Design of Narrow Band Filters Part 1

E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Design of Narrow Band Filters Part 1 Thomas Buch Institute of Communications Engineering University of Rostock Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 1

E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Description of Compact Filters N-Circle coupled Filter Resonant circuits and measures of coupling Ladder Network Circuits, continued fraction Normalization, LP-Design Specifications Relation between impedance and transfer function Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 2

Problem of Narrow Bandpass Filters E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Problem HF band filters are narrow band systems with BP << 1 Realization Possibilities: 1. Realization as reactance four-terminal network with source and load termination. Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 3

Problem of Narrow Bandpass Filters E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Pole Quality Factor j ω Requirement: Q L >> Q p with: Q L : Quality factor of the Inductance (coil) Q p = ω 0i 2σ i : Pole quality factor Reactance condition for HF application heavily or not fulfillable. For HF circuits only with crystal filters reachable ( Q 10 4 ). 3 σ i ω 0i σ Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 4

Problem of Narrow Bandpass Filters E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Consideration of the Losses 2. HF filters with consideration of the losses in the elements: 1 as multi-stage amplifiers with mutually detuned resonant circuits. 2 Compact filter with coupled resonant circuits. The realization of narrowband filters is examined with compact filters in the following. Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 5

Description of Compact Filters E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Coupled Resonant Circuits Assumption of an inductive coupling (other couplings later) circuit with lossy Parallel Resonant Circuits. M 1,2 M n 1, n L 1 L n C 1 R 1 Rn Cn U n I 0 Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 6

Description of Compact Filters E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Conversion of Sources I 0 C 1 U 0 C 1 The following relation applies to the conversion of a current source into a voltage source nearby the band center frequency: U 0 I 0 jω m C 1 Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 7

Description of Compact Filters E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Modified Network The following voltage transfer network results: M 1, 2 I 1 L 1 L 2 L n U 0 C 1 I 2 C 2 I n C n U n R 1 R 2 R n M 1,2 is the mutual inductance Now the investigation of the individual circuit sections followed: Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 8

Description of Compact Filters E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Impedance of Resonant Circuit All resonant circuits are coordinated with the same resonant frequency! with Z = R + jωl + 1 [ ( ω jωc = R 1 + jq ω )] m ω m ω Q = ω ml R = 1 ω m C R = Quality factor Z = R (1 + jq V ) with: V = ω ω m ω m ω = Detuning Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 9

Description of Compact Filters E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Parameter of Resonant Circuit Further parameters: d = 1 Q : Damping factor B = ω d2 ω d1 : Bandwidth Normalized Parameter: = B ω m : δ = d : = Ω = V : 1 Q Relative bandwidth Normalized damping factor Relative detuning, Normalized frequency Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 10

Description of Compact Filters E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Normalized Lowpass Impedance Function With these parameters becomes: Z = R δ (δ + jω) = Impedance function of the individual resonant circuit change jω s = u + jω = Z(s) = R δ (δ + s) Normalized lowpass impedance function Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 11

Description of Compact Filters E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Couple Parameter k ij = ω m M ij = Coupling measure with the dimension of a resistor (Ω) δ i δ j Coupling factor x ij = k ij = R i R j (normalized coupling measure) Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 12

Description of Compact Filters E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Total circuit M n-1,n U 0 Z 1 M 1,2 Z 2 M 2,3 Z 3 Z n This network can be analyzed by the Kirchhoff Law Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 13

Description of Compact Filters E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Mesh current matrix U 0 I 1 0. = I 2 Z. 0 I n Mesh current matrix with consideration only the coupling of neighbor resonant circuits. mit: Z = Z 1 jk 12 0... 0 0 jk 12 Z 2 jk 23... 0 jk 23 Z.. 3...... jk n 1,n 0 0 0... jk n 1,n Z n Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 14

Description of Compact Filters E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Input Impedance It is: D = Determinant of the mesh current matrix D 11 = sub-determinant with cancellation of the 1. Line and 1. Then applies: Column Z e = D D 11 = Input Impedance of the Network Z e (s) = Rational Fractional Function = Continued fractions arrangement = Results in elements values Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 15

Description of Compact Filters E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Continued Fractions Expansion: (first kind) It is for example: 1. step: X(s) = s4 + 10s 2 + 9 s 3 + 4s (s 4 + 10s 2 + 9) : (s 3 + 4s) = s + 6s2 + 9 s 3 + 4s s 4 + 4s 2 6s 2 + 9 further division in 2. step etc. = Function of a Reactance one port = s + 1 s 3 + 4s 6s 2 + 9 Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 16

