EE 508 Lecture 7. Degrees of Freedom The Approximation Problem

Transcription

1 EE 508 Lecture 7 Degrees of Freedom The Approxmaton Problem

2 vew 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

3 vew 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

4 vew from Last Tme pedance 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 -

5 vew from Last Tme pedance Scalng pedance 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

6 vew from Last Tme Typcal approach to lowpass flter desgn. Obtan normalzed approxmatng functon. Synthesze crcut to realze normalzed approxmatng functon 3. Denormalze crcut obtaned n step 4. pedance scale to obtan acceptable component values

7 Degrees of Freedom V IN R C Vo Ts VO VIN RCs T jω /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 ω 0 RC Performance/Cost strongly affected by how degrees of freedom n a desgn are used!

8 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+ R Q R V IN R 0 C C R 3 R R 3 VOLP INT INT A Popular Second-Order Lowpass Flter

9 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

10 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 QR 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?

11 Example: Desgn a nd order lowpass Butterworth flter wth 3dB passband attenuaton, a dc gan of 5, and a 3dB bandedge of 4KHz QR 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 + Settng R=C=R 3 = obtan the followng normalzed crcut

12 Example: Desgn a nd order lowpass Butterworth flter wth 3dB passband attenuaton, a dc gan of 5, and a 3dB bandedge of 4KHz Q V IN V OLP INT INT A Popular Second-Order Lowpass Flter n T s = Must now set Q Q s +s + Now we can do frequency scalng ω 0n= C L C/ω o L/ω o C= /(π 4K) = 39.8uF

13 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

14 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

15 vew from Last Tme Flter Concepts and Termnology Frequency scalng Frequency Normalzaton pedance 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

16 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

17 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

18 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))

19 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

20 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 ao- aω + a4ω j aω - a3ω + a5ω 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

21 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

22 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 )?

23 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!

24 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)

25 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

26 Observaton: If z s a zero (pole) of H A (ω ), then z, z*, and z* are also zeros (poles) of H A (ω ) z -z* z -z z* z -z z z z z* Thus, roots come as quadruples f off of the axs and as pars f they lay on the axs

27 Observaton: If z s a zero (pole) of H A (ω ), then z, z*, and z* are also zeros (poles) of H A (ω ) z -z* z Proof: -z z* Consder an even polynomal n ω wth real coeffcents Pω At a root, ths polynomal satsfes the expresson Pω placng ω wth ω, we obtan m m m P ω a -ω a ω a ω n * * n call (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 Snce a s real for all I, t thus follows that m * aj ω 0 0 ω* s a root of Pω

28 Theorem: If H A (ω ) s a ratonal fracton wth real coeffcents wth no poles or zeros of odd multplcty on the real axs, then there exsts a real number H 0 such that the functon H0 s-jz s-jz... s-jzm TAM s s-jp s-jp... s-jp n s a mnmum phase ratonal fracton wth real coeffcents that satsfes the relatonshp AM A T jω H ω where {z, z, z m } are the upper half-plane zeros of H A (ω ) and exactly half of the real axs zeros, and where where {p, p, p n } are the upper half-plane poles of H A (ω ) and exactly half of the real axs poles. Example: Roots of H A (ω ) Roots that Appear n T AM (s)

29 H A ω m ω+z ω+z... ω+z ω-p ω-p... ω-p ω+p ω+p... ω+p H ω-z ω-z... ω-z n 0 m If nverse exsts n Example: T AM s 0... m s-jp s-jp... s-jp H s-jz s-jz s-jz n Roots of H A (ω ) Pole Doublets Roots that appear n T AM (s) Example: Inverse does not exst because zeros are of odd multplcty on the real axs

30 H A ω m ω+z ω+z... ω+z ω-p ω-p... ω-p ω+p ω+p... ω+p H ω-z ω-z... ω-z n 0 m If nverse exsts n T AM s 0... m s-jp s-jp... s-jp H s-jz s-jz s-jz n Roots of H A (ω ) Roots that appear n T AM (s) Rotate roots by 90 o Roots of T AM (s)

31 H A ω m ω+z ω+z... ω+z ω-p ω-p... ω-p ω+p ω+p... ω+p H ω-z ω-z... ω-z n 0 m If nverse exsts n Observatons: T AM s Coeffcents of T AM (s) are real 0... m s-jp s-jp... s-jp H s-jz s-jz s-jz n If x s a root of H A (ω ), then jx s a root of T AM (s) Multplyng a root by j s equvalent to rotatng t by 90 o cc n the complex plane Roots of T AM (s) are obtaned from roots of H A (ω ) by multplyng by j Roots of T AM (s) are upper half-plane roots and exactly half of real axs roots all rotated cc by 90 o If a root of H A (ω ) has odd multplcty on the real axs, the nverse mappng does not exst Other (often many) nverse mappngs exst but are not mnmum phase (These can be obtaned by reflectng any subset of the zeros or poles around the magnary axs nto the RHP)

32 H A ω m ω+z ω+z... ω+z ω-p ω-p... ω-p ω+p ω+p... ω+p H ω-z ω-z... ω-z n 0 m If nverse exsts n T AM s 0... m s-jp s-jp... s-jp H s-jz s-jz s-jz n o 90 mnmum phase o 90 non mnmum phase

33 All pass functons (and factors) o 90 Pole-zero cancellaton T A (s)= All pass T A (s) o 90 Must not allow cancellatons to take place n H A (ω ) to obtan all-pass T A (s) Must keep upper HP poles and lower HP zeros n H A (ω ) to obtan all-pass T A (s) All-pass T A (s) s not mnmum phase

34 End of Lecture 7