active filter, freq profiles

8/13/2019 Active Filter, Freq Profiles

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PROFESSORS NOTES vers 1.1

24.1 ACTIVE FILTERS AND FREQUENCY PROFILES:

All circuits have a frequency response characterized by the mix of reactive components and active devices. In someinstances, we merely analyze the frequency character of the circuit, and assess it to see if it will accomplish our signalprocessing requirement. But in other instances we desire to command the circuits frequency behavior, and definethe frequency character, or fr equency profile of the circuit.

In most instances we define frequency profiles in terms of amplitude pass functions. There are four basic types:

1) Lowpass2) Highpass3) Bandpass4) Band stop

These are represented by figure 24.11.

|T|

|T||T|

log10 f log 10 f

log10 f log10 f

(dB)

(dB)(dB)

lowpass

bandpass

highpass

bandstop

|T|(dB)

Figure 24.11: Basic frequency profiles

Intermediate and highfrequency profiles use small components, capacitances in pF, inductances in H. Compo-nents are small and compact. Profiles are readily accomplished by means of judiciouslyconstructed RLC networks.The principal concern in highfrequency profiling is the ef fect of the circuit parasitics on the poles and zeros. Parasit -ics include leakage paths, wiring inductances, and fringing capacitances. Frequency profiles in excess of 500 MHzmay use resonant cavities or artificial crystals to define their frequency character.


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At lower frequencies, typically associated with audio systems, biological interface systems, or feedback control sys-tems, components may be large and cumbersome, and therefore techniques have been developed in which active driv -ers are used to replace components or reconfigure a circuit into one which may be cast into integratedcircuit form.These types of circuits are called active filters . An active filter is a frequencyresponsive network driven by one ormore active drivers. The typical driver is an opamp, or one of its cousins. The frequencyresponsive network consistsof resistances and frequencyactive components. Typically, the active driver is used to eliminate one of the more

cumbersome types of frequencyactive components, such as the inductances.

For these profiles, one or more characteristic frequencies are usually necessary to define the stop and passbandedges. Edge and amplitude definitions are represented by figure 24.12, for the four basic profiles

Figure 241.2. Edge and amplitude definition of the four basic frequency profiles.

Note that for bandpass and bandstop profiles, we must provide parameters to characterize both edges, unless the pro-files are symmetric. For frequency profiles of order greater than 2, the passband will (usually) include a rippleamplitude, as illustrated by figure 241.3.


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T able 24.11: Characteristics of the biquadratic pass functions

Coefficients (Normalized) function |T| response

N2 = 0, N 1 = 0T

20

s 2 s 0 Q 20

N0 0

N1 = 0, N 0 = 0

N2 0

T s2

s 2 s 0 Q 20

N2 = 0, N 0 = 0

N1 0

T s 0 Q

s 2 s 0 Q 20

N1 = 0T

s 2 20s 2 s 0 Q 20N2 0, N 0 0

T s 2 s 0 Q 20s 2 s 0 Q 20N2 0, N 0 0

N1 0

Type

lowpass

highpass

bandpass

bandstop

allpass

Note that if we make the choice N 1 = 0 /Q, then we have, for the bandpass function,

(24.13)T s 0 Q

s 2 s 0 Q 20

which is a normalized form of the bandpass function since |T| = 1 when = 0 . This is a convenient form to illustratethe meaning of the qualityfactor Q . Its response is identified by figure 24.13.


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Figur e 24.13. Biquadratic bandpass function

If we analyze equation (24.13) to find the frequency at which |T | 1 2 (= 3dB level), we find that there are foursolutions:

(24.14)02 Q0

2 Q 1 4 Q 2

Of these solutions, the only ones which are greater than zero are:

10

2 Q0

2 Q 1 4 Q 2 and 2

0

2 Q0

2 Q 1 4 Q 2

The difference, , represents the resonance width at 3dB, and will be:

2 10

Q

so that the quality factor Q represents the sharpness of the quadratic resonance peak, as

Q 0 (24.15)

The resonance peak also manifests itself for the lowpass and highpass quadratic functions when Q 1 2 . Thesketches in table 24.11 indicate this behavior .


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EXAMPLE 24.11: We usually have interest in singleamplifier biquadratic circuits since they yield relatively sim -ple building blocks for use in series profiles. The SallenKey circuit shown by figure 24.14 is such an example.

C2

Figur e 24.14: SallenKey singleamplifier biquad. This is a biquadratic lowpass circuit.

