## Abstract

The generalized analytical quadrature filter from a set of interferograms with arbitrary phase shifts is obtained. Both symmetrical and non symmetrical algorithms for any order are reported. The analytic expression is obtained through the convolution of a set of two-frame algorithms and expressed in terms of the combinatorial theory. Finally, the solution is applied to obtain several generalized tunable quadrature filters.

## 1. Introduction

A myriad of phase stepping algorithms have been reported since the 70’s [114]. However, an analytical generalized expression for arbitrary phase steps is not reported yet. In a previous work [12], it was proved that the design of a quadrature filter can be solved as geometrical problem from any desired conditions. However, the same procedure was reported as a new method in [13,14]. In a recent work, from a set of two-frame filters a quadrature filter is obtained and several novel tunable algorithms were reported [15]. Therefore, in this paper the generalized analytical expression for a quadrature filter for arbitrary phase steps is finally obtained instead the numerical approach reported in [16]. Thus, this paper is presented as follows: in chapter 2, a general non rotated (symmetric) two-frame algorithm is obtained, and through the rotation matrix the aliased rotated filter (non symmetric algorithm) is derived. In chapter 3, a formalism to obtain any quadrature filter through the convolution of a set of two-frame filters is introduced, and some particular cases for the rotated and non rotated filters are obtained. In chapter 4, the quadrature filters obtained are generalized through the combinatorial theory and two particular cases for symmetric and non symmetric algorithms are reported. Finally in chapter 5, the usefulness of our formalism is used to obtain several novel tunable algorithms.

## 2. The two-frame algorithm in phase shifting interferometry (PSI)

#### 2.1 The symmetric two-frame filter

In phase shifting interferometry an intensity frame $Ik$ measured at time $tk$ is given by

where α is the frequency carrier or usually named phase shift, $b(x,y)$ is the amplitude for the position (x,y) and $φ(x,y)$ is the phase to be estimated, meanwhile $a(x,y)$ is the background [115]. In a previous work [12], for $t1=1/2$, $t2=1/2$ and $a(x,y)=0$ the corresponding phase $φ(x,y)$ is obtained from the following tunable two-frame algorithmwhere, $I→2=[I1I2]T$is the intensity column vector formed by the two frames. Then, from [12,15] the Fourier transform $H(ω)$ of this filter that cuts off any frequency $ω=α$ is:
$H(ω)=−2sin[(ω−α)/2].$
Finally, from [15] the numerator and denominator that describe the quadrature filter are:In others words, through a two-frame filter it is possible to cut off any specific frequency α over the frequency axes because it Eq. (3) satisfies the quadrature condition $H(ω=α)=0$. As shown in [12,15], for $a(x,y)≠0$, the quadrature condition $H(ω=0)=0$ must be also satisfied to cancel the DC component and more frame are required to recover the desired phase. That is, two or more individual filters are required to satisfy both quadrature conditions.

#### 2.2 The discrete temporal convolution

For two individual algorithms their corresponding individual phases $φ1$, $φ2$ are given by:

where, and are the column vectors with n and m frames respectively. On the other hand, $N1$and $D1$ are two [nx1] row vectors, likewise $N2$ and $D2$ are two [mx1] row vectors too. Therefore, the temporal impulse responses $h1(t)$ and $h2(t)$ for both filters are expressed bywhere, $δ→n$ and $δ→m$ are the [nx1] and [mx1] column vectors expressed in terms of the impulse functions such as:Therefore, the combined filter between $h1(t)$ and $h2(t)$ is obtained through the following convolution h(t)=h1(t)*h2(t), where ∗ denotes the temporal discrete convolution. Then, from Eq. (6) the temporal response $h(t)$ becomeswhere, $δ→n∗δ→m=δ→m+n−1$. From the above expression a new (n + m-1) resultant filter is obtained, and from Eq. (8) the desired phase $φ(x,y)$ is recovered asForemost, from Eq. (9) the vector $I→n+m−1$ is dropped to obtain an easier way to evaluate the combined filter.
$ND=(N1D1)∗(N2D2)=N1∗D2+D1∗N2D1∗D2−N1∗N2.$
The discrete convolution between the X and Y vectors with n and m orders is obtained from $Z(k)=∑p=1m+n−1X(p) Y(k−p+1)$ for k = 1,2 ... m + n-1. The coefficients of the product are given by the convolution of the original coefficient sequences X and Y, extended with zeros where necessary to avoid undefined terms and this is the well known Cauchy product of two polynomials.

#### 2.3 The tunable rotated two-frame algorithm

The numerator $Nθ$ and denominator $Dθ$ and of a rotated filter is expressed as [11]

$(Nθ Dθ )=(cos(θ)−sin(θ)sin(θ)cos(θ))(ND)=(Ncos(θ)−Dsin(θ)Nsin(θ)+Dcos(θ)),$
where θ is the rotation angle. That implies that a filter N/D rotated and angle $θ=π/2$has an associated or aliased filter –D/N. Then, from Eqs. (4),11), the rotated two-frame filter for an arbitrary θ is given byTherefore, the numerator and denominator of this rotated filter becomes
$Nθ Dθ =[cos(θ+α/2),−cos(θ−α/2)][sin(θ+α/2),−sin(θ−α/2)].$
Consequently, the corresponding phase for this rotated two-frame algorithm is
$tan(φ)=Nθ I→2 Dθ I→2=cos(θ+α/2)I1−cos(θ−α/2)I2sin(θ+α/2)I1−sin(θ−α/2)I2,$
and the Fourier transform $Hr(ω)$ of this filter becomes
$Hr(ω)=−2exp(iθ )sin[(ω−α)/2].$
For $ω=α$ the quadrature condition $Hr(α)=H(α)=0$ is satisfied. That is, both rotated and non rotated filters $Hr(ω)$, $H(ω)$ cut off the same frequency α. Then, Eq. (13) for a specific angle $θ=α/2$, the ratio between numerator and denominator becomes
$Nα/2 Dα/2 =[cosα−1][sinα0].$
Thus, for an arbitrary frequency $αk$ the individual phase $φk$ recovered is

$tan(φk)=NkI←2DkI←2=[cosαk−1]I←2[sinαk0]I←2=I1cosαk−I2I1sinαk.$

## 3. The design of an M order quadrature filter

In a previous work [15], it was proved that a quadrature filter order M can be obtained from a set of M-1 frequencies $αk$ and a convolution algorithm, which implies that a generalized expression for any order with arbitrary conditions can be achieved.

