A new look at the Discrete Fourier Transform The missing Fourier transform The two-modular Fourier transform of binary functions Yi Hong (Monash University, Melbourne, Australia) Emanuele Viterbo (Monash University, Melbourne, Australia) Jean-Claude Belfiore (Telecom ParisTech) Emanuele Viterbo The two-modular Fourier transform
A new look at the Discrete Fourier Transform The missing Fourier transform The Discrete Fourier Transform N samples of a discrete-time signal (real or complex) form the time-domain vector x = ( x [ n ]) N − 1 n =0 The discrete Fourier transform (DFT) of x is the frequency-domain vector X = ( X [ k ]) N − 1 k =0 N − 1 � x [ n ] e − 2 π nk X [ k ] = k = 0 , . . . N − 1 N n =0 The inverse discrete Fourier transform (IDFT) of X is N − 1 x [ n ] = 1 � X [ k ] e 2 π nk n = 0 , . . . N − 1 N N k =0 Emanuele Viterbo The two-modular Fourier transform
A new look at the Discrete Fourier Transform The missing Fourier transform The Discrete Fourier Transform The vector x = ( x [ n ]) N − 1 n =0 gives the N samples of a time-domain function f : Z → C If f is periodic by N samples then f : Z N → C (assumption for DFT to provide the discrete spectrum) The time axis Z N = { 0 , 1 , . . . N − 1 } is an additive group G = ( Z N , +) with addition mod N then f : G → C In the frequency domain X = ( X [ k ]) N − 1 k =0 represents the transform of f as a periodic function ˆ f : G → C The frequency axis has the same additive group structure G = ( Z N , +) Emanuele Viterbo The two-modular Fourier transform
A new look at the Discrete Fourier Transform The missing Fourier transform The Discrete Fourier Transform The DFT matrix F = { e − 2 π nk N } N − 1 n,k =0 is a unitary matrix such that x T = 1 X T = Fx T N F H X T The vectors x and X are a two representations of the signal x [ n ] in different coordinate systems, defined by the time basis and frequency basis. ▼ ❇ ✻ ❇ ❇ ❇ s x [ n ] x ✏ ✏ ❇ ✏ ✏ ❇ ❇ ✏ ✏ X ❇ ✏ ✏ frequency ✶ ✏ ✏✏✏✏✏✏✏✏ ❇ ❇ ❇ ❇ ❇ ✲ time Emanuele Viterbo The two-modular Fourier transform
A new look at the Discrete Fourier Transform The missing Fourier transform One-dimensional group representation The Abelian group G = ( Z N , +) with addition mod N admits the following one-dimensional representations χ k ( n ) = e − 2 π nk χ k : G → S k ⊂ C N where k = 0 , . . . , N − 1 and � � N , · · · , e − 2 π ( N − 1) k 1 , e − 2 π k N , e − 2 π 2 k S k = N The representation χ k is a group homomorphism transforming the addition mod N in G into the multiplication of N -th roots of unity in S k , i.e., for any a, b ∈ G χ k ( a + b ) = χ k ( a ) χ k ( b ) since e − 2 π ( a + b ) k = e − 2 π ak N e − 2 π bk N N Emanuele Viterbo The two-modular Fourier transform
A new look at the Discrete Fourier Transform The missing Fourier transform Example Z 6 k S k = { χ k ( g ) , g ∈ G = { 0 , 1 , 2 , 3 , 4 , 5 }} Ker ( χ k ) , G/ Ker ( χ k ) 0 { 1 } { 0 , 1 , 2 , 3 , 4 , 5 } , { 0 } 0 4 5 { 1 , e − 2 π 6 , e − 4 π 6 , e − 6 π 6 , e − 8 π 6 , e − 10 π 6 } 1 { 0 } , { 0 , 1 , 2 , 3 , 4 , 5 } 3 0 2 1 4 { 1 , e − 4 π 6 , e − 8 π 6 } 2 { 0 , 3 } , { 0 , 2 , 4 } 0 2 3 { 1 , − 1 } { 0 , 2 , 4 } , { 0 , 3 } 3 0 2 { 1 , e 4 π 6 , e 8 π 6 } 4 { 0 , 3 } , { 0 , 2 , 4 } 0 4 2 1 { 1 , e 2 π 6 , e 4 π 6 , e 6 π 6 , e 8 π 6 , e 10 π 6 } 5 { 0 } , { 0 , 1 , 2 , 3 , 4 , 5 } 3 0 4 5 Emanuele Viterbo The two-modular Fourier transform
A new look at the Discrete Fourier Transform The missing Fourier transform Example Z 6 (cont.) g ∈ G 0 1 2 3 4 5 χ 0 ( g ) 1 1 1 1 1 1 ψ 0 e − 2 π e − 4 π e − 6 π e − 8 π e − 10 π χ 1 ( g ) 1 ψ 1 6 6 6 6 6 e − 4 π e − 8 π e − 4 π e − 8 π χ 2 ( g ) 1 1 ψ 2 6 6 6 6 χ 3 ( g ) 1 − 1 1 − 1 1 − 1 ψ 3 e 4 π e 8 π e 4 π e 8 π χ 4 ( g ) 1 1 ψ 4 6 6 6 6 e 2 π e 4 π e 6 π e 8 π e 10 π χ 5 ( g ) 1 ψ 5 6 6 6 6 6 ψ 0 . . F = . ψ 5 Emanuele Viterbo The two-modular Fourier transform
A new look at the Discrete Fourier Transform The missing Fourier transform DFT using representations as Fourier basis The representations χ k for k = 0 , . . . , N − 1 are all inequivalent. Some are one-to-one and some are many-to-one and the images can be associated with subgroups of G We can formally rewrite the DFT as � X [ k ] = � x , ψ k � = x [ g ] χ k ( g ) k = 0 , . . . , N − 1 g ∈ G The complex vectors ψ k = [ χ k ( g )] g ∈ G form the discrete Fourier basis vectors Each representation provides a “lens” through which we observe the time-domain signal x [ n ] . Emanuele Viterbo The two-modular Fourier transform
