Data Unfolding with Wiener-SVD Method arXiv:1705.03568 Tang, a,1 X. - PowerPoint PPT Presentation

Data Unfolding with Wiener-SVD Method arXiv:1705.03568 Tang, a,1 X. Li, b,1 X. Qian, a,2 H. Wei, a C. Zhang a W. a Physics Department, Brookhaven National Laboratory, Upton, NY, USA b State University of New Y ork at Stony Brook, Department of Physics and Astronomy, Stony Brook, NY, USA xiaoyue.li@stonybrook.edu DPF – FNAL, Aug. 2 nd , 2017 Xiaoyue Li DPF2017 - FNAL 1

Outline § The unfolding problem § Wiener filter in digital signal processing § SVD unfolding with Wiener filter § Data unfolding example § Cross-section § Reactor neutrino flux § Summary and recommendation Xiaoyue Li DPF2017 - FNAL 2

The data unfolding problem § The problem: 1200 1200 Smeared distribution True distribution Data 1000 1000 𝑔 "#$%& (𝑦) = +𝑆 𝑦 𝑧 𝑔 .&/$ 𝑧 d𝑧 800 800 600 600 § 𝑆 𝑦 𝑧 : response function 400 400 200 200 § 𝑛 3 = ∑ 𝑆 35 𝑡 5 5 0 0 0 2 4 6 8 10 12 14 16 18 20 0 2 4 6 8 10 12 14 16 18 20 § 𝑆 35 = 𝑄 observed in bin 𝑗 true value in bin 𝑘) § Why not do 𝒕 = 𝑆 GH I m ? § Due to statistical fluctuation, the unfolded spectrum from direct inversion of response matrix bares no resemblance to the true spectrum. § Introduce additional constraints. § E.g. trade bias for smoothness § Bayesian analysis Xiaoyue Li DPF2017 - FNAL 3

The data unfolding problem § The problem: 1200 1200 Smeared distribution True distribution Data 1000 1000 𝑔 "#$%& (𝑦) = +𝑆 𝑦 𝑧 𝑔 .&/$ 𝑧 d𝑧 800 800 600 600 § 𝑆 𝑦 𝑧 : response function 400 400 200 200 § 𝑛 3 = ∑ 𝑆 35 𝑡 5 5 0 0 0 2 4 6 8 10 12 14 16 18 20 0 2 4 6 8 10 12 14 16 18 20 § 𝑆 35 = 𝑄 observed in bin 𝑗 true value in bin 𝑘) § Why not do 𝒕 = 𝑆 GH I m ? § Due to statistical fluctuation, the unfolded spectrum from direct inversion of response matrix bares no resemblance to the true spectrum. § Introduce additional constraints. § E.g. trade bias for smoothness § Bayesian analysis Xiaoyue Li DPF2017 - FNAL 4

The data unfolding problem § The problem: 1200 1200 Smeared distribution True distribution Data 1000 1000 𝑔 "#$%& (𝑦) = +𝑆 𝑦 𝑧 𝑔 .&/$ 𝑧 d𝑧 800 800 600 600 § 𝑆 𝑦 𝑧 : response function 400 400 200 200 § 𝑛 3 = ∑ 𝑆 35 𝑡 5 5 0 0 0 2 4 6 8 10 12 14 16 18 20 0 2 4 6 8 10 12 14 16 18 20 § 𝑆 35 = 𝑄 observed in bin 𝑗 true value in bin 𝑘) § Why not do 𝒕 = 𝑆 GH I m ? 2000 2000 True distribution True distribution Unfolded Data Unfolded Data § Due to statistical fluctuation, the unfolded 1500 1500 spectrum from direct inversion of response matrix 1000 bares no resemblance to the true spectrum. 1000 500 § Introduce additional constraints. 500 0 § E.g. trade bias for smoothness 0 0 2 4 6 8 10 12 14 16 18 20 0 2 4 6 8 10 12 14 16 18 20 § Bayesian analysis Xiaoyue Li DPF2017 - FNAL 5

