EE 508 Lecture 6 Degrees of Freedom The Approxmaton Problem
Revew from Last Tme Desgn Strategy Theorem: A crcut wth transfer functon T(s) can be obtaned from a crcut wth normalzed transfer functon T n (s n ) by denormalzng all frequency dependent components. C L C/ω o L/ω o
Revew from Last Tme Frequency normalzaton/scalng The frequency scaled crcut can be obtaned from the normalzed crcut smply by scalng the frequency dependent mpedances (up or down) by the scalng factor Component denormalzaton by factor of ω 0 Normalzed Component Denormalzed Component R R C C/ω o L L/ω o Other Components Unchanged Component values of energy storage elements are scaled down by a factor of ω 0
Revew from Last Tme Example: Desgn a V-V passve 3 rd -order Lowpass Butterworth flter wth a 3-db band-edge of K rad/sec and equal source and load termnatons. 3 (from the BW approxmaton whch wll be dscussed later:) T s = s +s +s+ R S L L 3 V OUT V IN C R L Flter archtecture L =H L 3 =H V OUT V IN C =F Normalzed flter C L C/θ L/θ Ts =K s 3 +s +s+ L =mh L 3 =mh V OUT V IN C =mf Denormalzed flter 9 0 Ts =K s + 0 s 0 s 0 3 3 6 9
Revew from Last Tme Example: Desgn a V-V passve 3 rd -order Lowpass Butterworth flter wth a band-edge of K Rad/Sec and equal source and load termnatons. L =mh L 3 =mh V OUT V IN C =mf 9 0 Ts =K s + 0 s 0 s 0 3 3 6 9 Is ths soluton practcal? Some component values are too bg and some are too small!
Flter Concepts and Termnology Frequency scalng Frequency Normalzaton Impedance scalng Transformatons LP to BP LP to HP LP to BR
Revew from Last Tme Impedance Scalng Impedance scalng of a crcut s acheved by multplyng ALL mpedances n the crcut by a constant R C L A θr C/θ Lθ θa for transresstance gan A for dmensonless gan A/θ for transconductance gan
Revew from Last Tme Impedance Scalng Theorem: If all mpedances n a crcut are scaled by a constant θ, then a) All dmensonless transfer functons are unchanged b) All transresstance transfer functons are scaled by θ c) All transconductance transfer functons are scaled by θ -
Revew from Last Tme Example: Desgn a V-V passve 3 rd -order Lowpass Butterworth flter wth a band-edge of K Rad/Sec and equal source and load termnatons. L =mh L 3 =mh V OUT V IN C =mf 9 0 Ts =K s + 0 s 0 s 0 3 3 6 9 Is ths soluton practcal? Some component values are too bg and some are too small! Impedance scale by θ=000 R C L θr C/θ θl K L =H L 3 =H V OUT V IN C =uf K 9 0 Ts =K s + 0 s 0 s 0 3 3 6 9 Component values more practcal
Typcal approach to lowpass flter desgn. Obtan normalzed approxmatng functon. Synthesze crcut to realze normalzed approxmatng functon 3. Denormalze crcut obtaned n step 4. Impedance scale to obtan acceptable component values
Degrees of Freedom V IN R C Vo Ts VO VIN RCs T jω 0.707 /RC ω n Crcut has two desgn varables: {R,C} Crcut has one key controllable performance characterstc: If ω 0 s specfed for a desgn, crcut has desgn varables constrant Degree of Freedom ω0 RC Performance/Cost strongly affected by how degrees of freedom n a desgn are used!
Example: Desgn a nd order lowpass Butterworth flter wth 3dB passband attenuaton, a dc gan of 5, and a 3dB bandedge of 4KHz Note: We have not dscussed the Butterworth approxmaton yet so some detals here wll be based upon concepts that wll be developed later T BWn = 5 s + s+ ω 0 = Q 0.707 R Q R V IN R 0 C C R 3 R R 3 VOLP INT INT A Popular Second-Order Lowpass Flter
Example: Desgn a nd order lowpass Butterworth flter wth 3dB passband attenuaton, a dc gan of 5, and a 3dB band edge of 4KHz R Q R V IN R 0 C C R 3 R R 3 VOLP ω = 0 INT INT R R C C A Popular Second-Order Lowpass Flter R R C C 0 T s = R C R R C C s +s + Q= Q R Q 7 desgn varables and only two constrants (gnorng the gan rght now) Crcut has 5 Degrees of Freedom! RR C C
Example: Desgn a nd order lowpass Butterworth flter wth 3dB passband attenuaton, a dc gan of 5, and a 3dB band edge of 4KHz R Q R V IN R 0 C C R 3 R R 3 VOLP If C =C =C and R =R =R 0 =R, ths reduces to INT INT A Popular Second-Order Lowpass Flter T s = RC R s +s + RQ RC RC R Q R V IN R C R C R 3 R 3 V OLP INT INT A Popular Second-Order Lowpass Flter How many degrees of freedom reman?
