ECEN 2420 Wireless Electronics for Communication Spring 201203-03-12 P. Mathys

Using Filter Tables

1 Normalized Filters

The figure below shows the structure of a lowpass ladder filter of order n with a normalizedresistive load of RL = 1 Ω.

VS

V

RS

R

C1

C

C3

C

Cn

CRL1.0

L2

L

Ln

L

RL1.0

n even n odd

Note that, depending on whether the filter order n is even or odd, the last element is eithera horizontal L branch (n even) or a vertical C branch (n odd).

The values of the individual capacitors and inductors depend on the filter type, the filterorder n, the value of the source resistor RS, and, of course, the cutoff or reference frequencyof the lowpass filter (LPF), ωr = 2πfr. The computation of these values is somewhatcomplicated. To simplify the design of filters, tables for the L and C have been computedfor a normalized reference frequency of ωr = 1 rad/s.

For Butterworth filters of order n with RS = 1 the normalized susceptances and reactancesai (also called immittances) are given by

ai = 2 sin((2i− 1)π

2n

), i = 1, 2, . . . , n ,

where i is the component index. The table below shows these values for n ≤ 7.

Order RS C1 L2 C3 L4 C5 L6 C7

a1 a2 a3 a4 a5 a6 a7

1 1.0 2.00002 1.0 1.4142 1.41423 1.0 1.0000 2.0000 1.00004 1.0 0.7654 1.8478 1.8478 0.76545 1.0 0.6180 1.6180 2.0000 1.6180 0.61806 1.0 0.5176 1.4142 1.9319 1.9319 1.4142 0.51767 1.0 0.4450 1.2470 1.8019 2.0000 1.8019 1.2470 0.4450

Example: The figure below shows the schematic of a normalized Butterworth LPF of ordern = 3.

1

VS

AC 1

RS

1

RL1

C1

1

C3

1

L2

2

.ac dec 200 0.01 1

N003

The system function H(s) = VO(s)/VS(s), where VO is the voltage across the load resistorRL, of this filter for general component values is

H(s) =RL

RS + RL

· 1

RSRLL2C1C3

RS + RL

s3 +RSC1 + RLC3

RS + RL

L2s2 +

RSRL(C1 + C3) + L2

RS + RL

s + 1

.

Note: H(s) is a generalized version of the frequency response H(jω) that is computedin phasor notation as H(jω) = VO/VS. The complex variable s in H(s) is of the forms = σ+jω, i.e., it has a real part σ added to the jω variable. The advantage of using s insteadof just jω is that the roots of the denominator polynomial in H(s) can be computed anddisplayed in the complex s-plane, which gives some additional insight for the classification ofdifferent filter types. Another important property of H(s) is that it is the Laplace transformof the unit impulse response h(t) of the circuit with input vS(t) and ouput vO(t). Thefrequency response of a system described by h(t)⇔ H(s) is obtained by evaluating H(s) ats = jω.

With the component values shown in the schematic above, the system function of the nor-malized (ωr = 1 rad/s, RS = RL = 1) Butterworth LPF of order n = 3 is

H(s) =1

2

1

s3 + 2s2 + 2s + 1=

1

2

1

(s2 + s + 1)(s + 1).

The poles of H(s) (i.e., the roots of the denominator polynomial) lie in the left half of thecomplex s plane on a circle with radius ωr and are spaced 180/3 = 60 apart as shown inthe plot below.

2

−1 −0.8 −0.6 −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

0.4

0.6

0.8

1

Res

Ims

Poles of n=3 Butterworth Filter

The frequency response H(jω) is obtained by evaluating H(s) at s = jω. The magnitudeof H(jω) in dB is shown in the following graph. The -3 dB frequency of the filter is ωr = 1rad/s, corresponding to fr = ωr/(2π) = 0.159 Hz.

10−2

10−1

100

−60

−50

−40

−30

−20

−10

0

Mag

nitu

de [d

B]

Butterworth, n=3, Normalized, V(n003)

f [Hz]

Note that the maximum of the magnitude is -6 dB (since RL/(RS +RL) = 1/2) and thereforethe -3 dB frequency is the frequency at which the magnitude has decreased to -9 dB.

