4.5 Signal Flow Graphs

Transcription

1 3/9/009 4_5 ignl Flow Grphs.doc / 4.5 ignl Flow Grphs Reding Assignment: pp Q: Using individul device scttering prmeters to nlze comple microwve network results in lot of mess mth! Isn t there n esier w? A: Yes! We cn represent microwve network with its signl flow grph. HO: IGNAL FLOW GRAPH Then, we cn decompose this grph using set of stndrd rules. HO: ERIE RULE HO: PARALLEL RULE HO: ELF-LOOP RULE HO: PLITTING RULE It s sort of grphicl w to do lger! Let s do some emples: EXAMPLE: DECOMPOITION OF IGNAL FLOW GRAPH EXAMPLE: IGNAL FLOW GRAPH ANALYI Jim tiles The Univ. of Knss Dept. of EEC

2 3/9/009 4_5 ignl Flow Grphs.doc / ignl Flow grphs cn likewise help us understnd the fundmentl phsicl ehvior of network or device. It cn even help us pproimte the network in w tht mkes it simpler to nlze nd/or design! HO: THE PROPAGATION ERIE Jim tiles The Univ. of Knss Dept. of EEC

3 3/3/009 ignl Flow Grphs present /6 ignl Flow Grphs Consider comple 3-port microwve network, constructed of 5 simpler microwve devices: n is the scttering mtri of ech device, nd is the overll scttering mtri where of the entire 3-port network. Q: Is there n w to determine this overll network scttering mtri from the individul device scttering mtrices? n Jim tiles The Univ. of Knss Dept. of EEC

4 3/3/009 ignl Flow Grphs present /6 A: Definitel! Note the wve eiting one port of device is wve entering (i.e., incident on) nother (nd vice vers). This is oundr condition t the port connection etween devices. Add to this the scttering prmeter equtions from ech individul device, nd we hve sufficient mount of mth to determine the reltionship etween the incident nd eiting wves of the remining three ports in other words, the scttering mtri of the 3-port network! Q: Yikes! Wouldn t tht require lot of tedious lger! A: It sure would! We might use computer to ssist us, or we might use tool emploed since the erl ds of microwve engineering the signl flow grph. ignl flow grphs re helpful in (count em ) three ws! W - ignl flow grphs provide us with grphicl mens of solving lrge sstems of simultneous equtions. Jim tiles The Univ. of Knss Dept. of EEC

5 3/3/009 ignl Flow Grphs present 3/6 W We ll see the signl flow grph cn provide us with rod mp of the wve propgtion pths throughout microwve device or network. If we re ping ttention, we cn glen gret phsicl insight s to the inner working of the microwve device represented the grph. W 3 - ignl flow grphs provide us with quick nd ccurte method for pproimting network or device. We will find tht we cn often replce rther comple grph with much simpler one tht is lmost equivlent. We find this to e ver helpful when designing microwve components. From the nlsis of these pproimte grphs, we cn often determine design rules or equtions tht re trctle, nd llow us to design components with (ner) optiml performnce. Q: But wht is signl flow grph? A: First, some definitions! Jim tiles The Univ. of Knss Dept. of EEC

6 3/3/009 ignl Flow Grphs present 4/6 Ever signl flow grph consists of set of nodes. These nodes re connected rnches, which re simpl contours with specified direction. Ech rnch likewise hs n ssocited comple vlue. 0.7 j -j Q: Wht could this possil hve to do with microwve engineering? A: Ech port of microwve device is represented two nodes the node nd the node. The node simpl represents the vlue of the normlized mplitude of the wve incident on tht port, evluted t the plne of tht port: ( = ) V z z n Z + n n np 0n Likewise, the node simpl represents the normlized mplitude of the wve eiting tht port, evluted t the plne of tht port: Vn ( zn = znp ) n Z 0n Jim tiles The Univ. of Knss Dept. of EEC

7 3/3/009 ignl Flow Grphs present 5/6 Note then tht the totl voltge t port is simpl: ( = ) = ( + ) 0 V z z Z n n np n n n The vlue of the rnch connecting two nodes is simpl the vlue of the scttering prmeter relting these two voltge vlues: ( = ) V z z n Z + n n np 0n mn ( = ) V z z m Z m m mp 0m The signl flow grph ove is simpl grphicl representtion of the eqution: m = mn n Moreover, if multiple rnches enter node, then the voltge represented tht node is the sum of the vlues from ech rnch. Jim tiles The Univ. of Knss Dept. of EEC

