (4.46) before we can evaluate the theta func-tion.

Another practical benefit of the Jacobi theta functions is that they provide a means forefficient computation of the Jacobi elliptic functions. Let

v = x

2K

then by [BE55, Sec. 13.20]

sn(x;k) = 1pk

θ1(v ;k)

θ4(v ;k)(4.47)

31

Also k can be uniquely determined by q with the following equations

pk = θ2

(0, q

)θ3

(0, q

) pk ′ = θ4

(0, q

)θ3

(0, q

)The other Jacobi functions can be expressed similarly

cn(x;k) =√

k ′

k

θ2(v ;k)

θ4(v ;k)dn(x;k) =

pk ′ θ3(v ;k)

θ4(v ;k)

Moreover from [BE55, Sec. 13.20] we have

K = π

2θ2

3(0;k) K ′ = K

πlog

(1

q

)

32

5 Elliptic Rational Function

The elliptic rational function is the approximating function used for the elliptic filter, andthe key to understanding the elliptic filter lies with this function. There are many equivalentways of defining and formulating the elliptic rational function, but all of which require useof the Jacobi elliptic functions. It is also known as the Chebyshev rational function [Dan74],or the Chebyshev-Blashke product [NT13]. The mathematics behind this function date backto Zolotarev in 1877 [Zol77] a student of Chebyshev, however it wasn’t until 1958 when Cauerused Zolotarev’s ideas to design the elliptic filter for signal processing [Cau58]. As such some-times the elliptic filter is called the Zolotarev-Cauer filter.

In fact these various sources all provide different definitions and derivations of the ellipticrational function, which don’t appear to be equivalent. This caused a lot of confusion, as thefunctions they derived were used for the same purposes and had the same properties, butother than that, it appeared as if there were no other connections. For example consider inLutovac’s text [LTE01, Sec. 12.6] the elliptic rational function is defined to be

Rn(k, x) = cd( u

M;λ

)x = cd(u;k) (5.1)

where M is a scaling factor given in Theorem 4.5. However in Daniels’ text [Dan74, Sec. 5.12],the elliptic rational function is derived to be

Rn(k, x) = r1x

n−12∏

r=1

x2 − sn2(2r

n K ;k)

x2 − [k sn

(2rn K ;k

)]−2 if n odd (5.2)

Rn(k, x) = r2

n2∏

r=1

x2 − sn2(2r−1

n K ;k)

x2 − [k sn

(2r−1n K ;k

)]−2 if n even (5.3)

Where r1 and r2 are normalizing constants chosen so that Rn(k,1) = 1.The two functions don’t seem to be equivalent, and even after applying the nth degree

transformation to Lutovac’s form, the resulting rational function looks similar to Daniels’function, but not quite algebraically identical. These are standard textbooks that an engineerwould use to learn about signal processing (or to use as a reference), and it can be frustratingseeing such apparent discrepancies with no clear resolution. So where do these differencescome from, and what exactly is the elliptic rational function? Our goal for this chapter willbe to define and construct the elliptic rational function, and establish the connections be-tween the various results given from different authors. We will begin by examining some ofthe problems Zolotarev proposed and solved, and we will define the elliptic rational functionfrom the solution to one of his problems.

5.1 Statement of the Problems

We define the deviation of a function g (x) from a function f (x) on some interval I as

supI

∣∣ f (x)− g (x)∣∣

33

Consider the following problems adapted from [Akh70, Sec. 50].Problem A : Find the rational function y(t ) =ϕ(t )/ψ(t ) (withϕ andψ polynomials of degree

at most n) that deviate the least from the function

sgn t = −1 : t < 0

1 : t > 0

on the intervals

[−1/κ,−1] [1,1/κ] (with 0 < κ< 1 )

Problem B : Consider rational functions z(x) = f (x)/g (x) (with f and g polynomials of de-gree at most n) that satisfy |z(x)| ≥ 1 on the intervals (−∞,−1/k] and [1/k,∞) (with 0 < k < 1). Find the one that deviate the least from 0 on [−1,1].

The last problem we will consider isProblem C : Of all real rational functions

Y (X ) =p

XΨ(X )

Φ(X )

where Ψ and Φ are polynomials of degree r , find the one that deviates the least from 1 on[1,1/κ2] with 0 < κ< 1.

We now show that problems A, B and C are actually equivalent, that given a solution to one,we can construct a solution to the other. The following discussion is adapted from [Akh70,Sec. 50].

First suppose that z(x) = f0(x)/g0(x) is a solution to Problem B. We can see that f0(x) mustbe a polynomial of degree n, because if it were less than n consider z = kx f0(x)/g0(x). z isalso a rational function with the degree of the numerator and denominator at most n and wealso have the inequality |z(x)| ≥ 1 on the intervals (−∞,−1/k] and [1/k,∞). However

max[−1,1]

|z(x)| = max[−1,1]

∣∣∣∣kxf0(x)

g0(x)

∣∣∣∣≤ k max[−1,1]

∣∣∣∣ f0(x)

g0(x)

∣∣∣∣< max[−1,1]

|z(x)|

contradicting our assumption that z(x) is a solution to problem B.From the statement of the problem we see that on the intervals (−∞,−1/k] and [1/k,∞),

min |z| = 1. Now letmax[−1,1]

|z(x)| = m (5.4)

We see that m < 1, shown by the function z(x) = kx which solves problem B in the case n = 1.Let

y(t ) = 1−m

1+m

z(t )−pm

z(t )+pm

(5.5)

t = 1+pk

1−pk

xp

k −1

xp

k +1(5.6)

κ=(

1−pk

1+pk

)2

(5.7)

34

We show that y(t ) solves problem A. Now if −1 ≤ x ≤ 1 then it is easy to check that t rangesfrom −1/κ to −1. Similarly for |x| ≥ 1/k then 1 ≤ t ≤ 1/κ. Now we have for the first interval

m = maxx∈[−1,1]

|z(x)| = maxt∈[−1/κ,−1]

|z(t )|

And on this interval we have

y − sgn t = y +1 = 2(z +mp

m)

(1+m)(z +pm)

We note that y +1 is an increasing function since

d y

d z= 2

pm

(z +pm)2

> 0

therefore

max[−1/κ,−1]

|y +1| = 2(m +mp

m)

(1+m)(m +pm)

