Break_11 : Analog Butterworth Filters 1. Introduction (a) There are many kinds of analog filters, where a group of filters employs passive discrete components such as resistors (R), capacitors (C) and inductors (L), and the other group uses a solid device such as a SAW (Surface Acoustic Wave) filter for microwave.

The former group are called LRC filters.

In the frequencies range from 1 Hz to 1 MHz, the inductor itself gets quite bulky and uneconomical to manufacture.

In this case, active filters using operational amplifier (op-amp) are widely employed.

The analog filter in this area may be viewed as a counter-part of the digital filter.

Both of them have

similarities and the analog filters are placed prerequisite for the digital filters. (b) As we know, filters are equivalent to the linear time-invariant (LTI) systems.

The circuit theory

begins with the impulse response of the system (circuits), and the filter output is determined by the convolution of the input with the impulse response.

It is expressed by the equation as

y(t)= ∫0∞ h()x(t-)d where x(t) is the input to the filter, h(t) denotes the impulse response, and y(t) is the filter output. If we apply the Laplace transform to h(t), we can obtain the transfer function of the filter as H(s)= ∫0∞ exp(-st)h(t)dt The variable s is expressed by s=jΩ+, and H(s) generally has a form of rational functions as H(s)=∑k=0M βksk/ ∑k=0N αksk =(β0+β1s+β2s2+…+βMsM)/(α0+α1s+α2s2+…+αNsN),

N≥M

The transfer function H(s) can be decomposed such as H(s)=(β0/α0)∏k=1M(s-zk)/∏k=1N(s-pk) where zk and pk denote the zeros and poles, respectively. s-plane, the frequency response is obtained as H(Ω).

If we evaluate H(s) on the jΩ axis of the

The following constraints must be imposed with

regard to the location of poles and zeros in the s-plane. * All poles of H(s) should be placed in the left half of the s-plane in order for the filter to be stable. * All complex zeros and poles of H(s) must occur in complex-conjugate pairs in order for the filter coefficients to be real. 2. Fundamentals of LPF (a) The most simple lowpass filter (LPF) is the passive RC network illustrated below. R Vout

Vin =C

≡ (common) Its transfer function in the Ω (angular frequency) domain is H(Ω)=1/(1+jΩCR) Let us introduce the cutoff frequency (-3 dB corner frequency) defined by Ωc=1/RC Then, the transfer function can be rewritten as H(Ω)=1/[1+j(Ω/Ωc)] The magnitude response is, by simply taking its absolute value

|H(Ω)|=1/[1+(Ω/Ωc)2]1/2 where for frequencies (Ω/Ωc)>>1, the roll-off is 20 dB/decade. Consider the s-plane, where the frequency axis is jΩ. function in the s-domain is obtained as H(s).

If we replace jΩ in H(Ω) by s, the transfer

Then, we have

H(s)=H(Ω)|s=jΩ=1/(1+s/Ωc)=Ωc/(Ωc+s) (b) The simplest LRC lowpass filter circuit can be illustrated as below. L

R Vout

Vin =C

≡ (common) The transfer function is H(Ω)=1/(1+jΩCR-Ω2LC) In a similar manner to part (a), let us introduce the other cutoff frequency defined by Ω0=1/(LC)1/2 Then, the transfer function can be rewritten as H(Ω)=1/[1+j(Ω/Ω0)R(C/L)1/2-(Ω/Ω0)2] By replacing jΩ by s, the transfer function in the s-domain is H(s)=H(Ω)|s=jΩ=1/[1+(s/Ω0)R(C/L)1/2+(s/Ω0)2]=Ω02/[Ω02+sΩ0R(C/L)1/2+s2] (c) Let us introduce an active filter using a positive feedback amplifier illustrated below. +K =

C Vout

Vin R

R

=C ≡

where K denotes the gain of the amplifier.

The transfer functions is

H(Ω)=(1/RC)2/[-Ω2+jΩ(3-K)/RC+(1/RC)2] By introducing the cutoff frequency Ωc=1/RC, it can be rewritten as H(Ω)=Ωc2/[-Ω2+jΩ(3-K)Ωc+Ωc2] The transfer function in the s-domain is H(s)=H(Ω)|s=jΩ=Ωc2/[Ωc2+s(3-K)Ωc+s2] This is similar to the transfer function of LRC circuit.

