S-55 3120 Lecture Slides 6

8/9/2019 S-55 3120 Lecture Slides 6

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S-55.3120 Passive Filters

Lecture 6: Elliptic (Cauer) filtersJarmo Virtanen

|Page 1 / 26 | S-55.3120 Lecture 6: Elliptic (Cauer) filters

Copyright c 2010 Jarmo Virtanen



A! Aalto University

School of Scienceand Technology

Elliptic (Cauer) filters

|S 21(jω)|2 = 1

1 +

P 2n(ω)

Q 2p(ω)

Q p(ω) = (ω2 − ω22)(ω2 − ω2

4)(...)(ω2 − ω2p )








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Math tables of elliptic functions

◮ Complete elliptic integral of the first kind (K ):

K = K (m = k 2) and K ′

(m ) = K (1−m )

◮ Jacobi elliptic sine function (sn): sn(u , k ) = sn(ε\α)

ε = u K · 90o, α = arcsin(k )

◮ Nome function q :

q = q (m ) = q (k 2) = e−πK ′

/K

if k 2 << 1, then q (k 2) ≈ k 2

16






0 1 1/k ω

1

1

1 + ε21

1 + ε2

/k 20

|S 21( j ω)|2 = 11 + ε2R2

n (ω)

0

|Rn

| ≤1, when 0

≤ω

≤1

|Rn | ≥ 1/k 0, when ω ≥ 1/k






Elliptic filter

|S 21|2 =

1

1 + ε2R2n (ω)

Equiripple both on pass and stop bands:

dRn

dω= 0, when

|Rn (ω)| = 1, except when |ω| = 1|Rn (ω)| = 1/k 0, except when |ω| = 1/k

⇒ dRn

dω= C n

[1− R2n ][1 − k

2

0R2n ]

[1− ω2][1 − k 2ω2]

Solution is obtained with the elliptic integrals






Design formula of an elliptic filter

1. integrate over the pass band (0 ≤ ω ≤ 1, Rn alternates n times between 0...1 and 0...

−1)

n

1 0

dRn [1 −R2

n ][1− k 20R2n ]

= C n

1 0

dω [1 − ω2][1 − k 2ω2]

⇒ nK (k 0) = nK 0 = C n K (k ) = C n K

2. integrate 1 ≤ ω ≤ 1/k (Rn : 1 → 1/k 0):

1/k 0 1

dRn √ ...

= C n

1/k 1

dω√ ...

⇒ jK (k ′

0) = jK ′

0 = jC n K (k ′

) = jC n K ′






Design formula of an elliptic filter (cont)

Combining the two equations from the previous slide gives us thedesign formula for the elliptic filters:

nK ′

K = K

′

0

K 0

What is needed:

◮ selectivity factor k (sharpness of the filter)

◮ stop band attenuation k 0 (at ω = 1/k )

◮ K = K (k ), K 0 = K (k 0), K ′

= K (k ′

) and K ′

0 = K (k ′

0)

⇒ order n






Design formula of an elliptic filter (cont)

Nome function q = e−πK ′

/K , q 0 = e−πK ′

0/K 0 :

nK ′

K = K

′

0

K 0⇒ q n = q 0

Attenuation at pass and stop bands:

Ap = 10 lg[1 + ε2] and As = 10 lg[1 + ε2/k 20 ] ⇒Approximation: if k 0 << 1, then k 20 ≈ 16q 0 = 16q n

As ≈ 10{lg[10Ap/10 − 1] − n · lg(q ) − 1.2}

Design specifications determine As, Ap and k (→ q ), the order n of the filter is obtained from the equation above








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Pass band minima and stop band maxima

Pass band minima (|Rn = 1|), i.e., largest attenuation at passband:

n odd: n even:

ω̂i = sn( (2i − 1)K n

, k ) ω̂i = sn( 2iK n

, k )

i = 1, 2, ..., (n − 1)/2 i = 0, 1, 2, ..., (n /2) − 1

Stop band maxima (|Rn | = 1/k 0), i.e, smallest attenuation at

the stop band: ω̃i = 1k ̂ωi



Ell fil |S (j )|2


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Elliptic filter |S 21(jω)|2

It can be shown, that

Rn (ω) = sn(z , k 0), where

z =

nK 0

K sn−1(ω, k ), n odd

z =

nK 0

K sn−1

(ω, k ) + K 0, n even

Rn (ω) = 1

k 0Rn ( 1

k ω)

|S 21( j ω)|2 = 1

1 + ε2R2n (ω)



|S |2



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Example: Determine expression of |S 21|2 for an elliptic filter,whose parameters are k = 1/2, n = 3 and ripple, i.e., themaximum value of the reflection coefficient at the pass band is20% (

→Ap

≈0.1773 dB).

