Chapter V. Microwave Filters

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1 hapter V Microwave Filters otet 5. Filter Desig by the Isertio oss Method ( 8.3) 5. Filter Trasformatios ad Implemetatio ( ) 5.3 ow-pass ad Bad-Pass Filter Desigs ( )

2 5. Filter Desig by the Isertio oss Method ( 8.3) The isertio loss method allows a high degree of cotrol over the passbad ad stopbad amplitude characteristics, with a systematic way to sythesize a desired respose for a filter. Power oss Ratio P R Power available from source P = = ic = Power delivered to load P Γ ( ) I = Isertio oss = log P Γ( ) is a eve fuctio of Γ ( ) = M ( ) R M( ) + N( ) Substitute ito the expressio of P : P R = + M ( ) N( ) R load

3 Maximally flat respose: this characteristic is also called the biomial or Butterworth respose, ad is optimum i the sese that it provides the flattest possible passbad respose for a give filter order. N PR = + k c where N is the order of the filter ad c is the cut-off agular frequecy. Why is it called "maximally flat"? Because the first (N ) deratives of PR are zero at =. At = c, the power loss ratio at the passbad edge is + k. N For >> c, PR k, which icreases at the rate of N db/decade. c Equal-ripple respose: a hebyshev polyomial T N (x) is used to specify the isertio loss of a filter, which results i a sharper cutoff at the expese of passbad ripple. PR = + k TN c where N is the order of the filter ad c is the cut-off agular frequecy. 3

4 At =, the power loss ratio at the passbad edge is + k. c N For large x, TN( x) ( x), so for >> c, the power loss ratio becomes P R N k T k c 4 c N But the power loss ratio for the equal-ripple case is ( flat respose., which also icreases at the rate of N db/decade. N ) / 4 greater tha the maximally 4

5 ow-pass Filter Prototypes adder-type circuits begiig with a series elemet (case A) Rs Rs R s 3 R R R N = N = N = 3 R s 3 R s R 4 R N = 4 N = 5 adder-type circuits begiig with a shut elemet (case B) Gs G Gs G Gs 3 G N = N = N = Gs 3 G Gs 3 5 G N = 4 N = 5 5

6 ase A Zi = j + // R j z i = Normalized iput impedace = ase B Y y i i = j ( / Rs ) + + = j g + + = j + ( / ) 3 3 R / R i j ( Rs)( R/ Rs) g j g g j + G = j Gs + + = j g + + Y = Normalized iput admittace = G G / G g 3 3 s Z R j ( G )( G / G ) j g g s s i s s s R s R s = Z i z i Gs N = = g Y i G s = y i = g N = N = = g = g N = R R = g3 G G = g3 6

7 For both cases z y z z y y Γ= = Γ =ΓΓ = = z y z y R * * i i * i i i i * * i + + i i + z i + + i + yi 4 ( 3) ( 3 3) 3 Γ 4g 3 P = = + g + g g + g g g g + g g g d order maximally flat low-pass filter prototype c 4g3 PR = + k = + ( g ) + ( g g + g g g g ) + g g g k k g =, g =, g = 3 c c d order equal-ripple low-pass filter prototype 4 PR = + k T = + k c c c 4 = + ( g3) + ( gg3 + g ggg3) + ggg3 4g For k =.3493 (.5 db), g =, g =, g3 =.984 c c 3.3. For k = (3dB), g =, g = 5339, g c c = 7

8 For Nth order low-pass filter prototypes R s = gr s = gr 3 3 s = gr 5 5 s = g / Rs 4 = g4 / Rs R = g R N+ s = g / Gs 4 = g4 / Gs G = g G Gs = ggs 3 = g3gs 5 = g5gs N+ s Maximally flat low-pass filter prototypes N k g = g, =,, iiin, gn+ = gn+ = c g Table 8

9 Equal-ripple low-pass filter prototypes g g,,, N, g g = = iii N+ = N+ c g Table ( k =.3493) g Table ( k = ) 9

10 Example: A maximally flat low-pass filter is to be desiged with a 3 db cutoff frequecy of 8 GHz ad a miimum atteuatio of db at GHz. How may filter elemets are required? Solutio: 3 db cutoff k =, db atteuatio P = N N PR = + k = + N 8 c 8 R Atteuatio vs. ormalized frequecy for maximally flat filter prototypes