8 3/3/009 ignl Flow Grphs present 6/6 For emple, the signl flow grph: 3 3 is grphicl representtion of the eqution: = Now, consider two-port device with scttering mtri : o tht: = = + = + Jim tiles The Univ. of Knss Dept. of EEC

9 3/3/009 ignl Flow Grphs present 7/6 We cn thus grphicll represent two-port device s: Now, consider cse where the second port is terminted some lod Γ L : Γ L We now hve et nother eqution: ( = ) =Γ ( = ) V z z V z z + P L P =Γ L Jim tiles The Univ. of Knss Dept. of EEC

10 3/3/009 ignl Flow Grphs present 8/6 Therefore, the signl flow grph of this terminted network is: Γ L Now let s cscde two different two-port networks Γ L Here, the output port of the first device is directl connected to the input port of the second device. We descrie this mthemticll s: = nd = Jim tiles The Univ. of Knss Dept. of EEC

11 3/3/009 ignl Flow Grphs present 9/6 Jim tiles The Univ. of Knss Dept. of EEC Thus, the signl flow grph of this network is: Q: But wht hppens if the networks re connected with trnsmission lines? A: Recll tht length of trnsmission line with chrcteristic impednce Z 0 is likewise two-port device. Its scttering mtri is: 0 0 j j e e β β = Thus, if the two devices re connected length of trnsmission line: L Γ L Γ 0 Z

12 3/3/009 ignl Flow Grphs present 0/6 = e = e j β j β so the signl flow grph is: e j β e j β Γ L Note tht there is one (nd onl one) independent vrile in this representtion. This independent vrile is node. This is the onl node of the sfg tht does not hve n incoming rnches. As result, its vlue depends on no other node vlues in the sfg. From the stndpoint of sfg, independent nodes re essentill sources! Jim tiles The Univ. of Knss Dept. of EEC

13 3/3/009 ignl Flow Grphs present /6 Of course, this likewise mkes sense phsicll (do ou see wh?). The node vlue represents the comple mplitude of the wve incident on the one-port network. If this vlue is zero, then no power is incident on the network the rest of the nodes (i.e., wve mplitudes) will likewise e zero! Now, s we wish to determine, for emple:. The reflection coefficient Γ in of the one-port device.. The totl current t port of second network (i.e., network ). 3. The power sored the lod t port of the second () network. In the first cse, we need to determine the vlue of dependent node : Γ = in For the second cse, we must determine the vlue of wve mplitudes nd : I = Z 0 Jim tiles The Univ. of Knss Dept. of EEC

14 3/3/009 ignl Flow Grphs present /6 And for the third nd finl cse, the vlues of nodes nd re required: P s = Q: But just how the heck do we determine the vlues of these wve mplitude nodes? A: One w, of course, is to solve the simultneous equtions tht descrie this network. From network nd network : = + = + = + = + From the trnsmission line: = e = e j β j β And finll from the lod: = Γ L Jim tiles The Univ. of Knss Dept. of EEC

15 3/3/009 ignl Flow Grphs present 3/6 But nother, EVEN BETTER w to determine these vlues is to decompose (reduce) the signl flow grph! Q: Huh? A: ignl flow grph reduction is method for simplifing the comple pths of tht signl flow grph into more direct (ut equivlent!) form. Reduction is rell just grphicl method of decoupling the simultneous equtions tht re descried the sfg. For instnce, in the emple we re considering, the sfg : e j β e j β Γ L Jim tiles The Univ. of Knss Dept. of EEC

16 3/3/009 ignl Flow Grphs present 4/6 might reduce to: j 8 0. e π j j 0. From this grph, we cn directl determine the vlue of ech node (i.e., the vlue of ech wve mplitude), in terms of the one independent vrile. = 0. = 06. = j 0... j π 8 = 005 = 0e = 03. = 0. Jim tiles The Univ. of Knss Dept. of EEC

17 3/3/009 ignl Flow Grphs present 5/6 And of course, we cn then determine vlues like:.. 0. Γ = = = I in j π 0 e = = Z Z P s ( 03. ) ( 0. ) = = Q: But how do we reduce the sfg to its simplified stte? Just wht is the procedure? A: ignl flow grphs cn e reduced sequentill ppling one of four simple rules. Q: Cn these rules e pplied in n order? A: No! The rules cn onl e pplied when/where the structure of the sfg llows. You must serch the sfg for structures tht llow rule to e pplied, nd the sfg will then e ( little it) reduced. You then serch for the net vlid structure where rule cn e pplied. Eventull, the sfg will e completel reduced! Jim tiles The Univ. of Knss Dept. of EEC

