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05/03/09

05/03/09. Dónde nos quedamos la última vez...?. Revisar código del fit y corregir los posibles errores. Restringir el intervalo de masa en torno al pico de masa del Bs. Fijar la fracción de señal en el fit (Fsig = 0.5 , S/B = 1) ‏

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05/03/09

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  1. 05/03/09

  2. Dónde nos quedamos la última vez...? • Revisar código del fit y corregir los posibles errores. • Restringir el intervalo de masa en torno al pico de masa del Bs. • Fijar la fracción de señal en el fit (Fsig = 0.5 , S/B = 1)‏ • Generar datos y visualizar la asimetría (funcion de t): ( N(Bs) – N(Bsbar) ) / ( N(Bs) + N(Bsbar) )‏ • Ajustar esos datos a la función de probabilidad y visualizar (sobre los datos) la predicción de la p.d.f. • Comprobar que los parámetros que salen del ajuste son consistentes con los que se meten de input.

  3. Fsig=1, No Time Acc., No Time Res., w = 0 Nt = 5000 NBs = NBsbar = 2000 btim = 0 ps^3 sigmat = 0 ps

  4. Fsig = 1, No Time Acc., No Time Res., w = 0 NO. NAME VALUE NEGATIVE POSITIVE 1 Rpa 9.44164e-02 -3.99102e-03 4.15076e-03 OK 2 Rpe 7.98491e-02 -5.69817e-03 5.86271e-03 X 3 del1 2.72011e-02 -5.28318e-02 5.27332e-02 OK 4 del2 3.14916e+00 -5.21984e-02 5.71066e-02 OK 5 gama 1.04347e+00 -2.53945e-02 2.57942e-02 OK 6 gamd 5.21238e-02 -5.59795e-02 5.55491e-02 X 7 phis 1.60644e-01 -2.31968e-02 2.32869e-02 X phi_s = 9.20423472797 + 1.33424364804 - 1.32907969852

  5. Fsig = 1, Time Acc., Time Res., w = 0 Nt = 5000 NBs = NBsbar = 2000 btim = 0.027 ps^3 sigmat = 0.045 ps

  6. Fsig = 1, Time Acc., Time Res., w = 0 NO. NAME VALUE NEGATIVE POSITIVE 1 Rpa 9.39432e-02 -4.12677e-03 4.20634e-03 OK 2 Rpe 8.76708e-02 -6.20614e-03 6.53852e-03 X 3 del1 5.48024e-02 -7.66503e-02 7.61579e-02 OK 4 del2 3.17901e+00 -6.76180e-02 6.71483e-02 OK 5 gama 1.01369e+00 -2.16648e-02 2.19685e-02 OK 6 gamd 1.91410e-01 -5.13327e-02 5.07411e-02 OK 7 phis 2.39055e-01 -3.17366e-02 3.26277e-02 X phi_s = 13.6968317029 + 1.86942730543 - 1.81837220075

  7. Fsig = 0.5, Time Acc., Time Res., w = 0 Nt = 5000 NBs = NBsbar = 2000 btim = 0.027 ps^3 sigmat = 0.045 ps

  8. Fsig = 0.5, Time Acc., Time Res., w = 0 NO. NAME VALUE NEGATIVE POSITIVE 1 Rpa 1.03226e-01 -7.38658e-03 7.62333e-03 OK 2 Rpe 7.44554e-02 -9.55611e-03 1.04784e-02 X 3 del1 -6.85116e-02 -1.33218e-01 1.35722e-01 OK 4 del2 3.11361e+00 -1.10850e-01 1.11274e-01 OK 5 gama 1.01437e+00 -3.35596e-02 3.42083e-02 X 6 gamd 3.27944e-01 -8.22682e-02 8.05971e-02 X 7 phis 1.66163e-01 -4.46237e-02 4.66493e-02 OK phi_s = 9.52042211315 + 2.67280587273 - 2.55675114589

  9. Fsig = 0.5, Time Acc., Time Res., w = 0.3 Nt = 5000 NBs = NBsbar = 2000 btim = 0.027 ps^3 sigmat = 0.045 ps

  10. Fsig = 0.5, Time Acc., Time Res., w = 0.3 NO. NAME VALUE NEGATIVE POSITIVE 1 Rpa 9.26899e-02 -8.11812e-03 8.46227e-03 OK 2 Rpe 9.20385e-02 -1.19684e-02 1.27354e-02 OK 3 del1 -1.51052e-01 -2.28346e-01 2.38222e-01 OK 4 del2 3.35116e+00 -2.05974e-01 2.06901e-01 OK 5 gama 1.00895e+00 -3.58098e-02 3.62170e-02 OK 6 gamd 1.16082e-01 -8.53322e-02 8.49316e-02 OK 7 phis 2.73578e-01 -1.20337e-01 1.27374e-01 OK phi_s = 15.6748769252º + 7.29796991171º - 6.89480233219º

  11. Conclusiones & Tareas • El código se comporta bien. • Los parámetros que se extraen del ajuste son consistentes con el input. • Aumentar el número de datos generados por experimento (N=10 000). • Realizar ~ 100 experimentos para estudiar las “pull distribution” de los parámetros extraídos del fit.

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