11 I last example, how may filter elemets are required if a equal-ripple low-pass filter is used istead? k= R N N c 8 P k T T = + = + N 4.5 db ripple 3 db ripple Atteuatio vs. ormalized frequecy for equal-ripple filter prototypes

12 Example: Desig a maximally flat low-pass filter prototype with a 3 db cutoff frequecy of GHz, impedace of 5 Ω, ad at least 5 db isertio loss at 3 GHz. ompare with a 3 db equal-ripple desig havig the same order. Solutio: 3 db cutoff k =, 5 db atteuatio PR = 3.6 N N 3 PR = + k = N 5 c For a 5th order maximally flat desig,the g values are g=.68, g =.68, g3 =, g4 =.68, g5 =.68, g6 =. ase A: PF prototype begiig with a series elemet g N 5 krs 9 c π g N 5 k.68 9 crs π g N 5 3 krs c π = = =.459 =.459 H = = =.575 =.575 pf 5 9 = = = = H

13 R g N 5 4 k crs π g N 5 5 krs 5 9 c π = g6rs = 5 = 5 g N 5 k.68 = = cr 9 s π 5 g N 5 krs 9 c π g N 5 3 k 3 9 crs π 5 g N 5 4 krs 4 9 c π g N 5 5 k.68 5 = = cr 9 s π 5 6 s = = =.575 =.575 pf = = =.459 =.459 H ase B: PF prototype begiig with a series elemet 3 = 9.84 =.984 pf = = = = H = = = 3.83 = 3.83 pf = = = = H 3 = 9.84 =.984 pf G = g G = R = 5 3 5

14 g For a 5th order 3 db equal-ripple desig, the values are g = 3.487, g =.768, g = 4.538, g =.768, g = 3.487, g =. ase A: PF prototype begiig with a series elemet gr s = = =.3853 = H 9 c π g.768 = = =. =. pf 9 crs π 5 gr 3 s = = = = 8.57 H 9 c π g = = =. =. pf 9 crs π 5 gr 5 s = = =.3853 = H 9 c π R = g R = 5 = 5 6 s ase B: PF prototype begiig with a shut elemet = = = 5.54 = 5.54 pf 5 4 g 9 crs π

15 grs c π g crs π g4rs c π g crs π = g6gs R 9 = = = 3.3 = 3.3 H 9 = = = 7.3 = 7.3 pf 5 = = = 3.3 = 3.3 H = = = 5.54 = 5.54 pf 5 G = = 5 5 maximally flat desig 5Ω.459 H H.459 H 5Ω H H.575 pf.575 pf 5Ω.984 pf 3.83 pf.984 pf 5Ω equal-ripple desig 5Ω H 8.57 H H 5Ω 3.3 H 3.3 H. pf. pf 5Ω 5.54 pf 7.3 pf 5.54 pf 5Ω 5

16 db(s) ompariso betwee maximally-flat ad equal-ripple resposes dφ d( S I (db) = db( S) = log( S ) (rad)) Group delay = τd = = d d.5 - Group delay (s) maximally flat equal-ripple - -3 maximally flat equal-ripple Frequecy (GHz) Frequecy (GHz) iear phase respose: I some applicatios, it is importat to have a liear phase respose for a filter to avoid sigal distortio. A liear phase characteristic ca be achieved with the followig phase respose: N φ ( ) = A + p c where φ ( ) is the phase of the voltage trasfer fuctio of the filter, ad p is a costat. A related quatity is the group delay, defied as 6

17 N dφ τd = = A + p(n + ) d c which shows that the group delay for a liear phase filter is a maximally flat fuctio. (Note: the first (N ) deratives of τ are zero at =.) d iear phase low-pass filter prototypes Filters havig a maximally flat group delay, or a liear phase respose, ca be desiged usig the followig ormalized value table: g Table 7

18 5. Filter Trasformatios ad Implemetatio ( ) Filter Trasformatios Respose: ow-pass to high-pass trasformatio PF HPF c c N N PF HPF c R c Maximally flat: PR = + k P = + k PF HPF Equal-ripple: P c R = + k TN PR = + k TN c Prototype: PF HPF + PF TO HPF X g R X g R PF HPF c = = s = s = = c c g PF HPF c g = + = + = = + = c Rs Rs + c+ 8 + B B