Both become identical if we set K as

(3-K)=(C/L)1/2R Thus, we can realize the same characteristic as the LRC lowpass filter without using the inductor L. (d) Consider again the LRC lowpass filter described in part (b). in this filter defined by Q=1/Ω0CR Then, the transfer function can be rewritten as H(Ω)=Ω02/[Ω02+jΩ(Ω0/Q)-Ω2] The magnitude response is |H(Ω)|=1/{[1-(Ω/Ω0)2]2+(Ω/Ω0)2(1/Q)2}1/2 The transfer function in the s-domain is

Let us introduce the quality factor Q

H(s)=Ω02/[Ω02+s(Ω0/Q)+s2] These functions represent the general 2nd-order lowpass filter, which magnitude response varies according to the value of Q such as shown below. 20

Magnitude Response of N=2 LPF, Q=0.5- 1: 2-. 5--

15

20*abs[H(F)] dB

10

5

0 -5

-10

-15

-20 0

0.5

1 1.5 2 Normalized Frequency, F/Fo

2.5

3

Figure_ba01 (The roll-off is 12 dB/octave for higher frequencies.) The transfer function of the 2nd-order lowpass filter has two poles at the roots of denominator. When Q=0.7071=1/21/2=1/1.414, it becomes the 2nd-order Butterworth LPF. (e) For designing active filters, the 2nd-order filter is an important subclass of the more general filters described in part 1(b).

In other words, the 2nd-order filter is usually used as basic building blocks

for realizing higher-order filters.

For the purpose of optimization, the total transfer functions has a

form of cascade of the 2nd-order sections.

In particular, the all-pole filter is commonly used for

designing the higher-order LPF, which has the form such as H(s)=1/{[1+a1s+b1s2][1+a2s+b2s2]…[1+aNs+bNs2]}=1/∏k=1N(1+aks+bks2) where the passband DC gain is one, and ak and bk are the filter coefficients, which determine the complex pole locations for each of the Nth-order lowpass filters. One of problems in the higher-order filters is the saturation of the individual stage.

Therefore, the

cascaded filters need to be placed in the order of rising Q values. The transfer function for the k-th stage filter of the cascaded 2nd-order LPF is Hk(s)=1/(1+aks+bks2) For a 1st-order filter, the coefficient b is always zero (b1=0), thus yielding H1(s)=1/(1+a1s) A filter with an even order number consists of 2nd-order stages only, while filters with an odd order number include an additional 1st-order stage at the beginning. The other types of filter, such as highpass, bandpass and bandstop filters can be realized by using the frequency transformation.

For example, the transfer function of highpass filter is obtained from

that of lowpass filter by just replacing s by 1/s, which corresponds to replacing the resistor by capacitor and the capacitor by resistor.

The general transfer function of highpass filter is then

H(s)=1/∏k=1N(1+ak/s+bk/s2) with the numerator being the passband magnitude. (f) The criteria for optimizing the frequency response rely on one of the following requirements. (i) Maximum passband flatness (ii) Immediate passband-to-stopband transition

(iii) Linear phase response For that purpose, there are three types, Butterworth, Chebyshev and Bessel filters.

Their filter

coefficients are available in many literatures. The features for the three types are Butterworth

Chebyshev

Bessel

(a) Passband flatness :

maximum

ripple

flat

(b) Sharpness at the transition :

intermediate

maximum

very bad

(c) Linear-phase :

up to mid-frequency

very bad

perfect

Notice that an analog LPF is necessary both before AD convertor and after DA convertor.

Since the

Butterworth LPF approximates the linear-phase characteristic up to some frequency in the passband, it is often used even in the digital circuitry. With regard to AD convertors and analog anti-aliasing filters, ΔΣ (delta-sigma) AD convertor chips for audio-band signals are widely available and much more preferable. oversampling and removes the need for complex analog filters.

The technology uses

It is inherently linear, and provides

16-bit accuracy with low cost. (g) The Butterworth LPF are all-pole filters characterized by the magnitude response |H(Ω)|=1/[1+(Ω/Ωc)2N]1/2 It provides maximum passband flatness. Ωc at equally spaced points.

The poles of H(s) in the s-plane occur on a circle of radius

For example, the pole position of 5th-order Butterworth filter is shown

below. Pole positions of N=5 Butterworth filter 1 0.8 0.6

Imag j(F/Fc)

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 -1

-0.8

-0.6

-0.4

-0.2

0 0.2 Real F/Fc

0.4

0.6

0.8

1

Figure_ba03 (Five poles are located in the left half of the s-plane.) The Chebyshev filters are all-pole filters that exhibit equi-ripple behavior in the passband and a monotonic characteristic in the stopband. Chebyshev polynomial.