#### 3.1 The general theory of phase shifting interferometry

The estimated phase ϕ of a quadrature filter order M is expressed as [115]

where is the column vector containing the frames, while, $N=[b1b2...bM]$ and $D=[a1a2...aM]$ are the numerator and denominator row vectors that describe the filter’s behavior. Fortunately, by using the characteristic polynomial $P(x)$ of the filter defined as [8]:
$P(x)=∑k=1M(ak+i bk)xk−1.$
Then, there are two sets $Ak$ and $Bk$ with M-1 values such that $P(x)$ can be factorized as
$P(x)=∏k=1M−1(xk−Ak−i Bk).$
From the fundamental theorem of the algebra, this polynomial above have exactly M-1 roots $xk=Ak+i Bk$; for k = 1, 2, … M-1. Thus, for each roots an angle $αk$ can be obtained as
$αk=tan−1(Bk/Ak).$
As showed in [15], each value $αk$ has the geometric interpretation as an interception over the frequency axis. That is, the Fourier transform of the filter has a zero at frequency $αk$. As a consequence, for each frequency $αk$ an individual Fourier transform $Hk(ω)$ that satisfies $Hk(αk)=0$ is obtained. Therefore, for the M-1 interceptions $αk$ the Fourier transform $H(ω)$ that meets all the desired conditions is given by:
$H(ω)=∏k=1M−1Hk(ω−αk).$
That is, the Fourier transform $H(ω)$ of a quadrature filter order M is expressed as the product of a set of M-1 individual filters $Hk(ω−αk)$. Then, the temporal response of the resultant filter $h(t)$ is obtained through the inverse of the Fourier transform of the filter $h(t)=ℑ−1[H(ω)]$ and expressed as:
$h(t)=ℑ−1[H1(ω)H2.(ω)...HM−1(ω)]=h1(t)∗h2(t)∗...hM−1(t)=Ωk=1M−1hk(t).$
For simplicity, the symbol Ω is introduced as a convolution operator that denotes the temporal convolution of a set of M-1 individual filters $hk(t)$. Then, each filter $hk(t)$ has an individual time response given by the two $Nk$and $Dk$ [2x1] row vectors such asThus, through the operator Ω, the temporal response of any quadrature filter is expressed aswhere, N and D are the vectors defined in Eq. (18). Then, from Eq. (10) the numerator and the denominator of an algorithm is expressed asBy using Eq. (13) into Eq. (26) the filter is expressed as
$ND=Ωk=1M−1{[cos(θk+αk/2),−cos(θk−αk/2)][sin(θk+αk/2),−sin(θk−αk/2)]}.$
In other words, in the same way that a polynomial with order M-1 is expressed as the product of M-1 first order polynomials, the Fourier transform of a quadrature filter order M is expressed as the product of M-1 individual filters. Where for each filter a two-frame filter is associated. Therefore, through the convolution of this set of M-1 two-frame filters showed above, the desired filter is obtained. Additionally, from Eqs. (15),23) the Fourier transform of the quadrature filter can be recovered asFor $σ=0$, $H(ω)$ is a non rotated filter or symmetric algorithm, because σ becomes the real quantity
$H(ω)=(−2)M−1∏k=1M−1sin[(ω−αk)/2].$
In this work, this case is so called symmetric algorithm and from Eq. (27) resultsOtherwise, for any real value $σ≠0$ an aliased or rotated filter is recovered and it is called non symmetric algorithm. Therefore, by using the specific angle $θ=α/2$ this rotated filter becomes
$NrDr=Ωk=1M−1{[cosαk−1][sinαk0]}.$
At this point, three goals are achieved by the introduced formalism. The first one is to obtain an M order filter from a set of M-1 arbitrary frequencies (interceptions). The second goal is to analyze the behavior of any given algorithm. Finally, the last goal is to improve a known algorithm adding new properties through the convolution algorithm.

#### 3.2 The tunable non symmetric three-frame algorithm

From Eqs. (26) and (31), a three-frame filter is obtained from two phases $α1$, and $α2$ as

$ND=[cosα1,−1][sinα1,0]∗[cosα2,−1][sinα2,0]=[sinα2,0]∗[cosα1,−1]+[cosα2,−1]∗[sinα1,0][sinα1,0]∗[sinα2,0]−[cosα1,−1]∗[cosα2,−1],$
because the convolution of two vectors is simply the product of both polynomials. This way, the use of impulse functions has been eliminated, and from Eq. (10) the filter obtained is
$ND=[sin(α1+α2),−sin(α1)−sin(α2),0][−cos(α1+α2),cos(α1)+cos(α2),−1].$
Then, this filter shifted an angle $π/2$ is the following aliased filter expressed by
$Nπ/2Dπ/2=[cos(α1+α2),−cos(α1)−cos(α2),1][sin(α1+α2),−sin(α1)−sin(α2),0].$
After that, from [12,15] by using the essential quadrature conditions $α2=0$ and $α1=α$in Eq. (34) a novel filter is obtained and given by
$ND=[sin(α),−sin(α),0][−cos(α),cos(α)+1,−1].$
Finally, the desired phase recovered from this tunable rotated three-frame algorithm is
$tan(φ)=sin(α)(I1−I2)−cos(α)(I1−I2)+(I2−I3).$
It should be noticed that this result was obtained without the usage of the Fourier formalism or temporal response algebra. In the same way, from Eq. (11), the estimated phase for this tunable filter shifted $π/2$ is given by
$tan(φ)=cos(α)(I1−I2)−(I2−I3)sin(α)(I1−I2),$
by using $α=π/2$ in Eqs. (37) and (38), both famous three-frame algorithms are recovered [4].