19 ow-pass to badpass trasformatio PF BPF c Δ where =, = fractioal badwidth Δ = N N PF BPF R c Δ PF BPF N PR = + k TN ( ) c Δ Maximally flat: P = + k P = + k ( ) R Equal-ripple: PR = + k T Prototype: Respose: PF TO BPF PF BPF X g R X g R PF BPF = = s = s = c Δ g PF BPF g = + = + = = + c Rs Δ Rs + c Δ Δ,, +, c + = + Δ c c+ Δ B B = = = 9

20 ow-pass to badstop trasformatio PF BSF Δ c where =, = fractioal badwidth Δ = N N PF BSF R c PF BSF N PR = + k TN Δ( ) c Maximally flat: P = + k P = + k Δ( ) R Equal-ripple: PR = + k T Prototype: PF BSF = = s =Δ s = c g PF BSF g = + = + =Δ = + c Rs Rs + cδ,,, c + Δ + = + = cδ c+ Δ X g R X g R B B = = PF BSF + Respose: PF TO BSF + +

21 c c Δ Δ c c Δ c Δ c Δ c c Δ c c Δ Δ

22 c = c + + = + c = Δ c Δ = c + + = Δ c = Δ c = Δ c + + = Δ c + + = Δ c + + Δ = PF Prototype HPF Prototype BPF Prototype BSF Prototype Summary of filter trasformatios for -ladder prototypes

23 Example: Desig a badpass filter prototype havig a.5 db equal-ripple respose with N=3. The ceter frequecy is GHz, the badwidth is %, ad the impedace is 5 Ω. Solutio: For a 3rd order.5 db equal-ripple desig, the g values are g=.5963, g =.967, g3 =.5963, g4 =. 9 The value of c ca be arbitrary. Thus, set that c = = π. The elemet values for a PF prototype begiig with a series elemet are =.7 H, = 3.49 pf, 3 =.7 H, R = 5 Ω. 9 From low-pass to badpass trasformatio ( Δ =., = π ), the correspodig badpass filter prototype elemets are 9 c.7 8 = = =.7 = 7 H Δ. Δ. 3 = = =.99 =.99 pf 9 9 c (π ).7 Δ. = = = 7.6 =.76 H 9 c (π ) 3.49 c 3.49 = = = 3.49 = 34.9 pf Δ. 3

24 9 c = = =.7 = 7 H Δ. Δ c 3 (π ).7 = = =.99 =.99 pf The elemet values for a PF prototype begiig with a shut elemet are = 5.8 pf, = 8.73 H, 3 = 5.8 pf, R = 5 Ω. From low-pass to badpass trasformatio, the correspodig badpass filter prototype elemets are Δ. = = = 4.99 =.499 H 9 (π ) 5.8 c c 5.8 = = = 5.8 = 5.8 pf Δ. 9 c = = = 8.73 = 87.3 H Δ. Δ c (π ) 8.73 = = =.9 =.9 pf Δ. = = = 4.99 =.499 H 3 9 c 3 (π ) 5.8 c = = = pf Δ. = 4

25 db(s) BPF prototypes ase A ase B 5Ω 7 H.99 pf 7 H.99 pf 5Ω 87.3 H.9 pf.76 H 34.9 pf 5Ω.499 H 5.8 pf.499 H 5.8 pf 5Ω Frequecy resposes Isertio loss 5 Group delay - Group delay (s) Frequecy (GHz) Frequecy (GHz) 5

26 Filter Implemetatio The lumped-elemet desig i the filter prototypes geerally works well at low frequecies, but two problems arise at microwave frequecies. First, lumped elemets such as iductors ad capacitors are available oly for a limited rage of values ad are difficult to implemet at microwave frequecies. Secod, at microwave frequecies the distaces betwee filter compoets is ot egligible. I cotrast, trasmissio-lie sectios are geerally suitable for use over microwave frequecies ad therefore very useful for microwave filter implemetatio. Richard s trasformatio The trasformatio is used for a iductor to a λ/8 short-circuited stub ad for a capacitor to a λ/8 ope-circuited stub. c l= λc /8 ta β jx = jc= jz l = jz Z = l= λc /8 ta β jbc = jc = jy l = jy Z = c Z = c Z = c 6

27 Kuroda s idetities The four Kuroda idetities use redudat trasmissio lie sectios to achieve a more practical microwave filter implemetatio by performig ay of the followig operatios:.physically separate trasmissio lie stubs.trasform series stubs ito shut stubs, or vice versa. 3.hage impractical characteristic impedaces ito more realizable oes = + Z / Z 7