Nodal analysis at v1

and v+

gives:

v1(G 1 G 2 sC 1) vOsC 1 v I G 1 v G 2 0

v (G 2 sC 2) v1G 2 0

Using vo = K v + , ( where K = 1 + R B /R A ), gives:

vOv I K

G 1G 2C 1C 2

s 2 sG 2(1 K )

C 2G 1 G 2

C 1G 1G 2C 1C 2

(24.16)

Typically we let C 1 = C 2 = C and G1 = G2 = G , in which case we get:

vOv I K 2

0s2 s 0(3 K ) 20

(24.17)

where 0 = G/C = 1/RC . Note that (24.17 is of the lowpass form.

For this case we see that Q = 1/(3 K), and it is necessary that the feedback ratio R B /R A < 2 for stability.

Since the ratio G2 /C 2 is consistent throughout equation (24.16), we may taper the SallenKey biquad by selectingC 2 = C 1 = C and G2 = G1 = G. This modification does not change the characteristic frequency 0 , but willchange the form of the expression for quality factor to tapered form

Q 1 (2 a K ) (24.18)


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EXAMPLE 24.12 A more general circuit form, the 2integrator loop, can produce most, if not all of the basic biqua -dratic functions. For this reason, it also may be called a statevariable filter. The generalized twointegrator loopis shown by figure 24.15.

0 s

0 s

v1

vi

1

0s v0

20

s2 v0v0

n2

+1/Q

v2

Figur e 24.15: Twointegrator loop, general schematic.

where the integrator component usually is the inverting Miller integrator,

0 s =

The basic feedback circuit consists of two integrator elements and one invertersumming element in a series loop.Analysis of the circuit of figure 24.15 shows that the output of the summing circuit will be:

(24.19)v01Q

0s v0

20

s2 v 0 n 2v i

for which, collecting like terms in v0 and vi .

(24.110)v0 s 20

Q s 20 n 2s

2v i

Resolving equation (24.110), we see that the transfer function from vi to vo with then be of a highpass form:

(24.111)v0vi

n 2s2

s2 s 0 Q 20

From equation (24.110) and the relationship between stages we see that:

v1vi

0s

v0vi bandpass v 2

20

s 2v0v i lowpass

which is why we identify this type of circuit as a statevariable filter , since it provides the three basic biquadratic func -tions, lowpass, highpass, and bandpass. Other biquadratic functions, such as the notch and the allpass can becreated from sums of these functions, usually by means of an extra summing element.

For example, if we add an extra amplifer to perform the sum

v3 v iv1

n 2Q v i 1 1n 2Q

v1v i

then the function

v3vi

s2 20s2 s 0 Q 20

is created. This is the notch, or bandstop function.


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2. R1 can then be adjusted to define Q without changing

0 . Then Q = R 1 /R.3. R3 can then be adjusted to define amplitude without affecting either Q or

0 . Then |A peak | = R 1 /R3

1. R2 may be adjusted to set

0 . Typically , we let C 1 = C 2 = C and R2 = R 4 = R. Then

0 = 1/RC

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Resolving the transfer function in terms of v2 and v1 , (eliminating v2 ) we get

v2v1

G 4G 3 C 1C 2s2 sG 1 C 1 G 2G 4 C 1C 2

(24.114)

which is lowpass. At node v2 , using equation (24.111), we get

v2v1sG 3 C 1

s2 sG 1 C 1 G 2G 4 C 1C 2(24.115)

which is bandpass.

The TowThomas circuit is well accepted because it has a good tuning algorithm. The algorithm is as follows:

We can add a fourth opamp as an inverter/summing stage, such that

v4 (v1 v2) v1 1 v2

v1then, using equation (24.113), we get

v2v1

s2 s(G 1 G 3) C 1 G 2G 4 C 1C 2s2 sG 1 C 1 G 2G 4 C 1C 2

(24.115)

If G1 = G 3 , then equation (24.115) is of the form of a notch filter function. If G3 = 2G 1 , then equation (24.115)is of the form of an allpass function.

24.2 TRANSFORMATION AND RESCALING OF FREQUENCY PROFILES:

In defining a frequency profile, it is usually appropriate to mark a frequency about which the rest of the profile willfall. This frequency typically will be at a symmetry point, a corner, a peak, or a valley . However, this frequency maybe elected from anywhere within the profile, since it is only used as a reference point.

