8. Active Filters - 2. Electronic Circuits. Prof. Dr. Qiuting Huang Integrated Systems Laboratory

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1 8. Active Filters - 2 Electronic Circuits Prof. Dr. Qiuting Huang Integrated Systems Laboratory

2 Blast From The Past: Algebra of Polynomials * PP xx is a polynomial of the variable xx: PP xx = aa 0 + aa 1 xx + + aa nn xx nn Coefficients The highest power of xx in PP xx with non-zero coefficient is called the degree of PP xx. cc 1 is a root of PP xx cc kk is a root of PP xx PP xx = xx cc 1 QQ nn 1 xx. has nn roots PP xx = xx cc 1 xx cc kk QQ nn kk xx. QQ nn 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. 2

3 Design of Filters The transfer function of a filter is a rational function. ( ) T s k bks + b s + + bs+ b Ns () = = n as + a s + + as+ a D() s n k 1 k n 1 n 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: 0 ( ) = T s Passive 5 th -Order Ladder FIlter b as + as + + as+ a

4 Design of Filters Filter performance improves with the order of the filter. Sharper response with higher nn 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. PB = pass-band TB = transition-band SB = stop-band 4

5 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. Magnitude (db) Chebyshev Low-pass resonse Max. Ripple = 0.1 db Passive 5 th -Order Ladder FIlter The design comprises an optimization process targeting minimizing the maximum deviation from the ideal filter response (minmax optimization). 5

6 Approximation Problem 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. 6

7 Realization Problem: Filter Synthesis We can represent the transfer function obtained from the mathematical approximation as: TT ss = bb kk ss kk + bb kk 1 ss kk bb 1 ss + bb 0 aa nn ss nn + aa nn 1 ss nn aa 1 ss + aa 0 = NN(ss) DD(ss) = bb 1,2 ss 2 + bb 1,1 ss + bb 1,0 bb 2,2 ss 2 + bb 2,1 ss + bb 2,0 bb kk + bb kk 2,2ss2 2,1ss + bb kk 2,0 aa 1,2 ss 2 + aa 1,1 ss + aa 1,0 aa 2,2 ss 2 + aa 2,1 ss + aa 2,0 aann 2,2ss2 + aann 2,1ss + aann 2,0 7

8 Realization Problem: Filter Synthesis We can represent the transfer function obtained from the mathematical approximation as: TT ss = bb kk ss kk + bb kk 1 ss kk bb 1 ss + bb 0 aa nn ss nn + aa nn 1 ss nn aa 1 ss + aa 0 = NN(ss) DD(ss) = KK 1 ss 2 + ωω z1 2 ss + ωω QQ z1 KK 2 ss 2 + ωω z2 2 ss + ωω z1 QQ z2 KKnn z2 2 ss 2 + ωω o1 2 ss + ωω QQ o1 ss 2 + ωω o2 2 ss + ωω o1 QQ o2 ss 2 + o2 Can be implemented as second-order filter biquads ss 2 + ωω o nn 2 QQ o nn 2 ωω z nn 2 QQ z nn 2 2 ss + ωω nn z 2 2 ss + ωω nn o 2 Filter 1 Filter 2 Filter nn 2 8

9 Realization Problem: Filter Synthesis Example: Synthesize a 4 th -order band-pass filter with the normalized transfer function: TT BP ss = 0.01KKss 2 ss ss ss ss+1 Solution: Use 2 nd -order biquads (if the required filter order is odd, add a single 1 st -order section as well). Remember, a BPF can be realized using the cascade of a HPF and a LPF. TT BP = TT HP. TT LP Sallen-Key HPF Sallen-Key LPF TT BP (ss) = KK 1 ss 2 ss ss KK 2 ss ss

10 Realization Problem: Filter Synthesis 2 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 +20ss ss 3 +9ss and draw the corresponding circuit diagram. 10

11 Realization Problem: Filter Synthesis 2 Example: Synthesize the transfer function TT ss = ss4 +20ss ss 3 +9ss and draw the corresponding circuit diagram. Solution: Perform long division to obtain: Cauer I Realization: 1 11ss TT ss = ss + 1ss 3 9ss 11ss ss 11ss ss ss 11ss ss + 11 ss ss ss ss 64 ss Ω Series L ss 3 + 9ss ss ss ss 4 + 9ss ss ss 3 + 9ss ss Ω 1 ss Parallel C ss Ω ss ss 11 11ss ss ss 1111 Ω : Inductance Ω 1 : Capacitance ss Ω 1 Series L Parallel C 11