28 Prove the followig equivalet trasmissio-lie circuits for illustratig Kurada idetity (a). Proof: cosβl jzsiβl jωz A B = j ta βl j = Z D siβl cos βl jω ( + ) Ω HS Z Z +Ω Z Z Z O.. shut stub Uit elemet T-lie where Ω= ta βl Z jω cos βl j si βl Z ( Z Z) A B j taβl + D = = RHS j jω Z si βl cos βl +Ω Ω Z Z Z Uit elemet T-lie S.. series stub 8

29 HS =+ Z /Z A B A B D = D RHS Q.E.D. Example: Desig a low-pass filter for fabricatio usig microstrip lies. The specificatios are: cutoff frequecy of 4 GHz, third order, characteristic impedace of 5 Ω, ad a 3 db equal-ripple characteristic. Solutio: Step : For a 3rd order 3 db equal-ripple desig, the g values are g= , g =.77, g3 = , g4 =. The elemet values for a PF prototype begiig with a series elemet are g = Rs R s = c c g = =.77 crs crs gr 3 s R s = = umped-elemet low-pass filter prototype c c i ormalized form (by settig R s = ad R = g4rs = Rs c = ) 9

30 Step : Use Richards trasformatio. For a iductor to a S.. series stub, Z = c= Rs. For a capacitor to a O.. shut stub, Z = = =.45 Rs.77 / R c Step 3: Add uit-elemet trasmissio lies at eds of filter. These redudat elemets do ot affect the magitudes of S-parameters for the filter sice they are matched to the source ad load impedace ( R = R = Z ). s s l = λ /8 at c l = λ /8 at c Step 4: Use Kurodas idetity to covert the series stubs to shut stubs. Z R = + s.99 Z = R = s l = λ /8 at c 3

31 Step 5: Impedace ad frequecy scale the circuit. This ivolves multiplyig the ormalized characteristic impedaces by 5 Ω ad choosig the lie ad stub legths to be λ/8 at 4 GHz. l = λ /8 at c Trasmissio-lie circuit of fial filter Amplitude resposes of fial filter Microstrip fabricatio of fial filter 3

32 S(dB) S(dB) PF desig usig Kuroda s idetity simulatio vs. experimet Rs = 5 Ω = π.4 GHz c rd 3 -order.5 db equal ripple Sim Mea utoff freq..35 GHz. Ghz ( Sim ) ( Mea ) -4 ( Sim ) ( Mea ) Frequecy (GHz) Frequecy (GHz) 3

33 Impedace ad admittace iverters The operatio of impedace ad admittace iverters is to form the iverse of the load impedace or admittace. These iverters ca be used to trasform series-coected elemets to shut-coected elemets, or vice versa. The iverse trasformatio procedure is commoly see i badpass ad badstop filter implemetatio. (K iverters) (J iverters), A B ± jk D = ± j/ K A B ± j/ J D = ± jj (a) Operatio of impedace ad admittace iverters, (b) Implemetatio as quarter-wave trasformers 33

34 (K iverters) (J iverters) (c) Implemetatio usig trasmissio lies ad reactive elemets (d) Implemetatio usig capacitor etworks 34

35 5.3 ow-pass ad Bad-Pass Filter Desigs ( ) Approximate equivalet circuits for short trasmissio lie sectios l Z, β = Z Z Z Z Z βl, (short trasmissio lie) jzcot βl jzcsc βl [ Z] = jzcsc βl jzcot βl T-equivalet circuit [ Z ] Z Z = Z Z βl θ= βl jzθ Z Z = jz cot βl + jz cscβl = jz ta = j / / θ= βl = jysi βl jyθ = j Z Zθ Y Z θ = = τ, = = Yτ Equivalet circuit where τ represets the propagatio delay time of trasmissio lie. 35

36 For a electrically short trasmissio lie ( θ = βl ), the T-equivalet circuit of trasmissio lie ca be approximately show as θ = βl Z Zθ Zτ = Zθ Zτ = Yθ Yτ = For a high-impedace (High-Z) lie with Z, = Yθ /. The equivalet circuit ca be reduced to a series iductace: θ = βl High-Z ie Z Zθ Zτ = For a low-impedace (low-z) lie with Z, = Zθ /. The equivalet circuit ca be reduced to a shut capacitace: θ = βl ow-z ie Z Y θ Yτ = 36