For those profiles that have been standardized, for which placement of poles and zeros are defined according to a spe -cific mathematical criterion, tables are available, just a matter of tracking them down. These tables may either identifythe specific poles and zeros, or if a particular type circuit construct is used, may tabulate values of components. V aluesare tabulated in normalized form, i.e. for a reference frequency 0 = 1 r/s, and it is up to the user to rescale valuesaccording to his/her need.

In identifying the rescaling process we will direct our attention primarily to RC circuits since it is possible to recastmost of our circuits, regardless of complexity, into RC form by use of active filter techniques. The process of rescalingis relatively simple if only one type of frequencydependent component is present, which in this case would be capaci -tances. For RC circuits the frequency profile is defined in terms of the poles and zeros that are given entirely by RCtime constants.

Using normal network techniques, a transfer function T(s) for any given circuit is constructed in terms of either admit -tances or impedances, as indicated by figure 24.21. It can be a simple ratio, or it can be a complex mess, but thefrequency characteristics are entirely the result of the frequency characteristics of these basic admittance (or imped-ance) terms. Therefore we rescale the response T( ) by merely rescaling the individual frequencyactive terms, usinga uniform rescaling process.


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For example, consider the simple, singletime constant RC lowpass circuit indicated by figure 24.21a. We let thereference frequency be the 3dB corner frequency = 1/RC . And for the normalized case, for which 0 = 1 r/s, thewe might have normalized values of R and C to be R = 1 and C = 1 F .

R0 = 1

C0 = 1

0 = 1 r/s

T G 0

sC 0 G 0

R1 = 1

C1 = 1 F

1 = 1 Mr/s

T G 1

sC 1 G 1

1 r/s 1 Mr/s

YC = 0C0 = 1C1

(a) (b)

Figure 24.21 Frequency rescaling: Frequencydependent components (in this case, capacitances) are re-scaled to yield the same response at a new frequency .

We see that the same profile is created at a new frequency 1 , as indicated by figure 24.21(b) if we just rescale thefrequencydependent component C such that it has the same admittance behavior at the new corner frequency as itdid at 0 . If we increase the new frequency by a factor of 10, the capacitative admittance will increase by a factorof 10, just as it did when the corner was at 0 = 1 r/s.

By requiring that the admittance be the same at the new reference frequency 1 as it was at the old reference frequency

0 we then have

1C 1

0C 0 (24.21)

which will produce the same profile at new frequency 1 as it did at

0 , provided C 1 is rescaled by the factor

0 /

1 .

However, since = 1/RC we also realize that the frequency corner could also have been rescaled by an appropriaterescaling of the resistance(s). This rescaling can only be accomplished independently of the capacitance rescalingby mathematically removing the frequencydependence from the capacitances and passing it to the resistances. Thismathematics can be accomplished by examining the transfer function, and noting that if we rewrite the transfer func-tion of the RC lowpass function, and multiply it by the factor s/s, we can lift the frequency dependence from thecapacitances, e.g.

T (s)G 1

G 1 sC 11 sC 1

R1 1 sC 1ss

1 C 1sR 1 1 C 1

With transfer function and components thus changed, we may rescale to yet a new frequency 2 by means of

2 R2

1 R1 (24.22)

where R1 = R 0 . Equation (24.22) rescales only the resistance. Capacitances does not rescale during this process,so that C 2 = C 1 .

What is evident from this twostep process is that we may rescale both the frequencyactive components (capaci-tances) and the frequencypassive components (resistances) independently of one another by passing through an inter -mediate rescaling frequency 1 . The entire transformation takes us from an intiial frequency

0 to a final frequency


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R

C

C

R

250

2 , with the sole purpose of frequency

1 being to rescale one type of component, usually capacitances, to a moreconvenient magnitude scale, such as F, as indicated by figure 24.21. Continuing the example given with figure24.21, we may desire to have the same STC lowpass profile fall at 2 = 2 kr/s, for which we would rescale the valueof resistance, R0 , by equation (24.22) to the new value:

R 2

1

2 R 1

10 62 10 3

1 500

while the capacitance has (already been) rescaled to:

C 2 C 1

0

1C 0

110 6

1F 1 F

We can, of course, take this same process and apply it to a circuit in which we have many resistances and capacitances,for which the entire set of resistances and capacitances will be rescaled according to an initial frequency 0 , a finalfrequency f , and an intermediate frequency

M . The rescaling algorithm for RC circuits, in general, may be takenfrom equations (2421) and (24.22), as:

C f

0

M C i R f

M

f R i (24.23)

where {C i , Ri } are the initial values of the set of the set of resistances and capacitances, and {C f , R f } are the final values,and where M is the intermediate rescaling frequency. As indicated by the example, the rescaling frequency

M isusually chosen to rescale one type of component, usually capacitances, to a more convenient scale value. It may bedefined by taking one of the capacitances and defining a final value for this capacitance, in which case M will begiven by:

M

0C i1C f 1

(24.24)

where C i1 and C f1 represent the initial and final values of the selected capacitance.