Reflection zeros:

ω00 = 0 (always when n is odd)

ω01 = sn(2K

3 , 0.5) = sn(60o\300) ≈ 0.88103

Transmission zeros:

ω∞ = 1

k ω00

=

∞and ω1 =

1

k ω01 ≈2.2701

Parameter ε:

ε2 = 1

1 − |S 11|2 − 1 =

1

24





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Rn (ω):

Rn (ω) = H ω(ω2 − ω2

01)

ω2 − ω21

H is obtained from |Rn (1)| = 1 (cut-off frequency of the passband):

H = 1 · (1− ω2

01)

|1 − ω2

1| −1

≈18.5595

Transducer power gain:

|S 21|2 = 1

1 + ε2R2n (ω)

= 24(ω2 − 2.27012)2

24(ω2 − 2.27012)2 + 344.454ω2(ω2 − 0.881032)2



Elli ti filt A B d C t



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Elliptic filters: A-, B-, and C-type

◮ A-type

Even order → |S 21(∞)|2 = 1

1 + ε2/k 20

An elliptic filter having even order can not be realized with TL

or πL type low-pass filter because these filter topologies havetransmission zero at the infinity (S 21(∞) = 0). A frequencytransformation must be used to obtain realizable filter (B- orC-type).

Odd order can be realized with TL11 or πL00 type circuit,termination resistors are equal.





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◮ B-type

Transfer the largest transmission zero ω1 = 1/[k ω01] to theinfinity. The frequency transformation is:

ω̃2 = (1− k 2ω2

01)ω2

1

−k 2ω2

01ω2

k̃ = k · 1− ω2

01

1− k 2ω201

< k

Termination resistors: Rg

= RL





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◮ C-typeAs in the case of B-type, but transform also the smallest

reflection zero to the origin (ω01 → 0). The response lookslike the response of an A-type filter having order (n − 1), i.e.,S 21(0) = 1, S 21(∞) = 0. The frequency transformationused is:

ω̄2 = (ω21 − 1)(ω2 − ω201)(1− ω2

01)(ω21 − ω2)

k̄ = k ·

1 − ω201

1

−k 2ω2

01

< k̃ < k

Termination resistors: Rg = RL





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Example: Investigate change in the sharpness of A-, B- and C-typefilter, when Ap = 1 dB, As = 32.5 dB, n = 4 and k = 0.766.

ω01 = sn(K

4 , k ) = sn(22.5o\50o) = 0.4559

⇒ ω1 = 1/[0.766 · 0.4559] = 2.864

B-type: k̃ ≈ 0.950k C-type: k̄ = 0.9502k ≈ 0.903k



Elliptic filters: using filter tables



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Elliptic filters: using filter tables

Filter tables have types A and B (A odd, B even)

C 052030 = C Cauer

05 n

20 max |S 11|(%)

30 θ=arcsin(k )

Ap = −10 lg[1 − |S 11|2max]

Data in the tables: order n , Ap, As, modulus angle θ (A-type),sharpness ω̂s = 1/k (A) or ω̂s = 1/k̃ (B), transmission zerosω̂∞1,...

Low-pass filter: ω̂s = f s/ f p

High-pass filter: ω̂s = f p/ f s

Band-pass filter: ω̂s = B s/B p (B = B p)

Band-stop filter: ω̂s = B p/B s (B = B p)



Elliptic low pass filter



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Elliptic low-pass filter

0 2π f p 2π f s ω

−As

−Ap

0

|S 21|2

dB

ω̂s = f s

f p



Elliptic low-pass prototype filter



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Elliptic low-pass prototype filter

0 1 ω̂s ω̂

−As

−Ap

0

|S 21|2 dB



Elliptic band-pass filter



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Elliptic band pass filter

0 ωs1 ωs2ωp1 ωp2 ω

0

−Ap

−As

|S 21|2 dB

ω0

ω20 = ωs1ωs2 = ωp1ωp2B p = ωp2 − ωp1B s = ωs2 − ωs1

ω̂s = B s

B p

B p

B s



Low-pass–band-pass frequency transformation



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Low pass band pass frequency transformation

L

C ⇒

L1 C 1

C 2

L2

≡

L+

C +

L−

C −

C ± = C 1

2+ 2C 2 ∓

C 1C 2 1 +

C 1

4C 2 L+C − = L−C + = 1/ω2

0





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L C ⇒

L1 C 1L2

C 2

≡

L+ C +

L− C −L± =

L1

2

+ 2L2

∓ L1L2 1 + L1

4L2 L+C − = L−C + = 1/ω2

0

Transmission zeros at frequencies ω−

= 1/ L−C − and

ω+ = 1/ L+C +



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S-55 3120 Lecture Slides 6