A new look at the Discrete Fourier Transform The missing Fourier transform FFT using representations as Fourier basis Given a normal subgroup H ✂ G we define the quotient group G/H consisting of the coset leaders u of the cosets u + H The direct product of H and G/H is isomorphic to G i.e., G = { u + v | u ∈ H, v ∈ G/H } ≈ H × G/H All u ∈ H = Ker ( χ k ) are mapped to the same value χ k ( u ) = χ k (0) = 1 ∈ S k Then we can compute the DFT more efficiently as � � � X [ k ] = x [ g ] χ k ( g ) = x [ u + v ] χ k ( u + v ) g ∈ G v ∈ G/H u ∈ H �� � � = x [ u + v ] χ k ( v ) k = 0 , . . . , N − 1 v ∈ G/H u ∈ H Emanuele Viterbo The two-modular Fourier transform
A new look at the Discrete Fourier Transform The missing Fourier transform Known generalizations of the DFT concept f : G → C , where G can be an arbitrary group (not only Abelian): Fourier coefficients are complex matrices These generalizations make use of multi-dimensional representations of the group G with matrices over C 1 The inverse Fourier transform uses | G | Tr ( · ) the Trace operator of a matrix to get back to time domain scalar values of f . f : G → K , where K is a field of characteristic p and p does not divide | G | : Fourier coefficients are scalars in K since an “exponential” function can be defined using a primitive element α ∈ K . Emanuele Viterbo The two-modular Fourier transform
A new look at the Discrete Fourier Transform The missing Fourier transform The missing Fourier Transform for binary functions We consider a finite commutative ring R of characteristic p = 2 , e.g., F 2 [ X ] /φ ( X ) , where φ ( X ) is a binary-coefficient polynomial of degree m . Elements of R are represented by m -bit vectors (or polynomials of degree at most m − 1 ) and multiplications are computed by polynomial multiplication mod φ ( X ) . Let G = C n 2 be the additive group of F n 2 ( n bit vectors) We study binary functions f : G → R ( n bit to m bit) and their convolutions (group ring R [ G ] ) If p = 2 divides | G | = 2 n = N , the What does not work? inverse DFT term 1 /N is not defined in R and the Trace fails to work in the generalized inverse DFT. Emanuele Viterbo The two-modular Fourier transform
A new look at the Discrete Fourier Transform The missing Fourier transform The two-modular representations of G Let G = C 2 = { 0 , 1 } then a two-modular representation as 2 × 2 binary matrices over R , is given by the two matrices � 1 � 1 � � 0 1 E 0 = π 1 (0) = and E 1 = π 1 (1) = 0 1 0 1 The matrix entries are the ‘zero’ and ‘one’ element in R . The n -fold direct product of C 2 , G = C n 2 = C 2 × · · · × C 2 can be represented as the Kroneker product of the representations of C 2 , i.e., π n ( G ) � π 1 ( C 2 ) ⊗ · · · ⊗ π 1 ( C 2 ) Emanuele Viterbo The two-modular Fourier transform
A new look at the Discrete Fourier Transform The missing Fourier transform The two-modular representations of G = C 2 × · · · × C 2 Let the binary vectors b = ( b 1 , . . . , b n ) represent the elements of G with bitwise addition mod 2 (XOR). Then π n ( b ) = E b = π 1 ( b 1 ) ⊗ · · · ⊗ π 1 ( b n ) Example G = C 2 × C 2 � 1 0 0 0 � 1 0 1 0 � 1 1 0 0 � 1 1 1 1 � � � � 0 1 0 0 0 1 0 1 0 1 0 0 0 1 0 1 E 00 = E 01 = E 10 = E 11 = 0 0 1 0 0 0 1 0 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 Emanuele Viterbo The two-modular Fourier transform
A new look at the Discrete Fourier Transform The missing Fourier transform The two-modular Fourier basis for G = C 2 × · · · × C 2 Consider the two-modular representations ( 2 k × 2 k matrices) of the nested subgroups of G = C n 2 , H 0 = { 0 n } ✁ H 1 ✁ · · · ✁ H k ✁ · · · ✁ H n − 1 ✁ H n = G where H 1 ∼ = C 2 , H 2 ∼ = C 2 × C 2 , H 3 ∼ = C 2 × C 2 × C 2 , etc. The Fourier basis ‘vectors’ are made up of all the inequivalent two-modular representations π k ∼ H 0 Im ( π 0 ) = { 1 } = ∼ H 1 = Im ( π 1 ) = { E 0 , E 1 } ∼ H 2 Im ( π 2 ) = { E 00 , E 01 , E 10 , E 11 } = ∼ H 3 = Im ( π 3 ) = { E 000 , E 001 , E 010 , E 011 , E 100 , E 101 , E 110 , E 111 } . . . Emanuele Viterbo The two-modular Fourier transform
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