Note#43 Butterworth Filter - Note#43 Butterworth Filters...

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Unformatted text preview: Note #43. Butterworth Filters BOSE-2410 Signals & Systems Fall 2004 0. Outline 1. Introduction 2. Example. Derive 3"d order Butterworth filter equations. 3. Realization of filter as RCL circuit 4. Matlab plot 5. Frequency transformations for bandpass filters. 6. Reading assignment: 0&W. Sect. 9.7.5. Butterworth Filters. 1. Introduction. Practical filters are derived fi'om various approximations to an ideal brick wall filter. Butterwcrth filters are realized by means of a Taylor series approximation. The Taylor series approximation of a function f (x) at a point, x, using values at x = 0 2 fl!) . 1d ,9.” .1er (it dflx) is f(x)= f(0)+ O 3: Note that this equation is simply a power series, f (x) = K0 4- K,x + szz + .... See 0&W figure 6.16 (p.445) for typical specifications for filters. 2. Derivation of 3'11 order filter. Let’s approximate the desired brick wall frequency response, |H(jw)| , with the frequency response :1.1 + (1,5 Iio +llzo,.s'+.b,s2 +6353 I To ensure that |B( j w)| —> 0 as a) —> oo , make order of denominator > order of numerator, with ()3 at 0 . [Howl 2 . |BUwJI= Bowed—3L?" M2114 :5in can of a (physically realizable) rational transfer function, |B(s)|L_M , where 3(5) = The Taylor series expansiou of |B(jaJ)| about a) = 0 is _ _ _ d{B(ja))| 1d2|3(jm)j 13(Jw)|-{B(}0)i+TLw+-5Tlxw ...t... Strategy. To force |B(j(o)| to approximate |H(ja)) , pick the undetermined coefficients in 3(5) so that as many derivatives of |B(j(o)| are equal to zero as possible. This condition is called maximally flat. Unforttmately, taking derivatives of |B(ja))| is very cumbersome. Consequently, let’s use. [Bumlz = Humane) b0 +b1(ja’)+b2 Um): +b3 U“)? be +bI(—ja’) +br (Thy): +53 ("J-“03 do +d2aJ2 co + 62602 + c‘w‘ + cam“ Note that |B( j w)| 2 has no j-terms so derivatives are taken with respect to a real variable — much easier! Furthermore, the derivatives of {B( j w)| and {B(_;'r;o)|2 are similar since both are set to zero. Note the following relationships: B(s)B(—s) = B(jw)B(-jw) = BUw)B’(jw) = |B(J'w)12 . Conversely, if we express 800)) as B(ja>) = 12(0)) + 310:») , then 13ml” -- m) + mm) = (R(w)+ jI(co))('R(w)— fl(aa)) = B(S)B(—s)|,.1., . Expanding |B(ja))12 in a Taylor series results in . 2_ . 2 dIBUm)I2 2 _1_d2|B(jw)|2 4 |B(}a,)l —|B(JO)I +W (0 +2lm-‘mow +... and] d d =£9.+i[d2 -62 #0.](02 +—!--[—-IE.1‘—--c—2[ml2 —¢':2 Ana)“ +Ren1ainder co co co Forcing the derivatives to zero gives, do=co d2 =0;1 . c4=0 cfia)6 Thus the approximation, thus far is {.B(j¢v)|2 = 1—————fl-. c0 + czco + 050) m Strung. To force |B( j w)| to zero as a) —> Q“smoothly” expand |B(jao)|2 in a Taylor series about a) -> I. This will define the remaining undetermined coeflicient in |B( j .sz))|2 . w ...2_ 2 Following the usual procedure, let a) = l and expand ‘B[j 1] about v = 0. Thus, V Now force c2 = 0 , so that the approximation becomes , c 1 IMMI‘ = min— = 90 +5359 1 + 52.96 cu Redefine mknown coefficient as 59— = (:5 , so that |B(jm)|z = +- Compare this equation to n ‘ 1+ 31’] 0": equation 9.140 in 0&W with N=3. To find the poles of the transfer function, we need to write |B( j w)|2 in terms of 3(5). Since ——1—6, then B(s)B(—s) = —1— a. 1+ 3 1+ i] 0’: Jmc TEX/J Ftp-b retahMI-CI'S mac-s), N. am +M1lv» Mwafisl {-13 Mirror ‘ "page. 9‘ J W“ “‘a‘ “‘4 1‘“ ‘1 . Lap-“4f PEMC F9153: £292) 3v JCKM‘Z] 3. Realization of Filter as RCL Circuit. (1)3 53 + 2n);2 :2wfs+ a): ’ is realized by the circuit. Express the elements C1 ,02 ,L in terms of R and the cutoff fi'equency, we. 1? L. First find the circuit transfer function using the voltage divider law. This is done in two steps. First find V(s) and then Y(s). Verify that the third order Butterworth filter transfer function, 3(3) = 1 9" .. its _ (S) (S) 7. I l ”c H: +31. 