Electronic Circuitseleccirc/docs/restricted/lecture08.pdf · and degree 2 (quadratic) polynomials....

Prof. Dr. Qiuting HuangIntegrated Systems Laboratory

Electronic Circuits

8. Active Filters - 2

𝑃𝑃 𝑥𝑥 is a polynomial of the variable 𝑥𝑥:

ETH 2Integrated Systems Laboratory

Blast From The Past: Algebra of Polynomials*

𝑃𝑃 𝑥𝑥 = 𝑎𝑎0 + 𝑎𝑎1𝑥𝑥 + ⋯+ 𝑎𝑎𝑛𝑛𝑥𝑥𝑛𝑛⋯

Coefficients

The highest power of 𝑥𝑥in 𝑃𝑃 𝑥𝑥 with non-zero coefficient is called the degree of 𝑃𝑃 𝑥𝑥 .

𝑐𝑐1 is a root of 𝑃𝑃 𝑥𝑥𝑃𝑃 𝑥𝑥 = 𝑥𝑥 − 𝑐𝑐1 𝑄𝑄𝑛𝑛−1 𝑥𝑥 .

⋮𝑐𝑐𝑘𝑘 is a root of 𝑃𝑃 𝑥𝑥

𝑃𝑃 𝑥𝑥 = 𝑥𝑥 − 𝑐𝑐1 ⋯ 𝑥𝑥 − 𝑐𝑐𝑘𝑘 𝑄𝑄𝑛𝑛−𝑘𝑘 𝑥𝑥 .

has 𝑛𝑛 roots

𝑄𝑄𝑛𝑛−𝑘𝑘 𝑥𝑥 is an irreducible polynomial of degree 1 or 2.

has no more roots

A real polynomial 𝑃𝑃 𝑥𝑥 can be factorized into degree 1 (linear) and degree 2 (quadratic) polynomials.

The transfer function of a filter is a rational function.

The degree of the denominator is usually referred to as the order of the filter (𝑛𝑛).

𝑛𝑛 is determined by the number of reactive elements in the filter.

Example:


Design of Filters

( )1

1 1 01

1 1 0

( )( )

−−

−−

+ + + += =

+ + + +

k kk k

n nn n

b s b s b s b N sT sa s a s a s a D s

Passive 5th-Order Ladder FIlter

( ) 05 4

5 4 1 0

=+ + + +

bT sa s a s a s a

Filter performance improves with the order of the filter.

However, a high-order filter is costly.

Optimization process of a filter:

Find 𝑇𝑇(𝑠𝑠) that satisfies the requirements with lowest possible order.

Designing a high-order filter to provide a desired response is not a trivial problem.


Design of Filters

PB = pass-bandTB = transition-bandSB = stop-band

Sharper response with higher 𝑛𝑛


Design of Filters

Calibrating the circuit manually is not sufficient to achieve the task.

The specifications of the filter are dominated by its worst performer.

A systematic mathematical approach is required.

The design comprises an optimization process targeting minimizing the maximum deviation from the ideal filter response (minmax optimization).

Passive 5th-Order Ladder FIlterM

agni

tude

(dB) Chebyshev Low-pass resonse

Max. Ripple = 0.1 dB

There are several well-known approaches for optimizing the approximation problem stemming from mathematics; i.e. (Butterworth, Chebyshev, Bessel, Elleptic,…).

The design parameters and component values for the filter circuits realizing these responses are well-documented.


Approximation Problem

We can represent the transfer function obtained from the mathematical approximation as:


Realization Problem: Filter Synthesis

𝑇𝑇 𝑠𝑠 =𝑏𝑏𝑘𝑘 𝑠𝑠 𝑘𝑘 + 𝑏𝑏𝑘𝑘−1 𝑠𝑠 𝑘𝑘−1 + ⋯+ 𝑏𝑏1 𝑠𝑠 + 𝑏𝑏0𝑎𝑎𝑛𝑛 𝑠𝑠 𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝑠𝑠 𝑛𝑛−1 + ⋯+ 𝑎𝑎1 𝑠𝑠 + 𝑎𝑎0