The magnitude of H(s) is expressed by the Nth-order

The poles of H(s) occur on an ellipse with crossing -axis less than Ωc.

The Bessel filters are a class of all-pole filters, in which the denominator of H(s) is the N-th-order Bessel polynomial.

The poles of H(s) occur on a circular shape with crossing -axis greater than Ωc.

The Bessel filter has a large transition bandwidth, but its phase is linear within the passband. There are other types of filters that contains both poles and zeros in H(s). equi-ripple behavior in both the passband and the stopband.

The elliptic filters exhibit

For a given order and a given set of

specifications, an elliptic filter has the smallest transition bandwidth.

In the following sections, our treatment is focused on the Butterworth filters.

See the available

literatures for designing the other types of filters. 3. Butterworth LPF (a) The Nth-order Butterworth LPF provides maximum passband flatness by the magnitude response |H(Ω)|=1/[1+(Ω/Ωc)2N]1/2 The curve is plotted below, from which we can determine the order N to satisfy the requirement. 0

Magnitude Response of Nth-order Butterworth LPF, N=1- 3: 5-. 7--

-10 -20

20*abs[H(F)] dB

-30 -40 -50 -60 -70 -80 -90 -100 0

1

2

3

4 5 6 Normalized Frequency, F/Fc

7

8

9

10

Figure_ba02 (The roll-off is 6N dB/octave=20N dB/decade.) As we have described in part 2(e), the 2nd-order Butterworth LPF is the basic block for realizing higher-order filters. The transfer function of the single 2nd-order Butterworth LPF can be rewritten by using Ωc such as H(s)=Ωc2/[Ωc2+s(Ωc/Q)+s2]=1/[1+(s/Ωc)/Q+(s/Ωc)2] where Ωc specifies the cutoff frequency of the Butterworth LPF, and Q=0.7071=1/1.4142. Since H(s) is a function of (s/Ωc), let us introduce a new variable p defined by p=s/Ωc=jΩ/Ωc Then, we can obtain a new transfer function for the basic 2nd-order Butterworth LPF as H(p)=1/[1+(1/Q)p+p2] where p=1 corresponds to the cutoff frequency. (b) The transfer function of the Nth-order Butterworth LPF is then HN(p)=1/∏k=1N(1+akp+bkp2) Let us represent the denominator by D(p). HN(p)=1/DN(p) Consider the equally spaced poles on the circle.

Then, DN(p) is obtained by the following formulae.

N=1 : D1(p)=1+p N=even : DN(p)=∏k=1N/2{1+2psin[(2k-1)/2N]+p2} N=odd : DN(p)=(1+p)∏k=1(N-1)/2{1+2psin[(2k-1)/2N]+p2} Let us write out for N=1, 2, …,5. N=1 : D1(p)=1+p N=2 : D2(p)=1+2psin(/4)+p2=1+1.4142p+p2 N=3 : D3(p)=(1+p)[1+2psin(/6)+p2]=(1+p)(1+p+p2) N=4 : D4(p)=[1+2psin(/8)+p2][1+2psin(3/8)+p2]=(1+0.7654p+p2)(1+1.8478p+p2)