24.3 THE RCCR TRANSFORMA TION

As we investigate the set of pass functions, we find that the fundamental profile is lowpass . We can derive other passfunctions by means of frequency transformations .

One of the frequency transformations is a simple complementary transformation. Low and highpass functions arecomplementary about their 3dB corners, inasmuch as the 3dB corner represents the frequency at which the reactiveimpedance is equal to the nonreactive impedance. We can see the complementary behavior by comparing the RC(=singletimeconstant) (STC) circuit for low and highpass circuit forms:

Figur e 24.31: The RCCR transformation.


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Note that when |1/j C| = R, then the transfer function becomes

v2v1

11 2 p 21

12 (24.31)

where p1 = 1/RC . This frequency represents the corner frequency of the Bode magnitude plot, a reference frequen -cy at which the transfer function has the value 3 dB. As we increase the frequency , the magnitude of the transfer

function will change, corresponding to whether the reactive impedance decreases the ratio (lowpass) or increasesit (hipass).

RCCR conversion, in which a lowpass function is transformed to a highpass function with the same corner, mustfollow the impedance effects, referenced to that point at which they are equal, which is the 3dB level. If we desireto invert the impedance effects, we must change the gender of the components centered at the frequency at which theeffects balance. This transformation is achieved when the impedance of each reactive component is equal to the im-pedance of an equivalent nonreactive component, and conversely. The transformation must then occur for:

R k 1

3dBC k

C k 1

3dB Rk

(24.32)

(24.33)

where R, C represent the replacement values of resistance and capacitance, respectively .

24.4 RLC LADDER CIRCUITS

As indicated by section 24.3 we find that it is appropriate to base our circuits on the lowpass frequency profile and thenapply a transformation to take us to another profile. If we examine lowpass profiles in the general sense, we perceivelowpass filters as being those which suppress all frequencies above a bandedge. After we have surveyed at some of the basic filters, e.g. one with a single resistance and capacitance, we find that the bandedge may be relatively soft.Many times we would like the cutoff edge to be much more abrupt, or to have other features, such as uniform phaseshift of pulse signals above and below the edge.

We get more abrupt cutoff if we add more poles, via additional reactive components. Rolloff is accomplished bythe increase of the or der of the rolloff profile. The rolloff is approximately (20 dB/decade order ).

As an example we consider the 5thor der Chebyshev profile, which has a very abrupt cutoff, on the order of 100 dB/de -cade, accomplished by placing poles and zeros according to the Chebychev functions. We will not undertake any of the discipline and entertainment associated with network analysis of the circuit, but will assume that tabulated resultsfor placement of poles and zeros, as ordained by many previous analyses of this profile, are reasonable and accurate.

The Chebyshev profile and many other profiles can be straightforwardly implemented by means of the doublytermi-nated RLC ladder , as shown by figure 24.41. The figure shows a 5thorder ladder . Note that it has five (5) frequencyactive components. And since the inductances are in series and the capacitances in parallel, we should recognize thatthis configuration is of lowpass form. By judicious choice of component values, poles and zeros may be defined suchthat this circuit is a 5thorder Chebyshev profile.

Figur e 24.41. Example of (5thorder) doublyterminated ladder .

C3 C5

R1

R7

L4 L6L2


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It might be noted that RLC ladder circuits are favored for frequency profiling since the circuit is relatively insensitiveto small variations of component values, a statement that we will not prove, but merely accept these types of circuitsas a robust and are a convenient baseline. For active filter implementation we prefer that some component types, usual -ly the inductances, be converted to equivalent active forms. As we will see, the doublyterminated ladder lends itself readily to component transformations, and to transformations into other type pass functions,

As indicated by tables 24.41, and 24.42, doublyterminated RLC lowpass filter profiles are readily available in

tabulated form. The tables are invariably based on a normalized corner frequency of

0 = 1 r/s .