Butterworth filters have a maximally flat frequency response magnitude in the passband.Another class of filters, called Chebyshev filters, achieve a steeper transition from the pass-band of the filter to the stopband by allowing some ripple in the magnitude of the passband

3

frequency response. The immittance coefficients ci for Chebyshev filters of oder n can becomputed recursively from the immittance coefficients ai for Butterworth filters using thetwo intermediate quantities

α = 10RdB/10 − 1 , and β = sinh(tanh−1(1/

√1 + α)

n

),

where RdB is the maximum passband ripple in decibels. Using α, β, and ai, the ci arecomputed as

c1 =a1

β, and ci =

ai ai−1

ci−1

(β2 + sin2[(i− 1)π/n])

, i = 2, 3, . . . , n .

The ci for RdB = 0.2 dB and RS = 1 Ω are given in the following table for n ≤ 7.

Order RS C1 L2 C3 L4 C5 L6 C7

c1 c2 c3 c4 c5 c6 c7

1 1.0 0.43422 1.0 1.0378 0.67463 1.0 1.2275 1.1525 1.22754 1.0 1.3028 1.2844 1.9762 0.84685 1.0 1.3394 1.3370 2.1661 1.3370 1.33946 1.0 1.3598 1.3632 2.2395 1.4556 2.0974 0.88387 1.0 1.3723 1.3782 2.2757 1.5002 2.2757 1.3782 1.3723

Example: Normalized 0.2 dB ripple Chebyshev LPF of order n = 3 with RS = 1 Ω. Theschematic of the filter with component values taken from the table above is:

VS

AC 1

RS

1

RL1

C1

1.228

C3

1.228

L2

1.153

.ac dec 200 0.01 1

N003

4

The magnitude of the frequency response of this filter is shown next.

10−2

10−1

100

−60

−50

−40

−30

−20

−10

0M

agni

tude

[dB

]Chebyshev, R=0.2 dB, n=3, Normalized, V(n003)

f [Hz]

The 0.2 dB ripple in the passband is only visible by zooming in on the passband as shownbelow.

10−2

10−1

100

−7

−6.8

−6.6

−6.4

−6.2

−6

−5.8

−5.6

−5.4

−5.2

−5

Mag

nitu

de [d

B]

Chebyshev, R=0.2 dB, n=3, Normalized, V(n003)

f [Hz]

But what is very visible in the first graph is that now the -3 dB frequency is at f3 ≈ 0.2 Hz orat ω3 = 1.268 rad/sec (rather than at 1 rad/s as for the normalized Butterworth filter). Bytradition (mainly driven by mathematical considerations), the design or reference frequencyωr for Chebyshev lowpass filters is the highest frequency up to which the magnitude of thefrequency response in the passband stays within the maximum ripple specification.

To be able to compare Butterworth and Chebyshev filters of order n = 3, the componentsof the Chebyshev filter were scaled as shown below so that it has the same -3 dB frequencyas the Butterworth filter.

5

VS

AC 1

RS

1

RL1

C1

1.557

C3

1.557

L2

1.462

.ac dec 200 0.01 1

N003

The graph below shows a plot of the magnitude of the frequency response of the normalizedButterworth LPF of order 3 and the 0.2 dB ripple Chebyshev LPF of order 3 with a -3dB frequency ω3 = 1 rad/sec (corresponding to a Chebyshev design reference frequency ofωr = 0.788 rad/sec).

10−2

10−1

100

−60

−50

−40

−30

−20

−10

0

Mag

nitu

de [d

B]

Butterworth vs Chebyshev (R=0.2 dB, ωL=0.789 rad/s), n=3

f [Hz]

ButterworthChebyshev

The asymptotic (for large f) decrease of the magnitude of the frequency response is -60dB/decade (20 n dB/decade, in general) for both filters. But since the magnitude of the fre-quency response of the Chebyshev filter decreases more quickly initially, it gets an additional5-6 dB attenuation in the stopband. Looking at the roots of the denominator polynomial ofthe system function H(s), the main difference between Butterworth and Chebyshev filtersis that the poles of the former lie on a circle (with radius ωr), whereas the poles of the latterlie on an ellipse. This is shown in the following plot for the Butterworth and the Chebyshevexample filters considered above.

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−1 −0.8 −0.6 −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

0.4

0.6

0.8

1

Res

Ims

Poles of n=3 Butterworth (ωL=1) and Chebyshev (ω

L=0.7885) Filters

ButterworthChebyshev

The maximum ripple specification RdB for Chebyshev filters determines how wide (for smallRdB) or how narrow (for larger RdB) the ellipse is on which the pole locations lie.