#### 3.3 The tunable non symmetric four frame algorithm

From Eqs. (26), (31), and (33) the rotated four-frame filter for arbitrary conditions $α1,α2,α3$ is

$ND=[sin(α1+α2),−sin(α1)−sin(α2),0][−cos(α1+α2),cos(α1)+cos(α2),1]∗[cosα3,−1][sinα3,0]=[b1,b2,b3,b4][a1,a2,a3,a4].$
Then, from Eq. (10), the coefficients of the numerator and the denominator above become:
$b1=−cos(α1+α2+α3),$
$b2=cos(α1+α2)+cos(α1+α3)+cos(α2+α3),$
$b3=−cos(α1)−cos(α2)−cos(α3), b4=1,$
$a1=−sin(α1+α2+α3),$
$a2=sin(α1+α2)+sin(α1+α3)+sin(α2+α3),$
$a3=−sin(α1)−sin(α2)−sin(α3), a4=0.$
Thus, by using $a3=0$ in Eqs. (39)(44) the filter that cuts off the background is given by
$ND=[−cos(α1+α2),cos(α1+α2)+cos(α1)+cos(α2),−1−cos(α1)−cos(α2),1][−sin(α1+α2),sin(α1+α2)+sin(α1)+sin(α2),−sin(α1)−sin(α2),0],$
and from Eq. (11) the filter shifted $θ=π/2$ is given by
$ND=−[sin(α1+α2),−sin(α1+α2)−sin(α1)−sin(α2),sin(α1)+sin(α2),0][cos(α1+α2),−cos(α1+α2)−cos(α1)−cos(α2),1+cos(α1)+cos(α2),−1].$
By means of the values $α2=π$ and $α1=α$ from Eq. (46) the following filter is recovered
$ND=[cos(α),−1,−cos(α),1][sin(α),0,−sin(α),0].$
According to the reference [11], the above filter should be named as tunable four-frame in cross algorithm [15]. And the estimated phase is
$tan(φ)=N I→D I→=cos(α)(I1−I3)−I2+I4sin(α)(I1−I3).$
In the same manner, from Eqs. (11) and (48) the corresponding shifted tunable filter is
$tan(φ)=N I→D I→=−sin(α)(I1−I3)cos(α)(I1−I3)−I2+I4$
for $α=π/2$ in Eqs. (48) and (49), the well known four-frames filters are recovered [3,11].

#### 3.4 The tunable non symmetric five-frame algorithm

Likewise, the general five-frame filter for arbitrary conditions $α1,α2,α3,α4$ is obtained by

$ND=Ωk=14{[cosαk,−1][sinαk,0]}=[b1,b2,b3,b4,b5][a1,a2,a3,a4,a5].$
Then by using the Eq. (33) in Eq. (50) the filter is expressed as
$ND=[sin(α1+α2),−sin(α1)−sin(α2),0][−cos(α1+α2),cos(α1)+cos(α2),1]∗[sin(α3+α4),−sin(α3)−sin(α4),0][−cos(α3+α4),cos(α3)+cos(α4),1],$
and the coefficients of the filter in Eq. (50) become
$b1=sin(α1+α2+α3+α4),$
$b2=−sin(α1+α2+α3)−sin(α1+α2+α4)−sin(α1+α3+α4)−sin(α2+α3+α4),$
$b4=−sin(α1)−sin(α2)−sin(α3)−sin(α4), b5=0,$
$a1=cos(α1+α2+α3+α4),$
$a2=−cos(α1+α2+α3)−cos(α1+α2+α4)−cos(α1+α3+α4)−cos(α2+α3+α4),$
By using $α1=α$, $α2=π−α$, $α3=π$ and the quadrature condition [12] $α4=0$ in Eqs. (50)(59) the recovered filter is the well known five-step filter [24]
$tan(φ)=N I→5D I→5=−2sin(α) [0,1,0,−10] I→5[1,0,−2,0,1] I→5=2sin(α)(I2−I4)−I1+2I3−I5.$
From Eq. (50) for the conditions $α1=α2=α,$ $α3=π$and the quadrature condition [12] $α4=0$, the tunable five-frame algorithm class A is obtained as
$tan(φ)=N I→5D I→5=sin(2α) (I1−I3)−2sin(α) (I2−I4) −cos(2α) (I1−I3)+2cos(α) (I2−I4)−(I3−I5) .$
This filter is insensitive to the linear phase shift detuning error for any phase step α [15].

#### 3.5 The symmetric tunable three-frame algorithm

Repeating the procedure used above a symmetric three-frame filter is expressed as

and from the quadrature conditions $α2=0$ and $α1=α$ the well known three-frame filter is recovered as

$tan(φ)=[sin(α/2),0,−sin(α/2)] I→3[−cos(α/2),2cos(α/2),−cos(α/2)] I→3=tan(α/2)I1−I3−I1+2I2−I3.$

#### 3.6 The symmetric tunable four-frame algorithm

From Eq. (31) the symmetric four-frame filter is given by

Through the convolution algorithm Eq. (10) the coefficients are given by
$b1=−cos[(α1+α2+α3)/2], a1=−sin[(α1+α2+α3)/2],$
$b2=cos[(α1+α2−α3)/2]+cos[(α1−α2+α3)/2]+cos[(−α1+α2+α3)/2],$
$a2=sin[(α1+α2−α3)/2]+sin[(α1−α2+α3)/2]+sin[(−α1+α2+α3)/2].$
Then, for the quadrature condition [12] $α3=0$, the general four-frame filter becomes
$b1=−cos[(α2+α3)/2], b2=2cos[(α2−α3)/2]+cos[(α2+α3)/2]$
$a1=−a2=−sin[(α2+α3)/2].$
As we expected, for $α1=α2=π/2$, the four-frame filter class B is recovered [7,15].

#### 3.7 The symmetric and tunable five-frame algorithm

To conclude, from Eq. (30) the expression for a symmetric tunable five-frame filter is

$b1=−sin[(α1+α2+α3+α4)/2], a1=cos[(α1+α2+α3+α4)/2],$
This result comprises all result reported for any symmetric five-frame filter.

For the values $α1=α2=α,$ $α3=π$ and $α4=0$ the recovered filter is:

$tan(φ)=N I→5D I→5=[−cos(α)20−2,cos(α)]I→5−sin(α) [10−201]I→5=−cos(α)(I1−I5)+2(I2−I4)−sin(α)(I1−2I3+I5).$
According to [7] we named this filter above as tunable five-frame algorithm class A. Obviously, for the value $α=π/2$ becomes the filter reported in [7] and given by
$tan(φ)=N I→5D I→5=[020−2,0]I→5 −[10−201]I→5=2(I2−I4)−I1+2I3−I5.$
By using $α1=α2=α3=α$ and $α4=0$ into Eq. (70), the tunable five-frame filter obtained is
$ND=[−sin(3α/2)sin(3α/2)+3sin(α/2)0−sin(3α/2)−3sin(α/2)sin(3α/2)][cos(3α/2)−cos(3α/2)−3cos(α/2)6sin(α/2)−cos(3α/2)−3cos(α/2)cos(3α/2)].$
Again, according to [7], this filter above should be named as tunable five-frame algorithm class B. Therefore, for $α=π/2$ the following normalized filter reported in [7] is obtained

## 4. The general non symmetric quadrature filter M order

To express the solutions obtained above the combinatorial theory will be used below.

