Filters and Tuned Amplifiers

• Essential building block in many systems, particularly incommunication and instrumentation systems

• Typically implemented in one of three technologies: passive LCfilters, active RC filters, switched-capacitor filters

• Passive LC filters work well at high frequencies, however, at lowfrequencies the required inductors are large, bulky and non-ideal.Furthermore, inductors are difficult to fabricate in monolithic form andare incompatible with many modern assembly systems

• Active RC filters utilize opamps together with resistors and capacitorsand are fabricated using discrete, thick-film and thin-film technologies.The performance of these filters is limited by the performance of theopamps (e.g. frequency response, noise, offsets, etc.)

• Switched-capacitor filters are monolithic filters which typically offerthe best performance in terms of cost. Fabricated using capacitors,switches and opamps. Generally poorer performance compared topassive LC or active RC filters

Butterworth

• For a Butterworth filter the optimum is flat response in the passbandand steep slope soon after cutoff (maximally flat filter)

• The problem is similar to finding Fourier series of square wave. Thesolution lies in the use of Butterworth Polynomials

• 4 and 8 pole Butterworth filters common. 12 and 16 pole filtersconsidered fairly sophisticated

• Standard table can be used to find component values

Chebyshev

• In some applications the sharpness of the cutoff response is moreimportant than the passband flatness

• By adding higher resonant peaks it is possible to obtain sharper cutoffat the expense of peaks in the passband

• We also observe that the transmission decreasestowards -∞ as ω approaches ∞. Thus the filtermust have one more zero at s = ∞. In general, thenumber of zeros at s = ∞ is the difference betweenthe degree of the numerator polynomial, m, and thedenominator polynomial , n

• This is because as s approaches ∞, T(s) approachesam / s

n-m and thus is said to have n − m zeros at

s = ∞• For a filter circuit to be stable all its poles must lie

in the left half of the s plane and thus all the polesmust have negative real parts

• the figure on the next page shows typical pole andzero locations for the transmission functiondepicted in figure 11.3

• We have assumed this filter is of fifth order (n =5). It has two pairs of complex conjugate polesand one real axis pole

• All the poles lie in the vicinity of the passbandwhich gives the filter its high transmission atpassband frequencies

• The transfer function of the filter is of the form

( )( )T s

a s s

s b s b s b s b s bL L

( ) =+ +

+ + + + +4 2 1

2 22

2

54

43

32

21

10

ω ω

• As a further example consider the low pass filterbelow. We observe that there are no finite valuesof ω at which transmission is 0. Thus all zeros areat s = ∞. The filter transfer function is of the form

T sa

s b s b s bm

nn

nn

n( ) =+ + +−

−−

−1

12

20K

• Such a filter is known as an all-pole filter

Butterworth and Chebyshev FiltersButterworth Filter

The figure below shows the magnitude response of a Butterworth filter

• The filter exhibits a monotonically decreasing transmission with alltransmission zeros at ω = ∞, making it an all pole filter

• The magnitude function for an nth

order Butterworth filter with apassband edge ωP is given by

T j

p

n

p

p

( )

)

ω

ε ωω

ω ω

ωε

=

+

=

=+

1

1

1

1

2

2

2

at

T(j

The parameter ε determines the maximum variation in passband transmission, AMAX (i.e. passband ripple)

( )AMAX

AMAX

= +

= −

20 1

10 1

2

10

log

/

ε

ε

• At the edge of the stopband ω = ωS

A S

S

P

n

S

P

n

( ) log

log

ω

ε ωω

ε ωω

= −

+

= +

20 1

1

10 1

22

22

• This equation can be used to determine the filter order required,which is the lowest value of n that yields A(ωS) ≥ AMIN

• The poles of a Butterworth filter can be determined from thegraphical construction below

• The poles lie on a circle of radius ωP(1/ε)1/n

and are spaced at equalangles of π/n

• Since all poles have equal radial distance from the origin they allhave the same frequency

ω ω εο =

P

n1 1

• Once the poles are known the transfer funciton can be written as

( )( ) ( )T s

ks p s p s p

n

n

( ) =− − −

ω ο

1 2 K

where k is a constant equal to the required DC gain of the filter

The Chebyshev Filter• The figure below shows representative transmission functions for