N=5 : D5(p)=(1+p)[1+2psin(/10)+p2][1+2psin(3/10)+p2]=(1+p)(1+0.6180p+p2)(1+1.6180p+p2) In some literatures, the transfer functions in the s-domain are written by using Ωc such as below. N=1 : H(s)=Ωc/(s+Ωc) N=2 : H(s)=Ωc2/(s2+1.4142Ωcs+Ωc2) N=3 : H(s)=[Ωc/(s+Ωc)][Ωc2/(s2+Ωcs+Ωc2)] N=4 : H(s)=[Ωc2/(s2+0.7654Ωcs+Ωc2)][Ωc2/(s2+1.8478Ωcs+Ωc2)] N=5 : H(s)=[Ωc/(s+Ωc)][Ωc2/(s2+0.6180Ωcs+Ωc2)][Ωc2/(s2+1.6180Ωcs+Ωc2)] If we convert the above into H(Ω) by using s=jΩ, the absolute values of H(Ω) must be reduced to |H(Ω)|=1/[1+(Ω/Ωc)2N]1/2 It can be verified if we calculate one-by-one such as N=1 : H(Ω)=1/[(jΩ/Ωc)+1], |H(Ω)|=[1+(Ω/Ωc)2]1/2 N=2 : H(Ω)=1/[(jΩ/Ωc)2+1.4142(jΩ/Ωc)+1], |H(Ω)|=1/{[1-(Ω/Ωc)2]2+2(Ω/Ωc)2}1/2=[1+(Ω/Ωc)4]1/2 N=3 : H(Ω)=1/[(jΩ/Ωc)+1][(jΩ/Ωc)2+(jΩ/Ωc)+1], |H(Ω)|=1/[1+(Ω/Ωc)2]1/2{[1-(Ω/Ωc)2]2+(Ω/Ωc)2}1/2 =[1+(Ω/Ωc)6]1/2 N=4 : H(Ω)=1/[(jΩ/Ωc)2+0.7654(jΩ/Ωc)+1][(jΩ/Ωc)2+1.8478(jΩ/Ωc)+1], …, |H(Ω)|=[1+(Ω/Ωc)8]1/2 N=5 : H(Ω)=…, |H(Ω)|=[1+(Ω/Ωc)10]1/2 Thus, all the formulae for the Nth-order Butterworth LPF are equivalent. (c) Let us rewrite the transfer function of the basic 2nd-order Butterworth LPF expressed by p below. H(p)=1/[1+(1/Q)p+p2] Then, from the equations of DN(p), the values of Q of the 2nd-order LPF for each of the Nth-order Butterworth LPF are N=2 : for k=1 : Q=1/1.4142=0.71 N=3 : for k=1 : Q=1/1=1 N=4 : for k=1 : Q=1/0.7654=1.31,

for k=2 : Q=1/1.8478=0.54

N=5 : for k=1 : Q=1/0.6180=1.62,

for k=2 : Q=1/1.6180=0.62

Thus, the formulae for the equally spaced poles provide us with the values of Q in decreasing order. Therefore, by reversing the order of the 2nd-order LPF, we can place the basic 2nd-order sections in increasing order for the optimization. (d) Let us determine the -3dB corner frequencies of the partial filter in DN(p) one-by-one.

If we

substitute p=jΩ/Ωc into DN(p) and make the squared magnitude of the cascaded section equal to 2, and solve the equation with regard to (Ω/Ωc), then we can obtain the corner frequency of the partial filter by the form of (Ω/Ωc), where Ωc is the cutoff frequency of the overall filter. N=1 : |D1(p)|2=1+(Ω/Ωc)2=2, Ω/Ωc=1 N=2 : |D2(p)|2=[1-(Ω/Ωc)2]2+[1.4142(Ω/Ωc)]2=2, 2(Ω/Ωc)2=2, Ω/Ωc=1 N=3 : for (1+p), Ω/Ωc=1 : for (1+p+p2), [1-(Ω/Ωc)2]2+(Ω/Ωc)2=2, (Ω/Ωc)2=(1/2)(1+51/2)=(1/2)(1+2.236)=1.618, Ω/Ωc=1.272 N=4 : for (1+0.7654p+p2), [1-(Ω/Ωc)2]2+[0.7654(Ω/Ωc)]2=2, (Ω/Ωc)2=(1/2)3.8637=1.9319, Ω/Ωc=1.390 : for (1+1.8478p+p2), [1-(Ω/Ωc)2]2+[1.8478(Ω/Ωc)]2=2, (Ω/Ωc)2=(1/2)1.0351=0.5176, Ω/Ωc=0.7194 N=5 : for (1+p), Ω/Ωc=1 : for (1+0.6180p+p2), [1-(Ω/Ωc)2]2+[0.6180(Ω/Ωc)]2=2, (Ω/Ωc)2=(1/2)4.1907=2.0954, Ω/Ωc=1.4475 : for (1+1.6180p+p2), [1-(Ω/Ωc)2]2+[1.6180(Ω/Ωc)]2=2, (Ω/Ωc)2=(1/2)1.4754=0.7377, Ω/Ωc=0.8589

There are three partial filters for N=5.

Then, let us compute and plot each of the three frequency

responses together with the overall frequency response in one figure. They are N=5 : f1(Ω/Ωc)=1/[1+j(Ω/Ωc)], |f1(Ω/Ωc)|=1/[1+(Ω/Ωc)2]1/2 : f2(Ω/Ωc)=1/[1-(Ω/Ωc)2+j1.6180(Ω/Ωc)], |f2(Ω/Ωc)|=1/[(Ω/Ωc)4+0.6179(Ω/Ωc)2+1]1/2 : f3(Ω/Ωc)=1/[1-(Ω/Ωc)2+j0.6180(Ω/Ωc)], |f3(Ω/Ωc)|=1/[(Ω/Ωc)4-1.6181(Ω/Ωc)2+1]1/2 : |H(Ω/Ωc)|=1/[1+(Ω/Ωc)10]1/2 where f2(Ω/Ωc) and f3(Ω/Ωc) are exchanged to align them in increasing order.