#### 4.1 Combinatorial theory

For a list of any real frequencies $α1,α2,...αn$, the notation for sets is standard, and the items are separated by commas and enclosed by curly brackets W = {$α1,α2,...αn$}, meanwhile the parenthesis denotes a specific combination or arrangement. Then, let $W={α1,α2,...αn}rn$ be a set of different arrangements with r objects chosen from W. For a trivial example, $W={a,b,c,d}$ denotes a set formed by four different numbers a, b, c, and d. From this set W, only four combinations with three objects chosen from W can be made, and they are:

$W34={a,b,c,d}34={ (a,b,c),(a,b,d),(a,c,d),(b,c,d) }.$
In the same way, six different arrangements for two objects chosen from W are:
$W24={a,b,c,d}24={ (a,b),(a,c),(a,d),(b,c),(b,d),(c,d) }.$
The other cases are, $W14={a,b,c,d}14={ (a),(b),(c),(d) }$, and $W44={a,b,c,d}44={(a,b,c,d)}$. Finally, the arrangement with zero objects chosen from W is the empty set and it is written as: $W04={a,b,c,d}04={}$. And the number of total combinations is given by
$∑r=0nCnr=∑r=0nn!(n−r)!r!n=2n,$
where $Cnr$ is the number of r objects taken from n possibilities.

To express the filters shown in section 3, the usage of combinatorial algebra is required, and through this formalism any quadrature filter can be expressed in an easier and simple expression. Foremost, any filter is easily obtained computationally, even symbolically.

Let Σ be the sum operator, that over a set gives a new set where each element is the sum of each combination in the set. Then, by applying this operator to Eq. (79) the result is

$ΣW34=Σ{a,b,c,d}34={ (a+b+c),(a+b+d),(a+c+d),(b+c+d) }.$
Thus, applying a function over the expression above the following set is obtained:
$cos[Σ{a,b,c,d}34]={ cos(a+b+c),cos(a+b+d),cos(a+c+d),cos(b+c+d) }.$
The operations with a scalar value σ obeys the following algebra rule
$σ+(a,b,c,d)=(σ+a,σ+b,σ+c,σ+d).$
Combining Eq. (82) into Eq. (84) the set obtained is:
$σ+Σ{a,b,c,d}34={ (σ+a+b+c),(σ+a+b+d),(σ+a+c+d),(σ+b+c+d) }.$
Finally, combining all the rules above the scalar value obtained is:
$Σcos(σ+ΣW34)= cos(σ+a+b+c)+cos(σ+a+b+d)+cos(σ+a+c+d)+cos(σ+b+c+d) .$
That is, through this notation, a filter is expressed analytically in terms of combinatory theory in a simple expression that can easily be evaluated computationally.

#### 4.2 A general tunable non symmetric M-frame algorithm

As showed in section 3, all particular quadrature rotated filters have a well defined model, and from such pattern the generalized expression of an M order rotated filter is expressed as:

$ND=Ωk=1M−1{[cosαk−1][sinαk0]}=[b1b2b3...bM][a1a2a3...aM].$
Through the combinatory theory, each element of the numerator is expressed as:
$b1=∑sin[Σ{α1,α2,α3...,αM−1}M−1M−1]=−sin(σ),$
$b2=(−1)∑sin[Σ{α1,α2,α3...,αM−1}M−2M−1]=(−1)∑k=1M−1sin(σ−αk),$
and the last element can be expressed as
$bM=(−1)M+1∑sin[Σ{α1,α2,α3...,αM−1}M−MM−1]=(−1)M+1sin(0)=0,$
where the scalar σ is simply the sum of all cut off frequencies (interceptions) and given by
$σ=∑k=1M−1αk.$
That is, each element of the numerator $bk$ is expressed by:In the same way, each element of the denominator $ar$ is expressed by:
$a1=−∑cos[Σ{α1,α2,α3...,αM−1}M−1M−1]=−cos(σ),$
$a2=(−1)2∑cos[Σ{α1,α2,α3...,αM−1}M−2M−1]=(−1)2∑k=1M−1cos(σ−αk),$
and the last element is:
$aM=(−1)M∑cos[Σ{α1,α2,α3...,αM−1}M−MM−1]=(−1)Mcos(0)=(−1)M.$
Therefore, each element of the denominator $ar$ is generalized as:To conclude, from Eqs. (18), (92), and (96) the desired phase is recovered from the next algorithm
$tan(φ)=N I→MD I→M=∑k=1M{ (−1)k+1∑sin[Σ{α1,α2,... αM−1}M−kM−1] } Ik∑k=1M{ (−1)k∑cos[Σ{α1,α2,... αM−1}M−kM−1] } Ik,$
and from Eq. (11) the corresponding shifted π/2 filter is,
$tan(φ)=N I→MD I→M=∑k=1M{ (−1)k+1∑cos[Σ{α1,α2,... αM−1}M−kM−1] } Ik∑k=1M{ (−1)k+1∑sin[Σ{α1,α2,... αM−1}M−kM−1] } Ik.$
That is, the problem of designing a rotated quadrature filter from a set of arbitrary phase steps (any condition) has finally been solved in this paper. From [12], recall that for any filter in this paper $α1=0$ is the essential condition to obtain a quadrature filter.

#### 4.3 A general symmetric and tunable M-frame algorithm

In the same way, from Eq. (30) any M order symmetric algorithm is expressed by

However, two cases for odd and even order are presented. From section 3, for any M odd order symmetric algorithm, each element of the numerator and the denominator become:
$b1=∑sin[σ/2−Σ{α1,α2,α3...,αM−1}0M−1]=−bM=sin(σ/2),$
$a1=(−1)∑cos[σ/2−Σ{α1,α2,α3...,αM−1}0M−1]=−aM=−cos(σ/2),$
$b(m+1)/2=(−1)(m+3)/2∑sin[σ/2−Σ{α1,α2,α3...,αM−1}(M−1)/2M−1]=0.$
Therefore, each element of the numerator and the denominator are generalized as: As expected, for an M odd order filter the following symmetries $br=−bM−r+1$, $ar=aM−r+1$, and $b(M+1)/2=0$, are satisfied and the phase is recovered from the following algorithm
$tan(φ)=N I→MD I→M=∑k=1M{ (−1)k+1∑sin[σ/2−Σ{α1,α2,α3...,αM−1}k−1M−1] } Ik∑k=1M{ (−1)k∑cos[σ/2−Σ{α1,α2,α3...,αM−1}k−1M−1] } Ik.$
On the other hand, although the desired phase for an even M order case can also be obtained from the expression above, but the obtained filter will be a shifted π/2 aliased algorithm. However, from Eq. (11) the symmetric algorithm to recover the phase is obtained as:
$tan(φ)=N I→MD I→M=∑k=1M{ (−1)k+1∑cos[σ/2−Σ{α1,α2,α3...,αM−1}k−1M−1] } Ik∑k=1M{ (−1)k+1∑sin[σ/2−Σ{α1,α2,α3...,αM−1}k−1M−1] } Ik.$
That is, for an M even order filter the symmetries $br=−bM−r+1$and $ar=aM−r+1$are also satisfied. Where $α1=0$ is a necessary condition to recover a quadrature filter [12].