= 2p

m

1+m(5.8)

Similarly we have in the second interval

1 = minx≤−1/k

|z(x)| = mint∈[1,1/κ]

|z(t )|

On this interval

y −1 = −2z(m +pm)

(1+m)(z +pm)

By the same reasoning we see that y − 1 is a decreasing function so its maximum on thisinterval is

max[1,1/κ]

∣∣y − sgn t∣∣= max

[1,1/κ]

∣∣y −1∣∣= 2

pm

1+m(5.9)

So we found that the deviation of y(t ) from sgn t in the intervals required in problem A is

µ= 2p

m

1+m(5.10)

Also we see that µ is an increasing function of m in (0,1) since

dµ

dm= 1−mp

m(1+m)2

So since m was the smallest deviation for problem B, we see that µ is the smallest deviationfor problem A, and therefore y(t ) is a solution to problem A. Now conversely, given a solutionto problem A, we can apply the inverse transformations in (5.5), (5.6), (5.7) to find a solutionto problem B.

We now show that problem A and problem C are equivalent. Suppose Y (X ) =pXΨ(X )/Φ(X )

is a solution to problem C. Let X = t 2 so then the interval X ∈ [1,1/κ2] becomes t ∈ [−1/κ,−1]and [1,1/κ], and our function transforms to

Y = tΨ

(t 2

)Φ

(t 2

)

35

And so we have

maxX∈[1,1/κ2]

∣∣∣∣1−pXΨ (X )

Φ (X )

∣∣∣∣= maxt∈[−1/κ,−1]∪[1,1/κ]

∣∣∣∣∣sgn t − tΨ

(t 2

)Φ

(t 2

) ∣∣∣∣∣And we have arrived at the solution to problem A with n = 2r + 1 (the numerator ψ(t ) hasdegree 2r +1 and the denominator ϕ(t ) has degree 2r ). Similarly we can show the converse,that is given a solution to problem A we can apply the inverse transformation to obtain asolution to problem C. So now everything rests upon the solution to problem C.

5.2 Solution to Problem C

We will apply the following theorem due to Chebyshev [Akh70, Sec. 51]

Theorem 5.1 (Chebyshev). Let [a,b] be a finite closed interval, and let f (x) and s(x) be con-tinuous functions on this interval, with s(x) 6= 0. We consider expressions of the form

W (x) = s(x)q0 +q1x +·· ·+qn xn

p0 +p1x +·· ·+pm xm

with m,n given. Of these functions W (x) there exists one deviating the least from f (x) on [a,b].In particular if this function has the form

Q(x) = s(x)B(x)

A(x)= s(x)

b0 +b1x +·· ·+bn−νxn−ν

a0 +a1x +·· ·+am−µxm−µ

with 0 ≤µ≤ m,0 ≤ ν≤ n, am−µ 6= 0 and B(x)/A(x) is irreducible. Then the number of points of[a,b] at which f (x)−Q(x) takes its maximal value is not less than m +n +2−minµ,ν. Thisproperty completely characterizes the function Q(x).

The most important part of the theorem that we will make use out of is where it assertsthat | f (x)−Q(x)| achieves its maximum value m+n−d +2 times in the interval [a,b]. In factthe converse holds too, that is if we find a function Q(x) such that f (x)−Q(x) achieves itsmaximum deviation m+n−d+2 times in the interval [a,b], then Q(x) deviates the least fromf (x) in the interval [a,b]. This fact is what we will need to verify our solution.

Now we apply this to problem C, where the interval is [1,1/k2], f (x) = 1, s(x) = pX and

m = n = r . We present the solution in parametric form and show that this indeed solvesproblem C.

X = sn2(u;k) Y = 2λ

1+λ sn( u

M;λ

)(5.11)

with M given in Theorem 4.4.

Proof. Adapted from [Akh70, Sec. 51]. Let 4L and 2i L′ be the periods of sn(v ;λ), and 4K ,2i K ′

denote the periods of sn(u;k); we require L = K /M and L′ = K ′/((2r +1)M). By applying the

36

nth degree transformation (Theorem 4.4) we see that Y is indeed a rational function of therequired form.

Y = 2λ

1+λsn(u;k)

M

r∏α=1

1+ sn2(u;k)c2α

1+ sn2(u;k)c2α−1

with cα given by (4.34). By subbing in X = sn2(u;k) we see Y is of the required form

Y = 2λ

1+λ

pX

M

r∏α=1

1+X /c2α

1+X /c2α−1

Now consider the difference ∆(X ) = 1−Y as x runs along the interval [1,1/k2]. To do thiswe will discover the values of u that keep X in this interval, and then examine ∆(X ) for thesevalues.

Let u = K + i v , so

X = sn2(K + i v ;k) = cn2(i v ;k)

dn2(i v ;k)= 1

dn2(v ;k ′)

Let v increase from 0 to K ′. Then dn2(v ;k ′) decreases from 1 to dn2(K ′;k ′) = 1− (k ′)2 = k2,therefore X increases from 1 to 1/k2 as desired.

Now we check what happens to ∆(X ) in this interval

∆(x) = 1− 2λ

1+λ sn

(K + i v

M;λ

)= 1− 2λ

1+λ1

dn(v/M ;λ′)

Now when v increases from 0 to K ′, w = v/M increases from 0 to (2r +1)L′. We know thatdn(w ;λ′) is always between 1 and λ, also it obtains the maximum and minimum when w =0,L′,2L′, . . . ,2r L′, (2r +1)L′, and respectively dn(w ;λ′) = 1,λ,1, . . . ,1,λ. The values of ∆(X ) atthese points are

1−λ1+λ , −1−λ

1+λ , . . . , −1−λ1+λ

So 1−Y takes on its maximum value on [1,1/k2] with alternating signs 2r +2 successivetimes. Therefore Chebyshev’s Theorem tells us (note µ= ν= 0) that Y deviates the least from1 on the interval [1,1/k2].

Plots of the solutions to these problems can be found in the Appendix.

