2. SYNTHESIS OF ANALOG FILTERS

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1 ,. SYNTHESIS OF ANALO FILTERS 9. a) H BW ( j ) = where ε =.A max ε N A max = log Hj ( ) = ( ) ε log( ) A min = log Hj ( ) = ( ) ε log N N.A min ε = N.A max log.a min A max log H BW j ( ) = = H ( )H( ) ε = j = ε D ( )D( ) ε ε N N = -- N = ε j -- N j -- = e j j ( k )π N k = j ε -- N j = = e j ( k )π N e j k ( )π N. Amax = log ( ε) = log ( ρ ). For example, ρ = 5% <=> Amax =.8796 db --- N. A ( ) = log ε N we get r p = ε --- Ar ( p ) ε r - p N N N log ε ε = = log = ε --- N N = log ε = log ( ε ε ) = log.db.4 Hj ( ) = = ε, ε N N Hj ( ) j N = ( ) = ε F whih we reognize a a Taylor erie. Fx ( ) = F ( ) x Fn were x = F ( ) x!. Hene, the firt N derivative are zero for x =. ( )! ( ) x n n!.4 Let h(t) (t) be the impule tep repone, repetively, of a filter H(). Sale the frequeny with a fator k aording to = k. The impule repone of the frequeny aled filter i h' t () π = Hjk ( )e jt d = Hjk ( )e jk ( t k) = π -- d( k ) k --h t k k - Thu, both the time axi the original impule repone i divided by k. ' () t h' () t d t h t = = k - -- d t = h t k k - t - d k = t k - Thu, the time axi of the original tep repone i divided by k. dt n n x ao( ) in( ).6 We get = whih i of the form / at x =. However, the other extreme dx x value our for x <. We have in(y) = => y = k π for k = integer the extreme value (maxima minima) are obtained from n ao(x) = y = k π. Finally, we get x = o(π k/n). For example, a filter of order N = 5 we get the extreme value at x =, x = o(π /5) =.89699, x = o(π /5) =.9699, x = o(π /5) =.9699, x 4 = o(π 4/5) =.89699, x 5 =. The extreme value at x = ± are neither a maxima nor a minima. r p ε / N.7 The pole are aording to Eq. (.8) where = ε =.A max. We get ε = r p = krad/ the normalized pole: p =.5 j p = p = p = p =.5 j p = either from the Table or by uing [, Z, P] = BW_POLES(W, W, Amax, Amin, N) We denormalize the pole by multiplying with r p whih yield p = j krad/ p = r p p = krad/ p = r p p = j krad/ p = r p H ( ) = = We elet o that the filter get ( p )( p )( p ) ( r p ) ( r p r p ) the proper gain, for example, ain = at =. Hene, = r p N. A( ) = =.86 db A(4 ) = = 9.74 db log ε ( ) 6 log ε ( 4 ) 6.8 ε =.A max = , x = /ε = 6.557, ainh(x) = ln( x x ) = ln( x x = , a = ) inh = , N ln( x x b = ) = oh N Normalized pole aording to Equation (.) are p = i p = p = i denormalize by multiplying with. We get p = j krad/ p = krad/ p = j krad/ Alternatively we get the pole from the Table or by uing [, Z, P] = CH_I_POLES(W, W, Amax, Amin, N) H ( ) = We elet o that the filter get the proper gain, for example, ( p )( p )( p ) ain = at =. Hene, = We an either ue Eq. (.) or Eq (.7), but the former i impler to olve. We have at =

