Xb = (tdc1-tdc2)/2 for the bottom counter and Xt = (tdc3-tdc4)/2 for the top counter (1) tdc*=TDC*0.5where TDC is Le Croy 3377 TDC count (with 0.5 nsec bin width) . The plane angle was measured thru Xb and Xt as:

tan(theta) = (Xb - Xt)/h (2)where Xb,Xh were rewritten in cm using the length of the counter of 235 cm and the light signal travel time of 15 nsec. The vertical distance between counters in hodoscope h is 275.6 cm. The mean of angular distribution in

SUM = tdc3 + tdc4 (3)was used to estimate the time resolution of the trigger (per one PMT). But first because the time of the trigger as a TDC common stop depends from the muon coordinate Xb in the bottom counter such dependence should be taken into account. The SUM vs Xb distributions are presented in

SUM = tdc3 + tdc4 - 2*abs((tdc1 - tdc2)/2) (4)

SUM = tdc3 + tdc4 - 2*abs((tdc1 - tdc2)/2) - 2*v*h*(SQRT(1 + ((Xb-Xt)/h)**2) - 1) (5)The possible contribution of the signal amplitude to the trigger timing was ignored. We don't have control on it. Also the discriminator level was set low enough to make such contribution small. The SUM vs Xb distributions after corrections above are presented in

tdc3 = Tr - t3 tdc4 = Tr - t4 tdc1 = Tr - t1 tdc2 = Tr - t2 (6)where Tr, t1, t2, t3 and t4 are absolute time of the trigger and PMT 1 thru 4 signals. Therefore (4) becomes:

SUM = Tr - (t3 + t4) + t1 if trigger from t2 (Tr = t2) SUM = Tr - (t3 + t4) + t2 if trigger from t1 (Tr = t1) (7)and the variance of the SUM is:

D(SUM) = D(Tr - t3 - t4 + t1) = 4*D(t) (8)if we assume that all 4 time signals in given hodoscope have one and the same resolution. The resolutions obtained as sqrt[D(SUM)]/2 are given in

teren@fnal.gov

Last modified: Fri Apr 30 18:00:00 CST 1999