See the plot below.

Frequency responses, 1st : g-, 2nd : b:, 3rd : r-., Total : k--

10

5

Magnitude (dB)

0

-5

-10

-15

-20 0

0.2

0.4

0.6

0.8 1 1.2 1.4 Normalized Frequency, F/Fc

1.6

1.8

2

Figure_ba05 (The 1st-filter:green, 2nd:blue, 3rd:red, Total:black) From the plot, we can see that Q

Ω/Ωc

Roll-off

1

-6dB/oct

Odd (N=5) Q & Ω/Ωc are aligned in increasing order.

1st-filter

-

2nd-filter

0.62

0.859

-12dB/oct

3rd-filter

1.62

1.448

-12dB/oct

N=5 filter

-

1

-30dB/oct

Cascading

Q & Ω/Ωc are aligned in increasing order.

The following “Filter Coefficient Table” is the result of these computations for optimization, and the filter design relies on the table actually.

4. Filter Coefficient Table The following table shows the coefficients for the Butterworth filters, in which N is the filter order, k is the number of the partial filter, ak, bk are the filter coefficients, Qk is the quality factor of the partial filter.

Kk is the ratio of the corner frequency of a partial filter, Fck, to the corner frequency

of the overall filter, Fc. Notice that the partial filters are aligned in increasing order of Kk and Qk for optimization. N

k

ak

bk

1

1

1.0000

0.0000

1.000

-

2

1

1.4142

1.0000

1.000

0.71

3

1

1.0000

0.0000

1.000

-

2

1.0000

1.0000

1.272

1.00

1

1.8478

1.0000

0.719

0.54

2

0.7654

1.0000

1.390

1.31

1

1.0000

0.0000

1.000

-

2

1.6180

1.0000

0.859

0.62

3

0.6180

1.0000

1.448

1.62

1

1.9319

1.0000

0.676

0.52

2

1.4142

1.0000

1.000

0.71

3

0.5176

1.0000

1.479

1.93

1

1.0000

0.0000

1.000

-

2

1.8019

1.0000

0.745

0.55

3

1.2470

1.0000

1.117

0.80

4

0.4450

1.0000

1.499

2.25

1

1.9616

1.0000

0.661

0.51

2

1.6629

1.0000

0.829

0.60

3

1.1111

1.0000

1.206

0.90

4

0.3902

1.0000

1.512

2.56

1

1.0000

0.0000

1.000

-

2

1.8794

1.0000

0.703

0.53

3

1.5321

1.0000

0.917

0.65

4

1.0000

1.0000

1.272

1.00

5

0.3473

1.0000

1.521

2.88

1

1.9754

1.0000

0.655

0.51

2

1.7820

1.0000

0.756

0.56

3

1.4142

1.0000

1.000

0.71

4

0.9080

1.0000

1.322

1.10

5

0.3129

1.0000

1.527

3.20

4 5

6

7

8

9

10

Kk=Fck/Fc

Qk

5. Actual Design of the 5th-order Butterworth LPF with cutoff frequency Fc=2400 Hz. (a) The coefficients for the 5th-order Butterworth LPF are obtained from the table as ak

bk

Filter 1 :

a1=1

b1=0

Filter 2 :

a2=1.6180

b2=1

Filter 3 :

a3=0.6180

b3=1

Let us determine the each partial filter by using Fc=2400 Hz. The design is conducted by specifying the capacitor values firstly, then calculating the required resistor values. (b) There are two candidates for the first filter. Let us begin with the 1st-order inverting LPF. The schematic is illustrated below. C || R

R –

Vin

Vout + ≡ The transfer function of the circuit is H(s)=-1/(1+ΩcCRs) If we select C=22 nF (E24, 2% tolerance), then R calculates to R=1/2FcC=1/2•2.4•103Hz•22•10-9F=3.0144 kΩ→3.01 kohm (E96, 1% tolerance) Notice that the value of C with 2% tolerance is firstly specified, then the value of R with 1% tolerance is selected due to commercially available capacitors and resistors. (c) The 1st-order unity-gain LPF is illustrated below. – R

Vout +

Vin =C ≡

With C=22 nF (E24, 2% tolerance), R=3.01 kohm (E96, 1% tolerance) Comparing the both circuits, the input impedance of inverting LPF is pure resistance, however, that of unity-gain LPF is the combination of R and C.

