is the cut off frequency in rads.

Transcription

1 0 ELETRIAL IRUITS 9. HIGH RDER ATIVE FILTERS (With Tle) Introdution Thi development explin how to deign Butterworth (Mximlly Flt) or heyhev (Equl Ripple) Low P, High P or Bnd P tive filter. Thi tretment ue tle of polynomil for the type of funtion. All of the exmple tret thee two type nd only rief referene i mde of other type uh Beel (liner Phe). However, the deign proedure for the other type will e imilr. We trt with the Low P filter etion where peifi filter my hve mny ded tge, follow with High p nd lt the Bnd P. Deign of Low P Ative Filter Setion We hooe to mke the t order low p pive imply n R etion. Figure illutrte the form. Figure R low p Eqution give the trnfer funtion for Figure: V V UT IN R () The deign reltionhip for Figure i follow: R, where i the ut off frequeny in rd. The nd order etion i tive. The tndrd form hoen to implement i given y Figure.

2 Figure nd order tive implementtion A een in Figure the gin of the op mp onfigurtion i one. The other ontrint we implement i tht the reitor R, R re hoen equl. Under thee ondition the iruit gin expreion i given y Eqution. V V UT IN R R () Eqution i ompred gint the tndrd form of the nd order low p funtion given y Eqution 3: V V UT IN (3) In deigning vlue for Figure, & will e otined from the filter polynomil nd thu re ued to determine thoe vlue. Eqution 4 give the deign expreion for Figure : R : Deigner hoen R (4) R (4) Eqution 4 i otined y equting the oeffiient of like power of in the denomintor of Eqution nd 3. Any order filter n e otined y ding n even numer of Figure with Figure the finl tge if the order i odd numer. ne mut inure tht if Figure i ued tht it mut e the lt tge the input impedne of Figure will e loded y the output impedne of Figure. Additionlly, when the

3 order i odd the lt tge will e Figure nd thu the filter will hve non zero output impedne. When the order i even the output impedne will e from n op mp nd thu very low. Deign of High P Ative Filter Setion We hooe to mke the t order high p pive imply n R etion. Figure 3 illutrte the form. Figure 3 t order high p R The trnfer funtion for the firt order high p i given y Eqution 5. V V UT IN R R (5) The deign reltionhip for Figure 3 i follow: R, the ut off frequeny in rd. The nd order etion for high p i tive. The tndrd form hoen to implement i given y Figure 4.

4 3 Figure 4 nd order high p tive implementtion A een in Figure 4 the gin of the op mp onfigurtion i one. The other ontrint we implement i tht the pitor, re hoen equl. Under thee ondition the iruit gin expreion i given y Eqution 6. V V UT IN R R R R R (6) Eqution 6 i ompred gint the tndrd form of the nd order high p funtion given y Eqution 7: V V UT IN (7) In deigning vlue for Figure 4, & will e otined from the filter polynomil nd thu re ued to determine thoe vlue. Eqution 8 give the deign expreion for Figure 4: : Deigner hoen R (8) R R (8) Eqution 8 i otined y equting the oeffiient of like power of in the denomintor of Eqution 7 nd 6. Any order filter n e otined y ding n