Table 24.41: Table of doublyterminated RLC ladder values for normalized Butterworth lowpass re-sponse.

n C 1 C3 C5 C7 C9L2 L4 L6 L8 L10234567

8

910

1.414 1.4141.000 2.000 1.000

0.7654 1.848 1.848 0.76540.6180 1.618 2.000 1.618 0.61800.5176 1.414 1.932 1.932 1.414 0.51760.4450 1.247 1.802 2.000 1.802 1.247 0.4450

0.3902 1.111 1.663 1.962 1.962 1.663 1.111 0.3902

0.3473 1.000 1.532 1.879 2.000 1.879 1.532 1.000 0.34730.3129 0.9080 1.414 1.782 1.975 1.782 1.414 0.9080 0.31291.975

n L1 L3 L5 L7 L9C2 C4 C6 C8 C10

C2 C4

1

1

L3 LnL1

C3 C5

1

1

L4 LnL2

C1


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Table 24.42: Table of doublyterminated RLC ladder values for normalized Chebyshev lowpass response.

n C 1 C3 C5 C7L2 L4 L6

23

45

67

8

0.84304 0.622011.03156 1.14740 1.03156

1.10879 1.30618 1.77035 0.818071.14681 1.37121 1.97500 1.37121 1.14681

1.16811 1.40397 2.05621 1.51709 1.90280 0.861841.18118 1.42281 2.09667 1.57340 2.09667 1.42281 1.18118

1.18975 1.43465 2.11990 1.60101 2.16995 1.58408 1.94447 0.87781

n L 1 L3 L5 L7C2 C4 C6 C8

C2 C4

1

R2

L3 LnL1

C3C5

1

R2

L4L2

C1

3

5

7

1.5963 1.0967 1.5963

1.7058 1.2296 2.5408 1.2296 1.7058

1.7373 1.2582 2.6383 1.3443 2.6383 1.2582 1.7373

357

2.0236 0.9941 2.02362.1349 1.0911 3.0009 1.0911 2.13492.1666 1.1115 3.0936 1.1735 3.0936 1.1115 2.1666

L8 R2

0.737811.00000

0.737811.00000

0.737811.00000

0.73781

1.00000

1.00000

1.00000

1.00000

1.00000

1.00000

R2

(A) Ripple = 0.1 dB

(B) Ripple = 0.5 dB

(C) Ripple = 1.0 dB


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24.5 RLC:CRD TRANSFORMA TIONS:

Circuits such as that shown by figure 24.41 can be implemented in active form using only resistances and capaci-tances by executing a mathematical transformation of the circuit into one in which each type component has an equiva -lent. If we desire to transform an RL circuit into an RC equivalent, then it is necessary to multiply the the numeratorN(s) and the denominator D(s) of the transfer function by 1/s. This process is represented by figure 24.51.

Figur e 24.51 Transformation process in which the LR circuit is t ransformed into an equivalent RC circuit.

In this process we note that the L is transformed into an R and the R is transformed into a C . The same characteristicfrequency 0 results provided that the magnitude of R is the same as the magnitude of L and the magnitude of C isthe same as the magnitude of 1/R . This type transformation is of the form RL:CR

But when we have RLC circuits, the transformation must include all three type components and we cannot eliminatethe inductance L unless we define a new component derived from capacitance C . The technique is called theRLC:CRD transformation and is much like the RL:CR transformation, except C is transformed into an active equiva -lent = D . Otherwise for the transformation R is transformed into an equivalent C , L is transformed into an equivalent

R .For the RLC:CRD transformation we imply that R C , L R , and C D . The transformed circuit is representedby figure 24.52.

C7

R4C1 R6

D3 D5

R2

Figure 24.52. Same circuit as figure 24.41, (RLC doublyterminated ladder), transformed into CRD form.

If the same transformation as used for the LR:RC in which numerator and denominator components are are multipliedby 1/s is applied to and RLC circuit, then each capacitative impedance must be transformed into a component of fre-quency behavior:

1 sC 1 s 1 s 2C 1 s 2 D

T R R sL

T 1 sC

1 sC R

(a)

(b)

T R R sL

1 s

1 s R s

R s L

0 = R/L

0 = 1/RC

L RR 1/C

R

R

C

transformation:

L


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The strangelooking component, D , therefore depends on frequency as the square of s. This response is of the form of a frequencydependent negative resistance . It is naturally an active component, and may be implemented by means of the GIN (generalized immitance network) circuit, as shown by figure 24.53 (below).

Z IN

Figur e 24.53. GIN configured as frequency dependent negative resistance (FDNR)

Evaluate this thing and you will find that the GIN component has an admittance proportional to the square of the fre-quency. The magnitude of its admittance is

| Z D | 2 D 2C 2C 6 R5

This component will have the measure of a negative resistance with magnitude proportional to 2 .