Description of Compact Filters E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Resulting Element Values Result: 1 X(s) =s + 1 6 s + 1 12 5 s + 1 5 18 s L 1 = 1 1 2 L 3 = 1 5 5 C 2 = C 4 = 6 1 8 l 1 c 2 l 3 c 4 Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 17

Description of Compact Filters E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Continued Fractions Arrangement of Z e (s) If one divides the normalized input impedance of the total band filter circuit now into a continued fractions arrangement, one receives the following result after some simplifications: x 2 1,2 Z e (s) = s + δ 1 + R 1 /δ 1 x2,3 2 s + δ 2 + s + δ 3 +.. x 2. n 1,n s + δ n The interesting at the continued fractions representation is that it represents a direct connection between rationally broken input impedance function and the dimensioning of the elements. Now still the problem exists to derive from a given transfer function the input impedance. Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 18

Determination of the input impedance E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Doubly-terminated two-port Reactance Network For the realization of a given transfer function (or PZ-Map) of the BP the relations between transfer function H(s) and the input impedance Z e (s) must be determined. R 1 I 1 U0 U1 RVP U 2 R 2 Z e ( s) = R( s) + jx ( s) U2 P2 = R Starting point: Doubly-terminated two-port reactance network 2 2 Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 19

Determination of the input impedance E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Power Relations In the four-terminal reactance network no effective power is consumed, than: P 1 = U 1 I 1 P 2 = U 2 2 R 2 Under the condition of Power Fitting R 1 P max = U 0 2 4 R 1 U 0 R R 2 = 1 Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 20

Determination of the input impedance E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Transmission Function and with that: P 2 P max = U 2 2 4 R 1 U 0 2 R 2 It is then: H B = One therefore defines: H B (s) = 2 P2 R1 = 2 U 2 P max R 2 U 0 R1 U2(s) R 2 U 0 (s) = 2 R1 H(s) R 2 H B (s) : Transmission Function H(s) : Transfer Function Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 21

Determination of the input impedance E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Characteristic of H B (s) 1 H B (jω) ω H B (s) s=jω = H B (jω) 1 Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 22

Determination of the input impedance E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Power and Input Impedance In the four-terminal reactance network no effective power is consumed, than: P 1 = R {U 1 I1 } = P 2 { U1 2 } { U1 2 } (R(s) + jx(s)) P 1 = R Z e (s) = R R(s) 2 + X(s) 2 = P 1 = U 1 2 R Z e 2 = P 2 = U 2 2 R 2 U 2 2 = R 2 R Z e 2 U 1 U 1 2 R(s) R(s) 2 + X(s) 2 Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 23

Determination of the input impedance E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Power and Input Impedance Now we need the relation of the voltage ratio U 2 /U 0 and U 2 /U 1 : U 2 U 0 U 1 Z e = U 0 R 1 + Z e 2 = U 1 2 U 0 2 U 2 U 1 = R 2 R(s) R 1 + Z e (s) 2 With that we get a relation between Transmission Function and Input Impedance: Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 24

Determination of the input impedance E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Transmission Function Input Impedance From this, then follows: H B 2 = 4 R1 R 2 Reordering: U 2 U 0 = 4 R 1 R R 1 + Z e 2 Relation between transmission function and input impedance. 1 H B 2 = 1 4 R 1 R R 1 + Z e 2 = R 1 + Z e 2 4 R 1 R R 1 + Z e 2 Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 25

Determination of the input impedance E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Rearrangement of the Transmission Function Applies to the numerator: R 1 + Z e 2 4 R 1 R = R 1 + R + jx 2 4 R 1 R = (R 1 + R) 2 + X 2 4 R 1 R = R1 2 + R2 + 2 R 1 R 4 R 1 R + X 2 = (R 1 R) 2 + X 2 = R 1 R jx 2 = R 1 Z e 2 = Z e R 1 2 Becomes with that: 1 H B (s) 2 = R1 Z e (s) R1 + Z e (s) 2 = Z e (s) R1 Z e (s) + R1 2 Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 31

Determination of the input impedance E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Echo Transmission Function One defines: Echo Transmission Function H E (s) = Z e(s) R 1 Z e (s) + R 1 = Connections: (Composite return current coefficient) Z e (s) = R 1 1 + H E(s) 1 H E (s) H E (s) H E ( s) = 1 H B (s) H B ( s) Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 32

Summery E.U.I.T. Telecomunicación 2010, Madrid, Spain, 27.09 30.09.2010 Summery Reactance condition for HF Band Pass Filters heavily or not fulfillable. HF filter design with consideration of the losses in the elements is necessary. Use of coupled resonant circuits as compact filter. Relation between Input Impedance and device values by Continued Fractions Arrangement. Relation between Input Impedance and Transmission Function was derived. Th. Buch, Institute of Communications Engineering, University of Rostock, Germany 37