EE 508
Lecture 20
Sensitivity Functions• Comparison of Filter Structures
• Performance Prediction
C1
C2
R2R1
VIN
VOUTK
1 2 1 2
2
1 1 2 1 2 2 1 2 1 2
1
R R C CT s =K
1 1 1-K 1s +s + + +
R C R C R C R R C C
C1 C2
R2R1
VIN
VOUT-K
R3
R4
1 2 1 2
1 3 1 4 2 3 2 12 1 2
1 1 3 4 2 2 2 1 1 2 1 2
1
R R C CT s = -K
1+ R R 1+K + R R 1+ R R + R RR C1 1 1s +s 1+ + + 1+ +
R C R R C R C C R R C C
What causes the dramatic differences in performance between these two structures?
How can the performance of different structures be compared in general?
Equal R, Equal C, Q=10 Pole
Locus vs GBN
Revie
w f
rom
last
tim
e
Effects of GB on poles of KRC and -KRC Lowpass Filters
Im
Re
Over-order
pole
Actual “desired” poles
Desired poles
Review from last time
Dependent on circuit structure (for some circuits, also not dependent
on components)
Dependent only on components (not circuit structure)
1i
f ix
iN
dxdFS
F xN
k
Xi
R
C
VOUT
VIN
Consider:
1
T s = 1+RCs
0
0
ωT s =
s + ω
0
1ω =
R C
Revie
w f
rom
last
tim
e
i
f
xS
Theorem: If f(x1, ..xm) can be expressed as
where {α1, α2,… αm} are real numbers, then is not dependent
upon any of the variables in the set {x1, ..xm}
1 2
1 2 ... m
mf x x x
Proof:
i
i i
Xf
x xS Si
i
i
X i ix
i i
X xS
x X
ii
i
1i
i
X ix i
i
xS X
X
ii
ii
i
i
X
xSi
i
i
f
xS i
It is often the case that functions of interest are
of the form expressed in the hypothesis of the
theorem, and in these cases the previous claim is
correct
Review from last time
i
f
xS
Theorem: If f(x1, ..xm) can be expressed as
where {α1, α2,… αm} are real numbers, then is not dependent
upon any of the variables in the set {x1, ..xm}
1 2
1 2 ... m
mf x x x
Review from last time
Theorem: If f(x1, ..xm) can be expressed as
where {α1, α2,… αm} are real numbers, then the sensitivity terms in
are dependent only upon the circuit architecture and not dependent
upon the components and and the right terms are dependent only upon
the components and not dependent upon the architecture
1 2
1 2 ... m
mf x x x
1i
f ix
iN
dxdfS
f xN
k
Xi
This observation is useful for comparing the performance of two or more circuits
where the function f shares this property
Review from last time
Metrics for Comparing Circuits
Schoeffler Sensitivity
1i
f
xSm
i
Summed Sensitivity
Not very useful because sum can be small even when individual
sensitivities are large
1i
f
xSm
S
i
Strictly heuristic but does differentiate circuits with low sensitivities from those
with high sensitivities
Review from last time
Metrics for Comparing Circuits
1i
f
xSm
i
Often will consider several distinct sensitivity functions to consider
effects of different components
1i
f
xSm
i
i
f
RSR
All resistors
i
f
CSC
All capacitors
i
f
OA
All op amps
s
Review from last time
Homogeniety (defn)
A function f is homogeneous of order
m in the n variables {x1, x2, …xn} if
f(λx1, λx2, … λxn ) = λmf(x1,x2, … xn)
Note: f may be comprised of more than n variables
Review from last time
Theorem: If a function f is homogeneous of order m
in the n variables {x1, x2, …xn} then
i
nf
xi=1
= mS
, ,... , ,...1 2 n 1 2 nf x x x f x x xm
The concept of homogeneity and this theorem were
somewhat late to appear
Are there really any useful applications of this rather odd
observation?
Review from last time
Theorem: If all op amps in a filter are
ideal, then ωo, Q, BW, all band edges,
and all poles and zeros are homogeneous
of order 0 in the impedances.