18 3/3/009 ignl Flow Grphs present 6/6 Q:???? A: It s it like solving puzzle. Ever sfg is different, nd so ech will require different reduction procedure. It requires little thought, ut with little prctice, the reduction procedure is esil mstered. You m even find tht it s kind of fun! Jim tiles The Univ. of Knss Dept. of EEC

19 3/3/009 eries Rule present / eries Rule Consider these two comple equtions: = α = β where α nd β re ritrr comple constnts. Using the ssocitive propert of multipliction, these two equtions cn comined to form n equivlent set of equtions: ( ) ( ) = α = β = β α = αβ Now let s epress these two sets of equtions s signl flow grphs! The first set provides: α β = α = β While the second is: α αβ = α = αβ Jim tiles The Univ. of Knss Dept. of EEC

20 3/3/009 eries Rule present / Q: He wit! If the two sets of equtions re equivlent, shouldn t the two resulting signl flow grphs likewise e equivlent? A: Asolutel! The two signl flow grphs re indeed equivlent. This leds us to our first signl flow grph reduction rule: Rule - eries Rule If node hs one (nd onl one!) incoming rnch, nd one (nd onl one!) outgoing rnch, the node cn e eliminted nd the two rnches cn e comined, with the new rnch hving vlue equl to the product of the originl two. For emple, the grph: 03. j = 03. = j cn e reduced to: 03. = 03. = j 03. j 03. Jim tiles The Univ. of Knss Dept. of EEC

21 3/3/009 Prllel Rule present /6 Prllel Rule Consider the comple eqution: = α + β where α nd β re ritrr comple constnts. Using the distriutive propert, the eqution cn equivlentl e epressed s: ( α β ) = + Now let s epress these two equtions s signl flow grphs! The first is: α = α + β β With the second: α + β = ( α + β ) Jim tiles The Univ. of Knss Dept. of EEC

22 3/3/009 Prllel Rule present /6 Q: He wit! If the two equtions re equivlent, shouldn t the two resulting signl flow grphs likewise e equivlent? A: Asolutel! The two signl flow grphs re indeed equivlent. This leds us to our second signl flow grph reduction rule: Rule - Prllel Rule If two nodes re connected prllel rnches nd the rnches hve the sme direction the rnches cn e comined into single rnch, with vlue equl to the sum of ech two originl rnches. For emple, the grph: 0. = Cn e reduced to: 03. ( 03 0) =. +. = Jim tiles The Univ. of Knss Dept. of EEC

23 3/3/009 Prllel Rule present 3/6 Q: Wht out this signl flow grph? note direction! 0. Cn I rewrite this s: so tht (since =0.): ??? A: Asolutel not! NEVER DO THI!! Jim tiles The Univ. of Knss Dept. of EEC

24 3/3/009 Prllel Rule present 4/6 Q: Me I mde mistke. Perhps I should hve rewritten: note direction! 0. s this: = 0. so tht (since =5.3): ??? A: Asolutel not! NEVER DO THI EITHER!! Jim tiles The Univ. of Knss Dept. of EEC

25 3/3/009 Prllel Rule present 5/6 From the signl flow grph elow, we cn onl conclude tht = 0.3 nd = Using the series rule (or little it of lger), we cn conclude tht n equivlent signl flow grph to this is: = 0.06 = Q: Yikes! Wht kind of goof rnch egins nd ends t the sme node? 03. A: Brnches tht egin nd end t the sme node re clled self-loops. Jim tiles The Univ. of Knss Dept. of EEC

26 3/3/009 Prllel Rule present 6/6 Q: Do these self-loops ctull pper in signl flow grphs? A: Yes, ut the self-loop node will lws hve t lest one other incoming rnch. For emple: 006. = 0.06 j = 0.3 j 03. Q: But how do we reduce signl flow grph contining self-loop? A: ee rule 3! Jim tiles The Univ. of Knss Dept. of EEC

27 3/3/009 elf Loop Rule present /4 elf-loop Rule Now consider the eqution: = α + β + γ A little d of lger llows us to determine the vlue of node : = α + β + γ γ = α + β ( γ) = α + β α β = + γ γ The signl flow grph of the first eqution is: α γ = α + β + γ β Jim tiles The Univ. of Knss Dept. of EEC