12 Realization Problem: Filter Synthesis 2 Example: Synthesize the transfer function TT ss = ss4 +20ss ss 3 +9ss and draw the corresponding circuit diagram. Solution: Perform long division to obtain: Series L Series L Parallel C Parallel C Cauer I Realization: 1 TT ss = ss ss ss ss 12

13 Passive Filter Design 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 RR L ). Double termination minimizes losses and ensures maximum energy transfer. As the filter order nn increases, the quality factor increases. 13

14 From Passive to Active Consider the 5 th -order RLC ladder filter with double termination. Passive 5 th -Order RLC Low-pass Ladder FIlter 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). 14

15 Name currents and voltages for all components: + VV 11 + VV 33 + VV 55 VV 22 VV 44 VV 66 II 11 II 33 II 55 II 22 II 44 II 66 II 77 Derive current and voltage relations using KCL and KVL: VV 1 = VV i VV 2, VV 2 = II 2, VV sscc 3 = VV 2 VV 4, 1 VV 4 = II 4, VV 5 = VV 4 VV 6, VV sscc 6 = II 6, 3 sscc 5 VV o = VV 6, II 1 = VV 1 RR S, II 2 = II 1 II 3, II 3 = VV 3 ssss 2, II 4 = II 3 II 5, II 6 = II 5 II 7, II 5 = VV 5 ssss 4, II 7 = VV 6 RR L. Choose the relations between the voltage and the current. 15

16 Sketch the corresponding SFG using the equations VV 1 = VV i VV 2, VV 2 = II 2, VV sscc 3 = VV 2 VV 4, 1 VV 4 = II 4, VV 5 = VV 4 VV 6, VV sscc 6 = II 6, 3 sscc 5 VV o = VV 6, II 1 = VV 1 RR SS, II 2 = II 1 II 3, II 3 = VV 3 ssss 2, II 4 = II 3 II 5, II 6 = II 5 II 7, II 5 = VV 5 ssss 4, II 7 = VV 6 RR L. 16

17 Re-define the state-variables corresponding to the current nodes at the output of the branches with reactive elements to voltages. 17

18 Convert II 3 into VV 3. 18

19 Convert II 5 into VV 5. 19

20 Re-define the state-variables corresponding to the voltage nodes at the input of the vertical branches to currents. 20

21 Convert VV 1 into II 1. 21

22 Convert VV 3 into II 3. 22

23 Convert VV 5 into II 5. 23

24 Redefine the state variables with negative quantities on their left until there is a single negative branch in each loop. 24

25 Define VV 2A = VV 2. 25

26 Define II 3A = II 3. 26

27 Define VV 3A = VV 3. 27

28 Define II 4A = II 4. 28

29 Define VV 6A = VV 6. 29

30 Define VV oa = VV o. 30

31 Define II 7A = II 7. 31

32 Let the branches containing ss be in the form 1 ssss 32

33 Invert the sign of the positive reactive elements. 33

34 Eliminate the redundant state variables. 34

35 Recall (Lecture 4): The active integrator circuit VV out VV in = 1 ssssss 1 ssττ 35

36 From Passive to Active - Circuit Using the active integrator, we obtain a 5 th order active filter VV o = VV i 36

37 Blast From The Past: Algebra of Polynomials* If FF is a field and nn is a nonnegative integer, then a polynomial PP(xx) of degree nn over FF with aa ii FF for ii = 0,, nn, aa nn 0, and xx an indeterminate, is of the form: PP xx = aa 0 + aa 1 xx + + aa nn xx nn = nn ii=0 aa ii xx ii 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. 37

38 Blast From The Past: Algebra of Polynomials* 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 = aa aa nn xx nn is a complex polynomial then its conjugate is the polynomial PP xx = aa aa nn xx nn. That is, the conjugate is the polynomial whose coefficients aa ii are the conjugates of aa 0 of PP(xx). If ff(xx) R and ff zz 0 = 0, then ff zz 0 = 0; the complex roots of real polynomials come in conjugate pairs. 38

39 Blast From The Past: Algebra of Polynomials* Suppose ff(xx) C, with ff(xx) non-constant, then ff(xx) takes on a minimum value at some value zz 0 C, if ff(xx 0 ) 0, then ff(xx 0 ) is not the minimum value of ff(xx) and xx 0 zz 0. 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,