Page 1 EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 1 A/D DSP SC Filter Frequency Response â€¢ Sampled data (and hence SC) filter responses are periodic â€¢ CT and SD responses agree only for f << f s Ø Derive exact SD frequency response â€“ RC â€“ SC example â€“ z-transform â€“ SC integrator styles EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 2 A/D DSP RC / SC Bandpass phi1 phi1 phi1 phi2 phi2 Cs = 6.28pF -1M fo = 10kHz Q = 10 Cint = 10pF Cs = 6.28pF Cd = 628fF Cint = 10pF Cint = 10pF phi1 phi1 phi1 phi2 phi2 Cd = 628fF phi1 phi1 phi1 phi2 phi2 Cs = 6.28pF Vo_lp_sc Vo_bp_sc -1M Cs fs = 100kHz CLK1 RC / SC Biquad Comparison Vin V1 ac = 1V V1 ac = 1V phi1 phi1 phi2 phi2 phi1 6.28pF Periodic AC Analysis PAC_lin sweep from 0 to 200k (500 steps) PAC_lin sweep from 0 to 200k (500 steps) Periodic AC Analysis PAC_log log sweep from 10 to 200k (500 steps) PAC_log log sweep from 10 to 200k (500 steps) PAC_log log sweep from 10 to 200k (500 steps) PAC_log log sweep from 10 to 200k (500 steps) -1M -1M -1M -1M -1M -1M R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm Cint = 10pF Cint = 10pF Cint = 10pF Cint = 10pF Cint = 10pF R * Q = 15.92MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm Vo_bp_rc Vo_lp_rc Cint = 10pF Cint = 10pF Cint = 10pF Cint = 10pF Netlist ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ZOH1 T = 10us ZOH2 T = 10us phi1 Page 2 EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 1 A/D DSP SC Filter Frequency Response â€¢ Sampled data (and hence SC) filter responses are periodic â€¢ CT and SD responses agree only for f << f s Ø Derive exact SD frequency response â€“ RC â€“ SC example â€“ z-transform â€“ SC integrator styles EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 2 A/D DSP RC / SC Bandpass phi1 phi1 phi1 phi2 phi2 Cs = 6.28pF -1M fo = 10kHz Q = 10 Cint = 10pF Cs = 6.28pF Cd = 628fF Cint = 10pF Cint = 10pF phi1 phi1 phi1 phi2 phi2 Cd = 628fF phi1 phi1 phi1 phi2 phi2 Cs = 6.28pF Vo_lp_sc Vo_bp_sc -1M Cs fs = 100kHz CLK1 RC / SC Biquad Comparison Vin V1 ac = 1V V1 ac = 1V phi1 phi1 phi2 phi2 phi1 6.28pF Periodic AC Analysis PAC_lin sweep from 0 to 200k (500 steps) PAC_lin sweep from 0 to 200k (500 steps) Periodic AC Analysis PAC_log log sweep from 10 to 200k (500 steps) PAC_log log sweep from 10 to 200k (500 steps) PAC_log log sweep from 10 to 200k (500 steps) PAC_log log sweep from 10 to 200k (500 steps) -1M -1M -1M -1M -1M -1M R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm Cint = 10pF Cint = 10pF Cint = 10pF Cint = 10pF Cint = 10pF R * Q = 15.92MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm Vo_bp_rc Vo_lp_rc Cint = 10pF Cint = 10pF Cint = 10pF Cint = 10pF Netlist ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ZOH1 T = 10us ZOH2 T = 10us phi1 EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 3 A/D DSP Frequency response â€¢ CT bandpass has zero at fàinfinity â€¢ Where did that zero go in the SC filter? Note: f s /2 would be a reasonable place! RC Filter SC Aliases EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 4 A/D DSP Non-Inverting SC Integrator phi1a phi1a phi2b Cs 1pF Ci 1pF Gm = 50uS Vo1 Vo1 Vi VS1 1V 150kHz fs = 1MHz CLK1 Non-Inverting SC Integrator Transient Analysis to 3us phi1 phi2 phi1c phi1c phi2c Vo2 Vo2 Vo2 Vo2 Page 3 EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 1 A/D DSP SC Filter Frequency Response â€¢ Sampled data (and hence SC) filter responses are periodic â€¢ CT and SD responses agree only for f << f s Ø Derive exact SD frequency response â€“ RC â€“ SC example â€“ z-transform â€“ SC integrator styles EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 2 A/D DSP RC / SC Bandpass phi1 phi1 phi1 phi2 phi2 Cs = 6.28pF -1M fo = 10kHz Q = 10 Cint = 10pF Cs = 6.28pF Cd = 628fF Cint = 10pF Cint = 10pF phi1 phi1 phi1 phi2 phi2 Cd = 628fF phi1 phi1 phi1 phi2 phi2 Cs = 6.28pF Vo_lp_sc Vo_bp_sc -1M Cs fs = 100kHz CLK1 RC / SC Biquad Comparison Vin V1 ac = 1V V1 ac = 1V phi1 phi1 phi2 phi2 phi1 6.28pF Periodic AC Analysis PAC_lin sweep from 0 to 200k (500 steps) PAC_lin sweep from 0 to 200k (500 steps) Periodic AC Analysis PAC_log log sweep from 10 to 200k (500 steps) PAC_log log sweep from 10 to 200k (500 steps) PAC_log log sweep from 10 to 200k (500 steps) PAC_log log sweep from 10 to 200k (500 steps) -1M -1M -1M -1M -1M -1M R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm Cint = 10pF Cint = 10pF Cint = 10pF Cint = 10pF Cint = 10pF R * Q = 15.92MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm Vo_bp_rc Vo_lp_rc Cint = 10pF Cint = 10pF Cint = 10pF Cint = 10pF Netlist ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ZOH1 T = 10us ZOH2 T = 10us phi1 EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 3 A/D DSP Frequency response â€¢ CT bandpass has zero at fàinfinity â€¢ Where did that zero go in the SC filter? Note: f s /2 would be a reasonable place! RC Filter SC Aliases EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 4 A/D DSP Non-Inverting SC Integrator phi1a phi1a phi2b Cs 1pF Ci 1pF Gm = 50uS Vo1 Vo1 Vi VS1 1V 150kHz fs = 1MHz CLK1 Non-Inverting SC Integrator Transient Analysis to 3us phi1 phi2 phi1c phi1c phi2c Vo2 Vo2 Vo2 Vo2 EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 5 A/D DSP Transient Analysis EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 6 A/D DSP Difference Equation F 1 Q s(kT) = C s V i(kT) Q i(kT) = Q i(kT-T/2) F 2 Q s(kT+T/2) = 0 Q i(kT+T/2) = Q i(kT) + Q s(kT) = Q i(kT-T/2) + Q s(kT) With V o = Q i /C i and V i = Q s /C s à V o2(kT+T/2) = V o2(kT-T/2) + (C s /C i ) V i(kT) Page 4 EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 1 A/D DSP SC Filter Frequency Response â€¢ Sampled data (and hence SC) filter responses are periodic â€¢ CT and SD responses agree only for f << f s Ø Derive exact SD frequency response â€“ RC â€“ SC example â€“ z-transform â€“ SC integrator styles EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 2 A/D DSP RC / SC Bandpass phi1 phi1 phi1 phi2 phi2 Cs = 6.28pF -1M fo = 10kHz Q = 10 Cint = 10pF Cs = 6.28pF Cd = 628fF Cint = 10pF Cint = 10pF phi1 phi1 phi1 phi2 phi2 Cd = 628fF phi1 phi1 phi1 phi2 phi2 Cs = 6.28pF Vo_lp_sc Vo_bp_sc -1M Cs fs = 100kHz CLK1 RC / SC Biquad Comparison Vin V1 ac = 1V V1 ac = 1V phi1 phi1 phi2 phi2 phi1 6.28pF Periodic AC Analysis PAC_lin sweep from 0 to 200k (500 steps) PAC_lin sweep from 0 to 200k (500 steps) Periodic AC Analysis PAC_log log sweep from 10 to 200k (500 steps) PAC_log log sweep from 10 to 200k (500 steps) PAC_log log sweep from 10 to 200k (500 steps) PAC_log log sweep from 10 to 200k (500 steps) -1M -1M -1M -1M -1M -1M