RESCALING: As indicated by previous sections, tabulated frequency profiles, such as the Chebyshev are alwaysgiven for normalized at frequency 0 = 1 r/s and must be rescaled to the frequency of interest. For RLC:CRD trans -formations we might note that the resistances will be transformed into capacitances. In prototyping of circuits, we

often select a given value of capacitance, particularly if it is used more than once, and let it determine the scaling factor .It is convenient, so let us so do. Capacitance magnitudes will be transformed from a starting value of resistance accord -ing to C = 1/R . As capacitances, rescaling will take place after the transformation, for which

M 1

R i f C f

where M is the intermediate scaling frequency and f is the desired (final) characteristic frequency of the given cir-cuit. Using this value of scaling frequency all capacitances will scale according to

C f M

f

1 R i

(24.51)

Equation (25.51) may be used to get values for all of the capacitance in figure 24.42. Note that we must use f , thefrequency in rad/sec , since the profiles are always normalized in terms of the unity radian frequency 0 = 1 r/s.

Since L R , all inductances will be transformed and rescaled to resistances. Frequency rescaling must take placebefore before they are transformed into resistances, for which:

R f 0

M L i (24.52)


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Note that we must include resistances R A and R B to accommodate low frequencies and provide a DC path for the noninverting terminals of the opamps. The values selected are such that R >> 0 C and such that

v2v1 0.5

R B R A R2 R4 R B (24.45)

so that the transfer gain is 0.5 when f


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Figur e 24.57: Spreadsheet matrix for the 5thorder lowpass Chebyshev response

In the calculation matrix we did not identify the (intermediate) scaling frequency M . In this case it is defined by ourchoice of capacitance value, C = 1050 pF, for which, according to equation (24.51), we have:

M C f

f R1 1050 pF (2 10 4) 1 .066 mr s

where R1 = 1 , this case.

If everything has been done properly, a SPICE analysis of this circuit will look something like that shown by figure24.58


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Figure 24.58. 5thorder lowpass Chebyshev response

24.6: BANDPASS IMPLEMENT ATIONS FROM LOWPASS RLC LADDER

Because of its relative insensitivity to component variations, the doublyterminated RLC ladder is favored as a meansof defining higherorder polynomial profiles. The 3rdorder doublyterminated RLC ladder is represented by figure24.61

C3

R1

R5

L4L2

Figur e 24.61. Example of 3rdorder doublyterminated ladder .


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261

Y1 Y3 Y5

Z2 Z4 Z6

TY1 TZ2 TY3 TY5TZ4 TZ6V1V 6

V1 V6RLC Ladder

Leapfrog equivalent

V4V2

V2+

+

V4

Figure 24.64. Leapfrog techniques for simulation of a (RLC) ladder

For a ladder construction, as indicated by figure 24.64a, network analysis gives us the set of equations:

I 1 Y 1(V 1 V 2)

V 2 Z 2( I 1 I 3)

I 3 Y 3(V 2 V 4)

V 4 Z 4( I 3 I 5)

I 5 Y 5(V 4 V 6)

(24.61a)

(24.61b)

(24.61c)

(24.61d)

(24.61e)

Each of these equations represents a transfer function relationship, and can be expressed as such:

V I 1 T Y 1(V 1 V 2)

V 2 T Z 2(V I 1 V I 3)

V I 3 T Y 3(V 2 V 4)

V 4 T Z 4(V I 3 V I 5)

V I 5 T Y 5(V 4 V 6)

(24.62a)

(24.62b)

(24.62c)

(24.62d)(24.62e)

In each case we see that the righthand side of these transfer functions makes use of a simple difference of two inputs.To our great satisfaction we see that each of these inputs is available on the lefthand side, with exception of only V 1 ,which is THE input.

If we instead write the difference as addition of a negative, then we may find it to be advantageous to set up our set of equations accordingly. For example:

V 2 T Z 2[( V I 1) V I 3] (24.63b)


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provides us one of the inputs ( V 2 ) that is needed in equation (24.62a).