37 Stepped-Impedace (High-Z, ow-z) ow-pass Filters PF prototype + microstrip layout Z h Microstrip PF, τ Zl, τ + l = Z τ = Z ( w ) ε c l h h h eff Yl τ + + = + = εeff Zl( wl) c where wh : width of high-z microstrip lie wl : width of low-z microstrip lie ε : effective dielectric costat of high-z microstrip lie ε l l eff eff h l + h : effective dielectric costat of low-z microstrip lie : legth of the th sectio microstrip lie : legth of the (+)th sectio microstrip lie c : speed of light (3 m/sec) 8 l 37

38 Example: Desig a stepped-impedace low-pass filter havig a maximally flat respose ad a 3 db cut-off frequecy of.5 GHz. It is ecessary to have more tha 3 db isertio loss at 4. GHz. The filter impedace is 5 Ω; the highest practical lie impedace is 5 Ω, ad the lowest is Ω. 3 db cut-off k =, 3 db isertio loss P = R R = + N N 4 = + c.5 g 6.576,.44, 3.938, 4.938, 5 P k N For a 6th order maximally flat desig, the values are g = g = g = g = g =.44, g6 =.576, g7 =. The elemet values of low-pass filter prototype begiig with a shut elemet: =.659 pf, = 4.5 H, 3 =.46 pf, 4 = 6.49 H, 5 =.8 pf, 6 =.648 H, R = 5 Ω τ = = Z l =.659 = 6.59 = 6.59 ps Y l Yl τ = = = 3 = 3 ps Zh 5 τ = = Zl=.46 =.46 = 4.6 ps 38

39 db(s) τ τ τ 4 4 Zh Yl = = = 4.99 = 4.99 ps 5 9 = = Z =.8 =.8 = 8. ps l = = =.99 =.99 ps Z 5 h Stepped-impedace implemetatio Z = R = Ω 5 s τ Z l = Ω Z h = 5Ω Z l = Ω Z h = 6.59 ps τ = 3 ps τ 3 = 4.6 ps τ 4 = 5Ω Z l = Ω = 4.99 ps τ = 8. ps 5 Z h = 5Ω τ =.99 ps 6 Z = R = Ω lumped elemet High-Z, ow-z Group delay (s).5.4 lumped elemet High-Z, ow-z Frequecy (GHz) Frequecy (GHz) 39

40 Example: I last example, if the stepped-impedace low-pass filter is implemeted usig microstrip lies developed o a.6 mm thick FR4 substrate (ε r =4.5), show the fial microstrip layout dimesios. Desig for a 5 Ω microstrip lie: Z εr + εr A = + (.3 + ) = + (.3 + ) = ε + ε A r.544 r w_5 8e 8e = = =.88 w.544 _5 =.88 h=.88.6 = 3. (mm) h A e e Desig for a 5 Ω microstrip lie: Z εr + εr A = + (.3 + ) = + (.3 + ) = ε + ε A r 4.38 r w_5 8e 8e = =.8 w 4.38 _5 =.8 h=.8.6 =.73 (mm) h A e e εr + εr εeff _5 = + = + =.95 h w.73 _5 4

41 Desig for a Ω microstrip lie: 377π 377π B = = = 7.96 Z ε 4.5 r w_ ε.6 l( ) r = B B + l( B ) +.39 h π εr εr = 7.96 l( 7.96 ) + l(7.96 ) +.39 π = w = h= = 4.7 (mm) ε eff _ εr + εr = + = + = 4.63 h w _ The legth of each microstrip lie sectio is calculated as l l c τ = = ε 4.63 eff _5 8 4 = 9.8 =.98 mm c τ = = = 5.7 = 5.7 mm ε.95 eff 4

42 l l l l c τ = = = 3.66 = 3.66 mm ε 4.63 eff 8 c τ = = = 7. = 7. mm ε.95 eff 8 c τ = = =.68 =.68 mm ε 4.63 eff c τ = = ε.95 eff 8 3 =.93 =.93 mm Fial microstrip layout dimesios.98 mm 3.66 mm.68 mm 4.7 mm 3. mm.73 mm.73 mm 3. mm 4.7 mm 5.7 mm 7. mm.93 mm 4

43 S (db) S(dB) R Z Stepped-impedace PF desig simulatio vs. experimet s l th = 5 Ω, c = π.4 GHz = Ω, Z = 5 Ω h 7 -order,.5 db equal ripple Sim Mea utoff frequecy.4 GHz.3 GHz Frequecy (GHz) Frequecy (GHz) 43