=)𝑁𝑁(𝑠𝑠)𝐷𝐷(𝑠𝑠

=𝑏𝑏1,2𝑠𝑠2 + 𝑏𝑏1,1𝑠𝑠 + 𝑏𝑏1,0 𝑏𝑏2,2𝑠𝑠2 + 𝑏𝑏2,1𝑠𝑠 + 𝑏𝑏2,0 … 𝑏𝑏𝑘𝑘

2,2𝑠𝑠2 + 𝑏𝑏𝑘𝑘

2,1𝑠𝑠 + 𝑏𝑏𝑘𝑘

2,0

𝑎𝑎1,2𝑠𝑠2 + 𝑎𝑎1,1𝑠𝑠 + 𝑎𝑎1,0 𝑎𝑎2,2𝑠𝑠2 + 𝑎𝑎2,1𝑠𝑠 + 𝑎𝑎2,0 … 𝑎𝑎𝑛𝑛2,2𝑠𝑠

2 + 𝑎𝑎𝑛𝑛2,1𝑠𝑠 + 𝑎𝑎𝑛𝑛

2,0

We can represent the transfer function obtained from the mathematical approximation as:



Filter 1 Filter 2≡ ⋯ Filter 𝑛𝑛2

𝑇𝑇 𝑠𝑠 =𝑏𝑏𝑘𝑘 𝑠𝑠 𝑘𝑘 + 𝑏𝑏𝑘𝑘−1 𝑠𝑠 𝑘𝑘−1 + ⋯+ 𝑏𝑏1 𝑠𝑠 + 𝑏𝑏0𝑎𝑎𝑛𝑛 𝑠𝑠 𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝑠𝑠 𝑛𝑛−1 + ⋯+ 𝑎𝑎1 𝑠𝑠 + 𝑎𝑎0

=)𝑁𝑁(𝑠𝑠)𝐷𝐷(𝑠𝑠

𝐾𝐾1 𝑠𝑠2 + 𝜔𝜔z1𝑄𝑄z1

𝑠𝑠 + 𝜔𝜔z12 𝐾𝐾2 𝑠𝑠2 + 𝜔𝜔z2

𝑄𝑄z2𝑠𝑠 + 𝜔𝜔z2

2 …𝐾𝐾𝑛𝑛2

𝑠𝑠2 +𝜔𝜔z𝑛𝑛2𝑄𝑄z𝑛𝑛2

𝑠𝑠 + 𝜔𝜔z𝑛𝑛2

2

𝑠𝑠2 + 𝜔𝜔o1𝑄𝑄o1

𝑠𝑠 + 𝜔𝜔o12 𝑠𝑠2 + 𝜔𝜔o2𝑄𝑄o2

𝑠𝑠 + 𝜔𝜔o22 … 𝑠𝑠2 +𝜔𝜔o𝑛𝑛2𝑄𝑄o𝑛𝑛2

𝑠𝑠 + 𝜔𝜔o𝑛𝑛2

2

=

Can be implemented as second-order filter biquads

Example: Synthesize a 4th-order band-pass filter with the normalized transfer function:

𝑇𝑇BP 𝑠𝑠 = 0.01𝐾𝐾𝑠𝑠2

𝑠𝑠4+0.109773𝑠𝑠3+2.01103𝑠𝑠2+0.109773𝑠𝑠+1 Solution: Use 2nd-order biquads (if the required filter order is odd, add a

single 1st-order section as well). Remember, a BPF can be realized using the cascade of a HPF and a LPF.

𝑇𝑇BP = 𝑇𝑇HP.𝑇𝑇LP

𝑇𝑇BP(𝑠𝑠) =𝐾𝐾1𝑠𝑠2

𝑠𝑠2 + 0.057342𝑠𝑠 + 1.09365.

𝐾𝐾2𝑠𝑠2 + 0.0524317𝑠𝑠 + 0.914373



Sallen-Key HPF Sallen-Key LPF≡

Alternatively, the synthesis of a transfer function can be achieved by a successive removal of its poles.

The removal of a pole corresponds to the extraction of a network element (inductor or capacitor).

This process can be accomplished systematically by long division.

Example: Synthesize the transfer function 𝑇𝑇 𝑠𝑠 = 𝑠𝑠4+20𝑠𝑠2+64𝑠𝑠3+9𝑠𝑠

and draw the corresponding circuit diagram.