6,6 319) '- [24-29(5) d 3? SJ;*§E,+SL' SC: 3 I l ">ch; , E19. ‘V _ gut. New ”Ems: 34;) ,z; 4%.” qéfliu) + ,7 / E0 5“ + _\, I gag—l 31s) 3: 2-2 I fl) 1 _ .-—--. :- J, -- l + SLL (-2,, EH.) SL. +562. 4. MATLAB Plot Magnitude Plot of 3th order Butterworth Filter n=3; z__p_k=='The zeros, poles and gain constant'; [z,p,k]5buttap(n), pause; num_den='The numerator and denominator coefficients'; [nmn,den]=zp2tflz,p,k), pause; w=[0:.01 :5]; [mag phase]=bode(nmn,den,w); plot(w,mag); title(['Magnitude Plot of ' int25tr(n) 'th order Butterworth Filter'D; xlabelerequency'); ylabel(Magnitude'); Output atfirst pause: (hi t any key to continue executing) -0.5000 + 0.8660i -0.5000 - 0.3660i -l.0000 1.0000 Output at second pause; (hit any key to continue executing) num = 0 0 0 1.0000 den= 1.0000 2.0000 2.0000 1.0000 Example. Transform the 3rd order Butterworth filter in Sect. 3, with at, =1, i.e., 3(1)) = —3—?l—— , into a bandpass filter, H 3,, (s) , with bandwidth, B = 2 , and center frequency, p +2p +2p+l (00:5. Solution. The pole-zero diagram of B(p) is [r]=roots([l 2 2 1]); Sim=lreal(r)]; omega=[imag(r)]; plot(sigma. omega. 'X') hold ' x=[-l:.01:0]; y1=sqrt(1-X-"2); y2=-sq1't(1-x."2); plot(x,y1,'--',x,y2,'--' axis([—20 -1 1]) 3nd 1.2 4.3 4.6 -1.4 -1.2 -1 -o.s -o.e -o.4 41.2 and the poles axe > roots([1 2 2 1]) ans = -l.0000 -0.5000 + 0.8660i -0.5000 - 0.8660i 52 +w§ _ .5" +25 3 maps the poles of B(p) into the complex conjugate s Basically, the transfonnation p = poles in H a, (s) . That is, , _ 833 Ht“) = “Ply-'22” Wm __ 85’ ' s‘ + 455 + sss“ + zoss‘ + 207552 + 2500: +15625 The pole-zero diagram of H 3,, (s) is 6 [r]=roots([1 4 83 208 2075 2500 156251); sigma=[real(r)]; omega=limag(r)]; plot(sigma, omega, ‘x') where the roots are roots([l 4 83 208 2075 2500 15625]) 3115 5 -0.5857 + 5.9165i -0.5857 - 5.9165i -l.0000 + 4.8990i -1.0000 - 4.8990i -0.4l43 + 4.1845i -O.4143 - 4.184Si Thus the circle representing the pole locations for B(p) transforms into two circles representing the complex conjugate poles for H1909) . Also note the the order of the system has doubled. .92 +0)”2 32 +25 In effect the transformation, p = = maps BS 23 . . 52 + 25 2 pole p = —1 of B( p) mto complex conjugate poles —-1 = 23 or s + 2s+25 = 0 i.e., >> roots([1 2 25]) ans = -1.0000 + 4.8990i -l.0000 - 4.8990i 1 J3 1 «E s1+2s pole p = —-2—+ j T of B(p) into complex conjugate poles —§+ j—é— = 25 or s‘_+(1—jJ§)y+25 = 0 Le, >> roots([1 1-1.732i 25]) ms 1: 5 415857 +- 5.9165i 43.4143 - 4.1845i 1 .JS Similarly for pole p = —E—}? . l . The frequency re5ponse of B( p) = s 3—2—"_ 1 p +2p +2p+1 0.9 --------------------------- .en=[1 2 2 1]; —[0:.01:15]; 0‘3 """""""""""""" [mag,phase]=bode(num,dcn,w); m ..................................... ........................... .lot(w,mag) Note that the bandwidth is 1, i.e., we =1. 853 01133 Of H = —"‘—'—"—-—'—-———- is The frequency ”SP ’5‘“ (S) s“ + 435 + 835‘ + 20853 + 207532 + 2500: + 15625 1 Q9 ------ ----------------------------- -------------------------- - =[l 4 83 208 2075 2500 15625]; 0.3 ---------------------------- ;-----: ------- g- --------------------------- ------------------ —[0:.01:15]; [mag,phase]=bode(num,den,w); 0.707 ----------------------------------- -------------------------- o " 4.099 5 6.099 10 15 Solving the bandwidth equation, B a a)" ~— an and center frequency equation, me = a)“ a); , simultaneously for on results in (of + 2% — 25 = 0, the lower cutoff frequency of the passband, i.e., >> roots([l 2 -25]) ans = -6.0990 4.0990 FIG— Example. Use Matlab command, Ip2bp, to transform 1 . . the lowpass filter, 3(5) = m, w1th cutoff fi'equency (bandw1dth), m, =1, 8.93 , with s‘ + 4.95 + 835‘ + 208s3 + 2075.92 + 2500.: +5625 frequency, co, = 5, and bandwidth, B = 2. into a bandpass filter, H 3,, (s) = center Matiab Command. . The lp2bp command transforms lowpass analog prototype filters to bandpass filters. Syntax. [nudeent] = lp2bp(num,den,Wo,Bw) Description. lp2bp