Now we see that equation (24.63b) needs a term ( V I1 ). Therefore we might rewrite (24.62a) as:

( V I 1) T Y 1[V 1 ( V 2)] (24.63a)

If we continue with this approach through the whole list of transfer functions then we wil l have the set:

V I 1 T Y 1[V 1 ( V 2)] V 2 T Z 2[( V I 1) V I 3]

V I 3 T Y 3[( V 2) V 4]

V 4 T Z 4[V I 3 ( V I 5)] V I 5 T Y 5[V 4 ( V 6)]

(24.63a)

(24.63b)

(24.63c)

(24.63d)(24.63e)

This set of equations is exactly the equivalent of the circuit of figure 24.64b. Note that the transfer functions alternatein sign as we progress through the circuit.

The only aspect of leapfrog circuits that creates any complication is the signs of each of the impedance/admittanceterms T Zk or T Yk , respectively , in the string. In some cases, inverting stages must be inserted into the leapfrog string toensure correct sign.

The form of the RLC ladder when converted to bandpass, is of the form shown by figure 24.65.

R1

R5VI

VO

TY1

TZ2

TY3

Figur e 24.65. 3rdorder RLC ladder converted to bandpass form

The components enclosed by the dashed lines each represent an impedance or admittance term that can be realizedby a biquadratic circuit. The secondorder function for each of these component groupings is identified by Figure24.66. Note that, for the leapfrog realization, the series RLC string must relate to an admittance term whereas theparallel RLC, (LC if R not present) string must relate to an impedance term.


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R L C Y (1 L)s

s2 ( R L)s 1 LC

RC

L Z (1 C )s

s2 (1 RC )s 1 LC

CL Z

(1 C )ss2 1 LC

Figur e 24.66a. RLC series admittance function

Figur e 24.66b. RLC parallel impedance function

Figur e 24.66c. LC parallel lossless impedance function

Figure 24.66.

Subcircuits and equivalent (secondorder) impedance/admittance functions.

All of these RLC strings can be realized by an active circuit, with transfer function suitable for the leapfrog realizationby use of the DelyannisFriend singleamplifier biquad. The normalized Friend circuit ( 0 = 1 r/s) and equivalenttransfer function(s) are shown by figure 24.67. The two configurations presented are those suitable to the impedance/ admittance functions of figures 24.66a and 24.66c.

T (s)2kQs

s2 (1 Q)s 1

T (s) k (1 2Q 2) Q s

s2 1

Figur e 24.67a. Friend circuit and transfer function needed for RLC realizations

Figur e 24.67b. Friend circuit and transfer function needed for LC (lossless) realizations.

Figur e 24.67. DelyannisFriend biquadratic forms appropriate for realization of RLC circuits.


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If the input of either of these configurations is also used as a summing point, then the modification shown by figure24.68 must be made to the input node of the circuit. The transfer function is not affected as long as the parallel com -bination of the set of input resistances add up to R = 1 (or G = 1).

T (s)2kQs

s2 (1 Q)s 1

Figur e 24.68. Friend circuit with summing at input

Using the Friend circuit to realize impedance and admittance functions, the bandpass 3rdorder ladder given by figure

24.65 can be implemented in leapfrog form, as represented by figure 24.69.

Figure 24.69. Leapfrog Implementation of 3rdorder Chebychev bandpass

Note that the required impedance characteristics can be selected by appropriate choices of k 1 , k 2 , and k 3 , and Q1 , Q2 ,Q3. Since Q1 , Q 2 , Q 3 all define capacitances, it is convenient to to choose Q1 = Q 2 = Q 3 , which will make all capaci -tances equal. Note that the inverter/sum stage has arbitrary resistance values, the only requirement being that they

be equal.An example of a bandpass filter for f 0 = 10 kHz and BW = 5 kHz is represented by figure 24.612. The bandpass filteris developed from the 3rdorder 1dB ripple lowpass Chebyshev doublyterminated ladder shown by figure 24.610.and its bandpass realization, shown by figure 24.61 1.


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1

1

2.0236 2.0236

0.9941

Figur e 24.610. 3rdorder 1dB Chebyshev doublyterminated RLC ladder

VIVO

1.0 4.0472

1.9882

4.04720.2471 0.2471

1.00.503

R1 L2C2

C3 L3

C4L4

R5

Figur e 24.61 1. Bandpass realization of the normalized 3rdorder 1dB Chebyshev doublytermi -nated ladder . For these values 0 = 1 r/s and BW = 5kHz/10Khz = 0.5 . Resistances are in ohms and capaci -tances are in Farads. This realization was achieved by the component transformation shown by figure 24.62.

ANALYSIS: When we compare transfer functions in figure 24.67 to the circuit transfer functions in figure 24.66,we see that Q1 = L 2 /R1 = 4.0472/1.0 = 4.0472. Since L4 = L 2 and R5 = R 1 , then Q4 also = 4.0472.

