Lectures 29 and 30 BIQUADRATICS AND STATE SPACE OP AMP REALIZATIONS. I. Introduction
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1 EE-202/445, 3/18/ R. A. DeCarlo Lecures 29 and 30 BIQUADRATICS AND STATE SPACE OP AMP REALIZATIONS I. Inroducion 1. The biquadraic ransfer funcion has boh a 2nd order numeraor and a 2nd order denominaor: H (s) = b 0 s2 + b 1 s + b 2 s 2 + a 1 s + a 2 2. Review: Basic Invering Amplifier: V ou = Z f (s) Z in (s) V in = Y in (s) Y f (s) V in 3. Basic Invering Summer: v ou = G 1 v 1 G 2 v 2 G 3 v 3
2 EE-202/445, 3/18/ R. A. DeCarlo 4. Basic Invering Inegraor: v ou = G 1 v 1 G 2 v 2 G 3 v 3
3 EE-202/445, 3/18/ R. A. DeCarlo II. Conrollable Canonical 4 OP AMP Sae Space Realizaion of he Biquadraic TF: A CONCEPTUAL Block Diagram Developmen 1. Recall again he biquadraic srucure wih a 1 > 0 and a 2 > 0 o insure sabiliy (necessary and sufficien condiions for poles in he lef half complex plane): H (s) = V ou = b 0s 2 + b 1 s + b 2 V in s 2 = b 0s 2 + b 1 s + b 2 + a 1 s + a 2 = b 2 + b 1s + b 0s 2 where = s 2 + a 1 s + a 2. Since V ou (s) = H (s)v in (s), V ou (s) = b 2 V in (s) + b 1 s V in (s) + b 0 s2 V in (s) 2. Define he AUXILIARY EQUATION: V ou (s) = 1 V in (s) = 1 s 2 + a 1 s + a 2 V in (s) (a) Implicaion 1: s-domain auxiliary equaion ( s 2 + a 1 s + a 2 ) V ou (s) = V in (s) (b) Implicaion 2: s-domain oupu equaion V ou (s) = b 2 V in (s) + b 1 s V in(s) + b 0 s 2 V in = b 2 Vou + b 1 s V ou + b 0 s 2 V ou
4 EE-202/445, 3/18/ R. A. DeCarlo 3. OBJECTIVE: develop a flow diagram ha describes V ou (s) and hen enhance his flow diagram o obain V ou (s). There are wo criical s-domain equaions here: (a) (auxiliary equaion s-domain) and ( s 2 + a 1 s + a 2 ) V ou (s) = s 2 V ou (s) + a 1 sv ou (s) + a 2Vou (s) = V in (s) (b) (oupu equaion s-domain) V ou = b 2 Vou + b 1 s V ou + b 0 s 2 V ou 4. Inerpre s-domain equaions as differenial equaions in he - world. Recall, muliplicaion by s means differeniaion in he ime world. (The number of dos over he ime variables means he order of he ime derivaive.) (a) s 2 V ou (s) + a 1 s V ou (s) + a 2 Vou (s) = V in (s) implies ha in he -world: ˆv ou () + a 1 vou () + a 2 vou () = v in () or equivalenly ˆv ou () = v in () a 1 vou () a 2 vou () (b) V ou = b 2 Vou + b 1 s V ou + b 0 s 2 V ou implies ha in he -world: v ou () = b 2 vou () + b 1 vou () + b 0 vou () (c) Conclusion: ˆ vou () = v in () a 1 vou () a 2 vou () implies
5 EE-202/445, 3/18/ R. A. DeCarlo v ou = (b 2 b 0 a 2 ) v ou + (b 1 b 0 a 1 ) vou + b 0 v in 5. Summary: Two imporan relaionships resul: (a) Auxiliary Equaion in -world: ˆv ou () = v in () a vou 1 () a 2vou () (b) Oupu Equaion in -world: v ou = (b 2 b 0 a 2 ) v ou + (b 1 b 0 a 1 ) vou + b 0 v in 6. Block Diagram Inerpreaion of auxiliary equaion since we do no know he signs on he coefficiens and hence canno generae he acual op amp circui: Example 1: Realize H (s) = V ou (s) V in (s) = 0.1s s s 2 + 2s + 4 canonical realizaion. using he conrollable Sep 1. Time domain auxiliary equaion is:
6 EE-202/445, 3/18/ R. A. DeCarlo ˆv ou () = v in () 2 vou () 4 v ou () Sep 2. Inegrae boh sides and change signs o obain: vou () = vou ˆ (τ ) dτ = v in (τ )dτ 2 vou (τ ) dτ 4 [ v ou (τ )]dτ Sep 3. Realize equaion of sep 2: Sep 4. Recall v ou () = 0.7 v ou () vou () vou () and ˆv ou () = v in () 2 vou () 4 v ou () in which case