Example: Desgn a nd order lowpass Butterworth flter wth 3dB passband attenuaton, a dc gan of 5, and a 3dB bandedge of 4KHz R Q R V IN R C R C R 3 R 3 V OLP INT INT A Popular Second-Order Lowpass Flter T s = RC R s +s + RQ RC RC ω 0= RC RQ Q= R Normalzng by the factor ω 0, we obtan n T s = Q s +s + Lets now use up the two degrees of freedom n the crcut: Settng R=R 3 = obtan the followng normalzed crcut
Example: Desgn a nd order lowpass Butterworth flter wth 3dB passband attenuaton, a dc gan of 5, and a 3dB bandedge of 4KHz Settng R=R 3 = obtan the followng crcut R Q V IN C C V OLP INT INT A Popular Second-Order Lowpass Flter The two constrants become ω 0= RC C R R Q Q= RQ Ths leaves unknowns, R Q and C and two constrants (.e. no remanng degrees of freedom)
Example: Desgn a nd order lowpass Butterworth flter wth 3dB passband attenuaton, a dc gan of 5, and a 3dB bandedge of 4KHz n T s = Q s +s + ω 0n= Q N To satsfy the constrants, must now set RQ Q C 0.707 V IN V OLP INT INT A Popular Second-Order Lowpass Flter Now we can do frequency scalng C L C/ω o L/ω o C= /(π 4K) = 39.8uF
Example: Desgn a nd order lowpass Butterworth flter wth 3dB passband attenuaton, a dc gan of 5, and a 3dB bandedge of 4KHz Denormalzed crcut wth bandedge of 4 KHz.707 V IN 39.8uF 39.8uF V OLP INT INT A Popular Second-Order Lowpass Flter Ths has the rght transfer functon (but unty gan) Can now do mpedance scalng to get more practcal component values R θr C C/θ L θl A good mpedance scalng factor may be θ=000 R C K 39.8nF
Example: Desgn a nd order lowpass Butterworth flter wth 3dB passband attenuaton, a dc gan of 5, and a 3dB bandedge of 4KHz Denormalzed crcut wth bandedge of 4 KHz 707 K V IN K 39.8nF K 39.8nF K K V OLP Ths has the rght transfer functon (but unty gan) To fnsh the desgn, preceed or follow ths crcut wth an amplfer wth a gan of 5 to meet the dc gan requrements
Flter Concepts and Termnology Frequency scalng Frequency Normalzaton Impedance scalng Transformatons LP to BP LP to HP LP to BR It can be shown the standard HP, BP, and BR approxmatons can be obtaned by a frequency transformaton of a standard LP approxmatng functon Wll address the LP approxmaton frst, and then provde detals about the frequency transformatons
Flter Desgn Process Establsh Specfcatons - possbly T D (s) or H D (z) - magntude and phase characterstcs or restrctons - tme doman requrements Approxmaton - obtan acceptable transfer functons T A (s) or H A (z) - possbly acceptable realzable tme-doman responses Synthess - buld crcut or mplement algorthm that has response close to T A (s) or H A (z) - actually realze T R (s) or H R (z) Flter
The Approxmaton Problem The goal n the approxmaton problem s smple, just want a functon T A (s) or H A (z) that meets the flter requrements. Wll focus prmarly on approxmatons of the standard normalzed lowpass functon TLP j ω Frequency scalng wll be used to obtan other LP band edges Frequency transformatons wll be used to obtan HP, BP, and BR responses
The Approxmaton Problem TLP j T s =? A ω T A (s) s a ratonal fracton n s T s = m =0 n =0 as bs Ratonal fractons n s have no dscontnutes n ether magntude or phase response No natural metrcs for T A (s) that relate to magntude and phase characterstcs (dffcult to meanngfully compare T A (s) and T A (s))