Realization Problem: Filter Synthesis 2


and draw the corresponding circuit diagram. Solution: Perform long division to obtain:

𝑠𝑠 Ω𝑠𝑠3 + 9𝑠𝑠 𝑠𝑠4 + 20𝑠𝑠2 + 64

𝑠𝑠4 + 9𝑠𝑠2 111𝑠𝑠 Ω−1

11𝑠𝑠2 + 64 | 𝑠𝑠3 + 9𝑠𝑠

𝑠𝑠3 + 6411𝑠𝑠 121

35𝑠𝑠 Ω

3511𝑠𝑠 | 11𝑠𝑠2 + 64

11𝑠𝑠2 35704

𝑠𝑠 Ω−1

64 � 𝟑𝟑𝟑𝟑𝟏𝟏𝟏𝟏𝒔𝒔

𝑇𝑇 𝑠𝑠 = 𝑠𝑠 +1

𝑠𝑠3 + 9𝑠𝑠11𝑠𝑠2 + 64

𝑇𝑇 𝑠𝑠 = 𝑠𝑠 +11𝑠𝑠2 + 64𝑠𝑠3 + 9𝑠𝑠𝑇𝑇 𝑠𝑠 = 𝑠𝑠 +

11

11 𝑠𝑠 + 112135 𝑠𝑠 + 64

3511 𝑠𝑠


111 𝑠𝑠 + 1

11𝑠𝑠2 + 643511 𝑠𝑠


111 𝑠𝑠 +

3511 𝑠𝑠

11𝑠𝑠2 + 64


111 𝑠𝑠 + 1

12135 𝑠𝑠 +

13511 𝑠𝑠64



Ω : Inductance

Ω−1: Capacitance


111 𝑠𝑠 + 1

12135 𝑠𝑠 + 1

35704 𝑠𝑠

Cauer I Realization:

Series L

Parallel C

Series L

Parallel C

111

𝑠𝑠 Ω−1

12135

𝑠𝑠 Ω

35704

𝑠𝑠 Ω−1


and draw the corresponding circuit diagram. Solution: Perform long division to obtain:




111 𝑠𝑠 + 1

12135 𝑠𝑠 + 1

35704 𝑠𝑠

Cauer I Realization:

Series L

Parallel C

Series L

Parallel C

The previous technique is used to systematically synthesize ladder filter structures.

Low sensitivity to component mismatch is inherent to ladder filters.

LC ladder filter can be singly- or doubly-terminated (with or without 𝑅𝑅L).

Double termination minimizes losses and ensures maximum energy transfer.

As the filter order 𝑛𝑛 increases, the quality factor increases.ETH 13Integrated Systems Laboratory

Passive Filter Design

⋯

⋯

Consider the 5th-order RLC ladder filter with double termination.

The Signal Flowgraph (SFG) technique is used to convert the passive ladder circuit into an equivalent active circuit.

SFG is a special type of block diagram/directed graph where the variables are represented by nodes, while branches represent the relation between those variables.

SFG maps multiplications (weights) & additions (intersections).


From Passive to Active

Passive 5th-Order RLC Low-pass Ladder FIlter

Name currents and voltages for all components:

Derive current and voltage relations using KCL and KVL:

Choose the relations between the voltage and the current.


From Passive to Active - SFG

+ 𝑽𝑽𝟏𝟏 − + 𝑽𝑽𝟑𝟑 − + 𝑽𝑽𝟑𝟑 −𝑽𝑽𝟐𝟐 𝑽𝑽𝟒𝟒 𝑽𝑽𝟔𝟔

𝑰𝑰𝟏𝟏 𝑰𝑰𝟑𝟑 𝑰𝑰𝟑𝟑

𝑰𝑰𝟐𝟐 ↓ 𝑰𝑰𝟒𝟒 ↓ 𝑰𝑰𝟔𝟔 ↓ 𝑰𝑰𝟕𝟕 ↓

𝑉𝑉1 = 𝑉𝑉i − 𝑉𝑉2, 𝑉𝑉2 = 𝐼𝐼2𝑠𝑠𝐶𝐶1

, 𝑉𝑉3 = 𝑉𝑉2 − 𝑉𝑉4,

𝑉𝑉5 = 𝑉𝑉4 − 𝑉𝑉6, 𝑉𝑉o = 𝑉𝑉6,𝑉𝑉4 = 𝐼𝐼4𝑠𝑠𝐶𝐶3

, 𝑉𝑉6 = 𝐼𝐼6𝑠𝑠𝐶𝐶5

,

𝐼𝐼1 = 𝑉𝑉1𝑅𝑅S

, 𝐼𝐼2 = 𝐼𝐼1 − 𝐼𝐼3, 𝐼𝐼3 = 𝑉𝑉3𝑠𝑠𝑠𝑠2

,

𝐼𝐼4 = 𝐼𝐼3 − 𝐼𝐼5, 𝐼𝐼6 = 𝐼𝐼5 − 𝐼𝐼7,𝐼𝐼5 = 𝑉𝑉5𝑠𝑠𝑠𝑠4

, 𝐼𝐼7 = 𝑉𝑉6𝑅𝑅L

.

