Signal Processing for Charge Measurement OPTIMUM PROCESSOR signi noiseless processor T(s) i(t) C,+C n(o) System Transfer function A·T(s)/(s·(Cd+C)) System Impulse response function Vs(t)=A T(s)/(s(Cd+Cl) Output root mean square noise [VN211/2 Optimum Processor maximize p={Q·A(Cd+Ci)·MAX1(3)·T(s)/s)}/[2]2 Beijing January 2008 Luciano musa
Beijing, January 2008 Luciano Musa 11 Cd+Ci i(t) A noiseless signal processor T(s) n(w) f(t) System Transfer function AT(s) / (s (Cd+Ci )) System Impulse response function vs (t) = ʆ -1AT(s) / (s (Cd+Ci )) Output root mean square noise [vN 2 ] 1/2 Optimum Processor maximize r = {Q A/(Cd+Ci) MAX ʆ -1 (I(s) T(s) / s) } / [vN 2 ] 1/2 OPTIMUM PROCESSOR Signal Processing for Charge Measurement
Electronic Signal Processing R Vout= Xc+R j2lffc joc Low-pass(RC)flter Vout Vin 1+ RCo Example rc=0.5 Integrator s-transfer function SJo Impulse sponse function H( S)=1/(1+RCs Step function response h(t) t/rc h(sI RC Log-Log scale Beijing January 2008 Luciano musa
Beijing, January 2008 Luciano Musa 12 Electronic Signal Processing Example RC=0.5 s=jw R C Vout Vin Vin Xc R Xc Vout + = j fC j C Xc w 1 2 1 = = Vin RCj Vout + w = 1 1 Low-pass (RC) filter 1 2 3 4 5 0.5 1 1.5 2 Integrators-transfer function Impulse rsponse function H(s) = 1/(1+RCs) t RC t e / RC 1 h( ) − = Log-Log scale t f 0.01 0.05 0.1 0.5 1 5 10 0.05 0.1 0.2 0.5 1 1 2 3 4 5 0.2 0.4 0.6 0.8 1 Step function response |h(s)|
Electronic Signal Processing C R Vout Xc+r j2IIfc jo R Vout out RCjo-Vin 1+ RCio High-pass(Cr)filter Example RC=0.5 S=O Differentiator time function Hs)=RCs/(1+RCs h(t)=6(1)- -t/RC RC H(S) Step function response 0.1 Luciano musa
Beijing, January 2008 Luciano Musa 13 Electronic Signal Processing R C Vin Vout Vin Xc R R Vout + = j fC j C Xc w 1 2 1 = = Vin RCj RCj Vout w w + = 1 High-pass (CR) filter 1 2 3 4 5 -2 -1.5 -1 -0.5 0.5 1 Differentiator time function t RC t t e / RC 1 h( ) ( ) − = d − H(s) = RCs/(1+RCs) 0.01 0.05 0.1 0.5 1 5 10 0.05 0.1 0.2 0.5 1 |H(s)| Example RC=0.5 s=jw 1 2 3 4 5 0.2 0.4 0.6 0.8 1 Step function response
Electronic Signal Processing Combining one low-pass(RC) and one high-pass(Cr)filter: R Vin C R RCiO Vout= (1+RCjoC Example rc=o5 CR-RC S-transfer function CR-RC time function h(s)=RCs/(1+RCs)2 h(sI h(t=(1-t/RC)e URC .a Step function response 0.050.1 Log-Log scale f Luciano musa
Beijing, January 2008 Luciano Musa 14 Electronic Signal Processing R C Vout Vin 2 (1 RCjωC RCjω Vout + = R Vin C 1 Combining one low-pass (RC) and one high-pass (CR) filter : Example RC=0.5 s=jw CR-RC time function t/RC h(t) (1 t/RC) e − = − 1 2 3 4 5 -0.2 0.2 0.4 0.6 0.8 1 0.01 0.05 0.1 0.5 1 5 10 0.015 0.02 0.03 0.05 0.07 0.1 0.15 0.2 CR-RC s-transfer function h(s) = RCs/(1+RCs)2 Log-Log scale f |h(s)| 2 4 6 8 10 12 14 0.025 0.05 0.075 0.1 0.125 0.15 0.175 Step function response