Theorem: If all op amps in a filter are
ideal and if T(s) is a dimensionless transfer
function, T(s), T(jω), | T(jω) |, , are
homogeneous of order 0 in the impedances T jω
Review from last time
Theorem 1: If all op amps in a filter are
ideal and if T(s) is an impedance transfer
function, T(s) and T(jω) are homogeneous
of order 1 in the impedances
Theorem 2: If all op amps in a filter are
ideal and if T(s) is a conductance transfer
function, T(s) and T(jω) are homogeneous
of order -1 in the impedances
Review from last time
Corollary 1: If all op amps in an RC active
filter are ideal and there are k1 resistors and k2
capacitors and if a function f is homogeneous of
order 0 in the impedances, then
1 2
i i
k kf f
R Ci=1 i=1
S = S
Corollary 2: If all op amps in an RC
active filter are ideal and there are k1
resistors and k2 capacitors then1
i
kQ
Ri=1
S = 0
2
i
kQ
Ci=1
S = 0
Review from last time
Example
VIN
VOUTR1 R2
C1 C2
V1
1
Q
RSDetermine the passive Q sensitivities
2
Q
RS
1
Q
CS
2
Q
CS
OUT 2 211V sC +G V = G
1 1 2 IN 11 OUT 2sC +G +G =V VV G G
2
1 2 1 2 1 1 1 2 2 2
1T s =
s R R C C +s R C +R C +R C +1
0
1 2 1 2
1ω =
R R C C1 2 1 2
1 1 1 2 2 2
R R C CQ =
R C +R C +R C
1
-1/2 1/2
1 1 1 2 2 1 1 2 1 2 2 1 2 1 2 1 2 1 2Q 1
2R
1 1 1 2 2 2
1R C +R C +R C R R C C R C C - C +C R R C C
R2S •QR C +R C +R C
By the definition of sensitivity, it follows that
Example
VIN
VOUTR1 R2
C1 C2
V1
1
Q
RSDetermine the passive Q sensitivities
2
Q
RS
1
Q
CS
2
Q
CS
1
-1/2 1/2
1 1 1 2 2 1 1 2 1 2 2 1 2 1 2 1 2 1 2Q 1
2R
1 1 1 2 2 2
1R C +R C +R C R R C C R C C - C +C R R C C
R2S •QR C +R C +R C
1
21
Q 1 1 2
R
1 1 1 2 2 2
R C +CS
R C +R C +R C
Following some tedious manipulations, this simplifies to
Example
VIN
VOUTR1 R2
C1 C2
V1
Determine the passive Q sensitivities
1
21
Q 1 1 2
R
1 1 1 2 2 2
R C +CS
R C +R C +R C
Following the same type of calculations, can obtain
1
22
Q 2 2
R
1 1 1 2 2 2
R CS
R C +R C +R C
1
21
Q 1 1
C
1 1 1 2 2 2
R CS
R C +R C +R C
1
22
Q 2 1 2
C
1 1 1 2 2 2
C R RS
R C +R C +R C
2
i
kQ
Ci=1
S = 01
i
kQ
Ri=1
S = 0Verify
Could have saved considerable effort in calculations by using these theorems after
1
Q
RS
1
Q
CSand were calculated
Corollary 3: If all op amps in an RC
active filter are ideal and there are k1
resistors and k2 capacitors and if pk is any
pole and zh is any zero, then
11
k
i
kp
Ri=1
S = 12
k
i
kp
Ci=1
S =
11
h
i
kz
Ri=1
S = 12
h
i
kz
Ci=1
S = and
Corollary 3: If all op amps in an RC
active filter are ideal and there are k1
resistors and k2 capacitors and if pk is any
pole and zh is any zero, then
11
k
i
kp
Ri=1
S = 12
k
i
kp
Ci=1
S =
11
h
i
kz
Ri=1
S = 12
h
i
kz
Ci=1
S =
and
Proof:
It was shown that scaling the frequency dependent elements by a factor η divides
the pole (or zero) by η
Thus roots (poles and zeros) are homogeneous of order -1 in the frequency
scaling elements
Proof:
Thus roots (poles and zeros) are homogeneous of order -1 in the frequency
scaling elements
31 2
i i i
kk kp p p
R 1/C Li=1 i=1 i=1
S + S + S = 0
(For more generality, assume k3 inductors)
(1)132
i i
kkp p
C Li=1 i=1
S + S =
Since impedance scaling does not affects the poles, they are homogenous of
order 0 in the impedances
(2)
Since there are no inductors in an active RC network, is follows from (1) that
12
i
kp
Ci=1
S And then from (2) and the theorem about sensitivity to reciprocals that
11
i
kp
Ri=1
S
Corollary 4: If all op amps in an RC
active filter are ideal and there are k1
resistors and k2 capacitors and if ZIN is any
input impedance of the network, then
11 2
IN IN
i i
k kZ Z
R Ci=1 i=1
S - S =
Claim: If op amps in the filters
considered previously are not ideal but are
modeled by a gain A(s)=1/(s), then all
previous summed sensitivities developed for
ideal op amps hold provided they are
evaluated at the nominal value of =0
Sensitivity Analysis
If a closed-form expression for a function f
is obtained, a straightforward but tedious
analysis can be used to obtain the
sensitivity of the function to any
components
Closed-form expressions for T(s), T(jω), |T(jω)|, , ai, bi, can be
readily obtained
mmi
i ii=0 i=1
n ni
i ii=0 i=1
a s (s-z )T s = =K
b s (s-p )
f
x
f xS = •
x f
Consider:
T jω
Sensitivity AnalysisIf a closed-form expression for a function f is
obtained, a straightforward but tedious analysis
can be used to obtain the sensitivity of the
function to any components
Closed-form expressions for pi, zi, pole or zero Q, pole or zero
ω0, peak gain, ω3dB, BW, … (generally the most critical and
useful circuit characteristics) are difficult or impossible to
obtain !