28 3/3/009 elf Loop Rule present /4 While the signl flow grph of the second is: α β = + γ γ α γ β γ These two signl flow grphs re equivlent! Note the self-loop hs een removed in the second grph. Thus, we now hve method for removing self-loops. This method is rule 3. Rule 3 elf-loop Rule A self-loop cn e eliminte multipling ll of the rnches feeding the self-loop node ( sl ), where sl is the vlue of the self loop rnch. For emple: = j cn e simplified eliminting the self-loop. j 04. Jim tiles The Univ. of Knss Dept. of EEC

29 3/3/009 elf Loop Rule present 3/4 We multipl oth of the two rnches feeding the self-loop node : Therefore: = = ( 5. ) sl j 04. ( 5. ) And thus: 075. j = j Or nother emple: = 0.06 j = j 03. Jim tiles The Univ. of Knss Dept. of EEC

30 3/3/009 elf Loop Rule present 4/4 ecomes fter reduction using rule 3: = j 0.94 = 0.3 j Q: Wit minute! I think ou forgot something. houldn t ou lso divide the 0.3 rnch vlue =?? A: Nope! The 0.3 rnch is eiting the self-loop node. Onl incoming rnches (e.g., the j rnch) to the self-loop node re modified the self-loop rule! Jim tiles The Univ. of Knss Dept. of EEC

31 3/3/009 plitting Rule present /5 plitting Rule Now consider these three equtions: = α = β = γ 3 Using the ssocitive propert, we cn likewise write n equivlent set of equtions: = α = αβ = α 3 The signl flow grph of the first set of equtions is: β α 3 While the signl flow grph of the second is: γ αβ α γ 3 Jim tiles The Univ. of Knss Dept. of EEC

32 3/3/009 plitting Rule present /5 Rule 4 plitting Rule If node hs one (nd onl one!) incoming rnch, nd one (or more) eiting rnches, the incoming rnch cn e split, nd directl comined with ech of the eiting rnches. For emple: j = j = 03. = 0. 3 cn e rewritten s: j j 0. 3 = j = j 03. = 0. 3 Jim tiles The Univ. of Knss Dept. of EEC

33 3/3/009 plitting Rule present 3/5 Of course, from rule (or from rule 4!), this grph cn e further simplified s: j 03. j The splitting rule is prticulrl useful when we encounter signl flow grphs of the kind: 03. j j 0. 3 j 0. = j = j 03. = j 0. 3 Note this node hs two incoming rnches!! 0. Note this node hs onl one incoming rnch!! We cn split the -0. rnch, nd rewrite the grph s: ( ) j 0. j 0. Jim tiles The Univ. of Knss Dept. of EEC

34 3/3/009 plitting Rule present 4/5 Note we now hve self-loop, which cn e eliminted using rule #3: j j 0. Note tht this grph cn e further simplified using rule #. j 094. j 89. j 0. Q: Cn we split the other rnch of the loop? Is this signl flow grph: ( ) j Likewise equivlent to this one??: j j 0. j Jim tiles The Univ. of Knss Dept. of EEC

35 3/3/009 plitting Rule present 5/5 A: NO!! Do not mke this mistke! We cnnot split the 0.3 rnch ecuse it termintes in node with two incoming rnches (i.e., -j nd 0.3). This is violtion of rule 4. Moreover, the equtions represented the two signl flow grphs re not equivlent the two grphs descrie two different sets of equtions! It is importnt to rememer tht there is no mgic ehind signl flow grphs. The re simpl grphicl method of representing nd then solving set of liner equtions. As such, the four sic rules of nlzing signl flow grph represent sic lgeric opertions. In fct, signl flow grphs cn e pplied to the nlsis of n liner sstem, not just microwve networks. Jim tiles The Univ. of Knss Dept. of EEC

36 3/3/009 Emple Decomposition of ignl Flow Grph /3 Emple: Decomposition of ignl Flow Grphs Consider the sic -port network, terminted with lod Γ. L Γ L we wnt to determine the vlue: Γ L ( = P ) ( = ) V z z Γ =?? V z z + P In other words, wht is the reflection coefficient of the resulting one-port device? Q: Isn t this simpl? A: Onl if Γ L = 0 (nd it s not)!! o let s decompose (simplif) the signl flow grph nd find out! Jim tiles The Univ. of Knss Dept. of EEC