2 Frequency Scaling

An existing (linear) filter design can be scaled from one reference frequency to another bydividing all reactive elements (inductors and capacitors) by a frequency scaling factor kf ,defined as

kf =desired reference frequency

existing reference frequency.

Note that kf is a dimensionless number. Thus, the frequencies in the numerator and denom-inator must have the same dimensions, either both in radians per second or both in Hertz.

Example: Consider the following lowpass filter of order n = 2.

7

VS

AC 1

RS

R

RLR

C1

C

L2

L

Vo

+

-

The system function of this filter is

H(s) =RL

RS + RL

1RSL2C1

RS + RL

s2 +RSRLC1 + L2

RS + RL

s + 1.

The denominator of H(s) is a polynomial of degree 2 with 2 roots, s1 = −1/A and s2 = −1/B,where A and B are complex-valued numbers that can be computed from RS, RL, C1, andL2. Thus, H(s) can be expressed as

H(s) = K1

(As + 1) (Bs + 1),

where K = RL/(RS + RL). Now consider replacing s in H(s) by s/kf , i.e.,

old s =new s

kf

.

At s = jω this corresponds to the frequency transformation

old ω =new ω

kf

=⇒ new ω = kf old ω .

In terms of H(s) we can write

H(s/kf ) = K1

(As/kf + 1) (Bs/kf + 1)= K

1((A/kf ) s + 1

) ((B/kf ) s + 1

) ,

i.e., the new H(s/kf ) can be obtained from the old H(s) by simply changing the roots fromA and B to A/kf and B/kf . It is not too difficult to see that the same reasoning can beapplied to the more general case where the denominator (and possibly the numerator) ofH(s) has n roots.

In terms of the actual components that make up the filter, the change from s to s/kf affectsreactive components as follows

L s→ L (s/kf ) = (L/kf ) s , and1

C s→ 1

C (s/kf )=

1

(C/kf ) s,

and therefore

new L =old L

kf

, and new C =old C

kf

.

8

Example: Butterworth LPF of order n = 3 with -3 dB frequency fr = 1 kHz and RS =RL = 1 Ω (source and load resistors still normalized). We find that ωr = 2π 1000 = 6283.2rad/s. Thus,

kf =desired reference frequency

existing reference frequency=

6283.2

1.

From the Butterworth filter design table we find for n = 3 that

a1 = 1.0 , a2 = 2.0 , a3 = 1.0 ,

where a1 = “old” C1, a2 = “old” L2, and a3 = “old” C3. Dividing by kf = 6283.2 we thusfind the “new” element values

C1 = 159.15 µF , L2 = 318.3 µH C3 = 159.15 µF .

The corresponding schematic is shown below.

AC 1

VS

RS

1

RL1

C1

159.15µ

C3

159.15µ

L2

318.3µ

.ac dec 200 100 10k

N003

The magnitude of the frequency response of this filter is shown in the graph below.

102

103

104

−70

−60

−50

−40

−30

−20

−10

0

Mag

nitu

de [d

B]

Butterworth, n=3, ωL=6283.2 rad/s (f

L=1 kHz), V(n003)

f [Hz]

Clearly, the design specification of fr = 1 kHz (-3 dB frequency) is very well met.

9

3 Impedance Scaling

So far only the normalized load (and source) resistance RL = 1 Ω has been used. In generalRL will have a different value, e.g., 50 Ω for RF applications or RL = 600 Ω for telephonyand audio applications. To get an idea of how to scale filter designs for general RL values,consider again the second order LPF system function

H(s) =RL

RS + RL

1RSL2C1

RS + RL

s2 +RSRLC1 + L2

RS + RL

s + 1.

In general, the k−th coefficient in the denominator (and numerator, if applicable) polynomialof H(s) has dimension (sec/rad)k since s (especially for s = jω) has dimension (rad/sec)k

and H(s) is a dimensionless ratio (of output over input voltage). Clearly, terms that are justratios of RS and RL, such as RL/(RS +RL) are not changing H(s) as long as RS and RL arescaled by the same factor. But terms such as R C or L/R, which have dimension of sec/rad,would change H(s) if only the R part is scaled. However, R C and L/R remain constantif all the inductor values are scaled by the same factor as the resistance values and all thecapacitance values are changed by the inverse of the factor that is used for the resistancevalues. Define

kz =desired load impedance

existing load impedance.

