multiple-feedback band-pass filter band-stop (notch)...
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Multiple-Feedback Band-Pass Filter Band-Stop (Notch) Filter
". It,
..
Gain.12 1...".-2 I .•.'1.414
ft- ~ - 447.2 Hz
The notch filter is designed to block all frequencies that fall within its bandwidth.The circuit is made up of a high pass filter, a low-pass filter and a summingamplifier. The summing amplifier will have an output that is equal to the sum ofthe filter output voltages.
LowJ'lC'''6hu
Soommilll!••• pt;6crf,
C2 Rf----"N'v+V~
R) I cI
o---w
r I+~ VOu!
v.
1III
1-v
Hish po ••••
fillerL
"in
1 f;
Block diagram17
,,-,.•.,IDo-,.,.. );:)1""1
.3dB{
v,..1
[I t~
Frequency response18
Notch filter Transfer function HUm)
Vi • I
TransferFunction
H(jw)
H(j{J) = Vo(jro)V;(j{o)
H = Re(H) + jlm(H)
F_V
19
I. V.,
IHI = ~Re(H)2 + Im(H)2
LH = tan-I(IID(H)) Re(H)>-ORe(H)
LH =180° +tan-I(Im<H)) Re(H)-<ORe(H)
20
Frequency transfer function of filter HUm) Passive single pole low pass filter
(I) Low - Pass Filter
IH(jIV)1 = 1 r < f.
IH(j(9)I=O I>f.
(II) High - Pass Filter
IH(jm)1 = 0 f <f.IH(j(9)1 = If> f.
(Ill) Band - Pass Filter
IH(jm)1 = 1 fl, < f <fH
IH(j(9)I=O f <I, and f » Iii
1H(jm) = 1+ j(£R
I V~ = 1+ jaCR .
1m=~RC
~ IV.I=??
V =_I_V• 1+ j .
IVI- 1 IVI--1 IVI• - .Jl' + 1: I - .J2 .
(IV) Band - Stop (Notch) Filter
IH(jm~=O fL<f<j~
IH(j(~)1 = 1 f < IL and I> IH
R
~:; CT---2'°
[V) All-Pass (or phase -shift) Filter
IH(j(~~=1 for alii
has a specific phase response
v = Xc V.(I X('+R'
;(0(' VV -' I•• - _1 +R
jo)(;
1H(jm)= --, (J)
1+}-(J)o
_I_v.l+j(1)CR '
where
21
(i) ~ 0 => IVol = IVil ~ max. value(i) ~ 00 => IVol = 0 ~ min. value
Decibel (dB)(1) Power Gain in dB :
Iv.1Iv,L t "-tL...J2
Pill r:
A/dB) = 1OIOg.{i.)
OdB = 1010g,o(;:)- [!p 1-3dB = IOlog" \, •
+ 3dB = lOlOg,.(~~)
, ___ -1, \,(I), • (J)
IH(jCiJ)1
I
IT2 "------------------
Ie, = (1)" = RC (cut-off frequency) I i 'ar, • (I)
23
1m,,= RC
or
(J)o
H(s) = s +((Jowhere
s= jto
¢«(J) = - tall-'(:')
22
By Definition: dB = LOIOg,{1.)(2) Voltage Gain in dB: (p=V2/R)
V.1/1 Vom
A,.(dB) = 2010g,.( ::.)
OdB = 20 IOg".( ::)
[1 1-I'? .,
- 6dB = 20log,v -1',.
(21' )+6dB = 201og", v;~·
24
Cascaded System Bode Plot (single pole)H(j(9) = I =_1
1+J{£R 1+ J(;). 1
=> iH(Jm~= .L(:~}iH(j{"~. = 20Iog.IH(j"'i = 2010g.(11+(:)']
11()1l1
Ro Wv. ~o· C=L'0lOdE
A. = A" X Ar2 x A"J
A,. = 10xlOx 10= 103
A,(dB) = 20Iog,~(A,. x Ar, x A.,}
A,(dB) = 201og,JA,,)+ 20IoglO(AJ+ 20Iog",(A...)
A,(dB) = A"I(dB) + A.,(dB) + AjdB)
A. (dB) = 20dB + 20dB + 20dB
A,(dB) = 60dB
IH(j(9~ ",.-2010glO( ;)
For octave apart, {O = ~ IH(jm)1 ~ -6dB(J)) 1 (D-('=-(9] I IFor decade apart, () H (jOJ) ~ -2OdB
p:;"U
~
20dB lOdESinglepole low-pass filter
For (0)>(00
201og,~(lO)= 20dB201ogJI0')= 60dB
25
IH(j{u'll ~-20Ioo ({u)'IdB OH' (u.