37 3/3/009 Emple Decomposition of ignl Flow Grph /3 tep : Use rule #4 on node Γ L tep : Use rule #3 on node Γ L Γ L tep 3: And then using rule #: Γ L Γ L Γ L Γ L Γ L Γ L Jim tiles The Univ. of Knss Dept. of EEC

38 3/3/009 Emple Decomposition of ignl Flow Grph 3/3 tep 4: Use rule on nodes nd Γ L + Γ ΓL L ΓL ΓL Therefore: Γ = = Γ + Γ L L Note if Γ = 0, then L =! Jim tiles The Univ. of Knss Dept. of EEC

39 3/5/009 Emple ignl Flow Grph Anlsis /6 Emple: Anlsis Using ignl Flow Grphs Below is single-port device (with input t port ) constructed with two two-port devices ( nd ), qurter wvelength trnsmission line, nd lod impednce. j = λ 4 Z 0 Z 0 Γ L = port (input) port port port Where Z 0 = 50Ω. The scttering mtrices of the two-port devices re: = = Likewise, we know tht the vlue of the voltge wve incident on port of device is: ( = ) + V0 z zp j j = = Z V Jim tiles The Univ. of Knss Dept. of EEC

40 3/5/009 Emple ignl Flow Grph Anlsis /6 Now, let s drw the complete signl flow grph of this circuit, nd then reduce the grph to determine: ) The totl current through lod Γ L. ) The power delivered to (i.e., sored ) port. The signl flow grph descriing this network is: e j β e j β Γ L Inserting the numeric vlues of rnches: = j 5 j j 08. Jim tiles The Univ. of Knss Dept. of EEC

41 3/5/009 Emple ignl Flow Grph Anlsis 3/6 Removing the zero vlued rnches: = j 5 j j And now ppling splitting rule 4: 08. = j 5 j j Followed the self-loop rule 3: = j 5 j ( ) 04. = = j 08. Jim tiles The Univ. of Knss Dept. of EEC

42 3/5/009 Emple ignl Flow Grph Anlsis 4/6 Now, let s used this simplified signl flow grph to find the solutions to our questions! ) The totl current through lod Γ L. The totl current through the lod is: ( ) + ( = P ) ( = P ) I = I z = z L P V0 z z V0 z z = Z0 = Z0 = 50 Thus, we need to determine the vlue of nodes nd. Using the series rule on our signl flow grph: = j j j 04. From this grph we cn conclude: Note we ve simpl ignored (i.e., neglected to plot) the node for which we hve no interest! Jim tiles The Univ. of Knss Dept. of EEC

43 3/5/009 Emple ignl Flow Grph Anlsis 5/6 nd: j j j = = = 0. 5 ( ) = = 0. = 0 Therefore: ( ) I L = = = = 0 0. ma ) The power delivered to (i.e., sored ) port. The power delivered to port is: P = P P s + ( = ) ( = ) + P P V z z V z z = Z Z 0 0 = Thus, we need determine the vlues of nodes nd. Agin using the series rule on our signl flow grph: 035. = j 5 0. Agin we ve simpl ignored (i.e., neglected to plot) the node for which we hve no interest! Jim tiles The Univ. of Knss Dept. of EEC

44 3/5/009 Emple ignl Flow Grph Anlsis 6/6 And then using the prllel rule : = j 5 = Therefore: ( 5 ) = = j = j 0 nd: j 5 j P s = = = mw Jim tiles The Univ. of Knss Dept. of EEC

45 3/3/009 The Propgtion eries present /0 The Propgtion eries Q: You erlier stted tht signl flow grphs re helpful in (count em ) three ws. I now understnd the first w: W - ignl flow grphs provide us with grphicl mens of solving lrge sstems of simultneous equtions. But wht out ws nd 3?? W We ll see the signl flow grph cn provide us with rod mp of the wve propgtion pths throughout microwve device or network. W 3 - ignl flow grphs provide us with quick nd ccurte method for pproimting network or device. Jim tiles The Univ. of Knss Dept. of EEC

46 3/3/009 The Propgtion eries present /0 A: Consider the sfg elow: Note tht node is the onl independent node. This signl flow grph is for rther comple single-port (port ) device. j j we wish to determine the wve mplitude eiting port. In other words, we seek: = Γ in Using our four reduction rules, the signl flow grph ove is simplified to: j 3 4 j j 036. j Jim tiles The Univ. of Knss Dept. of EEC

