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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


i

f

xS


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



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


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


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


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


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?


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


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


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


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



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



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



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


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


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


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


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


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


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


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

ee 508 lecture 4 - class.ece.iastate.educlass.ece.iastate.edu/ee508/lectures/ee 508 lect 20 fall...

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