Chebyshev filters of even and odd order• The Chebyshev filter exhibits an equiripple response in the passband

and a monotonically decreasing transmission in the stopband• The total number of passband maxima and minima equals the order

of the filter• All the transmission zeros of the Chebyshev filter are at ω = ∞

making it an all pole filter• The magnitude of the transfer function of an nth order Chebyshev

filter with a passband edge ωP is given by

T j

n

T j

n

p

p

p

p

( )

cos cos

( )

cosh cosh

ωε ω

ω

ω ω

ωε ω

ω

ω ω

=+

≤

=+

≥

−

−

1

1

1

1

2 2 1

2 2 1

At the passband edge ω = ωP

T j( )ωε

=+

1

1 2

The parameter ε determines the passband ripple according to

( )( )

AMAX

X

= +

= −

10 1

10 1

2

10

log

/

ε

εΑ

ΜΑ

• The attenuation achieved by the Chebyshev filter at the stopbandedge (ω = ωS) is

( )A nSS

P( ) log cosh coshω ε ω

ω= +

−10 1 2 2 1

The poles of a Chebyshev filter are given by

Pkn n

jkn n

k p

p

= −−

+−

−ω πε

ω πε

sin sinh sinh

cos cosh sinh

2 1

2

1 1

2 12

1 1

1

The transfer function of the Chebyshev filter can be written as

( )( ) ( )T s

k

s p s p s pp

n

nn

( ) =− − −−

ωε 2 1

1 2 K

• The Chebyshev filter provides greater stopband attenuaton than theButterworth filter at the expense of passband ripple

Bessel (Tompson) Response

• The sharper the filter cutoff the worse the phase shift right beforecutoff (response after cutoff usually doesn’t matter as the signal isattenuated)

• Poor phase response results in unequal delay which results in poortransient response. For a low pass filter this would normally result inoutput ringing when given a high frequency response

• Bessel (Tompson) filter is a maximally flat filter which provides linearphase shift. (i.e. equal delay). A Bessel filter can be used tocompensate phase shifts introduced by other parts of the system

Filter Transfer Function• The filter transfer function can be written as the

ration of two polynomials

T sa s a s a

s b s bm

mm

mo

nn

no

( ) =+ + +

+ + +−

−

−−

11

11

K

K

• The degree of the denominator, n, is the filterorder. For the filter to be stable m ≤ n

• The numerator and denominator coefficients arereal numbers. The polynomials in the numeratorand denominator can be factored and T(s) can beexpressed in the form

T sa s z s z s z

s p s p s pm m

n

( )( )( ) ( )

( )( ) ( )=

− − −− − −

1 2

1 2

K

K

where the z’s are the zeroes and the p’s are the poles

• Each zero or pole can be a real or complexnumber. Complex zeroes and poles, however,must occur in conjugate pairs. Thus if -1 + j2 is azero then 1 + j2 is a zero

Frequency Transformations

• We can transform a low pass filter to a high passfilter using the following transformation

ps

=1

Third order high-pass Butterworth given by

T sk

s s s

k ss s s

( ) =

+

+

+=

+ + +12

12

11

2 2 13 2

3

3 2

• The high pass response will have the same cutofffrequency and passband flatness

• Usually there is a simple transformation of Cs, Rs

and Ls which will depend on the circuit topology

( )

( )

Low pass Band pass

ps

s

Low pass Band reject

ps

s

dB

dB

− →

=+

=

− →

=+

2 2

3

3

2 2

ωω

ω

ω

ω

οο

ο

Center Frequency

• Since in the filter stopband the transmission is required to be zero orsmall, the filter zeroes are usually placed on the jω axis at stopbandfrequencies

• This particular filter can be seen to have infiniteattenuation (zero transmission) at two stop bandfrequencies ωL1 and ωL2 the filter must have zerosat s = + jωL1 and s = + jωL2 however, sincecomplex zeros occur in conjugate pairs there mustalso be zeros at s = − jωL1 and s = − jωL2

