# Simulation of multiple Q dK2 errors(cont.2) - PowerPoint PPT Presentation

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 =1e-3(for 20cm Q) variable(for FD). Simulation of multiple Q dK2 errors(cont.2). dK2=2 sgn K1  /r (r=10mm, sgn= 1 : random for each Q ). dK2= + 2 K1  /r (r=10mm). dK2= - 2 K1  /r (r=10mm). Example: d FD=5e-4. Corr . It seems OK for d FD=5e-4.

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Simulation of multiple Q dK2 errors(cont.2)

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=1e-3(for 20cm Q)

variable(for FD)

### Simulation of multiple Q dK2 errors(cont.2)

dK2=2 sgn K1 /r

(r=10mm, sgn=1: random for each Q)

dK2=+2 K1 /r (r=10mm)

dK2=-2 K1 /r (r=10mm)

Example:dFD=5e-4

Corr

It seems OK for dFD=5e-4.

dK2=2 sgn K1 /r Sin[q], dSK2=2 sgn K1 /r Cos[q]

(r=10mm, sgn=1: random for each Q,0<q<2p:uniformly random)

### Simulation of multiple Q dK2 errors(cont.3)

Correction: roll(1mrad step) of SX+K2(1% step) of SX

dFD=1e-4

After cor.,

sy(95%CL)=44nm

Iteration of correction x2

 sy(95%CL)=47nm

Iteration of cor. is not effective

Cor.

dFD=5e-4, d20cmQ=5e-4

After cor.,

sy(95%CL)=225nm

SK2 of FD affects significantly

to sy. Tolerance of SK2 should

be much less than 1e-4.

Cor.

M.Woodley