2 / = A Chebyhev polynomial an, ε = T ( ) ε Hj ( ) = ε T H ( ) aording to tard mathematial hbook, be omputed reurively: T (x) =, T (x) = x, T n (x) = xt n (x) T n (x) for n =,, We get T (x) = xt (x) T (x) = x T (x) = xt (x) T (x) = x(x ) x = 4x x, T () = 4 =, T () = 4 6 = 6 T (4) = 4 4 = 44. We get Hj ( ) = = => A( ) = =.9 db ε T ( ) ε 6 log ε 6 A(4 ) = log = db ε ( 44 ).A min.9 We need to determine W. We have x = = Let y = W/W with.a max ln( x x ) = ln( y y ) N = we have N Iterating we get y = 5.58 W = 7.59 Mrad/. The following ript yield Amax =.; Amin = 4, W = 5*^6; W = 5.58*W; N = CH_ORDER(W, w, Amax, Amin) N = We elet N = ; [, Z, P] = CH_II_POLES(W, W, Amax, Amin, N) yield = e5 Z =.e7 * P =.e6 * i i i i with W = *W; Att = PZ ATT_S(, Z, P, W) we get Att = db with W = 4*W; we get Att =.9489 db. Trying with different W we get Amax =.; Amin = 4; W = 5*^6; W =.5*W N = CA_ORDER(W, W, Amax, Amin) W = 76 N = N = ; [, Z, P] = CA_POLES(W, W, Amax, Amin, N yield = e5 Z =.e7 * P =.e6 * i i i i W = *W A = PZ ATT_S(, Z, P, W) W = A = 4.87 db W = 4*W A = PZ ATT_S(, Z, P, W) W = A = db. a) N = => N = 5. We get ε = r p = Mrad/ either from the Table or by uing [, Z, P] = BW_POLES(W, W, Amax, Amin, N) p = j Mrad/ p = j.877 Mrad/ p =.7958 Mrad/ p4 = j.87 Mrad/ p5 = j Mrad/ log.a min A max Solving for A min : N = for A min => A min < 9. db log. A max =. db yield ε =.A max = r = π 6 p = ε N rad/. The pole are pk r π ( N k ) p π ( N k ) = o j in we get uing the following program N N N = 5; Amax =.; epilon = qrt(^(.*amax)-); epilon = ; % We elet epilon = W = *pi*^6; rp = W*epilon^(-/N); = 5; for k = :5 p = rp*(o(pi*(n*k-)/(*n)) i*in(pi*(n*k-)/(*n))) = *p; end p = e e6i p = e e6i p = e6 p4 = e e6i p5 = e e6i = e4. a) W = e; W = e; Amax =.; Amin = 5; % Speifiation W = linpae(, ^5, ); % Frequeny vetor, -axi t = linpae(, e-, ); % Time vetor %Butterworth % Determine filter order Nbutt = buttord(w, W, Amax, Amin, ''); % (MATLABS Signal Proeing Toolbox) Zbutt, Pbutt, butt] = buttap(nbutt); % Find pole zero epilon = qrt(^(.*amax) ); % Denormalize the pole zero with rp rp = W*epilon^( /Nbutt); Zbutt = Zbutt*rp; Pbutt = Pbutt*rp; butt = butt*(rp^nbutt); [Numbutt, Denbutt] = zptf(zbutt, Pbutt, butt); % Compute the tranfer funtion Hbutt = freq(numbutt,denbutt,w); % Compute the frequeny repone tgbutt = groupdelay(zbutt, Pbutt, butt, W, W); % Compute the group delay y = zpk(zbutt, Pbutt, butt); hbutt = impule(y,t); % Compute the impule repone butt = tep(y,t); % Compute the tep repone

3 e) We have aled the impule repone with a fator 5 x 4 in order to fit it into the ame plot a (t). -.8 The emi-irle ha the ame radii a the b edge -.6 x [rad/] τ g () [] x -4 x 5 5 zero at 5 zero at ) d) x [rad/] A() [db] figure(4); ubplot('poition', [ ]); PLOT_IMPULSE_RESPONSE_S(h*^-4, dira, t_axi) hold on PLOT_STEP_RESPONSE_S(_of_t, t_axi); [h, dira, t_axi] = PZ IMPULSE_RESPONSE_S(, Z, P, t); [_of_t, t_axi] = PZ STEP_RESPONSE_S(, Z, P, t); figure(); ymax = ^5; xmin = -4*^4; xmax = ^4; PLOT_PZ_S(Z, P, W, W, xmin, xmax, ymax) Tg = PZ T_S(, Z, P, W); figure(); ubplot('poition', [ ]); PLOT_T_S(W, Tg) or better ue our toolbox W = e; W = e; Amax =.; Amin = 5; % Speifiation W = linpae(, ^5, ); % Frequeny vetor w-axi t = linpae(, e-, ); % Time vetor N = BW_ORDER(W, W, Amax, Amin) % Selet an integer degree re-run the program N = 5; [, Z, P] = BW_POLES(W, W, Amax, Amin, N); Att = PZ ATT_S(, Z, P, W); figure(); ubplot('poition', [ ]); PLOT_ATTENUATION_S(W, Att) N = CH_ORDER(W, W, Amax, Amin) % Selet an integer degree N =... [, Z, P] = CH_II_POLES(W, W, Amax, Amin, N); [h, dira, t_axi] = PZ IMPULSE_RESPONSE_S(, Z, P, t); [_of_t, t_axi] = PZ STEP_RESPONSE_S(, Z, P, t); PLOT_IMPULSE_RESPONSE_S(h, dira, t_axi) PLOT_STEP_RESPONSE_S(_of_t, t_axi); or better ue our toolbox.5 a)% Chebyhev II filter Nheb = hebord(w, W, Amax, Amin, ''); % Determine filter order [Zheb, Pheb, heb] = hebap(nheb,amin); % Find pole zero Zheb = Zheb*W; % Denormalize the pole zero Pheb = Pheb*W; heb = heb*(w^(length(phe length(zhe)); [Numheb, Denheb] = zptf(zheb, Pheb, he; % Compute the tranfer funtion Hheb = freq(numheb,denheb,w); % Compute the frequeny repone tgheb = groupdelay(zheb, Pheb, heb, W, W); %Compute the group delay y = zpk(zheb, Pheb, he; hheb = impule(y,t); % Compute the impule repone heb = tep(y,t); % Compute the tep repone ) d) e) N = CH_ORDER(W, W, Amax, Amin) % Selet an integer degree N =... [, Z, P] = CH_I_POLES(W, W, Amax, Amin, N); [h, dira, t_axi] = PZ IMPULSE_RESPONSE_S(, Z, P, t); [_of_t, t_axi] = PZ STEP_RESPONSE_S(, Z, P, t); PLOT_IMPULSE_RESPONSE_S(h, dira, t_axi) PLOT_STEP_RESPONSE_S(_of_t, t_axi); or better ue our toolbox.4 a)% Chebyhev I filter Nheb = hebord(w, W, Amax, Amin, ''); % Determine filter order [Zheb, Pheb, heb] = hebap(nheb,amax); % Find pole zero Zheb = Zheb*W; % Denormalize the pole zero Pheb = Pheb*W; heb = heb*(w^nhe; [Numheb, Denheb] = zptf(zheb, Pheb, he; % Compute the tranfer funtion Hheb = freq(numheb,denheb,w); % Compute the frequeny repone tgheb = groupdelay(zheb, Pheb, heb, W, W); % Compute the group delay y = zpk(zheb, Pheb, he; % Compute the impule repone hheb = impule(y,t); heb = tep(y,t); % Compute the tep repone t [] x -..4 (t)

4 4 5 ) d) e) 5.6 a) % Cauer filter Na = ellipord(w, W, Amax, Amin, ''); % Determine filter orde [Za, Pa, a] = ellipap(na, Amax, Amin); % Find pole zero Za = Za*W; % Denormalize the pole zero Pa = Pa*W; a = a*(w^(length(pa) length(za))); [Numa, Dena] = zptf(za, Pa, a); % Compute the tranfer funtion Ha = freq(numau,denau,w); % Compute the frequeny repone tga = groupdelay(za, Pa, a, W, W); % Compute the group delay y = zpk(za, Pa, a); ha = impule(y, t); % Compute the impule repone a = tep(y, t); % Compute the tep repone or better ue our toolbox N = CA_ORDER(W, W, Amax, Amin) % Selet an integer degree N =... [, Z, P] = CA_POLES(W, W, Amax, Amin, N); [h, dira, t_axi] = PZ IMPULSE_RESPONSE_S(, Z, P, t); [_of_t, t_axi] = PZ STEP_RESPONSE_S(, Z, P, t); PLOT_IMPULSE_RESPONSE_S(h, dira, t_axi) PLOT_STEP_RESPONSE_S(_of_t, t_axi); ) d) e).7 %Plotting figure(); ubplot(4,, ) plot(w, *log(ab(hbutt))) axi([ e5 6 ]); ubplot(4,, ) plot(w, *log(ab(hhe)) axi([ e5 6 ]); ubplot(4,, ) plot(w,*log(ab(hhe)) axi([ e5 6 ]); ubplot(4,, 4) plot(w, *log(ab(ha))) axi([ e5 6 ]) x x x x 4 