47 3/3/009 The Propgtion eries present 3/0 Q: He, node is not connected to nthing. Wht does this men? A: It mens tht = 0 regrdless of the vlue of incident wve. I.E.,: Γ in = = 0 In other words, port is mtched lod! Q: But look t the originl signl flow grph; it doesn t look like mtched lod. How cn the eiting wve t port e zero? A: A signl flow grph provides it of propgtion rod mp through the device or network. It llows us to understnd often in ver phsicl w the propgtion of n incident wve once it enters device. We ccomplish this identifing from the sfg propgtion pths from n independent node to some other node (e.g., n eiting node). These pths re simpl sequence of rnches (pointing in the correct direction!) tht led from the independent node to this other node. Ech pth hs vlue tht is equl to the product of ech rnch of the pth. Jim tiles The Univ. of Knss Dept. of EEC

48 3/3/009 The Propgtion eries present 4/0 Perhps this is est eplined with some emples. One pth etween independent (incident wve) node nd (eiting wve) node is shown elow: We ll ritrril cll this pth, nd its vlue: j j 04. ( ) ( ) ( ) p = 0.5 j 0.4 j 0.5 = 0. Another propgtion pth (pth 5, s) is: 08. j j Jim tiles The Univ. of Knss Dept. of EEC

49 3/3/009 The Propgtion eries present 5/0 = 0.0 Q: Wh re we doing this? 5 ( 0.5) ( 0.4) ( 0.35) ( 0.8)( 0.5)( 0.8) ( 0.5) p = j j j j = j 4 ( 0.35)( 0.4)( 0.8) ( 0.5) 3 A: The eiting wve t port (wve mplitude ) is simpl the superposition of ll the propgtion pths from incident node! Mthemticll speking: = p Γ = = p n n in n n Q: Won t there e n wful lot of propgtion pths? A: Yes! As mtter of fct there re n infinite numer of pths tht connect node nd. Therefore: = pn Γ in = = pn Q: Yikes! Does this infinite series converge? n n A: Note tht the series represents finite phsicl vlue (e.g., Γ in ), so tht the infinite series must converge to the correct finite vlue. Jim tiles The Univ. of Knss Dept. of EEC

50 3/3/009 The Propgtion eries present 6/0 Q: In this emple we found tht Γ in = 0. This mens tht the infinite propgtion series is likewise zero: Γ = p = 0 Do we conclude from this tht ll propgtion pths re zero: in n n p = 0????? n A: Asolutel not! Rememer, we hve lred determined tht p = 0. nd p 4 = 0.0 definitel not zero-vlued! In fct for this emple, none of the propgtion pths p n re precisel equl to zero! Q: But then wh is: pn = 0??? n A: Rememer, the pth vlues p n re comple. A sum of non-zero comple vlues cn equl zero (s it pprentl does in this cse!). Jim tiles The Univ. of Knss Dept. of EEC

51 3/3/009 The Propgtion eries present 7/0 Thus, perfectl rtionl w of viewing this network is to conclude tht there re n infinite numer of non-zero wves eiting port : Γ = p where p 0 in n n n It just so hppens tht these wves coherentl dd together to zero: in Γ = p = n n 0 the essentill cncel ech other out! Q: o, I now pprecite the fct tht signl flow grphs: ) provides grphicl method for solving liner equtions nd ) lso provides method for phsicll evluting the wve propgtion pths through network/device. But wht out helpful W 3: W 3 - ignl flow grphs provide us with quick nd ccurte method for pproimting network or device.?? Jim tiles The Univ. of Knss Dept. of EEC

52 3/3/009 The Propgtion eries present 8/0 A: The propgtion series of microwve network is ver nlogous to Tlor eries epnsion: n d f( ) n f ( ) = ( ) n d n = 0 = Note tht there likewise is infinite numer of terms, et the Tlor eries is quite helpful in engineering. Often, we engineers simpl truncte this infinite series, mking it finite one: Q: Yikes! Doesn t this result in error? N n d f( ) f ( ) ( ) n d n = 0 = n A: Asolutel! The truncted series is n pproimtion. We hve less error if more terms re retined; more error if fewer terms re retined. The trick is to retin the significnt terms of the infinite series, nd truncte those less importnt insignificnt terms. In this w, we seek to form n ccurte pproimtion, using the fewest numer of terms. Jim tiles The Univ. of Knss Dept. of EEC