26
Bode plot (Two-pole)
IH(j(i))L.
Rl n,
V.I
0)----
aJOI
;;j:; aJQ2
lO(!J ){O 2(9x x • (o(1og::: I
I " i •
Active Filters (Part II)
Source: Analogue Electronic CircuitsDr.H.K.NgDepartment of Electrical EngineeringCity University of Hong Kong
Biquadratic function filtersOJ ?
S2 +~s+OJzS2 +cs +d _ K Qz
)- - OJ 2H(s - i +as+b i +~S+lOp
Qp
Realised by:
(I) Positive feedback
"I
(II) Negative feedback
Active Filters (Part II)
Contents
• Biquadratic function filters• Positive feedback active filter: VCVS• Negative feedback filter: IGMF• Butterworth Response• Chebyshev Response
(III) Band Pass
Biquadratic functions H(s)=K .s K
ss: +as+b s' + (I1F s+(O'
Qp r
(I) Low Pass (IV) Band Stop
H(s)=K 1 =K 1H(s) = K . s' + b = K s' + (11:
s' +l1s+b .v' +~s+l'J!s' + as + b . WI' ,s: +-S+W'
Q, rQp' r
(II) High Pass (V) All Pass
H(s)=K . s' =K s'• (iJ
H(s)=KS' =as+bs· --2.05+(0;
s' +as+b ,(up , K Q,S·+·-S+{U· s' +as+b s' +~-s+{O'Qr P o p-,
Low-Pass FilterVoltage Gun,1.5
I K=l,m.=11
VoIt~Glin(dB),o.~ e·:.....:'
·10.
0..5,·20" '.
2F~IeIl:y
3 4 0.
H(s) = s +12+_s Qp
Band-Pass FilterVoltageGin,-
1.5
I K=LN.=ll
2 4Fre(p.1eIl:y
6 8
·15
2Fre<pell:y
3
•2011'" 'q,;';oT' .·25
2
H(s) = s2 S
S +-+1Qp
4FrapJetl:y
High-Pass FilterVoItageGlin
r-- I '0.=1.5 I _ I1.5~ I ',K=l.m,.-l
I 0.= 1 " , , •I •.•..................... _.....I , 1
-----------0.51- •.•.•.•
\ .'...Q::~~ .
Voltagellin (dB)5, ,0,= 1.5" ' <,
1Q.=1 - - __0. I •.• ·-··,..~.~-~~~~=-:.:r:......_-:
\ .> (}=0.5. /\ 1
h / q,=_A 1 J2( ,I
I
.s
-10.
·15
"-""1.)\ .•.•.~ .•. ,~'
_ ,."
0' 6··'0. 1
2 3 ,I I I_ • "',' ,K-I· ••-tI, I44 '1
S2
H(s)= 2 !.-+1,. +-., 0-p
V~G!in5, 'I ,
Band-Stop FilterVol~~(dB)
• 1\, ~
IQ.= 1.5I \
I \'0,=1 \
1 .". \
1 ~I" " \. \ »:»", .:-:: ....
\. ~Q~QI"""-""'"Q' .....~" ,\, !t?s" "\~ -30 I K=I.m·=1' -. '':--'''~;:';'''~-~''''''---:-:-fQ,;, 'c!i' '." _." .._ " .._ .
Go.
VoltageGin (<13)
5, 19-1.50.1 I g.~1 I I
0.=-
~~J2v=> ..•..
Q=05' ~.:~...
I I': ,Q.=1.5
I 100
f
· ,K=I,&=I.O~=l!1 \ ~.:.\o.=1.5
, \\~\0.' I'
......... ,. .
84
Frecp.m:y6
2Frequen;y
I K= 1,((\.=I, <a=1 !
•10~ '.,q :
'l> •
]0'......
8 0.
,~,.\ .,.'8
H(s)= s2+222 Ss +-+1
Qp
.'.,.
1Q.= J2
~~"
4Frec:p.tn:y
3
-" _ ....