## 5. Some applications

The proposed formalism brings up a new viewpoint to design and to analyze quadrature filters. However, only some illustrative but novel examples are shown below.

#### 5.1 Improving a known algorithm

The proposed formalism can be used to improve a known filter into another with new desired properties. For example, from the well known five-frame filter [3,4] a new seven-frame filter can be obtained. From Eq. (19), the characteristic polynomial is given by [8]

$P(x)=∑k=15(ak+i bk)xk−1=−(x−1)(x+1)[x+cos(α)+isin(α)][x−cos(α)+isin(α)].$
From the roots of $P(x)=0$ and Eq. (21), the set of frequencies $W={0,π,α,π−α}$ is obtained. Then, the filter is improved by adding the two new frequencies $α5=β$, and $α6=π−β$. Therefore, from Eq. (30) the new seven-frame algorithm obtained is
$N D ={2sin(α)[010−10] [−1020−1] }∗{cos(β/2)[1−1]sin(β/2)[11]}∗{sin(β/2)[1−1]cos(β/2)[11]}.$
From Eq. (29) the corresponding Fourier transform of the filter becomes
$H(ω)=(−2)6sin(ω) [sin(ω)−sin(α)] [sin(ω)−sin(β)].$
Thus, by applying Eq. (10) the numerator and the denominator of the filter are:Therefore, the estimated phase recovered from this seven-frame algorithm is
$tan(φ)=N I→7 D I→7=−(I1−I7)+[3+4sin(α)sin(β)] (I3−I5)−2[sin(α)+sin(β)] (I2−2I4+I6) .$
This is, a novel wide-band tunable seven-frame filter was obtained. Then, for $β=0$ becomes the tunable filter, which is insensitive to the bias modulation error and given by
$tan(φ)=−(I1−I7)+3(I3−I5)−2sin(α) (I2−2I4+I6) .$
By using $α=π/2$ in the filter above, the algorithm reported in [10] is obtained as:
$tan(φ)=I1−3I3+3I5−I72I2−4I4+2I6.$
In the same way, from Eq. (111) by using the values $α=π/2$ and $β=π/6$ a wide-band seven-frame filter that practically cuts off all frequencies between is obtained as
$tan(φ)=I1−5I3+5I5−I7 3I2−6I4+3I6 .$
Finally, by using $β=α$, in Eq. (111) a novel seven-frame filter is recovered as
$tan(φ)=−(I1−I7)+[3+4sin2(α)](I3−I5)−4sin(α) (I2−2I4+I6) =(I1−I7)−[5−2cos(2α)](I3−I5)4sin(α) (I2−2I4+I6) ,$
and for the value $α=π/2$, the seven-frame algorithm reported in [7,8] is given by:

$tan(φ)=I1−7I3+7I5−I74 (I2−2I4+I6) .$

#### 5.2 Example of a tunable six-frame algorithm

The set $W={α1,α,α,α,α}$ where $α1=0$ corresponds to a quadrature filter that gives up the maximum insensitivity to the miscalibration error is given by

$ND=limα1→0 Ωk=14{[cosα−1][sinα0]}∗[cos(α1)−1][sin(α1)0].$
From Eq. (29) the corresponding Fourier transform of Eq. (117) for $α1=0$ is simply
$H(ω)=(−2)5exp(2α i)sin(ω/2)sin4[(ω−α)/2].$
From Eqs. (89) and (93) for the subset ${α,α,α,α}$ each element of this partial filter yields: From Eqs. (10), (117), (119), and (120) for $α1=0$ the desired numerator and denominator becomeAccording to the reference [7], the filter above should be named as tunable six-frame filter class B. Thus, by using $α=π/2$ the filter reported in [7] is yielded as:

$tan(φ)=N I→6D I→6=[1−1−661−1] I→6[04−4−440] I→6=I1−I2−6I3+6I4+I5−I64(I2−I3−I4+I5).$

#### 5.3 An example of a tunable eleven-frame algorithm

To conclude, an example of an eleven-frame algorithm is given by the following set $W={0,0,π,π,α,α,π−α,π−α,π+α,π+α}$. From Eq. (105) the numerator and the denominator of the filter are given by:

Then, by substituting $α=π/3$ the well known eleven-frame filter reported in [8,10] gives

## 6. Conclusions

The problem of designing any quadrature filter from a set of interferograms with arbitrary phase shifts has been solved. The problem is solved through the convolution of a set of two-frame filters and expressed in terms of the combinatory theory. Therefore, both the symmetric and the non-symmetric algorithms are reported. The introduced formalism is extremely versatile to obtain new quadrature filters from other known filters. Foremost, to illustrate the formalism, several tunable examples as five-frame, six-frame and eleven-frame algorithms are easily obtained. Additionally, the introduced expressions to recover the quadrature filters are easily evaluated computationally both numerically and symbolically through practically any mathematical software.

## Acknowledgments

This work is dedicated to the memory of P. A. Mosiño Alemán. The authors acknowledge the comments and help granted by MDD. J. J. Lozano in the revision of this work.

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### References

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#### Other (2)

H. Schreiber, J. H. Brunning, and J. E. Greivenkamp, “Phase shifting interferometry,” in Optical Shop Testing, D. Malacara ed., (John Wiley & Sons, Inc., 2007).

