Relating ξ to ∆ω(EE 221 Spring 2003 Midterm Project)Ramprasad Potluri∗April 2, 20031 IntroductionSuppose we have decided that we want to have an active bandpass filter (Figure 1), or aband-stop filter (Figure 2), with bandwidth ∆ω and center frequency ω0, and that thisfilter will have a quadratic pole (or quadratic zero).The transfer function of a bandpass filter has the form:(±)G01 + 2jξω/ω0+ (jω/ω0)2(1)The transfer function of a band-stop filter has the form:(±)G0©1 + 2jξω/ω0+ (jω/ω0)2ª(2)Here, G0> 0 and (±) means that the sign of G(jω) can be either positive or negative.This note helps us determine ξ given ωL, ωH, and ω0.2 Active bandpass filter: Quadratic poleFrom Equation (1), we have:|G| =G0r³1 −ω2ω20´2+³2ξωω0´2At ω = ω0and at ω = ωX(here, ω = ωXmeans that ω = ωLor ω = ωH), we have:|G||ω=ω0=G02ξ= Gmaxand |G||ω=ωX=G02√2ξ=Gmax√2(3)∗Department of Electrical & Computer Engineering, University of Kentucky, Lexington, KY, USA.Typeset in LATEX.1ωLω0ωHω Figure 1: Bode magnitude plot of bandpassfilter of the formG01+2jξωω0+hjωω0i2.ωLω0ωHω Figure 2: Bode magnitude plot of bandpassfilter of the form G0½1 + 2jξωω0+³jωω0´2¾.Thus,G0r³1 −ω2Xω20´2+³2ξωXω0´2=G02√2ξ=⇒µ1 −ω2Xω20¶2+µ2ξωXω0¶2= 8ξ2(4)=⇒ω4Xω40− 2ω2Xω20+ 1 +4ξ2ω2Xω20− 8ξ2= 0=⇒ω4Xω40+µ4ξ2− 2ω20¶ω2X+ 1 − 8ξ2= 0=⇒ ω4X+ 2(2ξ2− 1)ω20ω2X+ (1 − 8ξ2)ω40= 0=⇒ ω2X=−2(2ξ2− 1)ω20±q[2(2ξ2− 1)ω20]2− 4(1 − 8ξ2)ω402= −(2ξ2− 1)ω20±q4ξ4ω40− 4ξ2ω40+ ω40− ω40+ 8ξ2ω40= ω20− 2ξ2ω20± 2ω20ξpξ2+ 1= ω20− 2ξ2ω20µ1 ±r1 +1ξ2¶=⇒ ω2L= ω20− 2ξ2ω20µ1 +r1 +1ξ2¶(5)ω2H= ω20− 2ξ2ω20µ1 −r1 +1ξ2¶(6)=⇒ ω2H− ω2L= (ωH+ ωL)(ωH− ωL) = (ωH+ ωL)∆ωHere, ∆ω = ωH− ωL.2From Equation (5) and Equation (6), we have:ω2H− ω2L= 4ξ2ω20r1 +1ξ2Equating the last two equations above, we obtain:4ξ2ω20r1 +1ξ2= (ωH+ ωL)∆ω=⇒ ξ2r1 +1ξ2=(ωH+ ωL)∆ω4ω20LetK1=(ωH+ ωL)∆ω4ω20(7)Note that, K1is known because ωH, ωL, ∆ω, and ω0are known.Then we have:ξ4µ1 +1ξ2¶= K21=⇒ ξ4+ ξ2− K21= 0=⇒ ξ2=−1 ±p1 + 4K212=⇒ ξ = +s−1 +p1 + 4K212(8)≈s−1 +¡1 +12× 4K21¢2=r2K212=⇒ ξ ≈ K1Note that in the third to last equation above, we have used the approximation that(1 + δ)n≈ 1 + nδ (9)This approximation follows from Taylor’s series (please see the last section) and workswell when δ ¿ 1 (in particular, when δ ≤ 0.1).Thus, we have obtained the result:ξ ≈(ωH+ ωL)∆ω4ω20(10)Remark 1 : If the Bode magnitude plot is such that1√2Gmax= G0(please see Figure 1),then from Equation (3) we haveG02√2ξ= G0This means that ξ =12√2=√24=0.7072= 0.3535.3Remark 2 : When Bode magnitude plot is such that1√2Gmax= G0(as mentioned inRemark 1), or when1√2Gmax< G0, the filter is no longer a bandpass filter. It is a lowpass filter. Thus, it is meaningless to try to find ξ for such a case.Remark 3 : From Remark 1 and Remark 2, it follows that in our project, in case weuse active bandpass filters, the maximum value of ξ will be 0.3535.Remark 4 : If we find that 4K21> 0.1, it is b etter to use the exact expressions —Equation (7) and Equation (8) — to