Then, to convert a filter design from an existing to a desired load impedance, compute newR, L, and C values as

new R = kz old R , new L = kz old L , new C =old C

kz

.

Example: Convert the Butterworth LPF of order n = 3 with -3 dB frequency fr = 1 kHzwith RL = RS = 1 Ω (see previous example) to a filter with RS = RL = 600 Ω. In this casekz = 600 and the new values for C1, L2 and C3 become

C1 = C3 =159.15 µF

600= 265.25 nF , L2 = 600× 318.3 µH = 191 mH .

The resulting filter design is shown in the next schematic.

10

AC 1

VS

RS

600

RL600

C1

265.25n

C3

265.25n

L2

191m

.ac dec 200 100 10k

N003

The L and C values in this filter are now much more realistic and it is fairly easy to verify inLTspice that the filter has the designed frequency response magnitude with a -3 dB frequencyof 1 kHz.

4 Filter Transformation

Let sL = σL + jωL denote the complex frequency variable associated with a lowpass filter(LPF) design. To convert a LPF to a highpass filter (HPF), replace sL by 1/s in the lowpasssystem function HL(sL), i.e.,

sL =1

s, =⇒ at sL = jωL , s = σ + jω =

1

jωL

=⇒ ω = −1/ωL .

Thus, the frequency response H(jω) of the HPF is the same as the frequency responseof the LPF evaluated at sL = −j/ωL. In particular, the reference frequency ωr of theLPF is mapped to −1/ωr for the HPF. Note that, for filters with real-valued coefficients,|H(jω)| = |H(−jω)|, i.e., the magnitude of the frequency response has even symmetry in ω.

The substitution sL 7→ 1/s affects the C and L elements of the LPF as follows:

1

C sL

7→ s

C, and L s 7→ L

s.

That is, a capacitor C(L)i in the LPF becomes an inductor Li = 1/C

(L)i in the HPF and an

inductor L(L)i in the LPF becomes a capacitor Ci = 1/L

(L)i in the HPF.

Example: Chebyshev 0.2 dB ripple HPF of order n = 5 with reference cutoff frequency 2MHz and RL = 50 Ω. Start from the normalized Chebyshev LPF shown below.

11

AC 1

VS

RS

1

C1

1.339

C3

2.1661

C5

1.339

L2

1.337

L4

1.337

RL1

.ac dec 400 0.01 1

Next, apply the LPF to HPF transform to obtain

AC 1

VS

RS

1

RL1

C2

0.748

C4

0.748

L1

0.747

L3

0.462

L5

0.747

.ac dec 400 0.01 1

Finally, apply a frequency scaling factor of kf = 2π 2× 106 = 1.257× 107 and an impedancescaling factor of kz = 50 for the final denormalized HPF design shown in the schematicbelow.

12

AC 1

VS

RS

50

RL50

C2

1.19n

C4

1.19n

L1

2.97µ

L3

1.84µ

L5

2.97µ

.ac dec 400 100k 100meg

To convert a lowpass filter with design frequency ωr to a bandpass filter with center frequencyωc and bandwidth 2ωr, replace sL by (s2 + ω2

c )/(2s) in the lowpass system function HL(sL),i.e.,

sL =s2 + ω2

c

2 s=⇒ s2 − 2sL s + ω2

c = 0 =⇒ s = sL ±√

s2L − ω2

c = sL ± j√

ω2c − s2

L .

That is, for each value of sL, two values of s are created, one with the imaginary part shiftedup by

√ω2

c − s2L, and one with the imaginary part shifted down by

√ω2

c − s2L. Setting

sL = jωL, yields

s = σ + jω = jωL ± j√

ω2c − (jωL)2 = j

(ωL ±

√ω2

c + ω2L

)=⇒ ω = ωL ±

√ω2

c + ω2L .

In particular, considering only ω > 0, the design frequency ±ωr of the LPF maps into

±ωr +√

ω2c + ω2

r ≈ ωc ± ωr ,

where the approximation holds for narrowband BPFs for which ωc ωr. Thus, if ωr is the-3 dB frequency of the LPF, then the -3 dB bandwidth of the (narrowband) BPF is 2ωr.

The mapping sL 7→ (s2 + ω2c )/(2s) leads to the following conversion of all capacitors C in

the LPF:

1

Cs

1

Cs2 + ω2

c

2s

=1

Cs

2+

ω2c C

2s

=⇒

=⇒

•

•

CC

2

2

ω2c C

13

Similarly, sL 7→ (s2 + ω2c )/(2s) leads to a conversion of all inductors L in the LPF as shown

below.