5 4 even numer of Figure 4 with Figure 3 the finl tge if the order i odd numer. ne mut inure tht if Figure 3 i ued tht it mut e the lt tge the input impedne of Figure 4 will e loded y the output impedne of Figure 3. Additionlly, when the order i odd the lt tge will e Figure 3 nd thu the filter will hve non zero output impedne. When the order i even the output impedne will e from n op mp nd thu very low. Filter Polynomil The tle of polynomil for mximlly flt, Butterworth (Tle ) nd equl ripple, heyhev (Tle ) re provided. In oth tle only the 3d ripple ( ) i given. The Butterworth n e led to nother error y the following utitution proe: d Ripple d Ripple 0Log0 ( ), 0 0 (9) Ue Eqution 9 to onvert from d Ripple to error,. Upon otining the error, modify the 3dRipple polynomil to new ripple y plugging in the 3dRipple polynomil : N,where N i the order of the polynomil (0) Additionlly, Appix provide Mtl ode to to thi. The heyhev polynomil n not e led nd if nother mut e otined for the new dripple i deired new tle d Ripple. Appix Provide the Mtl ode to do tht. Tle give the Butterworth 3d polynomil for order N = to 0. D D.44 D3 ( )( (.5) ) D4 ( (.387) )( (.939) ) D ( )( (.3090) )( (.8090) 5 D6 ( (.588) )( (.707) )( (.9659) ) D ( )( (.5) )( (.635) )( (.900) 7 D8 ( (.95) )( (.5556) )( (.835) )( (.9808) ) D ( )( (.737) )( (.5000) )( (.7660) )( (.9397) 9 ) D0 ( (.564) )( (.4540) )( (.707) )( (.890) )( (.9877) ) Tle Butterworth 3d ripple polynomil Tle give the heyhev 3d polynomil for order N = to 0. ) )

6 5 D.004 D [(.35).777 D3 (.986)[(.493).9038 ] D [(.085).9465 ][(.056) 4 ].39 D5 (.775)[(.0548).9659 ][(.436).5970 ] D [(.038).9764 ][(.045).748 ][(.47) 6 ].66 D7 (.65)[(.08).987 ][(.0789).788 ][(.40).4373 ] D [(.06).9868 ][(.065).8365 ][(.090).5590 ][(.085) 8 D9 (.0983)[(.07).9896 ][(.049).870 ][( D 0 [(.038).995 ][(.040) Tle heyhev 3d ripple polynomil Now given tht the proper dripple the deign proe. Auming tht the exmple here in. e Butterworth Low P.8945 ][(.065).0753) ].6459 ] [(.094).7099 ][(.0788) [(.0873) polynomil tle re ville we n proeed with dripple will e 3d, Tle & will e ued for ll The Butterworth polynomil re ll ded nd order for n even or firt order with ded nd order for n odd. Eqution give the normlized firt order etion nd Eqution give the form of the typil normlized nd order etion ] ] ].4558 ] () () To deign Butterworth filter the order n, the dripple nd the utoff frequeny in rd re either peified or eleted. If the dripple i not 3d then the dt from Tle mut e onverted uing Eqution 9. For ny order n, the vriou etion of the polynomil for tht order will onit of Eqution or. Thee Eqution re onverted into polynomil tht re relized Figure or y the following utitution: (3) Thu Eqution nd eome:

7 6 (4) (5) The low p form tht Eqution 4 nd 5 led to re given y Eqution 6 nd 7: (6) (7) Thu for ny order n, the vriou etion in the Butterworth polynomil generte peifi Eqution 7 nd if n i odd ingle etion of Eqution 6. onider the following imple exmple. Exmple Given: n =3, 3d, Solution: From Tle the polynomil i: =000rd nd Butterworth low p D3 ( )( (.5) ) (8) Thi will require one etion of Eqution 6 nd one etion of Eqution 7. We plug in 000rd into Eqution 6 nd 000rd long with the =0.5 otined from Eqution 8 into Eqution (9)

8 7 000 (.5) 000 (0) We pik R in Eqution where =000rd. Let R =0k, then = 0.f. Thi pplie to the firt order of Figure. The eond order etion of Eqution 0 i implemented with Figure. The ytem funtion for Figure i given y Eqution. The deign proe elet R, nd uh tht Eqution eome Eqution 0. We pik R =0k, then uing Eqution 4 nd 4 we otin = 0.05f nd = 0.f. Thi pplie to the nd order of Figure. Figure 5 illutrte the omplete hemti. erve how the t order etion i ded on the output of the mplifier of the nd order etion eue of impedne loding effet. Note the omponent nme, (,, 3, R, R 3 ) in Figure 5, 6 re for SPIE nd my not follow the numering given y Eqution, 4, 4. Figure 5 3 rd rder Butterworth Low P Filter A PSPIE nlyi w performed uing voltge ontrolled oure with gin of for the op mp. Figure 6 illutrte the nlyi hemti.