53 3/3/009 The Propgtion eries present 9/0 Q: But how do we know which terms re significnt, nd which re not? A: For Tlor eries, we find tht s the order n increses, the significnce of the term generll (ut not lws!) decreses. Q: But wht out our propgtion series? How cn we determine which pths re significnt in the series? A: Almost lws, the most significnt pths in propgtion series re the forwrd pths of signl flow grph. forwrd pth \ˈfoṙ wərdˈ päth\ noun A pth through signl flow grph tht psses through n given node no more thn once. A pth tht psses through n node two times (or more) is therefore not forwrd pth. In our emple, pth is forwrd pth. It psses through four nodes s it trvels from node to node, ut it psses through ech of these nodes onl once: 044. j j Jim tiles The Univ. of Knss Dept. of EEC

54 3/3/009 The Propgtion eries present 0/0 Alterntivel, pth 5 is not forwrd pth: 08. j j We see tht pth 5 psses through si different nodes s it trvels from node to node. However, it twice psses through four of these nodes. The good news out forwrd pths is tht there re lws finite numer of them. Agin, these pths re tpicll the most significnt in the propgtion series, so we cn determine n pproimte vlue for sfg nodes considering onl these forwrd pths in the propgtion series: N pn n n= fp where p n represents the vlue of one of the N forwrd pths. p fp n Jim tiles The Univ. of Knss Dept. of EEC

55 3/3/009 The Propgtion eries present /0 Q: Is pth the onl forwrd pth in our emple sfg? A: No, there re three. Pth is the most direct: p = j j Of course we lred hve identified pth : p = j j Jim tiles The Univ. of Knss Dept. of EEC

56 3/3/009 The Propgtion eries present /0 And finll, pth 3 is the longest forwrd pth: 3 ( 0.5) ( 0.8)( 0.5)( 0.8) ( 0.5) p = j j = j = 0.08 ( 0.8) ( 0.5) j j Thus, n pproimte vlue of Γin is: Γ in = 3 n = p fp n = = Q: He wit! We determined erlier tht Γ in = 0, ut now our sing tht Γ in = Which is correct?? Jim tiles The Univ. of Knss Dept. of EEC

57 3/3/009 The Propgtion eries present 3/0 A: The correct nswer is Γ in = 0. It ws determined using the four sfg reduction rules no pproimtions were involved! Conversel, the vlue Γ in = ws determined using truncted form of the propgtion series the series ws limited to just the three most significnt terms (i.e., the forwrd pths). The result is esier to otin, ut it is just n pproimtion (the nswers will contin error!). For emple, consider the reduced signl flow grph (no pproimtion error): 04. j j 03. Ect FG 088. j j Compre this to the sme sfg, computed using onl the forwrd pths: j j 04. Appro. FG 036. j 036. j Jim tiles The Univ. of Knss Dept. of EEC

58 3/3/009 The Propgtion eries present 4/0 No surprise, the pproimte sfg (using forwrd pths onl) is not the sme s the ect sfg (using reduction rules). The pproimte sfg contins error, ut note this error is not too d. The vlues of the pproimte sfg re certinl close to tht of the ect sfg. Q: Is there n w to improve the ccurc of this pproimtion? A: Certinl. The error is result of truncting the infinite propgtion series. Note we severel truncted the series out of n infinite numer of terms, we retined onl three (the forwrd pths). If we retin more terms, we will likel get more ccurte nswer. Q: o wh did these pproimte nswers turn out so well, given tht we onl used three terms? A: We retined the three most significnt terms, we will find tht the forwrd pths tpicll hve the lrgest mgnitudes of ll propgtion pths. Q: An ide wht the net most significnt terms re? A: Yup. The forwrd pths re ll those propgtion pths tht pss through n node no more thn one time. The net most significnt pths re lmost certinl those pths tht pss through n node no more thn two times. Jim tiles The Univ. of Knss Dept. of EEC

59 3/3/009 The Propgtion eries present 5/0 Pth 4 is n emple of such pth: 044. There re three more of these pths (pssing through node no more thn two times) see if ou cn find them! After determining the vlues for these pths, we cn dd 4 more terms to our summtion (now we hve seven terms!): Γ in = 7 n = p n 035. ( p p p3) ( p4 p5 p6 p7) ( 0.036) ( ) = j j = = Jim tiles The Univ. of Knss Dept. of EEC