5.3 Elliptic Rational Function

Recall the properties of the approximating function P (ω) given in Section 1.3: |P (ω)| ≤ 1 in[−1,1], and outside this interval we want P (ω) to grow large. Notice the similarities to thesolution z(x) in problem B; in [−1,1] we have |z(x)| ≤ m, where

m = max[−1,1]

|z(x)|

37

and in the intervals (−∞,−1/k], [1/k,∞) (for 0 < k < 1) we have |z| ≥ 1. If we simply multiplyz(x) by 1/m, we have in [−1,1], |z(x)| ≤ 1 and in (−∞,1/k]∪ [1/k,∞), |z| ≥ 1/m. Since m isvery small, 1/m is very large and we have exactly what we need.

Definition 5.1. Let zn(k, x) be the solution to problem B with the degree of the numerator anddenominator at most n, the parameter 0 < k < 1 indicating the intervals (−∞,−1/k]∪[1/k,∞)where |zn(k, x)| ≥ 1, and with m = max |zn(k, x)| on [−1,1]. The elliptic rational function is:

Rn(k, x) = zn (k, x)

m(5.12)

The way we define the elliptic rational function is very similar to one way of defining theChebyshev polynomials. Consider the following problem:

Of all polynomials p(x) of degree n with leading coefficient 1, we desire the one which devi-ates the least from 0 on [−1,1].

The solution to this problem is

p(x) = 21−n cos(n arccos x) (5.13)

where the maximum deviation from 0 is ν = 21−n [Akh70, Sec. 52]. And we then define theChebyshev polynomial as Tn(x) = p(x)/ν = cos(n arccos x). We defined the elliptic rationalfunction in a similar fashion based off the solution to problem B.

Although, there is one slight problem with this definition. From the previous section wecan construct the solution to problem B from problem A and C, but only for odd degree n.It is easy to show that the solution to problem A must be an odd function, and so the degreeof the numerator and denominator is at most n, which is odd. Thus when we construct thesolution to problem B, based off the solution to A, we can only have a solution for when n isodd. So we seek to construct the solution to problem B which is not dependent on the parityof n.

Lemma 5.1. The solution to problem B can be represented parametrically as

zn(k, x) =λcd( u

M;λ

)x = cd(u;k) (5.14)

where L = K /(nM) and L′ = K /M as in the first principal nth degree transformation (Theorem4.5). The maximum deviation of z from 0 on [−1,1] is m =λ [NT13, Sec. 3.2.5].

Proof. First we prove that the maximum deviation from 0 on [−1,1] is m = λ. Let u rangefrom 0 to 2K . Then

x = cd(u;k) = sn(u +K ;k) (5.15)

At u = 0, x = snK = 1. As u increases to 2K , sn(u +K ) decreases to −1, and hence x ∈ [−1,1].Also

zn(k, x) =λcd( u

M;λ

)=λsn

( u

M+L;λ

)(5.16)

38

Consider when u = K , then z =λsn((n+1)L;λ) =λ,0,−λ,0 if n = 0,1,2,3 mod 4 respectively.Since elliptic sinus is absolutely bounded by 1 for real arguments, we see that max |z| = λ on[−1,1].

Also we can easily see that |zn(k, x)| ≥ 1 on the intervals (−∞,−1/k]∪ [1/k,∞). Let u = i vand have v range from K ′ to K ′+ i K , so

x = cd(i v ;k) = 1

dn(u;k ′)(5.17)

at v = K ′, dn(v ;k ′) = k so x = 1/k. As v runs to K ′+ i K , dn(v ;k ′) is decreasing untildn(K ′+ i K ;k ′) = 0, hence x increases from 1/k to +∞. For z we have

z =λcd

(i v

M;λ

)= λ

dn( v

M ;λ′) (5.18)

At v = K ′, we have

z = λ

dn(K ′

M ;λ′) = λ

dn(L′;λ′)= 1 (5.19)

And as v runs to K ′+ i K , dn(v/M ;λ′) will oscillate from 1 to −1, ending at v = K ′+ i K whichgives

dn

(K ′+ i K

M;λ′

)= dn(L′+ i nL;λ′) =λ,0,−λ,0

if n mod 4 = 0,1,2,3 respectively

Hence on [1/k,∞), |z| ≥ 1, and a similar argument will show the same for the interval (−∞,−1/k](let v range from K ′+2i K to K ′+3i K ).

Now to show that zn(k, x) solves problem B, instead we will show that yn(κ, t ) solves prob-lem A, with

yn(κ, t ) = 1−λ1+λ

z −pλ

z +pλ

(5.20)

t = 1+pk

1−pk

xp

k −1

xp

k +1(5.21)

κ=(

1−pk

1+pk

)2

(5.22)

These are the same relations discussed in Section 5.1, and by that previous discussion if weshow y solves problem A, then z solves problem B. We want to show that |y−sgn t | is minimaxon [−1/κ,−1]∪[1,1/κ]. Let x ∈ [−1,1], we found earlier this implies t ∈ [−1/κ,−1] (see Section5.1). Now consider the difference y − sgn t = y +1 on this interval.

y +1 = 2(z +λpλ)

(1+λ)(z +pλ)

=2(λcd

( uM ;λ

)+λpλ)(1+λ)

(λcd

( uM ;λ

)+pλ)

39

Since x ∈ [−1,1], then u ∈ [0,2K ], and so u/M ∈ [0,2nL]. The maximum deviation here is

µ= 2pλ

1+λAnd we reach this maximum whenever cd(u/M ;λ) =±1. This happens when u/M is an evenmultiple of L, i.e. u/M = 2 j L where j is an integer. Since LM = K /n, we see

u = 2 j

nK

for j = 0, ...,n. Therefore |y − sgn t | hits its maximum n +1 times on the interval [−1/κ,−1]. Asimilar argument will show that |y − sgn t | hits its maximum n +1 times on the other interval[1,1/κ] as well (let |x| ≥ 1/k). Therefore by Chebyshev’s Theorem yn(κ, t ) deviates the leastfrom sgn t on the required intervals.

From Lemma 5.1 and Definition 5.1 we have the following theorem.

Theorem 5.2 (The Elliptic Rational Function). Denote the periods of sn(v ;λ) as L,L′, withL = K /(nM) and L′ = K ′/M (as in Theorem 4.5).

Rn(k, x) = cd( u

M;λ

)x = cd(u;k) (5.23)

Often the notation ξ= 1/k is used, and this parameter ξ is called the selectivity factor. Sinceλ is determined by k,n the notation Ln(ξ) = 1/λ is often used, and we call Ln(ξ) the discrimi-nation factor [LTE01, Sec. 12.6]. We will avoid this notation since we have been denoting thecomplete elliptic integral with modulus λ as L, and this could cause confusion.