9 8 Figure 6 PSPIE nlyi hemti of Figure 5 Figure 7 illutrte the frequeny repone where mgnitude i plotted gint frequeny. The 3d point our t 59hz whih i 000rd. Figure 7 Mgnitude plot for Exmple

10 9 e heyhev Low P The heyhev polynomil re relized with Figure nd in very imilr mnner the Butterworth. An importnt differene i tht when the order n i even the over ll D gin mut e djuted to e n ttenution of the d. Thu if the 3d heyhev polynomil of Tle re ued with n even voltge divider mut e ued to djut the to gin to e 3d. A very imple pproh to ommodte thi requirement i to turn the firt reitor of Figure into voltge divider with the deired ttenution nd the prllel omintion equl to the firt reitor. Ripple Figure 8 voltge divider input to nd order tive etion The deign reltion for R nd R for Figure 8 re, given given y Eqution. dripple nd reitor R, A 0 R R R R d Ripple 0 R, R R R A R R RA RA A () In Tle d Ripple=3d therefore A. The other pet of the heyhev t nd nd order etion tht i different i tht they re not in the ommodting form of Eqution nd. The t order form nd nd order form for the heyhev polynomil re given y Eqution nd 3 repetively. ( ) () (3)

11 30 Eqution, the heyhev t order form i onverted to the low p t order y Eqution 4: (4) erve how the over ll peified for the t order etion eome. The nd order heyhev form, Eqution 3 i onverted into the low p y Eqution 5: ) ( ) ( (5) erve how the over ll peified for the heyhev nd order etion eome.nd whih w jut for the Butterworth i for the heyhev. Thu the t order heyhev etion for Figure i implemented vi Eqution eome: R (6) The nd order etion, of Figure, deign Eqution 4 nd 4 for heyhev eome: R R (7) ) ( R (7) For ny heyhev order n the t order etion, if n w odd, of the polynomil i implemented Figure with Eqution 4 nd 6. Eh of the nd order etion will e

12 3 implemented Figure with Eqution 5, 7 & 7. Additionlly, if n were even one of the nd order etion mut e of the form of Figure 8 to ommodte the d ttenution. Let do Exmple with the me pe of Exmple ut heyhev. Exmple Given: n =3, 3d, Solution: From Tle the polynomil i: =000rd nd heyhev low p Ripple D3 (.986)[(.493).9038 ] (8) Thi will require one etion of Eqution 4 nd one etion of Eqution 5. We plug in 000rd nd the ontnt =.986 into Eqution 4 nd 000rd long with the ontnt =.493 nd =.9038 otined from Eqution 8 into Eqution 5. Thi yield the trnfer funtion of the individul etion. Thee me ontnt long with 000rd nd R =0k re loded into Eqution 6, 7 & 7. Thi will omplete the deign for one etion of Figure nd one etion of Figure. Figure 9 illutrte the omplete hemti. Note the omponent nme, (,, 3, R, R, R 3 ) in Figure 9, 0 re for SPIE nd my not follow the numering given y Eqution, 4, 4. Figure 9 Shemti of Exmple Figure 0 give the PSPIE hemti for the nlyi tht w performed

13 3 Figure 0 PSPIE iruit for Exmple The plot of the mgnitude repone i given y Figure. erve tht the mgnitude i t out.7 t 59hz or 000rd. Figure mgnitude plot for Exmple

14 33 e 3 Butterworth High P Beue of the form of the Butterworth polynomil etion given y Eqution nd the high p eily trnform out of the low p of Eqution 6 nd 7 to eome Eqution 9 nd 30. (9) (30) Thu for ny order n, the vriou etion in the Butterworth polynomil generte peifi Eqution 30 nd if n i odd ingle etion of Eqution 9. onider the following imple deign exmple for 3 rd order Butterworth high p. Exmple 3 Given: n =3, 3d, =000rd nd Butterworth high p Solution: From Tle the polynomil i: ) (.5) )( ( 3 D (3) Thi will require one etion of Eqution 9 nd one etion of Eqution 30. We plug in 000rd into Eqution 9 nd 000rd long with the =0.5 otined from Eqution 3 into Eqution (3)