Sketch the corresponding SFG using the equations



𝑉𝑉1 = 𝑉𝑉i − 𝑉𝑉2, 𝑉𝑉2 = 𝐼𝐼2𝑠𝑠𝐶𝐶1

, 𝑉𝑉3 = 𝑉𝑉2 − 𝑉𝑉4,

𝑉𝑉5 = 𝑉𝑉4 − 𝑉𝑉6, 𝑉𝑉o = 𝑉𝑉6,𝑉𝑉4 = 𝐼𝐼4𝑠𝑠𝐶𝐶3

, 𝑉𝑉6 = 𝐼𝐼6𝑠𝑠𝐶𝐶5

,

𝐼𝐼1 = 𝑉𝑉1𝑅𝑅𝑆𝑆

, 𝐼𝐼2 = 𝐼𝐼1 − 𝐼𝐼3, 𝐼𝐼3 = 𝑉𝑉3𝑠𝑠𝑠𝑠2

,

𝐼𝐼4 = 𝐼𝐼3 − 𝐼𝐼5, 𝐼𝐼6 = 𝐼𝐼5 − 𝐼𝐼7,𝐼𝐼5 = 𝑉𝑉5𝑠𝑠𝑠𝑠4

, 𝐼𝐼7 = 𝑉𝑉6𝑅𝑅L

.

Re-define the state-variables corresponding to the current nodes at the output of the branches with reactive elements to voltages.



Convert 𝐼𝐼3 into 𝑉𝑉3′.



Convert 𝐼𝐼5 into 𝑉𝑉5′.



Re-define the state-variables corresponding to the voltage nodes at the input of the vertical branches to currents.



Convert 𝑉𝑉1 into 𝐼𝐼1′ .



Redefine the state variables with negative quantities on their left until there is a single negative branch in each loop.



Define 𝑉𝑉2A = −𝑉𝑉2.



Define 𝐼𝐼3A′ = −𝐼𝐼3′ .



Define 𝑉𝑉3A′ = −𝑉𝑉3′.



Define 𝐼𝐼4A = −𝐼𝐼4.



Define 𝑉𝑉6A = −𝑉𝑉6.



Define 𝑉𝑉oA = −𝑉𝑉o.



Define 𝐼𝐼7A = −𝐼𝐼7.



Let the branches containing 𝑠𝑠 be in the form 1𝑠𝑠𝑠𝑠



Invert the sign of the positive reactive elements.



Eliminate the redundant state variables.



Recall (Lecture 4): The active integrator circuit



𝑉𝑉out𝑉𝑉in

= −1𝑠𝑠𝑅𝑅𝑠𝑠

→ −1𝑠𝑠𝜏𝜏

Using the active integrator, we obtain a 5th order active filter


From Passive to Active - Circuit

𝑉𝑉o = −𝑉𝑉i

If 𝐹𝐹 is a field and 𝑛𝑛 is a nonnegative integer, then a polynomial 𝑃𝑃(𝑥𝑥) of degree 𝑛𝑛 over 𝐹𝐹 with 𝑎𝑎𝑖𝑖 ∈ 𝐹𝐹 for 𝑖𝑖 = 0,⋯ ,𝑛𝑛, 𝑎𝑎𝑛𝑛 ≠ 0, and 𝑥𝑥 an indeterminate, is of the form: 𝑃𝑃 𝑥𝑥 = 𝑎𝑎0 + 𝑎𝑎1𝑥𝑥 + ⋯+ 𝑎𝑎𝑛𝑛𝑥𝑥𝑛𝑛 = ∑𝑖𝑖=0𝑛𝑛 𝑎𝑎𝑖𝑖𝑥𝑥𝑖𝑖

Denoting the degree of 𝑃𝑃(𝑥𝑥) as deg 𝑃𝑃 𝑥𝑥 , let 𝑃𝑃 𝑥𝑥 ≠ 0,𝑄𝑄 𝑥𝑥 ≠ 0 ∈ 𝐹𝐹.