We can show that Rn(k, x) is a rational function of polynomials. Indeed, suppose n is even,as the case where n is odd is nearly identical. From the nth degree transformation, we cancompute the constants M and λ (see Theorem 4.5), as well as express Rn(k, x) as a rationalfunction of the required form. Proceeding

Rn(k, x) = cd(u/M ;λ) = sn(u/M +L;λ)

=n2∏

r=1

1− sn2(u;k)c2r−1

1−k2c2r−1 sn2(u;k)(5.24)

where

cr = sn2(

r K

n;k

)(5.25)

We want to express Rn(k, x) as a rational function of x, but what we have is a rational functionof elliptic sinus. The trick to do this is to express x = cd(u;k) in terms of elliptic sinus. Observe

x2 = cd2(u;k) = 1− sn2(u;k)

1−k2 sn2(u;k)

Solve for sn2(u;k),

sn2(u;k) = x2 −1

x2k2 −1

40

Substitute back into (5.24)

Rn(k, x) =n2∏

r=1

1− x2−1(x2k2−1)c2r−1

1−k2c2r−1x2−1

x2k2−1

(5.26)

=n2∏

r=1

x2k2c2r−1 −x2 +1− c2r−1

x2k2c2r−1 −x2k2c22r−1 +k2c2

2r−1 − c2r−1(5.27)

=n2∏

r=1

x2(k2c2r−1 −1

)+1− c2r−1

c2r−1(k2c2r−1 −1

)+x2k2c2r−1 (1− c2r−1)(5.28)

Note that in the solution to problem C, we used the second principal nth degree transfor-mation (Theorem 4.4), and the solution given above to problem B, we used the first principalnth degree transformation (Theorem 4.5). Since the solution to problem B follows from thesolution to problem C, where does the solution change from one nth degree transformationto the other?

I suspect the change happens with the mapping from problem A to problem B (from Sec.5.1). Now, if we take the solution given for problem C, and use the mappings we defined toobtain the solution to problem B, the resulting expression is significantly different from thesolution in Lemma 5.1; I do not know if it can be shown algebraically that these two expres-sions are equal.

The most common definition of the elliptic rational function coincides with Theorem 5.2[LTE01], and its also much simpler than the expression obtained from mapping the solutionto problem C to problem B and then dividing by the constant m = max |z| on [−1,1]. For thesereasons we use the solution from Lemma 5.1 to create the elliptic rational function.

Some plots of the elliptic rational function for n = 4,5 are included on the next page.

41

Figure 5.1: Plot of R4(k, x) with k = 0.7 on the intervals [−1,1] and [1,8], with horizontal linesat ±1/λ.

Figure 5.2: Plot of R5(k, x) with k = 0.7 on the intervals [−1,1] and [1,6] with horizontal linesat ±1/λ.

5.3.1 Connections Between Texts

Here we aim to understand the differences between the elliptic rational function given inDaniels’ text [Dan74] and in Lutovac’s text [LTE01], and in the paper by Tuen Wai Ng and Chiu

42

Yin Tsang [NT13]. We have already seen how Lutovac’s solution follows from the solution ofZolotarev’s problems, but we have not seen any connection to Daniels’ results yet. Comparethe rational function we derived in (5.28) to (5.3); they’re similar but not equivalent.

I believe the problem lies with the way the elliptic rational function was derived in Daniels’text [Dan74, Sec. 5.4, 5.8], in fact this approach seems to originate from Cauer [Cau58, p. 738-758]. Here Daniels sets up the following differential equations

du = MdRn√(1−R2

n)(1−λ2R2

n) (5.29)

= d x√(1−x2)

(1−k2x2

) (5.30)

And the solution is

x = sn(u +C1;k) (5.31)

Rn(k, x) = sn(u/M +C2;λ) (5.32)

with arbitrary constants C1,C2. Daniels arbitrarily sets the constant C1 = 0, and this leads tothe rational functions (5.2) (5.3) (here, Cauer has the same results with different notation).However if we set C1 = K , then x = cd(u;k) agreeing with our earlier results. Also when u = 0,then x = 1, and since we require Rn(k,1) = 1 [Dan74, p. 53],

1 = sn(C2;λ)

hence C2 = L, and we have Rn(k, x) = cd(u/M ;λ), as before.As for the paper by Tuen Wai Ng and Chiu Yin Tsang, they define the Chebyshev Blaschke

Product parametrically [NT13]

fn,k (x) =pλcd(nL w ;λ) x =

pk cd(K w ;k) (5.33)

Then they show z =pλ fn,k (x) solves a modified version of problem B. The same problem

except |z| ≥ 1 on(−∞,−1/

pk]∪

[1/p

k,∞), and z deviates the least from 0 on the interval[

−pk,p

k]

.

Substituting u = w/K and M = K /(nL) returns the same notation we’ve been using for theelliptic rational function, and we see this is really the same idea, just scaled accordingly tosuit their purposes.

5.4 Zeros and Poles of the Elliptic Rational Function

We can express the elliptic rational function in terms of its zeros and poles since it’s a rationalfunction of polynomials. Let x j be the j th zero, and xp, j be the j th pole, then

Rn(k, x) = r0

n∏j=1

x −x j

x −xp, j(5.34)

43

where r0 is a normalizing constant so that Rn(k,1) = 1 [Dan74, Sec. 5.3]. Before we derive thezeros and poles, we will make use of the following lemma.