mmi
i ii=0 i=1
n ni
i ii=0 i=1
a s (s-z )T s = =K
b s (s-p )
f
x
f xS = •
x f
Consider:
Bilinear Property of Electrical Networks
Theorem: Let x be any component or Op Amp time constant
(1st order Op Amp model) of any linear active network
employing a finite number of amplifiers and lumped passive
components. Any transfer function of the network can be
expressed in the form
0 1
0 1
N s +xN sT s =
D s +xD s
where N0, N1, D0, and D1 are polynomials in s that are not dependent upon x
A function that can be expressed as given above is said to be a bilinear
function in the variable x and this is termed a bilateral property of electrical
networks.
The bilinear relationship is useful for
1. Checking for possible errors in an analysis
2. Pole sensitivity analysis
C1
C2
R2R1
VIN
VOUTK
Example of Bilinear Property : +KRC Lowpass Filter
0
1 2 1 2
2 20
0
1 1 2 1 2 2 1 2 1 2 1 1 2 1 2 2 1 2 1 2
K
R R C CT s =
1-K1 1 1 1 1 1 1s +s + + + K s s +s + + +
R C R C R C R R C C R C R C R C R R C C
Consider R1
0
2 1 2
2 20
1 1 1 0 1 1 1
1 2 1 2 2 2 1 2 1 2 1 2 2 2 1 2
K
R C CT s =
1-K1 1 1 1 1 1 1R s +s +R +R + K s R s +s +R +R +
C R C R C R C C C R C R C R C C
001
2 1 2
2 20
0 1 0
1 2 1 2 1 2 1 2 2 1 2 2 2 1 2 2
KR
R C CT s =
1-K1 1 1 1 1 1 1s K s s + +R s +s + +K s s +s +
C R C C C R C C R C R C R C R C
VINR1
R2
C
VOUT
1
A s =s
V1
Example of Bilinear Property
1
1 1 2 IN 1 OUT 2
OUT 1
V G +G +sC = V G +V sC+G
V = -Vs
2
1 1 2 1 2 1 2
-RT s =
R +R R Cs+ s sCR R +R +R
Consider R1
02 1
2 1 2 2
-R RT s =
sR +R 1+R Cs+ s sCR +1
Consider τ
02
1 2 2 1 2
-RT s =
R 1+R Cs sR sR sCR +1
C1
C2
R2R1
VIN
VOUTK
Example of Bilinear Property : +KRC Lowpass Filter
0
2 2
2 20
02 2 2 2
K
R CT s =3-K 1 3 1
s +s + K s s +s +RC R C RC R C
Can not eliminate the R2 term
Equal R Equal C
3
0
2 2 2 2 2
0 0 0 0
KT s =
R C s K sC +R sC 3-K K C s + 1+K s
• Bilinear property only applies to individual components
• Bilinear property was established only for T(s)
Root SensitivitiesConsider expressing T(s) as a bilinear fraction in x
0 1
0 1
N s +xN s N sT s =
D s +xD s D s
Theorem: If zi is any simple zero and/or pi is any
simple pole of T(s), then
iz 1 i
x
ii
i
N zx
N zz
z
Sd
d
ip 1 i
x
ii
i
D px
D pp
p
Sd
d
and
Note: Do not need to find expressions for the poles or the zeros to find the pole
and zero sensitivities !
