Some results on convolution idempotents May 28, 2020 1 IIT Hyderabad, India 2 Stanford University 1 P.Charantej Reddy 1 Aditya Siripuram 1 Brad Osgood 2
Problem Statement then Motivation comes from sampling and Fuglede’s conjecture. . that vanish on , fjnd all idempotents and a set Given a positive integer Zero-set problem then • If • If Convolution idempotent (Defjnition) • For example: is support set an indicator, , where • We have or • The DFT coeffjcients are either 2 A mapping h : Z N → C N is a convolution idempotent if h ∗ h = h . Where Z N are integers modulo N and ∗ is circular convolution.
Problem Statement then Motivation comes from sampling and Fuglede’s conjecture. . that vanish on , fjnd all idempotents and a set Given a positive integer Zero-set problem then • If • If Convolution idempotent (Defjnition) • For example: is support set an indicator, , where • We have 2 A mapping h : Z N → C N is a convolution idempotent if h ∗ h = h . Where Z N are integers modulo N and ∗ is circular convolution. • The DFT coeffjcients are either 0 or 1 ⇒ ( F h ) 2 = F h h ∗ h = h =
Problem Statement • If Motivation comes from sampling and Fuglede’s conjecture. . that vanish on , fjnd all idempotents and a set Given a positive integer Zero-set problem then then Convolution idempotent (Defjnition) • If • For example: 2 A mapping h : Z N → C N is a convolution idempotent if h ∗ h = h . Where Z N are integers modulo N and ∗ is circular convolution. • The DFT coeffjcients are either 0 or 1 ⇒ ( F h ) 2 = F h h ∗ h = h = • We have h J = F − 1 1 J , where 1 J an indicator, J is support set
Problem Statement Zero-set problem Motivation comes from sampling and Fuglede’s conjecture. . that vanish on , fjnd all idempotents and a set Given a positive integer 2 Convolution idempotent (Defjnition) • For example: A mapping h : Z N → C N is a convolution idempotent if h ∗ h = h . Where Z N are integers modulo N and ∗ is circular convolution. • The DFT coeffjcients are either 0 or 1 ⇒ ( F h ) 2 = F h h ∗ h = h = • We have h J = F − 1 1 J , where 1 J an indicator, J is support set • If J = Z N then h = δ • If J = {} then h = 0
Problem Statement Convolution idempotent (Defjnition) • For example: Zero-set problem Motivation comes from sampling and Fuglede’s conjecture. 2 A mapping h : Z N → C N is a convolution idempotent if h ∗ h = h . Where Z N are integers modulo N and ∗ is circular convolution. • The DFT coeffjcients are either 0 or 1 ⇒ ( F h ) 2 = F h h ∗ h = h = • We have h J = F − 1 1 J , where 1 J an indicator, J is support set • If J = Z N then h = δ • If J = {} then h = 0 Given a positive integer N and a set Z ⊆ Z N , fjnd all idempotents h : Z N → C N that vanish on Z .
Problem Statement Convolution idempotent (Defjnition) • For example: Zero-set problem Motivation comes from sampling and Fuglede’s conjecture. 2 A mapping h : Z N → C N is a convolution idempotent if h ∗ h = h . Where Z N are integers modulo N and ∗ is circular convolution. • The DFT coeffjcients are either 0 or 1 ⇒ ( F h ) 2 = F h h ∗ h = h = • We have h J = F − 1 1 J , where 1 J an indicator, J is support set • If J = Z N then h = δ • If J = {} then h = 0 Given a positive integer N and a set Z ⊆ Z N , fjnd all idempotents h : Z N → C N that vanish on Z .
Motivation I: Sampling • Sampling is a well studied problem in Signal Processing • Traditional sampling with uniformly spaced samples can be ineffjcient while sampling signals with fragmented spectra. Figure 1: Example signal with two fragments, for . • In traditional setting average number of samples taken per second is at-least for the example signal in Figure 1. • Can we do better than this by using the frequency support information? 3
Motivation I: Sampling • Sampling is a well studied problem in Signal Processing information? • Can we do better than this by using the frequency support for the example signal in Figure 1. at-least • In traditional setting average number of samples taken per second is 3 ineffjcient while sampling signals with fragmented spectra. • Traditional sampling with uniformly spaced samples can be F f ( s ) s 0 1 2 3 Figure 1: Example signal with two fragments, for F = { 0 , 2 } .
Motivation I: Sampling • Sampling is a well studied problem in Signal Processing • Traditional sampling with uniformly spaced samples can be ineffjcient while sampling signals with fragmented spectra. • In traditional setting average number of samples taken per second is • Can we do better than this by using the frequency support information? 3 F f ( s ) s 0 1 2 3 Figure 1: Example signal with two fragments, for F = { 0 , 2 } . at-least 3 for the example signal in Figure 1.