The Approxmaton Problem T LP j Approach we wll follow: Magntude Squared Approxmatng Functons Inverse Transform A H ω T s A HA ω ω Collocaton Least Squares Pade Approxmatns Other Analytcal Optmzaton Numercal Optmzaton Canoncal Approxmatons Butterworth (BW) Chebyschev (CC) Ellptc Thompson
Magntude Squared Approxmatng Functons T jω = m =0 n =0 a b jω jω T s = m =0 n =0 as bs m a + a jω + a jω +...+ a jω T jω = b + b jω + b jω +...+ b jω T jω = o m n o n 4 3 5 ao- aω + a4ω +.. + j aω - a3ω + a5ω +... 4 3 5 bo- bω + b4ω +.. + j bω - b3ω + b5ω +... T jω = k k- akω + jω akω 0 k m 0 k m k even kodd k k- bkω + jω bkω 0 k n 0 k n k even kodd T jω = where F, F, F 3 and F 4 are even functons of ω F ω + jωf ω F ω + jωf ω 3 4
Magntude Squared Approxmatng Functons T s = m =0 n =0 as bs T jω = T jω = F ω + jωf ω F ω + jωf ω 3 4 F ω + ω F ω F3 ω + ω F4 ω Thus T jω s an even functon of ω It follows that T jω s a ratonal fracton n ω wth real coeffcents Snce T jω s a real varable, natural metrcs exst for comparng approxmatng functons to T jω
Magntude Squared Approxmatng Functons T s = m =0 n =0 as bs If a desred magntude response s gven, t s common to fnd a ratonal fracton n ω wth real coeffcents, denoted as H A (ω ), that approxmates the desred magntude squared response and then obtan a functon T A (s) T jω = H ω that satsfes the relatonshp A A H A (ω ) s real so natural metrcs exst for obtanng H A (ω ) H ω = A l =0 k =0 Obtanng T A (s) from H A (ω ) s termed the nverse mappng problem c ω d ω But how s T A (s) obtaned from H A (ω )?
Inverse mappng problem: T A s H A ω well defned A A H ω T jω TA s? HA ω Consder an example: T s s T s s A H ω ω Thus, the nverse mappng n ths example s not unque!
Inverse mappng problem: T A s H A ω A H ω T jω A TA s? HA ω Some observatons: If an nverse mappng exsts, t s not necessarly unque If an nverse mappng exsts, than a mnmum phase nverse mappng exsts and t s unque (wthn all-pass factors) The mappng from T A (s) to H A (ω ) ncreases order by a factor of Any nverse mappng from H A (ω ) to T A (s) wll reduce order by a factor of (wthn all-pass factors)
Example: H ω + A ω 4 ω + ω + T A s s+ s+ s+ Example: H ω - A ω 4 ω + ω +? Inverse mappng does not exst! It can be shown that many even ratonal fractons n ω do not have an nverse mappng back to the s-doman! Often these functons have a magntude squared response that does a good job of approxmatng the desred flter magntude response If an nverse mappng exsts, there are often several nverse mappngs that exst
Observaton: If z s a zero (pole) of H A (ω ), then z, z*, and z* are also zeros (poles) of H A (ω ) Im z -z* Im z Re Re -z z* Im Im z Re -z z Re Im z Im z Re Re z* Thus, roots come as quadruples f off of the axs and as pars f they lay on the axs
Observaton: If z s a zero (pole) of H A (ω ), then z, z*, and z* are also zeros (poles) of H A (ω ) Im z -z* Im z Re Re Proof: -z z* Consder an even polynomal n ω wth real coeffcents Pω At a root, ths polynomal satsfes the expresson Pω Replacng ω wth ω, we obtan m m m P ω a -ω a ω a ω 0 0 0 0 n * * n Recall (xy)*=x*y* and x x m aω 0 m aω 0 0 ω s a root of Pω Takng the complex conjugate of P(ω )=0 we obtan * m * m * m * * * j j P ω aω a ω a ω 0 0 0 0 Snce a s real for all I, t thus follows that m * aj ω 0 0 ω* s a root of Pω
End of Lecture 6