R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm Cint = 10pF Cint = 10pF Cint = 10pF Cint = 10pF Cint = 10pF R * Q = 15.92MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm Vo_bp_rc Vo_lp_rc Cint = 10pF Cint = 10pF Cint = 10pF Cint = 10pF Netlist ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ZOH1 T = 10us ZOH2 T = 10us phi1 EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 3 A/D DSP Frequency response â€¢ CT bandpass has zero at fàinfinity â€¢ Where did that zero go in the SC filter? Note: f s /2 would be a reasonable place! RC Filter SC Aliases EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 4 A/D DSP Non-Inverting SC Integrator phi1a phi1a phi2b Cs 1pF Ci 1pF Gm = 50uS Vo1 Vo1 Vi VS1 1V 150kHz fs = 1MHz CLK1 Non-Inverting SC Integrator Transient Analysis to 3us phi1 phi2 phi1c phi1c phi2c Vo2 Vo2 Vo2 Vo2 EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 5 A/D DSP Transient Analysis EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 6 A/D DSP Difference Equation F 1 Q s(kT) = C s V i(kT) Q i(kT) = Q i(kT-T/2) F 2 Q s(kT+T/2) = 0 Q i(kT+T/2) = Q i(kT) + Q s(kT) = Q i(kT-T/2) + Q s(kT) With V o = Q i /C i and V i = Q s /C s à V o2(kT+T/2) = V o2(kT-T/2) + (C s /C i ) V i(kT) EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 7 A/D DSP Laplace Transform V o2(kT+T/2) = V o2(kT-T/2) + (C s /C i ) * V i(kT) Laplace Transform à ) ( ) ( ) ( 2 2 2 2 s V C C e s V e s V i i s T s o T s o + = - + EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 8 A/D DSP z-Transform 1 1 1 1 inverting_ - non int_ 1 2 2 inverting_ - non int_ 1 2 2 2 2 2 2 2 2 1 ) ( ) ( ) ( 1 ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( 2 1 2 1 2 1 2 1 - - - - - - - - + - = = - = = + = + = = + = z z C C z V z V z H z z C C z V z V z H z z V C C z z V z V z V C C z z V z z V e z s V C C e s V e s V i s i o i s i o i i s o o i i s o o sT i i s T s o T s o Note: the derivation assumes that the output is taken at the end of phase 2. If, as is often the case, the output is used only at the end of the next phase 1, the numerator is z -1 . Ref: Oppenheim and Schafer, Discrete-Time Signal Processing, Chapter 4 Page 5 EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 1 A/D DSP SC Filter Frequency Response â€¢ Sampled data (and hence SC) filter responses are periodic â€¢ CT and SD responses agree only for f << f s Ø Derive exact SD frequency response â€“ RC â€“ SC example â€“ z-transform â€“ SC integrator styles EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 2 A/D DSP RC / SC Bandpass phi1 phi1 phi1 phi2 phi2 Cs = 6.28pF -1M fo = 10kHz Q = 10 Cint = 10pF Cs = 6.28pF Cd = 628fF Cint = 10pF Cint = 10pF phi1 phi1 phi1 phi2 phi2 Cd = 628fF phi1 phi1 phi1 phi2 phi2 Cs = 6.28pF Vo_lp_sc Vo_bp_sc -1M Cs fs = 100kHz CLK1 RC / SC Biquad Comparison Vin V1 ac = 1V V1 ac = 1V phi1 phi1 phi2 phi2 phi1 6.28pF Periodic AC Analysis PAC_lin sweep from 0 to 200k (500 steps) PAC_lin sweep from 0 to 200k (500 steps) Periodic AC Analysis PAC_log log sweep from 10 to 200k (500 steps) PAC_log log sweep from 10 to 200k (500 steps) PAC_log log sweep from 10 to 200k (500 steps) PAC_log log sweep from 10 to 200k (500 steps) -1M -1M -1M -1M -1M -1M R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm R = -1.592MOhm Cint = 10pF Cint = 10pF Cint = 10pF Cint = 10pF Cint = 10pF R * Q = 