Lemma 5.2.

cd(u + i K ′;k) = 1

k cd(u;k)(5.35)

Proof. Observe

cd(u + i K ′)= sn

(u +K + i K ′)

= snu cn(K + i K ′) dn

(K + i K ′)+ sn

(K + i K ′) cnu dnu

1−k2 sn2 u sn2 (K + i K ′)

= 1

k

cnu dnu

1− sn2 u

= dnu

k cnu= 1

k cdu

The zeros of the elliptic rational function happen when

Rn(k, x) = cd( u

M;λ

)= 0

Since cd(w ;λ) = 0 when w = L,3L, . . . , (2 j −1)L, we have that u j = (2 j −1)K /n for j = 1, . . . ,n.Therefore the zeros of the elliptic rational function are at

x j = cd

(2 j −1

nK ;k

)(5.36)

Now the poles are the solutions to

1

Rn(k, x)= 1

cd( u

M ;λ) = 0

This equation only has solutions for complex u, so let u = v + i K ′. Then

1

cd( u

M ;λ) = 1

cd( v

M + i L′;λ) =λcd

( v

M;λ

)= 0

We just solved this equation and found v j = (2 j −1)K /n. Hence the poles are at

xp, j = cd(v j + i K ′;k

)= 1

k cd(

2 j−1n K ;k

)= 1

kx j(5.37)

44

So we see how the zeros and poles are related to one another, also they come in pairs withequal magnitudes but opposite signs [LTE01, p. 532,533], that is x j =−xn− j+1. Hence we canwrite

Rn(k, x) = r1

n2∏

j=1

x2 −x2j

x2 − 1k2x2

j

if n is even (5.38)

Rn(k, x) = r2x

n−12∏

j=1

x2 −x2j

x2 − 1k2x2

j

if n is odd (5.39)

with r1,r2 normalizing constants so that Rn(k,1) = 1. Explicitly

r1 =n2∏

j=1

1− 1k2x2

j

1−x2j

r2 =n−1

2∏j=1

1− 1k2x2

j

1−x2j

This gives us another way of expressing the elliptic rational function as a ratio of polynomi-als, equivalent to applying the nth degree transformation. This form is useful since it is easyto identify the zeros and poles of the function.

5.5 The Elliptic Rational function for n = 1,2,3

Here we will derive explicit formulas for the elliptic rational function that avoid the use ofthe Jacobi elliptic functions. With these formulas and the nesting property, we can obtainexpressions for any order n = 2i 3 j . People have made algorithms that exploit the nestingproperty of the elliptic rational function for orders n = 2i 3 j , and these algorithms performsignificantly faster than the more traditional means of computing (i.e. using Theorem 5.2 or(5.28)) [LT05, p. 606, 607].

The nesting property of the elliptic rational function is as follows (see [LTE01, Sec. 12.7.1]).Denote the selectivity factor ξ = 1/k and the discrimination factor Ln(ξ) = 1/λ. We use thisnotation to differentiate between the discrimination factor of Rn and Rm .

Theorem 5.3 (Nesting Property).

Rmn(ξ, x) = Rm(Ln(ξ),Rn(ξ, x)

)(5.40)

So Rmn can be expressed as an mth degree elliptic rational function where the selectivityfactor is the nth degree discrimination factor, and the independent variable is Rn(ξ, x).

The case n = 1 is very simple, from the nth degree transformation it forces M = 1,λ = k.Hence

R1(k, x) = cd(u;k) = x (5.41)

45

The form of the 2nd degree elliptic rational function is

R2(k, x) =1− 1

k2x2j

1−x2j

x2 −x2j

x2 − 1k2x2

j

Where the two zeros are (see the Appendix for values of the Jacobi functions at K /2)

x j =±cd

(K

2;k

)=± cn

(K2 ;k

)dn

(K2 ;k

)=± 1p

1+k ′

Hence we have

R2(k, x) =1− 1

k2(1+k ′)

1− (1+k ′)x2 − (1+k ′)x2 − 1

k2(1+k ′)

Lutovac derives the simplified expression [LTE01, Sec. 13.2]

R2(k, x) = (k ′+1)x2 −1

(k ′−1)x2 +1(5.42)

Deriving a formula for the 3rd degree elliptic rational function requires a far lengthier (anddull) discussion. So instead we will skip to the results and refer the reader to Lutovac’s text forthe details [LTE01, Sec. 13.4].

Theorem 5.4. Let ξ= 1/k

R3(ξ, x) = x(x2 −x2

z )(1−x2p )

(x2 −x2p )(1−x2

z )(5.43)

where the pole xp is a function of ξ

x2p = 2ξ2

pG√

8ξ2(ξ2 +1)+12Gξ2 −G3 −p

G3(5.44)

with the auxiliary parameter G

G =√

4ξ2 + (4ξ2(ξ2 −1)

)2/3 (5.45)

and the zero xz can be found as a function of the pole by

x2z = x2

p (3−2x2p )+2xp

√(x2

p −1)3 (5.46)

46

6 The Elliptic Filter

Elliptic filters are equiripple in both the pass and stop bands; meaning the filter has equal lossmaximums in the passband, and equal loss minimums in the stopband. These filters can beexpressed in terms of Jacobi’s elliptic functions, hence the name elliptic filter. Sometimes thisfilter is called Zolotarev or Cauer filter, as Zolotarev was the man who pioneered the under-lying mathematics, and Cauer was the engineer who first applied Zolotarev’s theory to signalprocessing [Cau58] (see also [HW67] ).

6.1 The Elliptic Rational Function

Similar to how polynomials were the functions used to approximate the ideal lowpass transferfunction with the Butterworth and Chebyshev filters, for Zolotarev-Cauer filter we use a ratioof polynomials as the approximating function. The elliptic rational function Rn(k, x), is a ratioof polynomials of degree at most n; the parameter 0 < k < 1 is called the selectivity factor andit determines the intervals (−∞,−1/k]∪ [1/k,∞) where |Rn(k, x)| ≥ 1/λ and 1/λ= Rn(k,1/k).

Similar to the Chebyshev polynomial, we express the elliptic rational function parametri-cally. First denote the periods of sn(u;k) as 4K ,2i K ′ and the periods of sn(u;λ) as 4L,2i L′,with L = K /(nM) and L′ = K ′/M . That is K is the complete elliptic integral with modulus k,and L is the complete elliptic integral with modulus λ, and the constant M can be computedfrom Theorem 4.5. From Theorem 5.2

x = cd(u;k) Rn(k, x) = cd( u

M;λ

)(6.1)

Often engineers use the notation 1/k = ξ, and 1/λ = Ln(ξ) (since λ is determined by n,ξ)[LTE01, Sec. 12.6]. From the first principal nth degree transformation (Theorem 4.5) we canexpress Rn(k, x) as a rational function of x (see Sec 5.3), also we immediately have what engi-neers call the degree equation [Dan74, pg. 72] [LTE01, pg. 527,528]

n = K L′

K ′L(6.2)

Given the proper filter specifications, the elliptic integrals K ,K ′,L,L′ are determined, andfrom this equation the degree can be calculated.