6
Voltage Controlled Votage Source (VCVS)Positive Feedback Active Filter (Sallen-Key)
~z,
But,Z.J -,
r - = KiJZ. = K Z. +z.. - (2)
Substitute (2) into (I) gives
V v ",(Z:+Z')lfl 1 1)~+....t.._ -+-+-- =0Z, ZJ KZ, Z, Z, Z2 + Z,
or
V.V K
H=~= ZZ 1v, Z, (l-K)+--L.l+l+-(Z, +Z,)Z, Z,2, Z,(3)
By KCLat Va: i; +ij =i, = 0
where, v-vi =-'--'. z,
v-vi =-'-', Z,
Therefore, we getv -v v -T,' v~+~-~=O
Z, z, Z, +Z,
In admittance form:
r:t. = Z,+'Z,
Re-arrange into voltage group gives:
v v rill)-+-'--v. -+-+-- =0 (I)z, Z, ..\ 2, Z, Z, + Z,
H - ( )t t Y )";1'l+Y -+- +.:2..(1_K)+-'-4
.• )~ }~ 1~ )~)~
K(4)
• This configuration is often used as a low-pass filter, so aspecific example will be considered.
VCVS Low Pass Filter H(s)=K-1 we continue from equation (5),
Z, =R, Z~= R2I I
Z, = .-, - sC. '/(o('3 .1
I JZ, ::: .i{o(~'~- sC'~
H(s) =, ~s R,RoC,C. + sc,(R, + RJ+ sR,CI (1- K)+ I
K_lH(s) = R,~C,C,
s' + s[c.(R, +RJ+R,C,(I-K)]+[. l __]R,R,C,C. R,R,C,C.
K
In order to obtain the above response, we let:
(6)
Equating the coefficient from equations (6) and (5), it gives:
1 1 I 1(i)= ---- 0= -
p ~R,R:C,C.. ~R,C; ~R:C. -P IR,C~ + R2C'I_+(I_K)~RIC)IsC\ R,C) R:C.
Then the transfer function (3) becomes:KH(s) = ---,--- ----;- --:----
1+ sC.(R, + R,)+ sR,c.(I - K)+ s'R,Rl',C.K'
,(Op •s: +-s+(O;o.(5)
Now, K=I, equation (5) will then become,
1H(.r) = 1+.lC.(R, + R,)+/R,R/-",C.
Simplified Design (VCVS filter)I. =mR I: =R
I 1I,= =-
. j(u(nC) snC
H(s) = 11+sRC(m+l)+s2nmR2C2
Comparing with the low-pass response:I
H(s)=K----SZ +~!:.S+(()2o P
~p
It gives the following:
1(t)p = RC& Q=~
p -m+l
I Iz~=-=-
j(()C «:
Example (VCVS low pass filter)1
To design a low-pass filter with fo = 512Hz and Q = .,fiIIC
Let m = 1 ----i!f----Q,~Jmn~Jlxn~.r.:=J, -.:"iliTI,V' l
m+l 1+1 - .fi v _ -L . ,ill C_-.- 1 v
~ n=2 I' "o , 0
(cJ = 1 _ I 1I' RC-Jrnii - RC.Jl;2 = RC.fi = 27i(512H=)
200nF,---If----,
Choose C = lOOnF i2k1 : J'Tj'1 +'. i~~v, I <, ..•..~~
v.:~< [(IOnF =+- L~~~:~J1".
IThen R = 2,198n - 2.2kQ
What happen if n = I?
VCVS High Pass Filter VCVS Band Pass Filter
R3
Vo
I ~ II, 1+ Rj-f' R, "v.
In enj -, R4 - I v0
v.In R.
K'S' K_S_'
H(s) = R,C, = K'8., s ( 1 I 1 ( ) C I R, + R ,(0. 1
S +-- -:-+--+--..l-K +_~s~ I+··--~·-'- s: +~S+(J)C, R, R, R, R,c,/ R,R,R,C,c, Q, P
H(05) = KS's' +s(_l_+_l_+_l_(l_K))+ __ l_
R,C, R.C, R,C, R,R.C,e,
• (()r •S' +-05+(1);Qr
=-----
Infinite-Gain Multiple-Feedback (IGMF)Negative Feedback Active Filter
rearranging equation (3), it gives,
H=V,_Vi -
1Z,Zj
1(1 1 1 1) 1- -+-+-+- +--Z; Z, Zl 2j 2. 232.
1fT.~----~I~----~I------~Or in admittance form:
v; =0 1'~ =02, Z,~ V =--v ~~T =-...2.v
e Z)·'( :( Z~ (.'