The data unfolding problem § The problem: 1200 1200 Smeared distribution True distribution Data 1000 1000 𝑔 "#$%& (𝑦) = +𝑆 𝑦 𝑧 𝑔 .&/$ 𝑧 d𝑧 800 800 600 600 § 𝑆 𝑦 𝑧 : response function 400 400 200 200 § 𝑛 3 = ∑ 𝑆 35 𝑡 5 5 0 0 0 2 4 6 8 10 12 14 16 18 20 0 2 4 6 8 10 12 14 16 18 20 § 𝑆 35 = 𝑄 observed in bin 𝑗 true value in bin 𝑘) § Why not do 𝒕 = 𝑆 GH I m ? 2000 2000 True distribution True distribution Unfolded Data Unfolded Data § Due to statistical fluctuation, the unfolded 1500 1500 spectrum from direct inversion of response matrix 1000 bares no resemblance to the true spectrum. 1000 500 § Introduce additional constraints. 500 0 § E.g. trade bias for smoothness 0 0 2 4 6 8 10 12 14 16 18 20 0 2 4 6 8 10 12 14 16 18 20 § Bayesian analysis Xiaoyue Li DPF2017 - FNAL 6

The data unfolding problem § The problem: 1200 1200 Smeared distribution True distribution Data 1000 1000 𝑔 "#$%& (𝑦) = +𝑆 𝑦 𝑧 𝑔 .&/$ 𝑧 d𝑧 800 800 600 600 § 𝑆 𝑦 𝑧 : response function 400 400 200 200 § 𝑛 3 = ∑ 𝑆 35 𝑡 5 5 0 0 0 2 4 6 8 10 12 14 16 18 20 0 2 4 6 8 10 12 14 16 18 20 § 𝑆 35 = 𝑄 observed in bin 𝑗 true value in bin 𝑘) § Why not do 𝒕 = 𝑆 GH I m ? 2000 2000 True distribution True distribution Unfolded Data Unfolded Data § Due to statistical fluctuation, the unfolded 1500 1500 spectrum from direct inversion of response matrix 1000 bares no resemblance to the true spectrum. 1000 500 § Introduce additional constraints. 500 0 § E.g. trade bias for smoothness 0 0 2 4 6 8 10 12 14 16 18 20 0 2 4 6 8 10 12 14 16 18 20 § Bayesian analysis Xiaoyue Li DPF2017 - FNAL 7

Wiener filter in digital signal processing (1) § Deconvolution problem: Response function 𝑆(𝑢, 𝑢′) True signal 𝑇(𝑢) § 𝑁 𝑢 M = ∫ Q 𝑆 𝑢, 𝑢 M I 𝑇 𝑢 d𝑢 Signal: 200k electrons Single Electron Response -3 10 × V (mV) - GQ e 4000 0.15 § True signal 𝑇(𝑢) , measured signal 𝑁(𝑢 M ) , response 3000 0.1 function 𝑆(𝑢, 𝑢′) ≡ 𝑆(𝑢 − 𝑢′) 2000 0.05 § Fourier transform: 1000 0 𝑁 𝜕 = 𝑆(𝜕) I 𝑇(𝜕) → 𝑇 𝜕 = 𝑁 𝜕 /𝑆(𝜕) 0 0 20 40 60 80 100 0 20 40 60 80 100 Time ( s) Time ( s) µ µ → Inverse FFT 𝑇 𝜕 → 𝑇 𝑢 Simulated Measured Signal ADC § The response function 𝑆(𝜕) does not address 30 noise contributions to the measured signal. 20 Worse still, 𝑆(𝜕) is generally smaller at higher Measured signal, with statistical fluctuation 10 frequencies due to the shaping features of electronics, resulting in amplification of noises. 0 0 20 40 60 80 100 Time ( s) µ Xiaoyue Li DPF2017 - FNAL 8