Note: Do need the poles or zeros but they will generally be known by design
Note: Will make minor modifications for extreme values for x (i.e. τ for op amps)
Root SensitivitiesTheorem: If pi is any simple pole of T(s), then
ip 1 i
x
ii
i
D px
D pp
p
Sd
d
Proof (similar argument for the zeros)
0 1D s =D s +xD sBy definition of a pole,
iD p =0
i 0 i 1 iD p =D p +xD p 0\
Root Sensitivities
Re-grouping, obtain
But term in brackets is derivative of D(pi) wrt pi, thus
i 0 i 1 iD p =D p +xD p\
Differentiating this expression implicitly WRT x, we obtain
0 i 1 ii i
1 i
i i
D p D pp px D p 0
p x p x
0 i 1 ii
1 i
i i
D p D ppx D p
x p p
1 ii
i
i
D pp
D px
p
Root Sensitivities
1 ii
i
i
D pp
D px
p
Finally, from the definition of sensitivity,
ip 1 iix
ii i
i
D ppx x
D pp x p
p
S
Observation: Although the sensitivity expression is readily
obtainable, direction information about the pole movement is
obscured because the derivative is multiplied by the quantity pi
which is often complex. Usually will use either
or
which preserve direction information when working with pole or
zero sensitivity analysis.
Root Sensitivities
ip ip
xx
s
ip 1 iix
ii i
i
D ppx x
D pp x p
p
S
ip 1 iix
ii i
i
D ppx x
D pp x p
p
S
Root Sensitivities
ip
ipx
x s
i
iN
p 1 i
i
i p
D p
D p
p
x
s
ip
i i x
xp p
xS
Summary: Pole (or zero) locations due to component
variations can be approximated with simple analytical
calculations without obtaining parametric expressions for
the poles (or zeros).
Ideal
Componentsi ip p p
i
0 1D s D s x D s
where
and
Alternately,
C1
C2
R2R1
VIN
VOUTK
Example: Determine for the +KRC Lowpass Filter for equal R, equal C
0
0
KK s
1+K s
0
1 2 1 2
2 20
0
1 1 2 1 2 2 1 2 1 2 1 1 2 1 2 2 1 2 1 2
K
R R C CT s =
1-K1 1 1 1 1 1 1s +s + + + K s s +s + + +
R C R C R C R R C C R C R C R C R R C C
i
1
p
RS
0
2 1 2
2 20
0 1 0
1 2 1 2 1 2 1 2 2 1 2 2 2 1 2 2
K
R C CT s =
1-K1 1 1 1 1 1 1s K s s + +R s +s + K s s +s +
C R C C C R C C R C R C R C R C
0
2 1 2
2 0
1
1 2 1 2 2 1 2 2 1 2 2
K
R C CT s =
1-K1 1 1 1s + +R s +s +
C R C C R C C R C R C
ip 1 iix
ii i
i
D ppx x
D pp x p
p
S
0 1
0 1
N s +xN sT s =
D s +xD s
write in bilinear form
evaluate at τ=0
C1
C2
R2R1
VIN
VOUTK
Example: Determine for the +KRC Lowpass Filter for equal R, equal C
0
0
KK s
1+K s
i
1
p
RS
0
2 1 2
2 0
1
1 2 1 2 2 1 2 2 1 2 2
K
R C CT s =
1-K1 1 1 1s + +R s +s +
C R C C R C C R C R C
ip 1 iix
ii i
i
D ppx x
D pp x p
p
S
0 1
0 1
N s +xN sT s =
D s +xD s
02
1
2 1 2 2
1-K1D s =s +s +
R C R C
02 2 20
1 1 0
1 2 1 2 2 1 2 2 1 2 2
1-K ω1 1 1 1D s = s + +R s +s + R s +s +ω
C R C C R C C R C R C Q
21
2 0
p 2 1 2 2iR
0i ii
1-K1p +p +
R C R Cpx 1
ωp x pp
Q
S
C1
C2
R2R1
VIN
VOUTK
Example: Determine for the +KRC Lowpass Filter for equal R, equal C