6 figure(); ubplot(4,, ) plot(w, *log(ab(hbutt))) axi([ e4.5.]); ubplot(4,, ) plot(w, *log(ab(hhe)) axi([ e4.5.]); ubplot(4,, ) plot(w, *log(ab(hhe)) axi([ e4.5.]); ubplot(4,, 4) plot(w, *log(ab(ha))) axi([ e4.5.]) figure() ubplot(,,) zplane(zbutt, Pbutt) ubplot(,, ) zplane(zheb, Phe ubplot(,, ) zplane(zheb, Phe ubplot(,, 4) zplane(za, Pa) figure(4) ubplot(4,, ) plot(w, tgbutt) ubplot(4,, ) plot(w, tghe ubplot(4,, ) plot(w, tghe ubplot(4,, 4) plot(w, tga) x x - x x -4 x x -4 x 4 4 x Real part x 5 Real part x Imaginary part Imaginary part - x 4 x Real part x Real part x Imaginary part Imaginary part.5 x 4 x

5 7 figure(5) ubplot(4,, ) plot(t, hbutt) ubplot(4,, ) plot(t, hhe ubplot(4,, ) plot(t, hhe ubplot(4,, 4) plot(t, ha) x x x x - figure(6) ubplot(4,, ) plot(t, butt) ubplot(4,, ) plot(t, he ubplot(4,, ) plot(t, he ubplot(4,, 4) plot(t, a) x x x x -.8 = e5 when H max = hene = e5 = e6 The differene in order of the numerator denominator i N, hene, the rate of attenuation inreae i 6N db per otave (= doubling of the frequeny)..9 a) N =.686, elet N = S p, =.9587 ± j S p, = ± j S p, = ± j S p4, = ± j S p5, 9 = ± j S p6, 8 = ± j S p7 = p, = ± j p, = ± j p, = ± j p4, = ± j p5, 9 = ± j p6, 8 = ± j p7 = N = 7.5, N = 8 p,8 = ± j p,7 = ± j p,6 = ± j p4, 5 = ± j ) N = 7.5, N = 8 z,8 = ±j p,8 = ± j z,7 = ±j p,7 = ± j z,6 = ±j p,6 = ± j z4,5 = ±j p4,5 = ± j d) N = 5.659, N = 6 whih yield A min 76 db z,6 = ± j p,6 = ± j z,5 = ± j p,5 = ± j z,4 = ± j p,4 = ± j a) Amax = ; Amin = 5; W = ; W = ; NBW = BW_ORDER(W, W, Amax, Amin) NCH = CH_ORDER(W, W, Amax, Amin) NCA = CA_ORDER(W, W, Amax, Amin) yield NBW = NCH = NCA = The following program yield Amax = ; Amin = 5; W = ; W =.; NBW = ; while NBW > 4 NBW = BW_ORDER(W, W, Amax, Amin); W = W*.; end % Butterworth W W = W =.; NCH = ; while NCH > 4 NCH = CH_ORDER(W, W, Amax, Amin); W = W*.; end % Chebyhev I II W W = W =.; NCA = ; while NCA > 4 NCA = CA_ORDER(W, W, Amax, Amin); W = W*.; end % Cauer W W =.99765

6 9.We get uing the program below Butterworth Chebyhev I W = ^7; W = 5*^6; W = ^7; W = 5*^6; Amax = *log(.5^); Amin = 6; N = BW_ORDER(W, W, Amax, Amin) N = ; % Re-run the program after % eleting an integer order [, Z, P] = BW_POLES(W, W, Amax, Amin, N) xmax = *^6; xmin = *^7; ymax = 4*^7; PLOT_PZ_S(Z, P, W, W, xmin, xmax, ymax) N = = e7 P =.e7 * i i i i i i i i i i All zero at = Amax = *log(.5^); Amin = 6; N = CH_ORDER(W, W, Amax, Amin) N = ; % Re-run the program after % eleting an integer order [, Z, P] = CH_I_POLES(W, W, Amax, Amin, N) xmax = *^6; xmin = *^7; ymax = 4*^7; PLOT_PZ_S(Z, P, W, W, xmin, xmax, ymax) N = = e48 P =.e7 * i i i i i i All 7 zero at = Chebyhev II Cauer W = ^7; W = 5*^6; W = ^7; W = 5*^6; Amax = *log(.5^); Amin = 6; N = CH_ORDER(W, W, Amax, Amin) N = 7; % Re-run the program after % eleting an integer order [, Z, P] = CH_II_POLES(W, W, Amax, Amin, N) xmax = *^6; xmin = *^7; ymax = 4*^7; PLOT_PZ_S(Z, P, W, W, xmin, xmax, ymax) N = = e4 Z =.e7 * i i i i i i P =.e7 * i i i i i i Amax = *log(.5^); Amin = 6; N = CA_ORDER(W, W, Amax, Amin) N = 5; % Re-run the program after % eleting an integer order [, Z, P] = CA_POLES(W, W, Amax, Amin, N) xmax = *^6; xmin = *^7; ymax = 4*^7; PLOT_PZ_S(Z, P, W, W, xmin, xmax, ymax) N = = e4 Z =.e7 * i i i i P =.e7 * i i i i Note that zero at = ar not printed. The pole zero are hown below with Butterworth, Chebyhev I, Chebyhev II, Cauer filter left to the right. 