Ls Ls2 + ω2

c

2s=

Ls

2+

ω2c L

2s=⇒

=⇒L

L

2

2

ω2c L

Example: Butterworth BPF of order n = 6 with bandwidth 100 kHz, center frequencyfc = 1 MHz, and RS = RL = 50Ω. Start from a normalized Butterworth LPF of order n = 3with C1 = C3 = 1 F and L2 = 2 H. Then use frequency and impedance scaling factors

kf = 2π100, 000

2= 3.1416× 105 , and kz = 50 ,

to obtain the following LPF.

VS

AC 1

RS

50

RL50

C1

63.66n

C3

63.66n

L2

318.3µ

.ac dec 500 10k 1meg

14

Next, use ωc = 2π× 106 in sL = (s2 + ω2c )/(2s), replace all capacitors with parallel resonant

LC circuits (C ′ = C/2 and L′ = 2/(ω2c C)), and all inductors with series resonant LC circuits

(L′ = L/2 and C ′ = 2/(ω2c L)). This yields the following BPF.

AC 1

V1

R1

50

C1

31.83nL1

795.8n

L2

159.15µ

C2

159.16p

C3

31.83nL3

795.8n

R250

.ac lin 500 0.75meg 1.25meg

Example: Design of n = 4 Butterworth BPF with bandwidth 400 Hz, center frequencyfc = 4.915 MHz, and RS = RL = 120 Ω. Start from a n = 2 Butterworth LPF with design(-3 dB) frequency 200 Hz and use the scaling factors

kf = 2π 200 = 1256.6 , and kz = 50 .

The resulting schematic is shown below.

VS

AC 1

RS

120

C1

9.38µ

L2

135m

RL120

.ac lin 400 1m 1k

n003

15

Now apply the mapping s 7→ (s2+ω2c )/(2s) with ωc = 2π 4.915×106 to the capacitor and the

inductor to obtain a parallel combination of C1 = 4.69µF and L1 = 223.57353 pH (note that1/√

L1C1 = 2π 4, 915, 000) as replacement for the LPF capacitor and a series combinationof L2 = 67.64902 mH and C2 = 15.5 fF (note again that 1/

√L2C2 = 2π 4, 915, 000) as

replacement for the LPF inductor as shown in the following schematic.

AC 1

VS

RS

120

RL120

L1

223.57353p

L2

67.64902m

C1

4.69µ

C2

15.5f

.ac lin 400 4.914meg 4.916meg

n004

A problem with this circuit is the wide range over which the component values are scattered.A capacitor with 4.69µF can conceiveably be found, but no conventional capacitor with 15.5fF exists. Similarly, an inductor of 67.6 mH can be found (even though obtaining a highQ for such an inductor at 4.915 MHz would be a challenge), but 224 pH is definitely toosmall to be practical. Looking closely, one can see that the series resonant circuit can beimplemented using a quartz crystal. But that still leaves the problem of how to implementthe parallel resonant circuit. Fortunately, it is possible to convert the parallel resonant circuitinto a series resonant circuit, as shown in the next section.

5 Impedance Inverter

Sometimes it is necessary to convert a series resonant circuit into a parallel resonant circuitand vice versa. This can be done with an impedance inverter circuit, such as the “T” circuitshown below.

•

•

Zi →

jX

−jX

jX

ZL

The input impedance Zi of this circuit can be computed as

Zi = jX +−jX (jX + ZL)

−jX + jX + ZL

=jX ZL − (jX)2 − jX ZL

ZL

=X2

ZL

.

16

Example: Using an impedance inverter to convert the parallel resonant LC circuit in then = 4 Butterworth BPF with fc = 4.915 MHz and bandwidth 400 Hz to a series resonantcircuit that can be implemented using a quartz crystal.

VS

AC 1

RS

120

L11

3.9µ

L12

3.9µ

L1T

67.64902m

L21

3.9µ

L22

3.9µ

L2

67.64902m

C11

270p

C21

270p

C1T

15.5f

C2

15.5f

RL120

.ac lin 400 4.914meg 4.916meg

n010

Impedance Inverter, X=120 Impedance Inverter, X=120

Series LC Series LC

c©2012, P. Mathys. Last revised: 3-21-12, PM.

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