Comparing numerators for the transfer function of figure 24.67b to the impedance function of figure 24.66c, asneeded to realize stage T Z2 , we see that

1 C k 2(1 2Q 22) Q 2 (24.64)

We have flexibility of electing k 2 and then finding Q2 , or electing Q2 and then finding k 2 . As noted earlier, it is of considerable convenience to the qualification process to let Q2 = Q 1 = Q 4 = 4.0472, since this choice will give us acircuit for which all capacitances are equal. Using this choice for Q 2, then

k 21 C 3

(1 2Q 22) Q 21

1.9882 8.341 0.0603

with the choices of Q made equal, all capacitances will be equal. Furthermore, specification of Q will also let us speci -fy several of the resistances.

Other resistances are determined by requiring that numerator of the transfer function for figure 24.67a be equal tothe admittance numerator of figure 24.66a. This circuit is needed to realize stage T Y1. Stage T Y3 has identical re-quirements except that it does not need to be used as a summing point. This requirement gives

2Q 1k 1 1 L 2 1 4.0472 (24.65)

Since Q1 = 4.0472, we get k 1 = 0.0305. Likewise we get k 3 = 0.0305. Now we have defined all of the values of k and Q , which is sufficient to define all of normalized values of resistance and capacitance. The leapfrog circuit, withnormalized values of resistance and capacitance, is shown by figure 24.612.


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0.1235 65.5232.85

1.065

32.85

16.61

1.064

1 32.76

1.03

0.1235

0.1235

0.1235

0.1235

0.1235

65.52 65.5232.85

Figur e 24.612. Example normalized circuit values for C3 bandpass profile, 0 = 1, BW = 0.5.

Note that a summing inverter is embedded in between T Y1 and T Z2 to accommodate the necessary sign change. Notethat resistance r is arbitrary, and we do not need to rescale it to some inconvenient value.

The last step is to rescale capacitances and resistances such that the circuit is implemented at the required characteristicfrequency. RC rescaling, when set by a final value of capacitance C f , is accomplished by determining an intermediate -scaling frequency M such that

M (C i C f ) 0 (24.66)

Since the circuit is developed in terms of 0 = 1 r/s, then the intermediate frequency is just M = C i /C f . In the examplecase, for which we have elected C f = 1050 pF from initial value C i = 0.1234 F , we find M = 0.1175 Gr/s. The resist -ance values will then scale according to

R f R i( M f ) (24.67)

where f is the final desired frequency. If we apply this rescaling to the values in figure 24.612 we end up with thefinal design for an f 0 = 10 kHz and BW = 5 kHz, as given by figure 24.613. For convenience, we let r = 10 k sinceit is not necessary to scale this value.

1050

1050

1050

1050

1050

1050

122.861.4

1.99

61.4

31.0

1.99

1.87

122.8 122.8

61.4

1.93

61.4

Figure 24.613. EXAMPLE FINAL DESIGN: Bandpass implementation for 3rdorder 1dB Chebyshevdoublyterminated ladder when f 0 = 10 kHz and BW = 5 kHz. Since Q1 = Q2 = Q 3 , the scaling factor waschosen such that all capacitances would be 1050 pF. All resistance values are in k .


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RECIPE FOR ANALYSIS:

1. Starting with figure 24.610, execute an LPBP transformation and develop values for a figure of the formof 24.61 1.

2. From figure 24.611, develop the appropriate values for Q

, Q2 , Q3 and k 1 , k 2 , k 3 , such that all capacitanceswill be equal. Using these values implement the normalized design, of the form of figure 24.612.

3. Choose a value of capacitance. Then determine scaling frequency m and resistance values necessary todefine the final circuit, such as that given by figure 24.613.

Create a table showing the renormalization process and the rescaled values of the components. You may find that aspreadsheet is useful for executing this process. In the spreadsheet which I created, I arranged all of the final values of resistances and capacitances to fall in a column, which could then be exported and printed for use as an input file toPSPICE. An example of a spreadsheet generator is shown by figure 24.614.

Figure 24.614. Spreadsheet generator. The values shown are those used to develop the circuits shown byfigures 24.611, 24.612 and 24.613.


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4. Check analysis with SPICE. For the BP3C1 form, the output is shown by figure 24.615

Figur e 24.615. Typical SPICE output for 3C1 bandpass filter. Opamps were assumed to be ideal.


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