0
0
KK s
1+K s
1
2
0 0
2 2 2 20 0 1 10 0 0 0
2 2
K ωT s =
ω ω R Cs +s +ω K s s +s K Q +ω
Q Q R C
i
1
p
RS
ip 1 iix
ii i
i
D ppx x
D pp x p
p
S
0 1
0 1
N s +xN sT s =
D s +xD s
21
2 0
p 2 1 2 2iR
0i ii
1-K1p +p +
R C R Cpx 1
ωp x pp
Q
S
02
1 1 2 1 2 2 1 2 1 2
1-K1 1 1p +p + + + =0
R C R C R C R R C C
0
1 2 1 2
2 0
1 1 2 1 2 2 1 2 1 2
K
R R C CT s =
1-K1 1 1s +s + + +
R C R C R C R R C C
02
2 1 2 2 1 2 1 2 1 1
1-K1 1 1p +p + = - -p
R C R C R R C C R C
21
p i 1 2 1 2 1 1R
0i ii
1 1+ p
p R R C C R Cx 1
ωp x pp
Q
S
21
2
0p i 1 1R
0i 0i
1ω + p
p R Cx 1
ωp x ωp
Q
S
C1
C2
R2R1
VIN
VOUTK
Example: Determine for the +KRC Lowpass Filter for equal R, equal C
0
0
KK s
1+K s
i
1
p
RS
2
i
2
0p i 1 1x
0i 0i
1ω + p
p R Cx 1
ωp x ωp
Q
S
For equal R, equal C1
0ω = RC
21
2p 0 0iR
0i 0i
ω + pωpx 1
ωp x ωp
Q
S
21
p 0iR
0i
ω + ppx
ωp xp
Q
S
1
20 00
p
R20
ω ωω - 1-4Q
2Q 2Qω
1-4QQ
S
1
2
p
R 2
1 1Q- 1-4Q
2 2
1-4QS
C1
C2
R2R1
VIN
VOUTK
Example: Determine for the +KRC Lowpass Filter for equal R, equal C
0
0
KK s
1+K s
i
1
p
RS
For equal R, equal C
1
2
p
R 2
1 1Q- 1-4Q
2 2
1-4QS
Note this contains magnitude and direction information
ip
i i x
xp p
xS
For high Q 1 1 1
2 2 2 2 21
2
p
R 2
1Q -4Q
Q 1 j2
-4QS
jQ jj
j Q j
ip ix
i
px
p xS
0
1
0 5 0 5 1i
Rp ω . .
R
j
C1
C2
R2R1
VIN
VOUTK
Example: Determine for the +KRC Lowpass Filter for equal R, equal C
0
0
KK s
1+K s
i
1
p
RS
For equal R, equal C
For high Q
ip ix
i
px
p xS
0
1
0 5 0 5 1i
Rp ω . .
R
j
Was this a lot of work for such a simple result?
What is the value of this result?
Yes ! But it is parametric and still only took maybe 20 minutes
But it needs to be done only once for this structure
Can do for each of the elements
Understand how components affect performance of this circuit
Compare performance of different circuits for architecture selection
Could we have assumed equal R equal C before calculation? No ! Analysis would not apply (not bilinear)
Results would obscure effects of variations in individual components
Transfer Function Sensitivities
T s T jω
x xs=jω
S S
T jωT jω θ
x x xjθS S S θ T jω where
T jω T jω
x x=ReS S
IT jωθ
x x
1= m
θS S
Transfer Function Sensitivities
0 0
i i
i x i xT s
x
a s b s-
N s D s
S SS
i im na b
i i
mi
i
i=0
ni
i
i=0
a sN s
T s = D s
b s
If T(s) is expressed as
then
0 1
0 1
N s N sT s =
D s D s
x
x
If T(s) is expressed as
T s 0 1 0 1
x
0 1 0 1
x D s N s -N s D s
N s +xN s D s +xD sS
Band-edge Sensitivities
The band edge of a filter is often of interest. A closed-form expression for
the band-edge of a filter may not be attainable and often the band-edges
are distinct from the ω0 of the poles. But the sensitivity of the band-edges
to a parameter x is often of interest.
