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Unformatted text preview: Note #43. Butterworth Filters BOSE2410 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)) 1d23(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 jterms 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_d2B(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 Ftpb retahMICI'S
macs), 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) ' [2429(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: 22 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 polezero 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(1X"2);
y2=sq1't(1x."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 polezero 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 11.732i 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 = w2w1.
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, 113,,(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 polezero 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); llot(w,mag) 300 250 200 150 100 10 15
{:0 Aﬁ use2.. Impedance for a parallel resonant circuit with resonant ﬁequency = 2. The impedance is max at resonant ﬁ'cqucncY giXSh) :5 S‘
quL 2" +31. ﬁstt 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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