44 aoical coupled-lie circuits for filter applicatios 44

45 Te caoical coupled-lie circuits (cot) 45

46 Microstrip coupled-lie filter sectio (MFI i ibra) O.. Z θ, Z e o O.. [ Z ] j j ( Ze + Zo)cot θ ( Ze Zo)cscθ = j j ( Ze Zo)csc θ ( Ze + Zo)cotθ Proof: The derivatio begis with the Z matrix of a 4-port coupled lie sectio. V Z i i =, i,,3,4 I = I I= I3= I4= (ports,3,4 are ope circuited) Port excitatio Eve-mode excitatio Odd-mode excitatio θ θ θ O O.. = I Ze, Z + o Z e O I I V Z Z I I V Z Z I I V Z Z I I V Z Z et ot = + O.. et ot = + eb ob 3 = eb ob 4 = I I I O.. θ Z e Z e O.. O.. O.. eve top eve bottom O.. I I Bisectio of the circuit I I Z o 3 4 S.. θ Z o Zo O.. S.. O.. O.. odd top odd bottom O.. 46

47 et eb et eb = = e θ = = e ot ob ot ob = = ocot θ, = = o where Z Z jz cot, Z Z jz cscθ Z Z jz Z Z jz ( cotθ cotθ) = = e o = = 33 = 44 I I= I3= I4= ( cscθ cscθ) = = e o I I= I3= I4= cscθ V Z jz jz Z Z Z V Z jz jz ( cotθ cotθ) ( cscθ cscθ) = Z = Z = Z V Z jz jz Z Z Z 3 3 = = e + o = 3 = 4 = 4 I I= I3= I4= V Z jz jz Z Z Z 4 4 = = e + o = 4 = 3 = 3 I I= I3= I4= For a microstrip coupled-lie filter sectio, set I = I3 = to yield V= ZI+ Z4I4 V Z Z4 I = V4 = Z4I + Z44I 4 V 4 Z4 Z 44 I 4 j j ( Ze + Zo)cot θ ( Ze Zo)cscθ O.. [ Z ] =, Q.E.D. j j ( Ze Zo)csc θ ( Ze + Zo)cotθ θ Z, Z e o 3 4 MFI O.. 47

48 Image impedace The image impedaces ca help specify the passbad ad stopbad characteristics for a filter desig usig a cascade of two-port etworks. They are defied as follows. Zi = iput impedace at port whe port is termiated with Zi. Z i = iput impedace at port whe port is termiated with Z. i V AV + BI AV ( / I) + B AZi + B Zi = = = = AB I V + DI ( V / I) + D Zi+ D Zi = D V Z DV i = BI DZi + B = = BD Z I ( V AI) Z i = + i+ A A For a microstrip coupled-lie filter sectio, Z AB ZZ ( Ze Zo) ( Ze + Zo) cos i= = Z = D Z siθ Due to symmetry of ports ad, Z = Z. i i θ 48

49 Equivalet circuit of microstrip coupled-lie filter sectio O.. Z θ, Z e o O.. θ π / MFI Equivalet circuit Derive the coditios for equivalece: overtig [ Z] ito ABD matrix for a microstrip coupled-lie filter sectio yields Ze + Zo ( Z ) cot ( ) csc cos e + Z o θ Z e Z o θ θ A B Ze Zo j( Ze Zo)cscθ = D HS jsiθ Ze + Z o cosθ Ze Zo Ze Zo cosθ jzsiθ cosθ jzsiθ A B j/ J jsiθ jsiθ D = cosθ cos RHS jj θ Z Z (T-lie) (J iverter) (T-lie) cos θ ( JZ + )siθcos θ j( JZsi θ ) JZ J = j( si θ Jcos θ) ( JZ + )siθcosθ JZ JZ 49

50 omparig A or D as θ π / obtais Z + Z = + Z Z JZ e o JZ e o omparig B or as θ π / obtais e o Z Z = JZ Solvig () ad () for Z ad Z gives e o Ze = Z + JZ + ( JZ) (3) Zo = Z JZ + ( JZ) (4) oupled-ie Badpass Filters Z, Z e o () () A N+ sectio coupled-lie badass filter ca realize a Nth order badpass filter prototype. Z, Z en on Z, Z en + on + 5