H= V._Vi -
l~Y;Y5(~ +1~+Yj +Y.)+ Y3Y.
(I)
By KCLat v:, v-v v V V-V_'_.T. = ---L +-L.+_"_0z, Z2 Z, Z.
(2)~e ZI z, Z3 Z4 Zs
LP RI C2 R3 R4 Cs
HP CI R2 C3 C4 Rs
BP RI R2 C3 C4 Rs
substitute (1) into (2) gives
Vi +~V =-~V _ V, _lv _V. (3)z, ZIZ;' 2;Z2' Z; Z.2; • 2.
IGMF Band-Pass Filter Simplified design (IGMF filter)Band-pass: Hes) = K / +as+b
s
H(s)=
sCRI
To obtain the band-pass response, we let I 2C 2(,.2--+s-+.v ,R1R, R,
1 121 = l~ 22 = l~ Z; = }aX.:\ = sC
3
Z. = _1 _ 1}aC - :-(-' 2, = TJ• "'4 • .,
H(.<) =
.Iel
R,
Comparing with the band-pass response
H(s)=K--S
s' + {j)p s+{(/Qp r
2 C; +C.. 1 ( 1 1 )J C;C. +S~+ R; ~+ Rl Its gives,
(IJ= ] Q if'E f)l' C~R.Rs r = '2rt. H\(r)1' = -2Q~*This filter prototype has a very low
sensitivity to component tolerance whencompared with other prototypes,
Example (IGMF band pass filter) Butterworth Response (Maximally flat)To design a band-pass filter with fo = 512Hz and Q = 10
(0 =_I_=27r(512H:)r C~R,R.
C=lOO1F ~ ~Rs=9,66474[i
IfR;o =-_. =10-? 2 JR. Normalize to OJo = lrad/s
I A ~ 1Hi jto~I+(iJ2n
1
. iH(j"'l= N:.r Butterworth polynomials
B.(s) = s+l
B2 (s) = Sl +Ji8+1
B3(S)= s' +2s1 + 2s+ 1
= (8+ lXs2+8+ I)
B4(S) = S4 +2.61l +3.4ls2 +2.6Is+1
= (s~+ 0.77 s+ lXs1 + 1.85s +1)
E5(S)= S5 +3.24s~ +5,24s~ + 5.24s2 +3.24s + I
= (s+lXs2 +0.62s+ IXs2+ 1.62s+ I)
where n is the order
~ R. = 155.40 R, = 62,1700
--r-IOOIlF l ~62.170f!
~-- :j' - ,
ISH!! I_y.:--..,,"-.. i>----,"'" I! ~IIJOIIF !+/? I'
jl? 0I' •.•
In !.
Butterworth polynomials:
IH(j{!)~ = iB,,/iliJ)iWith similar analysis, we can choose the following values:
C=lOnF ~ =1,5541 and R, =621tOcn
'R, iutterwort 1-.1I D", P~""Iionsct ••••• ; ••••••••••••. I'! 'till "I .. "I"!,, .. ,,;. ,.... "..I",•.Second order Butterworth response
IHI_IStarted from the low-pass biquadratic function H(s)= K--=----
1 (V ~1 s +-2..s+liJ;For wp=1 K=l Q= .J2 Q, .
.101
(second order butterwoth polynomial)H(s),5' +..fis+1
IH(jw) = _w1 +..fij{!J+1
I
Vi(jw~= ~(I-(:J'r + (..fi(:J/1
IH(j{~~ = ,Jt-2aJ' +{,,' +2W2
IjH(jaJ~= ,jl+a'
I _. __ 1IH(ja~= ~ - ./t+ ((<JY'"1+ «J "
I I ~.~o I ••
rad/s
-20
Second order Butterworth filterKH(s) - ----;----,---;-----:----
- 1+ sC.(R. + RJ+ .~R,C,(I - K)+ s'R,R,C,C, S' + (i), S + w'I () ,o = --.=~--.== "",
_F ~R.C, + R,C, +(I-K) i!S.c,R,C, R.c.. V R,C,
Setting R)= R2 and C) = C2
Q _ 1 __ I,- .Ji +.Ji+(I-K)J1- 2+(1-K) 3-K
Now K = I + RJ RA
o =_1_ I =_1_-, 3-K (R ) RA3- 1+-L 2--RA R"
For Butterworth response:I I Io =- => Q =-=---F.fi P .J2 '>_ R.