15 (.5) 000 (33) We pik R in Eqution 5 where =000rd. Let R =0k, then = 0.f. The firt order i Figure 3. The eond order etion of Eqution 33 i implemented with Figure 4. The ytem funtion for Figure 4 i given y Eqution 6. The deign proe elet, R nd R uh tht Eqution 6 eome Eqution 33. We pik.f, then uing =0.5 in Eqution 8 nd 8 we otin R =0k nd R =5k for the nd order etion of Figure 4. Figure illutrte the omplete hemti. A with the low p, the t order etion i ded on the output of the mplifier of the nd order etion eue of impedne loding effet. Figure Butterworth high p for Exmple 3 Figure 3 give the PSPIE equivlent iruit tht w nlyzed. Agin the idel mplifier form w ued. The plot of the reult i given y Figure 4. Note the omponent nme, (,, 3, R, R, R 3 ) in Figure 3 re for SPIE nd my not follow the numering given y Eqution 5, 8, 8.

16 35 Figure 3 PSPIE iruit for Exmple 3 Figure 4 PSPIE frequeny repone for Exmple 3

17 36 e 4 heyhev High P Trnforming ny t order etion of the heyhev polynomil into the tndrd form for t order, Eqution 9 or ll of the nd order etion into the form of Eqution 30 i more omplited thn with the Butterworth. Eqution nd 3 re the form of heyhev t d nd order polynomil etion repetively. onider the t order etion firt: The gin mut e unity thu the finl reult: ( ) (34) Eqution 34 n e rerrnged to look like Eqution 9: (35) A een in Eqution 35 the effetive ut-off frequeny for the t order high p i. Thi n eily e implemented Figure 3 nd Eqution 5 where: R (36) The nd order etion follow imilr pproh. Strting with Eqution 3: ) ( Now plug in : nd normlize the gin reult in Eqution 37:

18 37 (37) From Eqution 37 oerve tht the tht the pe rdin frequeny: (38) Additionlly, n e extrted from Eqution 37 denomintor middle term: (39) From Eqution 39: (40) Uing Eqution 38 nd 40 plugged into Eqution 8 & 8, we n now otin deign reltionhip for the prmeter of Figure 4 for ny nd order etion in the heyhev polynomil: R R (4) R R R (4)

19 38 Where: R i reitor in Figure 4 high p tive tge R i reitor in Figure 4 high p tive tge i the mthed pitor, in Figure 4 i the polynomil oeffiient from Eqution 3 i the other polynomil oeffiient from Eqution 3 ne dditionl point need to e overed. Tht i when the order n i even the over ll D gin mut e djuted to e n ttenution of the d. Thu if the 3d heyhev polynomil of Tle re ued with n even voltge divider mut e ued to djut the to gin to e 3d. A very imple pproh to ommodte thi requirement i to turn the firt pitor of Figure into voltge divider with the deired ttenution nd the prllel omintion equl to the firt pitor. Figure 6 illutrte the nd order etion with n ttenution of the d uing the firt pitor. The following reltionhip implement thi onept: Ripple Ripple A 0 d Ripple 0, A ( A) A (43) Figure 6 heyhev nd order with n ttenution of the onider Exmple 4 high p heyhev 3 rd order t 000rd: dripple

20 39 Exmple 4 Given: n =3, 3d, =000rd nd heyhev high p Solution: From Tle the polynomil i: D3 (.986)[(.493).9038 ] (44) Thi will require one etion of Figure 3 nd one etion of Figure 4. We hooe ll the pitor to e 0. f. We plug in 000rd, 0.f nd =.986 into Eqution 36 nd ompute R. 986k for the t order etion of Figure 3. We plug 000rd, the eleted pitne 0.f long with the =0.493 into Eqution 4to ompute R. 493k in the nd order etion of Figure 4. We plug 000rd, the eleted pitne 0.f, =0.493 nd =.9038 into Eqution 4 to ompute R k (not the me R!) lo in Figure 4 the nd order etion. Figure 7 illutrte the omplete hemti. Figure 7 hemti of Exmple 4 heyhev high p Figure 8 illutrte the PSPIE equivlent iruit:

21 40 The PSPIE frequeny repone for Exmple 4 i given in Figure 9 Note the omponent nme, (,, 3, R, R, R 3 ) in Figure 8 re for SPIE nd my not follow the numering given y Eqution 36, 4, 4. Figure 9 Exmple 4 frequeny repone

22 4 Appix Mtl ode For Generlized Butterworth Polynomil %%%%% Generl Butterworth %%%%%%%%%%%% %%%% Pik order,n %%%%% %%%% Pik -d error AKA rho %%%%%%%%%%% %%%% It ompute formt poly BJ doe for he Tle %%%% %%%% It ompute Ho, to mke D gin %%%%%%% %%%% It hek the d' of rho %%%% N=3; %%% Filter order j=; phi=-90; d=-;%%%% d error rho=0^(d/0); ep=(rho^--)^.5; %%% get BW pole while phi<360 phi=-90+(80+(j-)*360)/(*n); phe(j)=phi; j=j+; M=j-; for i=:m Rel(i)=ep^-(/N)*o(phe(i)*pi/80); Img(i)=ep^-(/N)*in(phe(i)*pi/80); %%% Get right hlf plne pole j=; for i=:m if Rel(i)<0 RRel(j)=Rel(i); IImg(j)=Img(i); j=j+; R=RRel+j*IImg; MM=length(R); %%%%%% ompute Ho %%%%%% Ho=; for i=:mm Ho=Ho*(-R(i)); Ho=rel(Ho); %%%%%%%%% d hek %%%%%%%%%%%%%

23 4 P=; for i=:mm P=P*(j-R(i)); HM=Ho/P; HMq=(HM)^; dhk=0*log0(hmq); %%%%%%%%%%%%%%%%%%%%%%%%% if eil(n/)>n/ S(,)=(R(eil(N/))); (,)=0; for i=:eil(n/)- S(i+,)=-rel(R(i)); S(i+,)= (img(r(i))); if eil(n/)==n/ for i=:eil(n/) S(i,)=-rel(R(i)); S(i,)=(img(R(i))); %%%%%%%%%%%%%% utput %%%%%%%%% dhk %% A hek ompute peified d S %%%% output polynomil oeffiient in he formt Ho %%% Numertor term to fore D repone to unity

24 43 Appix Mtl ode For Generlized heyhev Polynomil %%%%% BJ Generl heyhev %%%%%%%%%%%% %%%% Pik order,n %%%%% %%%% Pik -d AKA rho %%%%%%%%%%% %%%% It ompute formt poly BJ doe for he Tle %%%% %%%% It ompute Ho, to mke D gin for N odd or down y (/(+ep^))^.5 for N even %%%%%%% %%%% It hek the d' of rho %%%% N=3; %% rder of filter d=-3; % d error ro=0^(d/0); ep=(ro^--)^.5; gm=((+(+ep^)^.5)/ep)^(/n); %%%% Get Pole for k=:n Rel=-(gm-gm^-)*.5*in((*k-)*pi/(*N)); Img= (gm+gm^-)*.5*o((*k-)*pi/(*n)); R(k)=Rel+j*Img; P=; for k=:n P=P*(-R(k)); P=(P); %%%% Set gin for N odd if eil(n/)>n/ Ho=P; S(,)=(R(eil(N/))); (,)=0; for i=:eil(n/)- S(i+,)=-rel(R(i)); S(i+,)= (img(r(i))); %%%% Set gin for N even if eil(n/)==n/ Ho=(/(+ep^))^.5*P; for i=:eil(n/) S(i,)=-rel(R(i)); S(i,)=(img(R(i))); %%%%% d hek %%%%%%% P=; MM=N;

25 44 for i=:mm P=P*(j-R(i)); HM=Ho/P; HMq=(HM)^; dhk=0*log0(hmq); %%%%%%%%%%%%% utput %%%%%%%%%%%%%%%% dhk S Ho