J. Schwider, “Advanced evaluation techniques in interferometry,” in Progress in Optics, E. Wolf ed., (Elsevier, 1990).

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### Equations (125)

$H ( ω ) = − 2 sin [ ( ω − α ) / 2 ] .$
$N D = ( N 1 D 1 ) ∗ ( N 2 D 2 ) = N 1 ∗ D 2 + D 1 ∗ N 2 D 1 ∗ D 2 − N 1 ∗ N 2 .$
$( N θ D θ ) = ( cos ( θ ) − sin ( θ ) sin ( θ ) cos ( θ ) ) ( N D ) = ( N cos ( θ ) − D sin ( θ ) N sin ( θ ) + D cos ( θ ) ) ,$
$N θ D θ = [ cos ( θ + α / 2 ) , − cos ( θ − α / 2 ) ] [ sin ( θ + α / 2 ) , − sin ( θ − α / 2 ) ] .$
$tan ( φ ) = N θ I → 2 D θ I → 2 = cos ( θ + α / 2 ) I 1 − cos ( θ − α / 2 ) I 2 sin ( θ + α / 2 ) I 1 − sin ( θ − α / 2 ) I 2 ,$
$H r ( ω ) = − 2 exp ( i θ ) sin [ ( ω − α ) / 2 ] .$
$N α / 2 D α / 2 = [ cos α − 1 ] [ sin α 0 ] .$
$tan ( φ k ) = N k I ← 2 D k I ← 2 = [ cos α k − 1 ] I ← 2 [ sin α k 0 ] I ← 2 = I 1 cos α k − I 2 I 1 sin α k .$
$P ( x ) = ∑ k = 1 M ( a k + i b k ) x k − 1 .$
$P ( x ) = ∏ k = 1 M − 1 ( x k − A k − i B k ) .$
$α k = tan − 1 ( B k / A k ) .$
$H ( ω ) = ∏ k = 1 M − 1 H k ( ω − α k ) .$
$h ( t ) = ℑ − 1 [ H 1 ( ω ) H 2. ( ω ) ... H M − 1 ( ω ) ] = h 1 ( t ) ∗ h 2 ( t ) ∗ ... h M − 1 ( t ) = Ω k = 1 M − 1 h k ( t ) .$
$N D = Ω k = 1 M − 1 { [ cos ( θ k + α k / 2 ) , − cos ( θ k − α k / 2 ) ] [ sin ( θ k + α k / 2 ) , − sin ( θ k − α k / 2 ) ] } .$
$H ( ω ) = ( − 2 ) M − 1 ∏ k = 1 M − 1 sin [ ( ω − α k ) / 2 ] .$
$N r D r = Ω k = 1 M − 1 { [ cos α k − 1 ] [ sin α k 0 ] } .$
$N D = [ cos α 1 , − 1 ] [ sin α 1 , 0 ] ∗ [ cos α 2 , − 1 ] [ sin α 2 , 0 ] = [ sin α 2 , 0 ] ∗ [ cos α 1 , − 1 ] + [ cos α 2 , − 1 ] ∗ [ sin α 1 , 0 ] [ sin α 1 , 0 ] ∗ [ sin α 2 , 0 ] − [ cos α 1 , − 1 ] ∗ [ cos α 2 , − 1 ] ,$
$N D = [ sin ( α 1 + α 2 ) , − sin ( α 1 ) − sin ( α 2 ) , 0 ] [ − cos ( α 1 + α 2 ) , cos ( α 1 ) + cos ( α 2 ) , − 1 ] .$
$N π / 2 D π / 2 = [ cos ( α 1 + α 2 ) , − cos ( α 1 ) − cos ( α 2 ) , 1 ] [ sin ( α 1 + α 2 ) , − sin ( α 1 ) − sin ( α 2 ) , 0 ] .$
$N D = [ sin ( α ) , − sin ( α ) , 0 ] [ − cos ( α ) , cos ( α ) + 1 , − 1 ] .$
$tan ( φ ) = sin ( α ) ( I 1 − I 2 ) − cos ( α ) ( I 1 − I 2 ) + ( I 2 − I 3 ) .$
$tan ( φ ) = cos ( α ) ( I 1 − I 2 ) − ( I 2 − I 3 ) sin ( α ) ( I 1 − I 2 ) ,$
$N D = [ sin ( α 1 + α 2 ) , − sin ( α 1 ) − sin ( α 2 ) , 0 ] [ − cos ( α 1 + α 2 ) , cos ( α 1 ) + cos ( α 2 ) , 1 ] ∗ [ cos α 3 , − 1 ] [ sin α 3 , 0 ] = [ b 1 , b 2 , b 3 , b 4 ] [ a 1 , a 2 , a 3 , a 4 ] .$
$b 1 = − cos ( α 1 + α 2 + α 3 ) ,$
$b 2 = cos ( α 1 + α 2 ) + cos ( α 1 + α 3 ) + cos ( α 2 + α 3 ) ,$
$b 3 = − cos ( α 1 ) − cos ( α 2 ) − cos ( α 3 ) , b 4 = 1 ,$
$a 1 = − sin ( α 1 + α 2 + α 3 ) ,$
$a 2 = sin ( α 1 + α 2 ) + sin ( α 1 + α 3 ) + sin ( α 2 + α 3 ) ,$
$a 3 = − sin ( α 1 ) − sin ( α 2 ) − sin ( α 3 ) , a 4 = 0.$
$N D = [ − cos ( α 1 + α 2 ) , cos ( α 1 + α 2 ) + cos ( α 1 ) + cos ( α 2 ) , − 1 − cos ( α 1 ) − cos ( α 2 ) , 1 ] [ − sin ( α 1 + α 2 ) , sin ( α 1 + α 2 ) + sin ( α 1 ) + sin ( α 2 ) , − sin ( α 1 ) − sin ( α 2 ) , 0 ] ,$
$N D = − [ sin ( α 1 + α 2 ) , − sin ( α 1 + α 2 ) − sin ( α 1 ) − sin ( α 2 ) , sin ( α 1 ) + sin ( α 2 ) , 0 ] [ cos ( α 1 + α 2 ) , − cos ( α 1 + α 2 ) − cos ( α 1 ) − cos ( α 2 ) , 1 + cos ( α 1 ) + cos ( α 2 ) , − 1 ] .$
$N D = [ cos ( α ) , − 1 , − cos ( α ) , 1 ] [ sin ( α ) , 0 , − sin ( α ) , 0 ] .$