Wiener filter in digital signal processing (1) § Deconvolution problem: Response function 𝑆(𝑢, 𝑢′) Response function 𝑆(𝜕) True signal 𝑇(𝑢) § 𝑁 𝑢 M = ∫ Q 𝑆 𝑢, 𝑢 M I 𝑇 𝑢 d𝑢 Response function in Frequency domain Signal: 200k electrons Single Electron Response -3 10 × V (mV) ) - GQ e ω 4000 R( 0.15 30 § True signal 𝑇(𝑢) , measured signal 𝑁(𝑢 M ) , response 3000 0.1 function 𝑆(𝑢, 𝑢′) ≡ 𝑆(𝑢 − 𝑢′) 20 2000 0.05 § Fourier transform: 10 1000 0 𝑁 𝜕 = 𝑆(𝜕) I 𝑇(𝜕) → 𝑇 𝜕 = 𝑁 𝜕 /𝑆(𝜕) 0 0 0 20 40 60 80 100 0 20 40 60 80 100 0 0.2 0.4 0.6 0.8 1 Time ( s) Time ( s) µ µ (MHz) ω → Inverse FFT 𝑇 𝜕 → 𝑇 𝑢 Simulated Measured Signal Data in Frequency domain ADC ) § The response function 𝑆(𝜕) does not address ω M( 2000 30 noise contributions to the measured signal. 1500 20 Worse still, 𝑆(𝜕) is generally smaller at higher Measured signal in Measured signal, with 1000 frequency domain 𝑁 𝜕 statistical fluctuation 10 frequencies due to the shaping features of 500 electronics, resulting in amplification of noises. 0 0 0 20 40 60 80 100 0 0.2 0.4 0.6 0.8 1 Time ( s) µ ω (MHz) Xiaoyue Li DPF2017 - FNAL 9

Wiener filter in digital signal processing (1) § Deconvolution problem: Response function 𝑆(𝑢, 𝑢′) Response function 𝑆(𝜕) True signal 𝑇(𝑢) § 𝑁 𝑢 M = ∫ Q 𝑆 𝑢, 𝑢 M I 𝑇 𝑢 d𝑢 Response function in Frequency domain Signal: 200k electrons Single Electron Response -3 10 × V (mV) ) - GQ e ω 4000 R( 0.15 30 § True signal 𝑇(𝑢) , measured signal 𝑁(𝑢 M ) , response 3000 0.1 function 𝑆(𝑢, 𝑢′) ≡ 𝑆(𝑢 − 𝑢′) 20 2000 0.05 § Fourier transform: 10 1000 0 𝑁 𝜕 = 𝑆(𝜕) I 𝑇(𝜕) → 𝑇 𝜕 = 𝑁 𝜕 /𝑆(𝜕) 0 0 0 20 40 60 80 100 0 20 40 60 80 100 0 0.2 0.4 0.6 0.8 1 Time ( s) Time ( s) µ µ (MHz) ω → Inverse FFT 𝑇 𝜕 → 𝑇 𝑢 Deconvoluted signal 𝑇(𝑢) Simulated Measured Signal Data in Frequency domain ADC ) § The response function 𝑆(𝜕) does not address ω Deconvoluton without filter 6 M( 10 × 2000 30 - e noise contributions to the measured signal. 1500 20 10000 Worse still, 𝑆(𝜕) is generally smaller at higher Measured signal, with Measured signal in 1000 frequency domain 𝑁 𝜕 statistical fluctuation 10 frequencies due to the shaping features of 0 500 electronics, resulting in amplification of noises. 0 -10000 0 0 20 40 60 80 100 0 0.2 0.4 0.6 0.8 1 Time ( s) µ ω (MHz) 0 20 40 60 80 100 Time ( s) µ Xiaoyue Li DPF2017 - FNAL 10