60 3/3/009 The Propgtion eries present 6/0 Note this vlue is closer to the correct vlue of zero thn ws our previous (using onl three terms) nswer of As we dd more terms to the summtion, this pproimte nswer will get closer nd closer to the correct vlue of zero. However, it will e ectl zero (to n infinite numer of deciml points) onl if we sum n infinite numer of terms! Q: The significnce of given pth seem to e inversel proportionl to the numer of times it psses through n node. Is this true? If so, then wh is it true? A: It is true (generll speking)! A propgtion pth tht trvels though node ten times is much less likel to e significnt to the propgtion series (i.e., summtion) thn pth tht psses through n node no more thn (s) four times. The reson for this is tht the significnce of given term in summtion is dependent on its mgnitude (i.e., p n ). If the mgnitude of term is smll, it will hve fr less ffect (i.e., significnce) on the sum thn will term whose mgnitude is lrge. Q: You seem to e sing tht pths trveling through fewer nodes hve lrger mgnitudes thn those trveling through mn nodes. Is tht true? If so wh? Jim tiles The Univ. of Knss Dept. of EEC

61 3/3/009 The Propgtion eries present 7/0 A: Keep in mind tht microwve sfg reltes wve mplitudes. The rnch vlues re therefore lws scttering prmeters. One importnt thing out scttering prmeters, their mgnitudes (for pssive devices) re lws less thn or equl to one! mn Recll the vlue of pth is simpl the product of ech rnch tht forms the pth. The more rnches (nd thus nodes), the more terms in this product. ince ech term hs mgnitude less thn one, the mgnitude of product of mn terms is much smller thn product of few terms. For emple: 3 j 0.7 = nd 0 j 0.7 = 0.08 In other words, pths with more rnches (i.e., more nodes) will tpicll hve smller mgnitudes nd so re less significnt in the propgtion series. Note pth in our emple trveled long one rnch onl: p = 0.44 Pth hs five rnches: Jim tiles The Univ. of Knss Dept. of EEC

62 3/3/009 The Propgtion eries present 8/0 p = 0. Pth 3 seven rnches: p 3 = 0.08 Pth 4 nine rnches: p 4 = 0.04 Pth 5 eleven rnches: p 5 = 0.0 Pth 6 eleven rnches: p 6 = 0.0 Pth 7 thirteen rnches: p 7 = Hopefull it is evident tht the mgnitude diminishes s the pth length increses. Jim tiles The Univ. of Knss Dept. of EEC

63 3/3/009 The Propgtion eries present 9/0 Q: o, does this men tht we should ndon our four reduction rules, nd insted use truncted propgtion series to evlute signl flow grphs?? A: Asolutel not! Rememer, truncting the propgtion series lws results in some error. This error might e sufficientl smll if we retin enough terms, ut knowing precisel how mn terms to retin is prolemtic. We find tht in most cses it is simpl not worth the effort use the four reduction rules insted (it s not like the re prticulrl difficult!). Q: You s tht in most cses it is not worth the effort. Are there some cses where this pproimtion is ctull useful?? A: Yes. A truncted propgtion series (tpicll using onl the forwrd pths) is used when these three things re true:. The network or device is comple (lots of nodes nd rnches).. We cn conclude from our knowledge of the device tht the forwrd pths re sufficient for n ccurte pproimtion (i.e., the mgnitudes of ll other pths in the series re lmost certinl ver smll). Jim tiles The Univ. of Knss Dept. of EEC

64 3/3/009 The Propgtion eries present 0/0 3. The rnch vlues re not numeric, ut insted re vriles tht re dependent on the phsicl prmeters of the device (e.g., chrcteristic impednce or line length). The result is tpicll trctle mthemticl eqution tht reltes the design vriles (e.g., Z 0 or ) of comple device to specific device prmeter. For emple, we might use truncted propgtion series to pproimtel determine some function: Γin( Z,, Z, ) 0 0 If we desire mtched input (i.e., ( ) Γ in Z0,, Z0, = 0) we cn solve this trctle design eqution for the (nerl) proper vlues of Z0,, Z0,. We will use this technique to gret effect for designing multi-section mtching networks nd multi-section coupled line couplers. 3 3 e e j θ e j θ j θ j jc sinθ e θ sin j jc θ e θ e j θ e e j θ e j θ sin j jc θ e θ sin j jc θ e θ sin j jc3 θ e θ sin j jc θ e θ sin j jc θ e θ sin j jc3 θ e θ j θ sin j jc θ e θ sin j jc θ e θ e j θ Jim tiles The Univ. of Knss Dept. of EEC e e j θ e j θ j θ sin j jc3 θ e θ sin j jc3 θ e θ e j θ 4 4 The signl flow grph of three-section coupled-line coupler.