4 x 7 zero at 4 x 7 7 zero at 4 x 7 zero at 4 x 7 zero at x x 7 x 7 x 7.. We have N = log ( τ ) log ( τ ) log( τ ) whih an be τ written N τ = log whih obviouly ( τ )( = log τ ) τ τ jτ = log jτ jτ H ( ) ( τ ) equal N => = where the real zero i at ( τ )( τ ) = τ the two real pole are at = τ = τ, the gain ontant i = Note that if the pole zero alternate on the real axi, a in thi ae, the orreponding tranfer funtion an be realized by uing a network oniting of only reitor apaitor Tranformation of the HP peifiation to the orreponding LP peifiation uing S = I / Ω = I / where we neglet the negative ign ine the peifiation of the magnitude funtion i ymmetri around =. We get A max = db, Ω = I /7 6 A min = 5 db, Ω = I / 6 We elet I = 7 Mrad/ to get a normalized LP peifiation in order to allow the ue of tard table whih are normalized to Ω = Ω = 7/ =.5. Neeary order aording the Eq.(.7) i

7 .A min log A.5 log max N = =.86. We elet here N =. The pole are obtained Ω log log(.5 ) Ω π ( k ) from S k R p in for k =,..N N π ( k ) = j o N R p ε / N where = = ε =.A max = all zero are at S =. We get S p = j S p = S p = j S p4 = j S p5 = S p6 = j We elet only the pole in the left-h half of the -plane S p = j S p = S p = j We map the LP pole to the orreponding HP pole uing = I = S a jb I ( ) where S = a jb yield a b p = j Mrad/ p = Mrad/ p = j Mrad/ Mapping the LP zero to the orreponding HP zero uing z = I /S z yield z =, n =,, ine there are zero at S = in the LP filter. The tranfer funtion of the LP i H LP S ( ) = = ( S σ ) S ( σ S r ) ( S.5576 )( S.5576 S. ) σ r Inerting = I /S give 9.97 where = H HP ( ) I σ I σ I r = = ( I σ )( I σ I r ) = ( )( ).8.9 The zero are z, = ±5 rad/, z,4 = ± ±j rad/, whih i OK, but the pole are: p,,,4 = ±±j p5 = 5 rad/. Hene, there are three pole in the right-h half of the -plane, whih make the filter untable. Furthermore, there are ix zero only five pole. Hene the filter i non-aual. The filter ha omplex oeffiient an therefore not be realized uing tard method... We elet I = = 5.5 Mrad/ get Ω = Ω = I / = Ω /Ω = The required order for the Cauer filter i N = 4.99 we elet N = 5. We modify the. W = e; W = 6e; Amax =.; Amin = 5; % HP filter! Tranform O = W^/W; O = W^/W; Nheb = hebord(o, O, Amax, Amin, ) %Order Chebyhev I Nauer = ellipord(o, O, Amax, Amin, ) %Order Cauer % Pole zero Chebyhev I [ZhebLP, PhebLP, heblp]= hebap(nheb, Amax); [rad/] x 6 A() [db] The attenuation for the HP filter i hown below i i i i i i i i Zhp =.e6 * Php =.e6 * The pole zero of the HP filter are hp = i i i i i i i i ZLP =.e7 * PLP =.e6 * program in Example.7 a hown below. W = 