6.2 The Transfer Function for the Elliptic Filter

The magnitude of the transfer function for the elliptic filter with a passband of [−1,1] is|H(iω)|2 = 1+(

εRn(k,ω))2; where ε is again the ripple factor constant. Then we extend this to

H(s)H(−s) = 1+(εRn

(ξ,

s

i

))2(6.3)

We need to find the zeros with negative real part, in the next section we will derive a generalformula to find these zeros for an nth degree filter (adapted from [LTE01, Sec. 12.8]). Follow-ing that we will discuss the cases n = 1,2,3, where the poles can be expressed without usingthe Jacobi elliptic functions [LTE01, Sec. 12.8.1].

47

6.3 Transfer Function Poles for Order n

Lemma 6.1. Let

xr = cd

(2r −1

nK ;k

)v = sn−1

(1p

1+ε2;λ′

)(6.4)

The appropriate zeros of H(s) (i.e. those with ℜ(sr ) < 0 for the normalized elliptic filter transferfunction are

sr =−sn

( vM ;k ′)cn

( vM ;k ′)√1−x2

r

√1−k2x2

r + i xr dn( v

M ;k ′)cn2

( vM ;k ′)+k2x2

r sn2( v

M ;k ′) (6.5)

with r = 1, . . . ,n [LTE01, Sec. 12.8].

Proof. This proof is adapted from [LTE01, Sec. 12.8]. From (6.3) we must solve

Rn(k,−i sr ) =± i

ε

⇐⇒ cd

(cd−1(−i sr ;k)

M;λ

)=± i

ε

Letcd−1(−i sr ;k)/M = ur + i vr (6.6)

Using the addition theorem we have

cd(ur + i vr ;λ) = sn(ur +L+ i vr ;λ)

= sn(ur +L)cn(i vr )dn(i vr )+ sn(i vr )cn(ur +L)dn(ur +L)

1−λ2 sn2(ur +L)sn2(i vr )

We dropped the modulus λ to reduce the amount of clutter in the above equation. Now weapply the imaginary transformations, where again we will drop the moduli with the knowl-edge that functions with an argument of u will have a modulus of λ, and functions with anargument of v will have a modulus of λ′.

cd(ur + i vr ;λ) = sn(ur +L)dn(vr )+ i sn(vr )cn(vr )cn(ur +L)dn(ur +L)

cn2(vr )+λ2 sn2(ur +L)sn2(vr )=± i

ε(6.7)

Now by equating the real parts of (6.7) we see (recall dn(t ) > 0,∀t ∈R)

sn(ur +L;λ) = 0 (6.8)

⇒cn(ur +L;λ) = dn(ur +L;λ) = 1

Equate the imaginary parts of (6.7) and simplifying with aid of the above equation (6.8), wehave

sn(vr )

cn(vr )=±1

ε

48

Square both sides

sn2(vr )

1− sn2(vr )= 1

ε2

⇒ sn(vr ;λ′) = 1p1+ε2

(6.9)

Notice that in (6.9), vr only depends on n,k,ε, and not at all on the subscript. Hence v1 =v2 = ·· · = vn = v and we have obtained (6.4). Proceeding from (6.8), we have

0 = sn(ur +L;λ) = cd(ur ;λ)

Since cd(w ;k) = 0 at odd multiples of K , we have that

ur = (2r −1)L (6.10)

Proceeding from (6.6) we have

−i sr = cd(M(ur + i v);k)

= cd

(2r −1

nK + i

v

M;k

)since M = K /(nL). Again we use the addition theorem to separate the real and imaginaryparts (we want the roots with negative real part), and again we omit the modulus with theunderstanding that functions with an argument of u will have modulus k and those with vwill have modulus k ′. We obtain

− i sr =sn

(2r−1n K +K

)dn

( vM

)+ i sn( v

M

)cn

( vM

)cn

(2r−1n K +K

)dn

(2r−1n K +K

)cn2

( vM

)+k2 sn2(2r−1

n K +K)

sn2( v

M

) (6.11)

Recall that the zeros of the elliptic rational function are according to (5.36)

xr = cd

(2r −1

nK ;k

)= sn

(2r −1

nK +K ;k

)r = 1, . . . ,n

Substituting xr into (6.11) yields the result (6.5).

6.4 Transfer Function Poles for Orders n = 1,2,3

The formulas for the transfer function poles we found in the previous section are valid for anyn, but they require extensive use of the Jacobi elliptic functions. In this section we wish toderive formulas for the transfer function poles without using the Jacobi elliptic functions, fordegrees n = 1,2,3. Currently algebraic formulas (without using elliptic functions) are knownfor the elliptic rational function with degree n = 1,2,3. We will avoid the case n = 3 as theformulas are much more complicated and not very enlightening, the details can be foundin [LTE01, Sec. 13.4] for those interested.

49

For n = 1 we have R1(k, x) = x, and one zero at x1 = 0. The transfer function poles are thesolutions to

R1(k,−i s) = ±i

ε

That is

−i s = i

ε

⇒ s1 = −1

ε

The transfer function is thusT (s) = α1

s − s1(6.12)

where

1

α1= H ′(s1) = 2ε2R1(k, s1) =−2ε

Hence

T (s) = −1

2ε(s + 1

ε

) (6.13)

For n = 2, denote ζ= sn(v/M ;k ′), and we rewrite (6.5) as

sr =−ζ

√1−ζ2

√1−x2

r

√1−k2x2

r + i xr

√1− (1−k2)ζ2

1−ζ2(1−k2x2

r) (6.14)

In Sec. 5.5 we derived the following formula for the second order elliptic rational function,

R2(k, x) = (k ′+1)x2 −1

(k ′−1)x2 +1(6.15)

The transfer function poles are the solutions to

R2(k,−i s) =± i

ε

Let’s focus on the positive root, we have

i

ε= (k ′+1)(−i s)2 −1

(k ′−1)(−i s)2 +1

⇒ s2 = −ε+ i

ε(1+k ′)− i (1−k ′)

Then the squared magnitude of the transfer function poles is

|s|2 =p

1+ε2√ε2(1+k ′)2 + (1−k ′)2

(6.16)

50

Now we will use (6.14) to derive another expression for the squared magnitude of the trans-fer function poles, and compare the two. The zeros of the second order elliptic rational func-tion are

x1,2 =± cn(K /2;k)

dn(K /2;k)=± 1p

1+k ′ =±p

1−k ′

k(6.17)