7 EE-202/445, 3/18/ R. A. DeCarlo v ou () = 0.1v in () 0.3 v ou () 0.1 vou () = 0.1v in () 0.3 v ou () 0.1 vou () which has realizaion: Par 2: The Observable Canonical Form by example only Example 2: Realize H (s) = V ou (s) V in (s) = 0.1s s s 2 + 2s + 4 canonical realizaion. using he observable Sep 1. Given a ransfer funcion, generae a differenial equaion in he inpu and oupu. H (s) = V ou (s) V in (s) = 0.1s s s 2 + 2s + 4 implies ( s 2 + 2s + 4)V ou (s) = ( 0.1s s + 0.7)V in (s) implies
8 EE-202/445, 3/18/ R. A. DeCarlo v ou () + 2 v ou () + 4v ou () = 0.7v in () + 0.3v in () + 0.1v in () Sep 2. Replace derivaive by he D-operaor in which D k = dk d k and D k is he k-fold inegral. Equivalenly D 2 v ou () + 2Dv ou () + 4v ou () = 0.7v in () + 0.3Dv in () + 0.1Dv in () D 2 v ou () = 0.1D 2 v in () + D[ 0.3v in () 2v ou ()]+ [ 0.7v in () 4v ou ()] Muliplying boh sides by D 2 yields wha one would call an inegral equaion. v ou () = 0.1v in () + D 1 [ 0.3v in () 2v ou ()]+ D 2 [ 0.7v in () 4v ou ()] Sep 3. Define an inermediae variable x 1 and realize resuling equaion as an op amp circui assuming x 1 is available i will be made available in sep 4. In general he inermediae variable has no physical meaning. (a) From sep 2, recall ha v ou () = 0.1v in () + D 1 [ 0.3v in () 2v ou ()]+ D 2 [ 0.7v in () 4v ou ()] which has a bunch of messy implici inegrals. (b) Define an inermediae variable x 1 = D 1 [ 0.3v in () 2v ou ()]+ D 2 [ 0.7v in () 4v ou ()]
9 EE-202/445, 3/18/ R. A. DeCarlo o cover up he implici inegrals. (c) Thus v ou () = 0.1v in () + x 1 or v ou () = 0.1v in () x 1 (d) Op amp realizaion assuming x 1 is available: Sep 4. Consruc a differenial equaion for x 1. (a) Recall x 1 = D 1 [ 0.3v in () 2v ou ()]+ D 2 0.7v in () 4v ou () (b) Muliply boh sides by D o obain [ ] x 1 = Dx 1 = 0.3v in () 2v ou () + D 1 [ 0.7v in () 4v ou ()] (c) To make life simpler by making he implici inegrals disappear define a second inermediae variable x 2 : (d) Thus x 2 = D 1 [ 0.7v in () 4v ou ()] x 1 = Dx 1 = 0.3v in () 2v ou () + x 2 (e) Subsiue for v ou () where from sep 3 v ou () = 0.1v in () + x 1.
10 EE-202/445, 3/18/ R. A. DeCarlo (f) I follows ha x 1 = 0.1v in () 2x 1 + x 2. Sep 5. Realizaion of (f) in sep 4. (a) Inegraing boh sides of x 1 = 0.1v in () 2x 1 + x 2 and muliplying by -1 yields: x 1 () = x 1 (τ )dτ = 0.1 v in (τ )dτ 2 [ x 1 (τ )]dτ x 2 (τ )dτ (b) This inegral equaion can be realized by he op amp inegraor circui below: Noice ha an addiional op amp is used o drive he op amp circui of sep 3 which requires x 1. Sep 6. Build an op amp circui o generae x 2. Consider ha x 2 = D 1 [ 0.7v in () 4v ou ()] implies ha x 2 () = Dx 2 () = 0.7v in () 4v ou () = 0.3v in () 4x 1 () afer subsiuing for v ou. This is equivalen o he inegral equaion
11 EE-202/445, 3/18/ R. A. DeCarlo x 2 () = x 2 (τ )dτ = 0.3 v in (τ )dτ 4 [ x 1 (τ )]dτ This equaion can be realized using he op amp circui below. Remarks: 1. Overall he realizaion requires 5 op amps. This can be reduced o 4. Redo he above developmen so as o reduce he oal number of op amps o My experience wih filer design classes is ha sudens who use he observable form above have beer luck for a working op amp circui han hose who use he conrollable form.
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