5.5*^6; W =.5*^6; % Requirement for the highpa filter Amax = -*log(-.5^); Amin = 45; WI = W; % We elet the tranformation angular frequeny WI = W Omega = W; % Omega = WI^/W = W Omega = WI^/W; NLP = CA_ORDER(Omega, Omega, Amax, Amin) % Synthei of lowpa filter (Cauer) NLP = 5; [LP, ZLP, PLP] = CA_POLES(Omega, Omega, Amax, Amin, NLP) [hp, Zhp, Php] = PZ HP_S(LP, ZLP, PLP, WI^) % Tranform LP to HP filter N = NLP; % LP HP filter ha the ame order ubplot('poition', [ ]); W = linpae(, ^7, ); H = PZ FREQ_S(hp, Zhp, Php, W); Att = MA ATT(H); PLOT_ATTENUATION_S(W, Att) axi([ ^7 6]); The normalized pole zero of the LP filter are LP = e5

8 ZhebLP = O*ZhebLP % Denormalize Chebyhev I PhebLP = O*PhebLP [Zheb, Pheb] = zphp(zheblp, PhebLP, w^) % Tranform to HP Chebyhev I %Pole zero Cauer [ZauerLP, PauerLP, auerlp]= ellipap(nauer, Amax, Amin); ZauerLP = O*ZauerLP % Denormalize Cauer PauerLP = O*PauerLP [Zauer, Pauer] = zphp(zauerlp, PauerLP, w^) % Tranform to HP Cauer %Plot pole zero figure; ubplot(,,) hold on plot(zheb, o ) plot(pheb, x ) axi image hold off title( Chebyhev I ) ubplot(,,) hold on plot(zauer, o ) plot(pauer, x ) axi image hold off title( Cauer ) % Determine the tranfer funtion Chebyhev I [NumCheb, DenCheb] = zptf(zheb, Pheb, ) % POOR ACCURACY RUTINES % Determine the tranfer funtion Cauer [NumCauer, DenCauer] = zptf(zauer, Pauer,) Rewrite the program above by intead uing the toolbox!. We have for a Butterworth filter H LP ( j ) = The orreponding HP filter i ε N obtained by the tranformation S = I, i.e., Ω I by eleting = we obtain = I H HP ( j ) = The two filter have the ame gain at the roover frequeny,, i.e., ε N H LP ( j ) = H HP ( j ) =.5 ε = H LP j ( ) H HP ( j ) N N Hene, we mut have, whih yield = N N = N N = The two filter, whih are power N = N N = N omplementary, an be ued a roover filter in an audio ytem..4 We have for the LP-BP tranformation S = I iω i I = Ω i I = Ω I = = I From Eq.(.5)we get τ ghp ( ) I = τglp ( Ω ) = τ glp ( Ω ) I τ glp ( Ω ) Hene, the mapping of the group delay of the LP filter ha two omponent. Sine a the group delay of 4 a typial lowpa filter ha a peak at the pab edge there will be two peak in the BP filter group delay; one at eah b edge. The eond omponent i weighted with the fator I /, whih i large for low frequenie. Hene, the peak at the lower b edge i larger than that at the higher b edge..5 Firt, we map the BP peifiation to the orreponding LP peifiation. The geometri requirement i in thi ae atified ine = = 5 (krad/) = = 5.5 = 5 (krad/) We get Ω = = =.5 krad/ Ω = = 5.5 =.5 krad/ Nomogram or MATLAB with Ω /Ω = 9.4 => N = Normalized pole: (Approximately!) S plp =.494 S plp =.47j.9659 S plp =.47 j.9659 Denormalization of Chebyhev I filter are done by multipliation with Ω =.5 krad/. Denormalized pole: S plp =.5 krad/ S plp =.675 j.448 krad/ S plp = j.448 krad/ three zero at infinity. (There are alway equal number of pole zero.) The LP tranfer funtion i H LP ( S ) = ( S S plp )( S S plp )( S S plp ) LP tobp tranformation