Let’s focus on the positive root x1. To use the formula (6.14), we evaluate

√1−x2

1 =√

1− 1

1+k ′ =√

k ′

1+k ′√1−k2x2

1 =√

1− (1−k ′) =p

k ′

So

s1 = −k ′ζ√

1−ζ2 − i√

1− (1−k2)ζ2(1−k ′ζ2

)p1+k ′

And the squared magnitude is

|s1|2 = 1+k ′ζ2(1−k ′ζ2

)(1+k ′)

(6.18)

Now set (6.16) equal to (6.18)

1+k ′ζ2(1−k ′ζ2

)(1+k ′)

=p

1+ε2√ε2(1+k ′)2 + (1−k ′)2

Solve for ζ

ζ= 2

(1+k ′)p

1+ε2 +√

(1−k ′)2 +ε2(1+k ′)2(6.19)

Now with (6.14) and (6.19), one can compute the transfer function poles without using theJacobi functions for n = 2.

For n = 3 we will just give the results on how to compute ζ. in (6.14) Let

b = 3√

1− (1−2k2

)2

c =√

1+b +b2

d = 1

2

(1

c

(1−2k2)+p

2+b +2c −1

)b1 = 9ε2 (

1+d +d 2)+ (1−d)3

a1 = 3ε√

3(1+ε2

)[ε2(1+2d)3 + (1+d)(1−d)3

]E = 3ε(1+d)

3√

a1 +b1 − 3√

a1 −b1 +1−d

Then

ζ= 1p1+E 2

(6.20)

51

6.5 The Elliptic Filter

Now that we have the proper roots, we are able to express the transfer function T (s) = 1/H(s)in partial fractions as

n∑r=1

αr

s − sr

To find the constants αr we need to differentiate

H(s)H(−s) = 1+ε2 cd2(

cd−1(−i s;k)

M;λ

)Using the formulas [Fun]

d

ducd(u;k) =

(k2 −1

)sn(u;k)

dn2(u;k)(6.21)

d

ducd−1(u;k) = sn

[cd−1(u;k);k

]u2 −1

(6.22)

Denote

w(s) = cd−1(−i s;k)

M

Thus

d

d sH(s)H(−s) = 2ε2(λ2 −1)

i M

cd(w(s);λ

)sn

(w(s);λ

)sn

(w(s);k

)dn2 (

w(s);λ)(s2 −1)

(6.23)

And we can then evaluate (6.23) at the transfer function poles to get the transfer functionT (s). Just evaluate

αr = 1

H ′(sr )(6.24)

Theorem 6.1 (The Elliptic Filter). Let xin(t ), xout(t ) represent the input (output) signal at timet respectively. With sr given by (6.5), and αr given by (6.24) and (6.23), the elliptic filter is then

xout(t ) =∫ ∞

0

n∑r=1

αr e sr τxin(t −τ)dτ (6.25)

6.6 Conclusions

Lets revisit the example filter specifications given at the end of Chapter 2, but now using theelliptic transfer function. We have the following lowpass elliptic filter requirements: Amax =0.1dB, Amin = 30dB and ωH /ωb = 1.3, what degree n is necessary [Dan74, p. 72]? For theelliptic filter, the minimal degree is n = 5, which is even smaller than the required degree forthe Chebyshev filter (8). In fact, the degree of the elliptic filter will always be smaller than theButterworth or Chebyshev filters [Dan74, Sec. 5.14].

The elliptic filter is by far the most mathematically complicated filters that we have lookedat, however for any practical purposes, modern computers can calculate the Jacobi elliptic

52

functions with ease. The elliptic filter has the sharpest transition region out of all the filterswe have seen thus far (for equivalent degree n). Meaning that areas of high attenuation (thestopband) occur almost immediately outside the passband. So if the frequencies in the in-terval [ωa ,ωb] are those that you want to remain relatively unaffected by the filter, and anyfrequency outside that interval to be removed (mostly) then the elliptic filter is best for thejob.

Figure 6.1: Plot of |T (iω)|2 for the elliptic approximation with n = 5,ε = 0.15,k = ξ−1 = 0.9 toshow how well it approximates the ideal lowpass transfer function. We can seethat this transfer function is equiripple in both the pass and stop-bands, and thatthe transition from the passband to the stopband is extremely quick.

53

Figure 6.2: Plots of |T (iω)|2 for the Butterworth, Chebyshev and elliptic filters superim-posed on the same axis. Here we can directly compare the different types ofapproximations.

54

7 Appendix

Period Parallelograms for the Jacobi Elliptic Functions

Figure 7.1: Period parallelogram for z = sn(u;k) evaluated at common values for u [Akh70,p. 207].

Figure 7.2: Period parallelogram for z = cn(u;k) evaluated at common values for u [Akh70,p. 207].

55

Figure 7.3: Period parallelogram for z = dn(u;k) evaluated at common values for u [Akh70,p. 207].

56

Differentiation formulas

d

dusnu = cnu dnu

d

ducnu =−snu dnu

d

dudnu =−k2 snu cnu

d

ducd(u;k) =

(k2 −1

)sn(u;k)

dn2(u;k)d

dusn−1(u;k) = 1

cn[sn−1(u;k);k

]dn

[sn−1(u;k);k

]d

ducn−1(u;k) = −1

cn[sn−1(u;k);k

]sn

[cn−1(u;k);k

]d

dudn−1(u;k) = −1

k2 cn[dn−1(u;k);k

]sn

[dn−1(u;k);k

]d

ducd−1(u;k) = sn

[cd−1(u;k);k

]u2 −1

[Akh70, p. 208], [Fun]

Special Values

snK

2= 1p

1+k ′

cnK

2=

pk ′

p1+k ′

dnK

2=p

k ′ [Akh70, p. 209]