15.92MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm R = 1.592MOhm Vo_bp_rc Vo_lp_rc Cint = 10pF Cint = 10pF Cint = 10pF Cint = 10pF Netlist ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ahdl_include "zoh.def" ZOH1 T = 10us ZOH2 T = 10us phi1 EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 3 A/D DSP Frequency response â€¢ CT bandpass has zero at fàinfinity â€¢ Where did that zero go in the SC filter? Note: f s /2 would be a reasonable place! RC Filter SC Aliases EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 4 A/D DSP Non-Inverting SC Integrator phi1a phi1a phi2b Cs 1pF Ci 1pF Gm = 50uS Vo1 Vo1 Vi VS1 1V 150kHz fs = 1MHz CLK1 Non-Inverting SC Integrator Transient Analysis to 3us phi1 phi2 phi1c phi1c phi2c Vo2 Vo2 Vo2 Vo2 EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 5 A/D DSP Transient Analysis EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 6 A/D DSP Difference Equation F 1 Q s(kT) = C s V i(kT) Q i(kT) = Q i(kT-T/2) F 2 Q s(kT+T/2) = 0 Q i(kT+T/2) = Q i(kT) + Q s(kT) = Q i(kT-T/2) + Q s(kT) With V o = Q i /C i and V i = Q s /C s à V o2(kT+T/2) = V o2(kT-T/2) + (C s /C i ) V i(kT) EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 7 A/D DSP Laplace Transform V o2(kT+T/2) = V o2(kT-T/2) + (C s /C i ) * V i(kT) Laplace Transform à ) ( ) ( ) ( 2 2 2 2 s V C C e s V e s V i i s T s o T s o + = - + EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 8 A/D DSP z-Transform 1 1 1 1 inverting_ - non int_ 1 2 2 inverting_ - non int_ 1 2 2 2 2 2 2 2 2 1 ) ( ) ( ) ( 1 ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( 2 1 2 1 2 1 2 1 - - - - - - - - + - = = - = = + = + = = + = z z C C z V z V z H z z C C z V z V z H z z V C C z z V z V z V C C z z V z z V e z s V C C e s V e s V i s i o i s i o i i s o o i i s o o sT i i s T s o T s o Note: the derivation assumes that the output is taken at the end of phase 2. If, as is often the case, the output is used only at the end of the next phase 1, the numerator is z -1 . Ref: Oppenheim and Schafer, Discrete-Time Signal Processing, Chapter 4 EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 9 A/D DSP Frequency Response jfRC f H f f fT jfT C C fT j C C fT j fT fT j fT C C z z C C z z C C z H f H s fT j fT e z e z e z jfT jfT jfT p p p p p p p p p p p p 2 1 ) ( or 1 for 2 1 sin 2 1 sin cos sin cos 1 1 1 ) ( ) ( RC int s int s int s sin cos int s 1 int s int_ni SC_ni 2 1 2 1 2 1 2 2 1 2 = << << Ëœ = + - + = - = - = = + = = - = - - = EECS 247 Lecture 10: SC Filter Frequency Response © 2002 B. Boser 10 A/D DSP Inverting SC Integrator fT j C C f H z z C C z H p sin 2 1 ) ( 1 ) ( int s SC_i 1 int s int_i 2 1 - = - - = - - Note: H int_i (z) assumes that the output is used at the end of phase 2. If, as is often the case, the output is used already at the end of phase 1, the numerator is 1. If, as is normally the case, circuit topologies are used where inverting and non-inverting SC integrators alternate, this is not an issue. Cs 1pF Ci 1pF Gm = 50uS Vi VS1 1V 150kHz fs = 1MHz CLK1 Inverting SC Integrator Transient Analysis to 3us phi2 phi2 Vo1 Vo1 Vo1 Vo1 Vo1 Vo1 Vo1 Vo1 phi1c phi1c phi1c phi1c phi1c phi1c phi1c phi1c phi2c phi2c phi2c phi2c Vo2 Vo2 Vo2 Vo2 Vo2 Vo2 Vo2 Vo2 Vo2 Vo2 Vo2 Vo2 Vo2 Vo2 Vo2 Vo2Read More

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