• Thus the numerator polynomial of this filter willhave the factors

( )( )( )( )s j s j s j s jL L L L+ − + −ω ω ω ω1 1 2 2

which can be written as

( )( )s sL L2

12 2

22+ +ω ω

For s = jω the numerator becomes

( )( )− + − +ω ω ω ω21

2 22

2L L

which is zero at ω = ωL1 and ω = ωL2

• The section cutoff frequencies and Q factors can be determined usingChebyshev Polynomials . Whereas for the Butterworth filter we onlyspecify the number of poles (zeroes), for a Chebyshev filter we specifythe number of poles (zeroes) and passband flatness. (i.e. a 0.5 dBChebyshev filter has a maximum peak 0.5 dB above the minimumvalley in the passband (Equal - Ripple filter)

Elliptical (Cauer) Response

• The filter cutoff response can be improved further by following a basiclow pass filter with a notch filter. The notch decreases the responsejust after cutoff

• To be effective the notch has to be narrow and as a result the overallresponse will start to increase just after the notch

• To eliminate this we add a number of notches until the original filtercurve has dropped low enough

• Filter is specified with three parameters: passband ripple, order,minimum stop band attenuation

• Due to the large number of options it is difficult to find tables

• Filters are generally linear circuits that can be represented as a twoport network

• The filter transfer function is given as follows

T j T sV sV s

o

i

( ) ( )( )( )

ω = ≡

• The magnitude of transmission is often expressedin decibels in terms of the gain function

G T j dB( ) log ( )ω ω= 20

• A filter shapes the frequency spectrum of the inputsignal, according to the magnitude of the transferfunction

• The phase characteristics of the signal are alsomodified as it passes through the filter

• Filters can be classified into a number of categoriesbased on which frequency bands are passedthrough and which frequency bands are stopped

• The above transmission characteristics are for ideal filters. Realistictransmission characteristics for a low pass filter are shown below

• Transmission of a low pass filter is specified by four parameters

1. Passband edge, ωP

2. Maximum allowed variation in passband transmission, AMAX

3. Stopband edge, ωS

4. Minimum required stopband attenuation, AMIN

• The ratio ωS / ωP is usually used to measure the sharpness of the filterresponse and is called the selectivity factor

• The more tightly one specifies a filter (i.e. lower AMAX, higher AMIN,ωP / ωS closer to unity) the resulting filter must be of higher order andthus more complex and expensive

• Common to refer to AMAX as the passband ripple

• The process of obtaining a transfer function that meets givenspecifications is known as filter approximation. Filter approximationis usually performed using computer programs or filter design tables.In simpler cases, filter approximation can be performed using closedform expressions

• The figure below shows transmission specifications for a bandpassfilter

General Notes

• Each capacitive / inductive element adds a pole or zero to thefrequency response

• By adding additional poles / zeroes we can increase the rolloff of thefilter response (e.g. two pole low-pass filter would have -40 dB /decade rolloff at high frequency)

• In adding poles we must take into account the loading of successivestages

• Addition of buffer removes impedance matching effects

• When capacitors are combined with inductors it is possible to makecircuits with very sharp frequency response

T ssC

R sLsC

s L C sRC

sQ

s

LC

QL

R L C RC

( ) =+ +

=+ +

=+ +

=

= =

1

11

1

1

1

2

2 2

ωω ω

ω

ω

ο

οο

ο

• Two pole rolloff (40 dB / decade)

• Q of circuit affects response

• When Q = 1/2 response is equivalent to two pole RC filter

• High Q pumps up gain below cutoff and increases slope after cutoff

• Single RC filter has large passband ripple and shallow rolloff

• Addition of RC stages increases rolloff, however passband flatness notaffected and using a large number of stages does not give sharp cutofffor the first 50 dB. 8 section RLC is better than 32 section RC

• By adjusting the Q of each section of a filter it is possible to achievebetter response

• For a two section filter we can increase the Q of the second stage sothat the peak fills in the rounded area just beyond cutoff of the firstsection

• The scheme improves the passband flatness just short of cutoff andallows us to reach ultimate slope just after cutoff

• How do we select the Q to give optimum performance and what is ouroptimum?