51 Desig procedure Step : Establish the equivalet etwork of the filter as θ π /. N + sectio coupled-lie badpass Step : Simplify the etwork i Step by usig the followig equivalet circuits. For trasmissio lies of electrical legth θ as θ π/, = Proof: The ABD matrix for the cascade of the above T-ework ad :- trasformer is show as jzcot θ ( jzcot θ) + jz cotθ + A B jz /si θ jz /siθ D = jz cotθ + jz /si θ jz /siθ 5

52 cosθ jzsi θ cosθ jzsi θ ABD matrix = = = jy si θ cosθ jy si θ cosθ for a θ T-lie For the T-etwork as θ π /, Z = π π = Z Proof: As θ π /, jz cotθ ad assume that θ = π / + dθ. jz jz jz = si θ si( π + dθ) dθ dθ l λ /4 π dθ π θ = βl = l = = = dθ = d = d u d u f λ d p jz jz as θ π / si θ π d p 5

53 The impedace of the parallel circuit j = + d j / / Z( ) = = d j+ j( + d) + j( + d) + ( + ) j j( + d) j / / = d d j( + d) + ( ) ( j+ ) + j d j j j For equivalece, the followig coditios are satisfied. j+ = Z j = (5) π / jz = π d d (6) j d π = Z j Apply the equivalet circuits derived i this step to simplify the etwork i Step. For example, a three-sectio coupled-lie filter has a equivalet etwork show as θ θ θ θ θ θ J 3 Z Z Z Z Z Z Z 9 Z 9 J 9 J where θ π / 53

54 This equivalet etwork ca also be show as θ J : J 3 Z Z 9 Z 9 Z 9 Remove the trasmissio lies of electrical legth θ ad : - trasformers because they dot affect the magitude of S-parameters for the coupled-lie badpass filters. The, the equivalet etwork ca reduce to : J θ Z J 9 J 9 J 3 9 Z Step 3: Use the J coverters to trasform betwee series-coected elemets ad parallel-coected elemets. J Z 9 G = J 3 Z J 9 Z i of J 3 iverter 54

55 Z J 9 J / J Z i of J iverter J / J R = J Z / J 3 Z Z Z i of J iverter Δ gz gz Δ J J = Z Δ g J G = J J Z / J J g Z Δ Secod-order badpass filter prototype Z 3 Δ J = (7) gz = (8) J Δ J = (9) gz Z Δ J g g J Z J Z Z 3 J = Δ J J = () () 55

56 Solvig (5)-() for J, =,,3, yields π Δ π Δ π Δ JZ =, J Z =, JZ = 3 g gg g I geeral, a N + sectio coupled-lie badpass filter ca realize a Nth-order badpass filter prototype with the followig desig formulas: π Δ π Δ π Δ π Δ JZ =, J Z =, JZ =, iii, J Z =, iii, 3 g gg gg3 g g π Δ J Z =, J Z = N π Δ N+ g g N N Ng g N+ From (3) ad (4), the eve-mode ad odd-mode impedaces for each microstrip coupled-lie filter sectio ca be calculated by Ze = Z + JZ + ( JZ) Zo = Z JZ + ( JZ) =,,, N + 56

57 Example: Desig a coupled-lie badpass filter with N=3 ad a.5 db equalripple respose. The ceter frequecy is. GHz, the badwidth is %, ad Z = 5 W. What is the atteuatio at.8 GHz? Solutio: For a 3rd order.5 db equal-ripple desig, the g values are g=.5963, g =.967, g3 =.5963, g4 =. The J values are calculated as π. π. JZ = =.337, JZ = = π. π. JZ 3 = =.87, JZ 4 = = The, the Ze ad Zo values are foud by Z e = Z4 e = 5 ( ) = 7.6 Ω Z o = Z4 o = 5 ( ) = 39.4 Ω Z e = Z3 e = 5 ( ) = Ω Zo= Zo3= 5 ( ) = Ω.8. From the BPF to PF trasformatio, = =. c Δ...8 PR = + k TN = +. T 3 (.) = =.8 db. c 57

58 S(dB) S(dB) oupled-ie BPF desig simulatio vs. experimet Hair-pi desig Frequecy (GHz) Dual-bad desig Frequecy (GHz) 58

Butterworth LC Filter Designer

Butterworth LC Filter Designer Butterworth LC Filter Desiger R S = g g 4 g - V S g g 3 g R L = Fig. : LC filter used for odd-order aalysis g R S = g 4 g V S g g 3 g - R L = useful fuctios ad idetities Uits Costats Table of Cotets I.