calculate ξ than Equation (10).3 Active bandstop filter: Quadratic zeroFrom Equation (2), we have:|G| = G0sµ1 −ω2ω20¶2+µ2ξωω0¶2At ω = ω0and at ω = ωX(here, ω = ωXmeans that ω = ωLor ω = ωH), we have:|G||ω=ω0= G02ξ = Gminand |G||ω=ωX=√2Gmin=√2G02ξ (11)Thus,G0sµ1 −ω2Xω20¶2+µ2ξωXω0¶2= G02√2ξ=⇒µ1 −ω2Xω20¶2+µ2ξωXω0¶2= 8ξ2This equation is the same as Equation (4). So, similar to Equation (10), we can write:ξ ≈(ωH+ ωL)∆ω4ω20(12)Remark 5 : If the Bode magnitude plot is such that√2Gmin= G0(please see Figure 2),then from Equation (11) we haveG02√2ξ = G0This means that ξ =12√2=√24=0.7072= 0.3535.Remark 6 : When the Bode magnitude plot is such that√2Gmin= G0(as mentionedin Remark 5), or when√2Gmin> G0, the filter is no longer a bandstop filter. It is a highpass filter. Thus, it is meaningless to try to find ξ for such a case.Remark 7 : From Remark 5 and Remark 6, it follows that in our project, in case weuse active bandstop filters, the maximum value of ξ will be 0.3535.44 One-dimensional Taylor’s seriesA Taylor series is a series expansion of a function about a point.The one-dimensional Taylor’s series expansion of the function f(x) about the point x0isas follows:f(x) = f(x0) + (x − x0)f0(x0) +(x − x0)22!f00(x0) + . . . +(x − x0)nn!f(n)(x0) + Rn.Here, Rnis a remainder term known as Lagrange remainder.Letting x − x0= ∆x, or x = x0+ ∆x, we obtain:f(x) = f(x0+∆x) = f(x0)+(∆x)f0(x0)+(∆x)22!f00(x0)+. . .+(∆x)nn!f(n)(x0)+Rn. (13)For a more detailed treatment of Taylor’s series, please see, for example,http://mathworld.wolfram.com/TaylorSeries.html.The function f(1 + δ) = (1 + δ)n, where n ∈ < (< is the set of real numbers), can beexpanded into a Taylor’s series by substituting x0= 1, ∆x = δ into Equation (13):(1 + δ)n= f(1) + (δ)f0(1) +δ22!f00(1) + . . . +δnn!f(n)(1) + RnNote that x = 1 + δ. So,f(x) = (1 + δ)n=⇒ f(1) = 1f0(x) = n(1 + δ)n−1=⇒ f0(1) = nf00(x) = n(n − 1)(1 + δ)n−2=⇒ f00(1) = n(n − 1). . . . . . . . .So, we have:(1 + δ)n= 1 + δn +δ22!n(n − 1) + . . . +δnn!n(n − 1)(n − 2) ···(n −(n −1)) + Rn= 1 + δn +δ22!n(n − 1) + . . . +δnn!n(n − 1)(n − 2) ···1 + Rn(14)When δ ¿ 1 (δ ≤ 0.1), we can use Equation (14) to perform quick calculations withoutusing a calculator.Consider the following example:5(1 + 0.1)1/2= 1 +12(0.1) +(1/2)(−1/2)2!(0.1)2+(1/2)(−1/2)(−3/2)3!(0.1)3+(1/2)(−1/2)(−3/2)(−5/2)4!(0.1)4+ . . .= 1 + 0.05 − 0.00125 + 0.0000625 − 0.0000039063 + . . .Here is MATLAB’s evaluation of the last expression above:EDUÀ format longEDUÀ 1 + 0.05 - 0.00125 + 0.0000625 - 0.0000039063ans =1.04880859370000EDUÀ 1 + 0.05 - 0.00125 + 0.0000625ans =1.04881250000000EDUÀ 1 + 0.05 - 0.00125ans =1.04875000000000EDUÀ 1 + 0.05ans =1.05000000000000For√1.1, a CASIO fx-82C calculator gave the value 1.0488088. For most practical pur-poses 1.049 is good enough. Using Equation (9), we get 1.05 which is pretty close toadequate.Clearly, if δ < 0.1, we can obtain even closer
View Full Document
Unlocking...