$tan ( φ ) = N I → D I → = cos ( α ) ( I 1 − I 3 ) − I 2 + I 4 sin ( α ) ( I 1 − I 3 ) .$
$tan ( φ ) = N I → D I → = − sin ( α ) ( I 1 − I 3 ) cos ( α ) ( I 1 − I 3 ) − I 2 + I 4$
$N D = Ω k = 1 4 { [ cos α k , − 1 ] [ sin α k , 0 ] } = [ b 1 , b 2 , b 3 , b 4 , b 5 ] [ a 1 , a 2 , a 3 , a 4 , a 5 ] .$
$N D = [ sin ( α 1 + α 2 ) , − sin ( α 1 ) − sin ( α 2 ) , 0 ] [ − cos ( α 1 + α 2 ) , cos ( α 1 ) + cos ( α 2 ) , 1 ] ∗ [ sin ( α 3 + α 4 ) , − sin ( α 3 ) − sin ( α 4 ) , 0 ] [ − cos ( α 3 + α 4 ) , cos ( α 3 ) + cos ( α 4 ) , 1 ] ,$
$b 1 = sin ( α 1 + α 2 + α 3 + α 4 ) ,$
$b 2 = − sin ( α 1 + α 2 + α 3 ) − sin ( α 1 + α 2 + α 4 ) − sin ( α 1 + α 3 + α 4 ) − sin ( α 2 + α 3 + α 4 ) ,$
$b 4 = − sin ( α 1 ) − sin ( α 2 ) − sin ( α 3 ) − sin ( α 4 ) , b 5 = 0 ,$
$a 1 = cos ( α 1 + α 2 + α 3 + α 4 ) ,$
$a 2 = − cos ( α 1 + α 2 + α 3 ) − cos ( α 1 + α 2 + α 4 ) − cos ( α 1 + α 3 + α 4 ) − cos ( α 2 + α 3 + α 4 ) ,$
$tan ( φ ) = N I → 5 D I → 5 = − 2 sin ( α ) [ 0 , 1 , 0 , − 1 0 ] I → 5 [ 1 , 0 , − 2 , 0 , 1 ] I → 5 = 2 sin ( α ) ( I 2 − I 4 ) − I 1 + 2 I 3 − I 5 .$
$tan ( φ ) = N I → 5 D I → 5 = sin ( 2 α ) ( I 1 − I 3 ) − 2 sin ( α ) ( I 2 − I 4 ) − cos ( 2 α ) ( I 1 − I 3 ) + 2 cos ( α ) ( I 2 − I 4 ) − ( I 3 − I 5 ) .$
$tan ( φ ) = [ sin ( α / 2 ) , 0 , − sin ( α / 2 ) ] I → 3 [ − cos ( α / 2 ) , 2 cos ( α / 2 ) , − cos ( α / 2 ) ] I → 3 = tan ( α / 2 ) I 1 − I 3 − I 1 + 2 I 2 − I 3 .$
$b 1 = − cos [ ( α 1 + α 2 + α 3 ) / 2 ] , a 1 = − sin [ ( α 1 + α 2 + α 3 ) / 2 ] ,$
$b 2 = cos [ ( α 1 + α 2 − α 3 ) / 2 ] + cos [ ( α 1 − α 2 + α 3 ) / 2 ] + cos [ ( − α 1 + α 2 + α 3 ) / 2 ] ,$
$a 2 = sin [ ( α 1 + α 2 − α 3 ) / 2 ] + sin [ ( α 1 − α 2 + α 3 ) / 2 ] + sin [ ( − α 1 + α 2 + α 3 ) / 2 ] .$
$b 1 = − cos [ ( α 2 + α 3 ) / 2 ] , b 2 = 2 cos [ ( α 2 − α 3 ) / 2 ] + cos [ ( α 2 + α 3 ) / 2 ]$
$a 1 = − a 2 = − sin [ ( α 2 + α 3 ) / 2 ] .$
$b 1 = − sin [ ( α 1 + α 2 + α 3 + α 4 ) / 2 ] , a 1 = cos [ ( α 1 + α 2 + α 3 + α 4 ) / 2 ] ,$
$tan ( φ ) = N I → 5 D I → 5 = [ − cos ( α ) 2 0 − 2 , cos ( α ) ] I → 5 − sin ( α ) [ 1 0 − 2 0 1 ] I → 5 = − cos ( α ) ( I 1 − I 5 ) + 2 ( I 2 − I 4 ) − sin ( α ) ( I 1 − 2 I 3 + I 5 ) .$
$tan ( φ ) = N I → 5 D I → 5 = [ 0 2 0 − 2 , 0 ] I → 5 − [ 1 0 − 2 0 1 ] I → 5 = 2 ( I 2 − I 4 ) − I 1 + 2 I 3 − I 5 .$
$N D = [ − sin ( 3 α / 2 ) sin ( 3 α / 2 ) + 3 sin ( α / 2 ) 0 − sin ( 3 α / 2 ) − 3 sin ( α / 2 ) sin ( 3 α / 2 ) ] [ cos ( 3 α / 2 ) − cos ( 3 α / 2 ) − 3 cos ( α / 2 ) 6 sin ( α / 2 ) − cos ( 3 α / 2 ) − 3 cos ( α / 2 ) cos ( 3 α / 2 ) ] .$
$W 3 4 = { a , b , c , d } 3 4 = { ( a , b , c ) , ( a , b , d ) , ( a , c , d ) , ( b , c , d ) } .$
$W 2 4 = { a , b , c , d } 2 4 = { ( a , b ) , ( a , c ) , ( a , d ) , ( b , c ) , ( b , d ) , ( c , d ) } .$
$∑ r = 0 n C n r = ∑ r = 0 n n ! ( n − r ) ! r ! n = 2 n ,$
$Σ W 3 4 = Σ { a , b , c , d } 3 4 = { ( a + b + c ) , ( a + b + d ) , ( a + c + d ) , ( b + c + d ) } .$
$cos [ Σ { a , b , c , d } 3 4 ] = { cos ( a + b + c ) , cos ( a + b + d ) , cos ( a + c + d ) , cos ( b + c + d ) } .$
$σ + ( a , b , c , d ) = ( σ + a , σ + b , σ + c , σ + d ) .$
$σ + Σ { a , b , c , d } 3 4 = { ( σ + a + b + c ) , ( σ + a + b + d ) , ( σ + a + c + d ) , ( σ + b + c + d ) } .$
$Σ cos ( σ + Σ W 3 4 ) = cos ( σ + a + b + c ) + cos ( σ + a + b + d ) + cos ( σ + a + c + d ) + cos ( σ + b + c + d ) .$