Therefore, if the output impedance of previous

stage is not sufficiently lower than R=3.01 kohm, then the 1st-order inverting LPF is preferable. There is another consideration for the two first filters, that is the output phase. yields 180 degrees phase difference between the input and the output. the phase relation.

The inverting LPF

The unity-gain LPF keeps

The linear-phase requirement does not accept the phase inversion.

(d) The second filter can be realized by the 2nd-order unity-gain LPF since the output impedance of either first filter is sufficiently low. The schematic of the 2nd-order unity-gain LPF is || –

C2 R1

R2

Vout +

Vin = C1 ≡

The transfer function of the 2nd-order unity-gain circuit above illustrated is H(s)=1/[1+ΩcC1(R1+R2)s+Ωc2C1C2R1R2s2] Recall the transfer function for a single stage 2nd-order LPF described in part 2 (e) as Hk(s)=1/(1+aks+bks2) The coefficient comparison between the two transfer functions yields ak=ΩcC1(R1+R2) bk=Ωc2C1C2R1R2 At the k-th stage filter for given C1 and C2, the resistor values for R1 and R2 are calculated through R2, 1=[akC2±(ak2C22-4 bkC1C2)1/2]/(4FcC1C2) Notice that R2 corresponds to + of the symbol ±.

In order to obtain real values under the square root,

C2 must satisfy the following condition C2 ≥C1(4bk/ak2) (e) Recall the values of coefficients for the Butterworth second filter, a2=1.6180 and b2=1.

If we select

C1=18 nF (E24, 2% tolerance), C2 must be C2 ≥C1(4b2/a22)=18•10-9F•4•1/1.6182=27.5 nF→33 nF (E24, 2% tolerance), above value With C1=18 nF and C2=33 nF, calculate the values for R1 and R2 through R1=[a2C2-(a22C22-4b2C1C2)1/2]/(4FcC1C2) R2=[a2C2+(a22C22-4b2C1C2)1/2]/(4FcC1C2) And obtain R1={[1.618•33•10-9-[(1.618•33•10-9)2-4•1•18•10-9•33•10-9] 1/2}/(4•2.4•103•18•10-9•33•10-9) =1.764•103=1.764 kohm→1.78 kohm (E96, 1% tolerance), closest value R2={[1.618•33•10-9+[(1.618•33•10-9)2-4•1•18•10-9•33•10-9] 1/2}/(4•2.4•103•18•10-9•33•10-9) =4.197•103=4.197 kohm→4.22 kohm (E96, 1% tolerance), closest value (f) Let us move on the third unity-gain LPF.

The calculation is identical to that of the second filter,

except that a2=1.6180 and b2=1 are replaced by a3=0.6180 and b3=1. Specify C1 as 6.8 nF (E24, 2% tolerance), and obtain C2 with C2 ≥C1(4b3/a32)=6.8•10-9F•4•1/0.6182=71.2 nF→82 nF (E24, 2% tolerance), above value With C1=6.8 nF and C2=82 nF, the values for R1 and R2 are R1={[0.618•82•10-9-[(0.618•82•10-9)2-4•1•6.8•10-9•82•10-9] 1/2}/(4•2.4•103•6.8•10-9•82•10-9) =1.92•103=1.92 kohm→1.91 kohm (E96, 1% tolerance), closest value R2={[0.618•82•10-9+[(0.618•82•10-9)2-4•1•6.8•10-9•82•10-9] 1/2}/(4•2.4•103•6.8•10-9•82•10-9) =4.12•103=4.12 kohm→4.12 kohm (E96, 1% tolerance), closest value

6. Other Considerations (a) The most important op-amp parameter is the unity-gain bandwidth. The higher-order filters include the highest-valued Q in the last section.

See the filter coefficient table. So, the selection of the

op-amp relies on if the last partial filter can realize the required frequency response at its peak frequency. (b) Resistors values should stay within the range of 1 kohm to 100 kohm. excessive current draw from the op-amp.

The lower limit avoids

The upper limit is to avoid excessive resistor noise.

Capacitor values can range from 1 nF to several μF. The lower limit avoids coming too close to parasitic capacitances. (c) The filter with an even order number consists of 2nd-order stages only. at the beginning.

So, it may need a buffer

The filter with an odd order number includes the 1st-order stage, that is either

inverting or non-inverting filter, which may act also as a buffer section. may be preferred by system designers.

So, the odd-order number