T jω
ωC ω
Want
Cω C
x
C
ω x
x ωS
Band-edge Sensitivities
T jω
ωC ω
Theorem: The sensitivity of the band-edge of a filter is given by the expression
CC
C
T jω
xω=ωω
x T jω
ωω=ω
SS
S
Band-edge Sensitivities
T jω
ωC ω
T jω
ω
Observe T jω T jω
ω ω
T jω
T jω T jω x xωω x ω
x
Proof:
Band-edge Sensitivities T jω
ωC ω
T jω
ω
T jωT jω T jω x x
ωω x ω
x
T jω
ω xT jωx
ω
T jω x
x T jωω ω
T jωx xω
ω T jω
T jω x
x T jωω x
T jωx ω ω
ω T jω
T jωω xx T jω
ω
SS
S
CC
C
T jω
xω=ωω
x T jω
ωω=ω
SS
S
Sensitivity Comparisons
Consider 5 second-order lowpass filters
(all can realize same T(s) within a gain factor)
R L
CVIN
VOUT
C1
C2
R1 R2K
VOUTVIN
VOUT
VIN
R3 R1 R2
C1
C2VIN
VOUT
R0
R1RQ
R4
R3R2
C1 C2
Passive RLC +KRC
Bridged-T Feedback
Two-Integrator Loop
(a) (b)
(c) (d)
Sensitivity Comparisons
Consider 5 second-order lowpass filters
(all can realize same T(s) within a gain factor)
-KRC Lowpass
C1 C2
R2R1
VIN
VOUT
R3
R4 R5
5
4
R-K = -
R
(e)
2
0
2 200
KωT s =
ωs +s +ω
Q
For all 5 structures, will have same transfer function within a gain factor
R L
CVIN
VOUT
OUT
2IN
1V LCT s = =
RV 1s +s +LCL
a) – Passive RLC
0
1ω =
LC
1 LQ =
R C
b) + KRC (a Sallen and Key filter)
C1
C2
R2R1
VIN
VOUT
R4
R5
5
4
RK = 1+
R
1 2 1 2
2 1 1 2 2 1 2 1 1
2 2 1 1 2 1 2 2 1 2 1 21 2 1 2
K
R R C CT s =
R C R C R C R C1 1s +s + + K +
R C R C R C R C R R C CR R C C
0
1 2 1 2
1ω =
R R C C1 1 2 2 1 2 1 1
2 2 1 1 2 1 2 2
1Q =
R C R C R C R C+ + K
R C R C R C R C
Case b1 : Equal R, Equal C
1 2R = R = R1 2C = C = C
0
1ω =
RC1
K = 3 -Q
2
0
2 200
KωT s =
ωs +s +ω
Q
Case b2 : Equal R, K=1
1
2
C1Q =
2 C1 2R = R = R
VOUT
VIN
R3 R1 R2
C1
C2
1
1 3 1 2
2 1 22 1 2
1 3 1 3 1 2 1 21 2 1 2
1
R R C CT s =
R RC R R 1s +s + + +
C R R R R R C CR R C C
If R1=R2=R3=R and C2=9Q2C1
2 2 2
1
2
2 2 2 2
1 1
1
9Q R CT s =
1 1s +s +
3Q RC 9Q R C
c) Bridged T Feedback
0
1 2 1 2
1ω =
R R C C
1
1 22 1 2
1 3 1 3
Q = R RC R R
+ +C R R R
4
3 0 2 1 2
2 4
Q 2 3 0 2 1 2
R 1
R R R C CT s = -
R1 1s +s +
R C R R R C C
0 1 2R = R = R = R 1 2C = C = C3 4R = R
2 2
2
2 2
Q
1
R CT s = - 1 1
s +s +R C R C
0
1ω =
RCQR =QR
VINVOUT
R0
R1RQ
R4
R3R2
C1 C2
d) 2 integrator loop
For:
40
3 0 2 1 2
R 1 =
R R R C C
Q 2
10 2
R C Q=
CR R
d) - KRC (a Sallen and Key filter)
C1 C2
R2R1
VIN
VOUT
R3
R4 R5
5
4
R-K = -
R
3 4 3 11 11 1 1
1 2 1 2
2 1 1 2 21 2
3 1 1 1 2 2 4 2 1 2 1 2
K
R R C CT s = -
1+ R R K + R R R R R RR Cs +s 1+ 1+ +
R R C C R C R C R R C C
3 4 3 11 11 1 2 2
0
1 2 1 2
1+ R R K + R R R R R Rω =
R R C C
3 4 3 11 1
1 1 1
1 1 2 2
1 2 1 2
1 2
3 1 1 1 2 2 4 2
1+ R R K + R R R R R R
R R C CQ =
R C1+ 1+
R R C C R C R C
Often R1=R2=R3=R4=R, C1=C2=C05+K
Q = 5
Stay Safe and Stay Healthy !
End of Lecture 20