$N D = Ω k = 1 M − 1 { [ cos α k − 1 ] [ sin α k 0 ] } = [ b 1 b 2 b 3 ... b M ] [ a 1 a 2 a 3 ... a M ] .$
$b 1 = ∑ sin [ Σ { α 1 , α 2 , α 3 ... , α M − 1 } M − 1 M − 1 ] = − sin ( σ ) ,$
$b 2 = ( − 1 ) ∑ sin [ Σ { α 1 , α 2 , α 3 ... , α M − 1 } M − 2 M − 1 ] = ( − 1 ) ∑ k = 1 M − 1 sin ( σ − α k ) ,$
$b M = ( − 1 ) M + 1 ∑ sin [ Σ { α 1 , α 2 , α 3 ... , α M − 1 } M − M M − 1 ] = ( − 1 ) M + 1 sin ( 0 ) = 0 ,$
$σ = ∑ k = 1 M − 1 α k .$
$a 1 = − ∑ cos [ Σ { α 1 , α 2 , α 3 ... , α M − 1 } M − 1 M − 1 ] = − cos ( σ ) ,$
$a 2 = ( − 1 ) 2 ∑ cos [ Σ { α 1 , α 2 , α 3 ... , α M − 1 } M − 2 M − 1 ] = ( − 1 ) 2 ∑ k = 1 M − 1 cos ( σ − α k ) ,$
$a M = ( − 1 ) M ∑ cos [ Σ { α 1 , α 2 , α 3 ... , α M − 1 } M − M M − 1 ] = ( − 1 ) M cos ( 0 ) = ( − 1 ) M .$
$tan ( φ ) = N I → M D I → M = ∑ k = 1 M { ( − 1 ) k + 1 ∑ sin [ Σ { α 1 , α 2 , ... α M − 1 } M − k M − 1 ] } I k ∑ k = 1 M { ( − 1 ) k ∑ cos [ Σ { α 1 , α 2 , ... α M − 1 } M − k M − 1 ] } I k ,$
$tan ( φ ) = N I → M D I → M = ∑ k = 1 M { ( − 1 ) k + 1 ∑ cos [ Σ { α 1 , α 2 , ... α M − 1 } M − k M − 1 ] } I k ∑ k = 1 M { ( − 1 ) k + 1 ∑ sin [ Σ { α 1 , α 2 , ... α M − 1 } M − k M − 1 ] } I k .$
$b 1 = ∑ sin [ σ / 2 − Σ { α 1 , α 2 , α 3 ... , α M − 1 } 0 M − 1 ] = − b M = sin ( σ / 2 ) ,$
$a 1 = ( − 1 ) ∑ cos [ σ / 2 − Σ { α 1 , α 2 , α 3 ... , α M − 1 } 0 M − 1 ] = − a M = − cos ( σ / 2 ) ,$
$b ( m + 1 ) / 2 = ( − 1 ) ( m + 3 ) / 2 ∑ sin [ σ / 2 − Σ { α 1 , α 2 , α 3 ... , α M − 1 } ( M − 1 ) / 2 M − 1 ] = 0.$
$tan ( φ ) = N I → M D I → M = ∑ k = 1 M { ( − 1 ) k + 1 ∑ sin [ σ / 2 − Σ { α 1 , α 2 , α 3 ... , α M − 1 } k − 1 M − 1 ] } I k ∑ k = 1 M { ( − 1 ) k ∑ cos [ σ / 2 − Σ { α 1 , α 2 , α 3 ... , α M − 1 } k − 1 M − 1 ] } I k .$
$tan ( φ ) = N I → M D I → M = ∑ k = 1 M { ( − 1 ) k + 1 ∑ cos [ σ / 2 − Σ { α 1 , α 2 , α 3 ... , α M − 1 } k − 1 M − 1 ] } I k ∑ k = 1 M { ( − 1 ) k + 1 ∑ sin [ σ / 2 − Σ { α 1 , α 2 , α 3 ... , α M − 1 } k − 1 M − 1 ] } I k .$
$P ( x ) = ∑ k = 1 5 ( a k + i b k ) x k − 1 = − ( x − 1 ) ( x + 1 ) [ x + cos ( α ) + i sin ( α ) ] [ x − cos ( α ) + i sin ( α ) ] .$
$N D = { 2 sin ( α ) [ 0 1 0 − 1 0 ] [ − 1 0 2 0 − 1 ] } ∗ { cos ( β / 2 ) [ 1 − 1 ] sin ( β / 2 ) [ 1 1 ] } ∗ { sin ( β / 2 ) [ 1 − 1 ] cos ( β / 2 ) [ 1 1 ] } .$
$H ( ω ) = ( − 2 ) 6 sin ( ω ) [ sin ( ω ) − sin ( α ) ] [ sin ( ω ) − sin ( β ) ] .$
$tan ( φ ) = N I → 7 D I → 7 = − ( I 1 − I 7 ) + [ 3 + 4 sin ( α ) sin ( β ) ] ( I 3 − I 5 ) − 2 [ sin ( α ) + sin ( β ) ] ( I 2 − 2 I 4 + I 6 ) .$
$tan ( φ ) = − ( I 1 − I 7 ) + 3 ( I 3 − I 5 ) − 2 sin ( α ) ( I 2 − 2 I 4 + I 6 ) .$
$tan ( φ ) = I 1 − 3 I 3 + 3 I 5 − I 7 2 I 2 − 4 I 4 + 2 I 6 .$
$tan ( φ ) = I 1 − 5 I 3 + 5 I 5 − I 7 3 I 2 − 6 I 4 + 3 I 6 .$
$tan ( φ ) = − ( I 1 − I 7 ) + [ 3 + 4 sin 2 ( α ) ] ( I 3 − I 5 ) − 4 sin ( α ) ( I 2 − 2 I 4 + I 6 ) = ( I 1 − I 7 ) − [ 5 − 2 cos ( 2 α ) ] ( I 3 − I 5 ) 4 sin ( α ) ( I 2 − 2 I 4 + I 6 ) ,$
$tan ( φ ) = I 1 − 7 I 3 + 7 I 5 − I 7 4 ( I 2 − 2 I 4 + I 6 ) .$
$N D = lim α 1 → 0 Ω k = 1 4 { [ cos α − 1 ] [ sin α 0 ] } ∗ [ cos ( α 1 ) − 1 ] [ sin ( α 1 ) 0 ] .$
$H ( ω ) = ( − 2 ) 5 exp ( 2 α i ) sin ( ω / 2 ) sin 4 [ ( ω − α ) / 2 ] .$
$a k = ( − 1 ) k ∑ cos [ Σ { α , α , α , α } 5 − k 4 ] = ( − 1 ) k ( 4 5 − k ) cos [ ( 5 − k ) α ]$