57

The First Principal nth Degree Transformation

Subjecting the periods to L = KnM and L′ = K ′

M , we have

cr = sn2( r

nK ;k

)λ= kn

bn/2c∏r=1

c22r−1 M =

bn/2c∏r=1

c2r−1

c2r

sn( u

M;λ

)= 1

Msn(u;k)

n−12∏

r=1

1− sn2(u;k)c2r

1−k2c2r sn2(u;k)if n odd

sn( u

M+L;λ

)=

n2∏

r=1

1− sn2(u;k)c2r−1

1−k2c2r−1 sn2(u;k)if n even

cn( u

M;λ

)= cn(u;k)

n−12∏

r=1

1− sn2(u;k)c2r−1

1−k2c2r sn2(u;k)if n odd

cn( u

M+L;λ

)=−λ

′

M

sn(u;k)

cn(u;k)

n2∏

r=1

1− sn2(u;k)c2r

1−k2c2r−1 sn2(u;k)if n even

dn( u

M;λ

)= dn(u;k)

n−12∏

r=1

1−k2c2r−1 sn2(u;k)

1−k2c2r sn2(u;k)if n odd

dn( u

M+L;λ

)=− λ′

dn(u;k)

n2∏

r=1

1−k2c2r sn2(u;k)

1−k2c2r−1 sn2(u;k)if n even

[Akh70, p. 213]

58

The Second Principal nth Degree Transformation

Subjecting the periods to L = KM and L′ = K ′

nM , we have

M =b n

2 c∏r=1

sn2(2r−1

n K ′;k ′)sn2

(2rn K ′;k ′) λ=

n∏r=1

θ24

(r K ′nK ;k ′

)θ2

4

((2r−1)K ′

2nK ;k ′)

cr =sn2

( rn K ′;k ′)

cn2( r

n K ′;k ′) δr = dn2( r

nK ′;k ′

)

sn( u

M;λ

)= 1

Msn(u;k)

b n2 c∏

r=1

1+ sn2(u;k)c2r

1+ sn2(u;k)c2r−1

cn( u

M;λ

)= cn(u;k)

n−12∏

r=1

1−δ2r sn2(u;k)

1+ sn2(u;k)c2r−1

if n odd

cn( u

M;λ

)= cn(u;k)dn(u;k)

∏ n2 −1r=1 1−δ2r sn2(u;k)∏ n

2r=1 1+ sn2(u;k)

c2r−1

if n even

dn( u

M;λ

)= dn(u;k)

n−12∏

r=1

1−δ2r−1 sn2(u;k)

1+ sn2(u;k)c2r−1

if n odd

dn( u

M;λ

)=

n2∏

r=1

1−δ2r−1 sn2(u;k)

1+ sn2(u;k)c2r−1

if n even

[Akh70, p. 214]

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Plots of the Jacobi Elliptic Functions for Real Values

Figure 7.4: Plot sn(x;k) for various values of k superimposed on top of each other.

60

Figure 7.5: Plot dn(x;k) for various values of k superimposed on top of each other.

61

Figure 7.6: Plot cn(x;k) for k = 0.9. Since |cn(x)| isn’t bounded by 1, superimposing the plotsover each other would require adjusting the scale, and then we wouldn’t be able tosee everything properly.

62

Figure 7.7: Plot cn(x;k) for k = 0.99.

63

Figure 7.8: Plot of sn(x;k) [black] and sin(π

2K x)

[red]. The frequency of sine is scaled so thatthe periods of sin x and sn x are both 4K .

64

Plots of the Jacobi Elliptic Functions for Pure Imaginary Values

Figure 7.9: Plot sn(i x;k) using Theorem 4.3.

65

Figure 7.10: Plot cn(i x;k) using Theorem 4.3.

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Plots of the Solutions to Zolotarev’s Problems A, B, and C

Figure 7.11: Plot of Y (X ) the solution to problem C with κ = 0.5 on the interval [1,1/κ2] =[1,4]. The maximum deviation µ= 0.0025.

67

Figure 7.12: Plot of y(t ) the solution to problem A with κ= 0.5 on the interval [1,1/κ] = [1,2].The maximum deviation µ= 0.0025.

68

Figure 7.13: Plot of z(x) the solution to problem B with k = 0.207 on the interval [−1,1]. Themaximum deviation m = 0.0006.

69

Figure 7.14: Plot of z(x) the solution to problem B with k = 0.207 on the interval [1,15]. Avertical line is at 1/k and horizontal lines at ±1 to show that for x ≥ 1/k, |z| ≥ 1.The maximum deviation m = 0.0006.

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References

[Akh70] N. I. Akhiezer, Elements of the theory of elliptic functions, AMS, 1970.

[BE55] H. Bateman and A. Erdélyi, Higher transcendental functions, McGraw-Hill, 1955.

[But30] Chebyshev Butterworth, Filter approximation theory, Engineer 7 (1930), 536–541.

[Cau58] Wilhelm Cauer, Synthesis of linear communcation networks, McGraw-Hill, 1958.

[Dan74] Richard W. Daniels, Approximation methods for electronic filter design, McGraw-Hill, 1974.

[Fun] Wolfram Functions, Inverse jacobi elliptic function cd, http://functions.wolfram.com/09.37.20.0001.02, [Online; accessed 28-July-2013].

[HW67] MC Horton and RJ Wenzel, The digital elliptic filter–a compact sharp-cutoff designfor wide bandstop or bandpass requirements, Microwave Theory and Techniques,IEEE Transactions on 15 (1967), no. 5, 307–314.

[LT05] Miroslav D. Lutovac and Dejan V. Tosic, Elliptic rational functions, The Mathemat-ica Journal, vol. 9, Wolfram Media, 2005, pp. 598–608.

[LTE01] Miroslav D. Lutovac, Dejan V. Tosic, and Brian L. Evans, Filter design for signal pro-cessing using MATLAB and Mathematica, Prentice Hall, 2001.

[NT13] TuenWai Ng and ChiuYin Tsang, Polynomials versus finite blaschke products,Blaschke Products and Their Applications (Javad Mashreghi and EmmanuelFricain, eds.), Fields Institute Communications, vol. 65, Springer US, 2013, pp. 249–273.

[PTVF07] William H. Press, Saul A. Teukolsky, William T. Vetterling, and Brian P. Flannery,Numerical recipes, Cambridge University Press, New York, NY, USA, 2007.

[Zol77] E.I. Zolotarev, Application of elliptic functions to questions of functions deviatingleast and most from zero, Izvestiya Imp. Akad. Nauk, 1877, Reprinted in his Col-lected works Vol. 2, Izdat. Akad. Nauk SSSR, Moscow, 1932, pp. 1-59. (Russian)Ibuch Fortschritte Math. 9, 343.

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