Courses

# SC Filter Frequency Response Notes | EduRev

## : SC Filter Frequency Response Notes | EduRev

``` Page 1

EECS 247 Lecture 10:  SC Filter Frequency Response © 2002 B. Boser   1
A/D
DSP
SC Filter Frequency Response
• Sampled data (and hence SC) filter responses
are periodic
• CT and SD responses agree only for f << f
s
Ø Derive exact SD frequency response
– RC – SC example
– z-transform
– 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
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
• Sampled data (and hence SC) filter responses
are periodic
• CT and SD responses agree only for f << f
s
Ø Derive exact SD frequency response
– RC – SC example
– z-transform
– 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
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
• CT bandpass has zero
at fàinfinity
• 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
• Sampled data (and hence SC) filter responses
are periodic
• CT and SD responses agree only for f << f
s
Ø Derive exact SD frequency response
– RC – SC example
– z-transform
– 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
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
• CT bandpass has zero
at fàinfinity
• 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
• Sampled data (and hence SC) filter responses
are periodic
• CT and SD responses agree only for f << f
s
Ø Derive exact SD frequency response
– RC – SC example
– z-transform
– 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
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
• CT bandpass has zero
at fàinfinity
• 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
• Sampled data (and hence SC) filter responses
are periodic
• CT and SD responses agree only for f << f
s
Ø Derive exact SD frequency response
– RC – SC example
– z-transform
– 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
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
• CT bandpass has zero
at fàinfinity
• 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
=
<< << ˜
=
+ - +
=
-
=
-
=
=
+ = =
-
=
-
-
=
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 Vo2
```
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!