Motivation I: Sampling • Sampling is a well studied problem in Signal Processing • Traditional sampling with uniformly spaced samples can be ineffjcient while sampling signals with fragmented spectra. • In traditional setting average number of samples taken per second is • Can we do better than this by using the frequency support information? 3 F f ( s ) s 0 1 2 3 Figure 1: Example signal with two fragments, for F = { 0 , 2 } . at-least 3 for the example signal in Figure 1.
Motivation I: Sampling (contd) Multi-coset sampling • Two samples every second sampled Figure 2: Top-left: Example sampling pattern, Top-right: Spectrum of signal and Bottom: Spectrum of sampled signal 4 • At 0 s, 0 . 25 s, 1 s, 1 . 25 s, 2 s, 2 . 25 s, . . . 0 1 2 t
Motivation I: Sampling (contd) Multi-coset sampling and Bottom: Spectrum of sampled signal Figure 2: Top-left: Example sampling pattern, Top-right: Spectrum of signal sampled 4 • Two samples every second • At 0 s, 0 . 25 s, 1 s, 1 . 25 s, 2 s, 2 . 25 s, . . . F f ( s ) s 0 1 2 t 0 1 2 3
Motivation I: Sampling (contd) Multi-coset sampling and Bottom: Spectrum of sampled signal Figure 2: Top-left: Example sampling pattern, Top-right: Spectrum of signal 4 • Two samples every second • At 0 s, 0 . 25 s, 1 s, 1 . 25 s, 2 s, 2 . 25 s, . . . F f ( s ) s 0 1 2 t 0 1 2 3 |F f sampled ( s ) | s 0 1 2 3
Multi-coset sampling in the space, the Fourier transform Applied Physics 24.12 (1953), pp. 1432–1436. 1 Arthur Kohlenberg. “Exact interpolation of band-limited functions”. In: Journal of . is in zero only when the frequency is non • For any signal • Non-uniform deterministic sampling techniques 1 . Figure 3: Example signal with two fragments, for • Spectrum is non zero at integer intervals • Location of spectrum is known • Spectrum is non zero only in +ve frequency • With out loss of generality we make following assumptions • Frequency support is known 5
Multi-coset sampling • For any signal 1 Kohlenberg, “Exact interpolation of band-limited functions”. . is in zero only when the frequency is non in the space, the Fourier transform . • Non-uniform deterministic sampling techniques 1 Figure 3: Example signal with two fragments, for • Spectrum is non zero at integer intervals • Location of spectrum is known • Spectrum is non zero only in +ve frequency • With out loss of generality we make following assumptions • Frequency support is known 5
Multi-coset sampling • Spectrum is non zero at integer intervals 1 Kohlenberg, “Exact interpolation of band-limited functions”. • Non-uniform deterministic sampling techniques 1 5 • Location of spectrum is known • Spectrum is non zero only in +ve frequency • With out loss of generality we make following assumptions • Frequency support is known F f ( s ) s 0 1 2 3 Figure 3: Example signal with two fragments, for F = { 0 , 2 } . • For any signal f in the space, the Fourier transform F f ( s ) is non zero only when the frequency s is in ∪ n ∈ F [ n, n + 1] .
Multi-coset sampling (contd) • Let us cosider the sampling pattern to be picked. second ofgset 6 ∞ � � p J ( t − kN ) , where p J ( t ) = δ ( t − m/N ) k = −∞ m ∈J . . . . . . . . . 0 1 2 t 1 N Figure 4: Sampling pattern: Samples are taken at every second, and at a 1 /N • And J ⊆ [0 , N − 1] and N ( > max F + 1) are parameters that need
Multi-coset sampling (contd) • From elementary Fourier analysis, the sampled signal has spectrum Figure 5: Shifted spectrum for 7 ∞ � F f sampled ( s ) = h J ( k ) F f ( s − k ) k = −∞ • here h J ( n ) is discrete Fourier transform of 1 J
Multi-coset sampling (contd) • From elementary Fourier analysis, the sampled signal has spectrum 7 ∞ � F f sampled ( s ) = h J ( k ) F f ( s − k ) k = −∞ • here h J ( n ) is discrete Fourier transform of 1 J F f ( s − k ) s 0 1 2 3 4 5 6 Figure 5: Shifted spectrum for k = 0
Multi-coset sampling (contd) • From elementary Fourier analysis, the sampled signal has spectrum 7 ∞ � F f sampled ( s ) = h J ( k ) F f ( s − k ) k = −∞ • here h J ( n ) is discrete Fourier transform of 1 J F f ( s − k ) s 0 1 2 3 4 5 6 Figure 5: Shifted spectrum for k = 0 , 1
Multi-coset sampling (contd) • From elementary Fourier analysis, the sampled signal has spectrum 7 ∞ � F f sampled ( s ) = h J ( k ) F f ( s − k ) k = −∞ • here h J ( n ) is discrete Fourier transform of 1 J F f ( s − k ) s 0 1 2 3 4 5 6 Figure 5: Shifted spectrum for k = 0 , 1 , 2 .
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