1. deg 𝑃𝑃 𝑥𝑥 𝑄𝑄 𝑥𝑥 = deg(𝑃𝑃 𝑥𝑥 ) + deg(𝑄𝑄 𝑥𝑥 )2. deg(𝑃𝑃 𝑥𝑥 ± 𝑄𝑄 𝑥𝑥 ) ≤ max(deg(𝑃𝑃 𝑥𝑥 ), deg(𝑄𝑄 𝑥𝑥 ))

If 𝑓𝑓 𝑥𝑥 ,𝑔𝑔(𝑥𝑥) ∈ 𝐹𝐹 with 𝑔𝑔(𝑥𝑥) ≠ 0 then 𝑔𝑔(𝑥𝑥) divides 𝑓𝑓 𝑥𝑥 , or 𝑔𝑔(𝑥𝑥) is a factor of 𝑓𝑓(𝑥𝑥), if there exists a polynomial 𝑞𝑞(𝑥𝑥) ∈ 𝐹𝐹 such that 𝑓𝑓 𝑥𝑥 =𝑞𝑞 𝑥𝑥 𝑔𝑔 𝑥𝑥 . If 𝑓𝑓(𝑥𝑥) ≠ 0 has no non-trivial, non-unit factors (it cannot be factorized into polynomials of lower degree); then 𝑓𝑓(𝑥𝑥) is an irreducible polynomial, or prime polynomial.



Hence, there exist unique polynomials 𝑞𝑞 𝑥𝑥 , 𝑟𝑟 𝑥𝑥 ∈ 𝐹𝐹 such that 𝑓𝑓 𝑥𝑥 = 𝑞𝑞 𝑥𝑥 𝑔𝑔 𝑥𝑥 + 𝑟𝑟(𝑥𝑥), where 𝑟𝑟 𝑥𝑥 = 0 or deg 𝑟𝑟 𝑥𝑥 < deg 𝑔𝑔 𝑥𝑥 . The polynomials 𝑞𝑞 𝑥𝑥 and 𝑟𝑟 𝑥𝑥 are called respectively the quotient and remainder.

If 𝑐𝑐 is a root of 𝑃𝑃(𝑥𝑥), then (x − 𝑐𝑐) divides 𝑃𝑃(𝑥𝑥), such that 𝑃𝑃 𝑥𝑥 =𝑥𝑥 − 𝑐𝑐 𝑄𝑄(𝑥𝑥) with deg 𝑄𝑄 𝑥𝑥 = deg 𝑃𝑃 𝑥𝑥 − 1.

An irreducible polynomial of degree greater than one over a field 𝐹𝐹 has no roots in 𝐹𝐹. Hence, a polynomial of degree 𝒏𝒏 in 𝑭𝑭 can have at most 𝒏𝒏 distinct roots.

If 𝑃𝑃 𝑥𝑥 = 𝑎𝑎0 + ⋯+ 𝑎𝑎𝑛𝑛𝑥𝑥𝑛𝑛 is a complex polynomial then its conjugate is the polynomial �𝑃𝑃 𝑥𝑥 = �𝑎𝑎0 + ⋯+ �𝑎𝑎𝑛𝑛𝑥𝑥𝑛𝑛. That is, the conjugate is the polynomial whose coefficients �𝑎𝑎𝑖𝑖 are the conjugates of 𝑎𝑎0 of 𝑃𝑃(𝑥𝑥).

If 𝑓𝑓(𝑥𝑥) ∈ ℝ and 𝑓𝑓 𝑧𝑧0 = 0, then 𝑓𝑓 ̅𝑧𝑧0 = 0; the complex roots of real polynomials come in conjugate pairs.



Suppose 𝑓𝑓(𝑥𝑥) ∈ ℂ, with 𝑓𝑓(𝑥𝑥) non-constant, then 𝑓𝑓(𝑥𝑥) takes on a minimum value at some value 𝑧𝑧0 ∈ ℂ, if 𝑓𝑓(𝑥𝑥0) ≠ 0, then 𝑓𝑓(𝑥𝑥0) is not the minimum value of 𝑓𝑓(𝑥𝑥) and 𝑥𝑥0 ≠ 𝑧𝑧0. Consequently, 𝒇𝒇(𝒙𝒙) has at least one complex root.

A complex polynomial completely factorizes into linear factors. Additionally, a real polynomial factorizes into degree 1 and degree 2 factors. Equivalently, the only irreducible real polynomials are linear and quadratic polynomials.

* Proofs can be found in: Fine, D. and Rosenberger, G. “The Fundamental Theorem of Algebra”, Springer series on undergraduate texts in Mathematics, Springer, 1997.



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