- R,
Bode plot (n-th order Butterworth)e.~ 10.-
I ' \1 IH(jw ~=-../=1 +=a=!)2=.
IndBfonn:
IH(j(tJ)1 = 20 log_l_.Jl + w2"
IHU(o)1 = -20 log(.JI + w2.)
~ -40c:&-00coa~~ -80
suppose (i) »1 -100
IH(jw)1 ",-20nlog(w)
For decade condition, w =10{o"
The Butterworth filter would bave(-20n)dB/decade
For 2nd order : -40 dB/decade
For 3rd order : -60 dB/decade·
For n - th order: - 20n dB/.decade
K'I ~ ~(i)
---------------------------1----------- ! 1 I '' .L ~, , i~J -~JTJ·c~t iu-----J','A • ./' R I
/'j".:C I .> ,'. I >~jl__: _
/J"tUrwoTtlt re$JHlMe
Therefore, we have 2 - R._ =.J2 = 1.414RA
We define Damping Factor (DF) as:
I 2 s, 4DF=-= --=1.41Qp R"
Values for the Butterworth responseB,(s)= .HI
B2(S)= S2 +-Jis+1
B3(S)= s' +2s2+2s+1= (s+IXs~+s+I]
Bl~)= S4 +2.6b3 +3.4ls= +2.61s +1 = (S2 +O.77s+1Xs= +1.85s +1)
B~(s)=s; +3.24s4 +5.24s3 +5.24s: +3.24s + 1= (s + lXs: +0.62s +IXs: + 1.62s + I)
Damping Factor (DF)
• The value of the damping factor required to produce desire responsecharacteristic depends on the order of the filter.
Roll-off 1st stage 2nd stage 3rd stage
Order dB/decade poles DF poles DF poles DF
I -20 I optional
2 -40 2 1.414
3 -60 2 1.000 1 1.000
4 -80 2 1.848 2 0.765
5 -100 2 1.000 2 1.618 I 0.618
6 -120 2 1.932 2 1.414 2 0518
• The DF is determined by the negative feedback network of the filtercircuit.
• Because of its maximally flat response, the Butterworth characteristicis the most widely used.
• We willlirnit our converge to the Butterworth response to illustratebasic filter concepts.
Forth order Butterworth FilterRJ = R~,CJ = C4RJ = R2,CJ = C2
C,O.OII'f C,O.OII'f
R,I.2 illR, 1.2ill
v'"
R. I.Sill R. 27 kO
DF=2_RB =1.848R,
R.-2...=0.152R.R8 = (0. 152)R.• = (0. 152)tOknRB = 1.52kn
DF=2- Rs =0.7651(.
RB = 1.235RAR3 = (1.235)R. = (1.235)22kfl
Rs = 27.17kO
Chebyshev Cosine Polynomials
Co {co)= 1c,{w)= coC2 (co)= 2co2 -1C3(co) = 4co3
- 3coC4 (w ) = 8co4
- 8co2 + 1
C, (co) = 16co5- 20co3 + Sco
CII(w) = 2coC
II_1 (co)- Cn-2 (W)
Chebyshev Response (Equal-ripple)IH(j(o~= I ! J( ,
l+£-C~ W
Where e determines the ripple andC; is the Chebyshev cosine polynomial defined as
]"(1.1]
,I I --- , ••, 2
Second order Chebychev ResponseIH(;a>~=--1
~1+&2C.~(a»
Example: 0.969dB ripple gives E = 0.5, CJ(!J) = 2(tl~ -I
IH2(a»= 2C'2({(»)
2 1+&"
1
Roots: SZ -1 ±.Jl=5 12 -'2±j
IH,(,)H,(-sj- (h~+j)(,' +~- j1I+0.51(2a>-IY
Ia>4 - (tl2 +1.25
Hi«(tIL", =H2(s}H2(-S)I
s~+s2+1.25
Roots of first bracketed term
O±m~-. - 2
rr':=±'r"2- }
=±[ HH+(-~)J-jHH-(-~)JJ= ±[0.566- jO.899]
RootsRoots of second bracketed term
O±~)s= '';-,\2- J )
2
=±J-~+ j
=±[ MH+(-~)J+j HI%-(-~))]= ±[0.566+ jO.899]
H:{s)H2{-S)= (5+0.556- j0899XS-0.556+/0.899) (5+0.566+ jO.899XS-O.556- '
I I Ior H 2 (s) = , .. ,. '" • . ... .._