I LP BP S The BP tranfer funtion i H BP ( S ) = I S plp I S plp I S plp S The bpa pole an be found from equation: S plp pbp = ± -- S plp 4 I.6 Firt, we map the BP peifiation to the orreponding LP peifiation uing S = I /. We have Amax = log(.9) =.4 db, A min = Amin = 5 db, The geometri requirement: Ι = = => 5 = 5 5 = 6. We elet to redue: = krad/. Ω = = 5 = 5 krad/ Ω = = 5 = 5 krad/ We elet I = 5 krad/ to get a normalized LP peifiation with Ω = Ω = 5/5 = 5. The order for the LP filter i aording to Eq.(.6).A min aoh A.5 aoh max N aoh( 8.59 ) = = =.57 aoh aoh( 5.4 ) aoh( 5 )

9 Note that we get two BP pole (zero) for every LP pole (zero). In the ame way we get for S p = = j j = j ( j ) = j ( o ( ) j in( )) = j = j p, = j ± where we mut add π rad ine the pole mut be in the lhp. The BP pole are omputed = atan = ( π ) S ( p 4 I ) = ( j 5.66 ) 6 8 = j 75 For example, for a pole, S p = j 5.66, we firt ompute the quare root. S I = =. To ompute the left h ide we proeed a follow. S -- ± S ( 4 I ) 4 Mapping of LP pole zero are done aording to an be rewritten.5.5 [rad/] x S I I whih A() [db].8. τ g () [] 4.6 where ha been eleted o that H LP () =. All LP zero are at S =. The attenuation group delay funtion for the lowpa filter are hown below. ( S σ ) S σ S ( r ) H LP S ( ) = = ( S.986 )( S.986 S ) b = Pole are S p = *i S p = S p = *i We have ε = ine ainh(x) = x x we get a = a = inh,,. b = oh ε =.A max --- N ainh -- ε --- N ainh -- ε S pk = Ω a k in ( ) π for k =,,..N. where N jω b o ( k ) π N where aoh ( x ) = x x. We elet here N = get the LP pole for the Chebyhev I filter 5 pbs, = 4.59 ± j.6989 rad/ (from the real LP pole) pbs,4 = ± j.5 rad/ pbs5,6 = ± j6.8 rad/ The zero of the SB filer are obtained from the zero of the LP filter, i.e., from S =. Hene, we get zero at z,, = ±j I = ±j4.46 = ± j π rad/ All together 6 zero. The tranfer funtion i Tranformation to a SB filter uing S I = , i.e., I I yield S S = ± I I elet to dereae to = / =.7688 rad/ I = = ine we do not want to hange the pab edge. We get Ω = Ω = Nomogram or MATLAB with Ω /Ω = => N = Normalized pole for a Chebyhev I filter are: S plp =.458 S plp, =.669 ± j Denormalizing with Ω = yield S plp = rad/ S plp, = ± j rad/ The ymmetry requirement yield: = (rad/) = (rad/) We Ω = I /( ) = I /( )π.9 We get Ω = I /( ) = I /( )π.8.7 x [rad/] A() [db].8. τ g () [] 4.6 H BP ( ) = ( )( )( ) The attenuation group delay i hown below An LP Chebyhev I have all zero at S =, eah reulting in a zero at = =. The numerator i therefore ine the denominator ha order 6. The tranfer funtion i j S p = j 5.66 yield p56, = j j yield the bpa pole p4, = j.79 6

10 ..4 a) ) d) e) 7 H BS ( ) ( ) = ( ). ( ) ( ) N = 6 z,, = p, = 87 ± j87 p4, = 59 p56, = ± j959 ± j977 N = 6 zz,, = p, = 99 p4, = 54 p56, = 94 N = 6 z = ± j857 ± j69 ± j x 4 Butterworth - - x 4 Chebyhev-II x x 4 Chebyhev-I - - x 4 Cauer x 4 z z45, = ± j889, = ± j4569 p, = 85 p4, = 576 p56, = 58 N = 4 z z4 ± j6678 ± j459 ± j8644, = ± j4946, = ± j8 p, = 876 ± j858 p4, = 95 ± j x x