transforms analog lowpass filter prototypes with a cutoff angular frequency of 1 rad/s into bandpass filters with desired bandwidth and center frequency. The transformation is one step in the digital filter design process for the butter, chebyl, chebyz, and ellip fimctions. lp2bp can perform the transfonnation on linear system representations in the transfer function form. The input system must be an analog filter prototype. Transfer Function Form (Polynomial). [numt,dent] = lp2bp(nurn,dcn,Wo,Bw) transforms an analog Iowpass filter prototype given by polynomial coefficients into a bandpass filter with center frequency W0 and bandwidth Bw. Row vectors onto and den specify the coefficients of the numerator and denominator of the prototype in descending powers of s. Scalars W0 and Bw specify the center frequency and bandwidth in units of radls. For a filter with lower band edge wI and upper band edge w2, use We = sqrt(w1"'w2) and Bw = w2-w1. lp2bp returns the frequency transformed filter in row vectors bt and at. Solution. num=[l]; den=[1 2 2 1]; Wo=5; Bw=2; [numt,dent] = lp2bp(num,clen,Wo,Bw) M. numt= 8.0000 —0.0000 0.0000 -0.0000 dent= 1.0e+004 * 0.0001 0.0004 0.0083 0.0208 0.2075 0.2500 1.5625 853 s‘ +435 +835J + 203:;3 +2073:2 +2soos+15625' ._p7... which agiees with our previous answer, 1-13,,(5) = Example. Filter lowpass to bandpass transformations. 2 2 Insight into circuits and why the bandwidth transformation has the form p = S + 030 BS A. Consider a 2“Cl order Butterworth lowpass filter, H p (s) = , with bandwidth=l (i.e., 1 5'2 + ~55 +1 we =1). We call'this the prototype Butterworth filter. The pole-zero plot of H p (s) is The circuit for realizing this filter is with H I,(s) = ILC l . Comparing thefltwo transfer functions gives, SZ+ES+E 1 J_ 1 —= 2 :9 C:— C JE i=1 :9 L=l= 2 LC C B. Change the bandwidth (i.e., the cutoff fiequency, 0),) of the lowpass filter to B. The transfonnation from the prototype filter H P (s) to the lowpass filter, H u. (s) with bandwidth, 3, is s] 1 B2 -__.___.—_- [ST J13] _52+J§BS+BZ —— + 2 — +1 3 B 1 1 1 J5 -——=32 2» L1: 2 =————=— 14:, (:13 [—1—]32 3 J53 But from theprototype circuit, C=—JI—E and L=J§. Thensubstimtingabove, CI =%,and LI =§° Summary To change the 2rlcl order Butterworth prototype filter flow a bandwidth of 1 to a bandwidth of B, simply decrease the prototype C & L by B. L T % 2:1 0 LONE!“ Bavdwn'dffa. = 4- LN?!“ EMA ”Edda *- B C. Now transfonn the lowpass Butterworth filter of bandwidth B to a bandpass Butterworth filter of bandwidth B, and center frequency, (00. Basic thinking: For lowpass filter - u allZfor w<B Iargeror w>B largeroraJ<B ~. =lZforoJ>B These conditions allow low fiequenoies to pass through the filter and the high frequencies attenuated. ‘--.—-—.-- 1 large Z at resonant fiequency, can, i.e., when a): = L C 2 I small Z at frequencies higher and lower than mo. We can see this resonant phenomena flow the following Matlab plots. Impedance for a series resonant circuit with resonant fiequency = 2. The impedance is zero at resonant frequency. L S"+ "Ef. s: +1.1. _. _s‘L¢-+l_ 5"" c %*9L*§'Z'T- 155 " s 400 350 -[0:.01:40]; imag,phase =bode(num,den,w); l-lot(w,mag) 300 250 200 150 100 10 15 {:0 Afi- use-2.. Impedance for a parallel resonant circuit with resonant fiequency = 2. The impedance is max at resonant fi'cqucncY- giXSh) :5 S‘ quL 2" +31. fist-t- ESE-+4- 60 20 25 30 35 40 [mag,phase] =bode(nmn,den,w); 20 10 Lt? Lo «spas: haum Need the following flansfonnations. 2 2 2 5L, __) SLI+__1__SL,C2+1=[S +£00]L 5C2- C25 33 1 —— .91. ‘L __) [3C1]( 2) EL; _ 1 SC] L411: 1+52L2Cl 3341;: C, 5 I BS where 2 1 1 wt) =——-—=-—-—-— L20. L16; We can think of these values as complex impedanoes which involve [ s2 +a1§ instead or writing 3L, , we write [ Bs *2!” 9+0): BS instead of 5. Thus )1. . (Note that the L’s are different!) ...
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