DZ201PLC控制的恒壓供水系統(tǒng),dz201plc,控制,節(jié)制,供水系統(tǒng) 長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 1 -附 錄集成單元積分的最優(yōu)近似值函數(shù)1.前言：積分混合電路在無(wú)線電頻率系統(tǒng)范圍中尋找應(yīng)用對(duì)象。這包括單獨(dú)的邊頻帶相應(yīng)位鑒別器、平衡放大器、衰減器、相位轉(zhuǎn)換器和可靠的幾種平衡混頻器。在為微波頻率中，他們經(jīng)常用分布電路實(shí)現(xiàn)。但是，在較底頻率中，他們將被集成。這篇論文認(rèn)為那些展現(xiàn)精確積分特性的電路，假設(shè)用一個(gè)集成單元實(shí)現(xiàn)。在這論文中，一個(gè)源于實(shí)現(xiàn)方案執(zhí)行近似值函數(shù)在文獻(xiàn)[3、4]中可以不再抑制這篇論文尋址，這還沒(méi)有回答問(wèn)題：給出一個(gè)操作運(yùn)算的頻率帶和振幅平衡規(guī)則。什么是最優(yōu)的解決方法，象它將被電路的最小階數(shù)完成。假設(shè)他將被右無(wú)源電路模式中最終實(shí)現(xiàn)，當(dāng)然盡管他將可能用有效單元實(shí)現(xiàn)函數(shù)。2.問(wèn)題的陳述：一般的混合電路如圖 1所示。假設(shè)電路是無(wú)損的，如果所有的節(jié)點(diǎn)是完美的匹配，那么，理論指示他必須是一個(gè)方向的配合者。電源附屬于一個(gè)節(jié)點(diǎn)，將被在兩個(gè)較遠(yuǎn)的節(jié)點(diǎn)分開(kāi)，剩余的節(jié)點(diǎn)被絕緣。在這篇論文中，假設(shè)激發(fā)的主要節(jié)點(diǎn)是節(jié)點(diǎn) 1，功率在節(jié)點(diǎn) 2和 4中被分開(kāi)，接待內(nèi) 3被絕緣，它已經(jīng)被記載積分連接器的移動(dòng)函數(shù)可以在 W項(xiàng)中用表達(dá)式描述：)(1)(2wFSi???3.在 F（W）是 W的奇數(shù)函數(shù)。一個(gè)相似的表達(dá)式可以因?yàn)?￥被分開(kāi)，用他的倒數(shù) F（W） 。F（W）是一個(gè)奇函數(shù)，他因?yàn)楦S￥或￥在 0頻率時(shí)是整體。假設(shè)這應(yīng)用于接點(diǎn) 4 ，因此 F（ 0）=0。這樣接點(diǎn)變成“通過(guò)”節(jié)點(diǎn)，接點(diǎn)變成“耦合”節(jié)點(diǎn)，在普通的用法中它要求對(duì)于討論的目的來(lái)說(shuō)，轉(zhuǎn)移函數(shù)結(jié)果在近似值等于功率因數(shù)除以指定的頻率范圍。因此，F(xiàn)（W）近似于乘以這個(gè)范圍。一個(gè) F（W）的理想特征在圖之中顯示在￥和￥分別是較底和較高頻率通頻帶極限。 I01234混合積分 I0圖一混合積分類長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 2 -1W1 Wu圖二 F（W）的理想特設(shè)計(jì)者的任務(wù)是求出這理想指數(shù)的近似值。它不能用有限值函數(shù)產(chǎn)生。研究的目的也是求出最優(yōu)的解決方案：考慮到方程描述的函數(shù)如下： ).1)((.)( 2222dWAwFdNn??假設(shè)分子的等級(jí)和字母的第一項(xiàng)不同，第二項(xiàng)的更高求出函數(shù)的階數(shù)。此外，自然頻率的值建議在特性因式分解中，他們是￥，￥，￥，￥等等，考慮到此刻 W從 0上升的曲線，在靠近 0點(diǎn)，在分子中因式的第一階將計(jì)算出引起 F（W）近似線性的上升。然而，隨著 W上升加深分母中的每一個(gè)三次項(xiàng)和分子將引起函數(shù)脈動(dòng)。它可以看作一半脈動(dòng)數(shù)等于函數(shù)的階數(shù)。已經(jīng)選擇了 F（W）的階數(shù)，任務(wù)是選擇自然頻率的數(shù)值和常數(shù)乘以 A，象最后的脈動(dòng)函數(shù)之間和形成關(guān)于整體的幾何極限的切線。長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 3 -一個(gè)典型的解決方案如圖 3所示。對(duì)于第五階的函數(shù)帶有一個(gè)寬頻無(wú)線電的100，在一個(gè) 0.1~10標(biāo)準(zhǔn)化的范圍，較底和較高的脈動(dòng)極限分別是被指定的 1/M或M倍。對(duì)剩余的常規(guī)保留在這[篇論文中圖 3第五階的合理函數(shù)4、F（W）的本質(zhì)的最優(yōu)驗(yàn)證： 讓 是 n階分母函數(shù)是根據(jù)第三)(wQPF?部分中的方法所得?，F(xiàn)在讓￥作為第二函數(shù)，這符合以至更優(yōu)于￥的近似值。整體超過(guò)指定的頻率范圍，也適應(yīng)與 F（W）作為 W的奇函數(shù)相同的約束條件。在通頻帶F（W）在較底與較高極限之間脈動(dòng)幾次，G（W）必然在相同的通頻帶與 F（W）相交至少幾次。當(dāng)來(lái)年感個(gè)函數(shù)是奇數(shù)的，那必須有幾個(gè)更進(jìn)一步的交叉在負(fù)數(shù)通頻帶，在那兩個(gè)函數(shù)的脈動(dòng)與第一個(gè)相交 2n+1次。換句話說(shuō)，表達(dá)式： )()()()( wSQRpwSRQpw??????必須有至少 2n+1個(gè)零點(diǎn)。當(dāng) P（W）或 Q（W）是 n階的，他跟隨 P（W）或Q（W）必須是至少 n+1階的。而切，因此 G（W）是比 F（W）更高階的。他所跟隨F（W）必須是最優(yōu)的和唯一的對(duì)于給定的階數(shù)和通頻帶。5、F（W）的幾何對(duì)稱性：考慮到幾何學(xué)上的頻率限制被應(yīng)用于整體，因此￥現(xiàn)在考慮函數(shù) F（W）在第三階段向幾派生的情況，也考慮函數(shù) F（1/W） ，這也是n階的。而且在 M和 1/M極限間震蕩幾次，首先達(dá)到 M，當(dāng) W從 0開(kāi)始上升，幾為階數(shù)，考慮函數(shù) 1/F（1/W）從 0開(kāi)始上升到 1/M。當(dāng) W從 0開(kāi)始上升到￥它也在兩個(gè)極限間震蕩與原始函數(shù) F（W）相同的次數(shù)。他的特性與 F（W）嚴(yán)密的相同，他的階數(shù)當(dāng)然也是 n而且這個(gè)函數(shù)已在第四部分中證明是唯一的，它服從于F(w)=1/F(1/w) 或 F(w) n為奇數(shù)（6）)1(w?現(xiàn)在考慮在 n為偶數(shù)時(shí)的情形。這時(shí)，F(xiàn)（1/W）的性質(zhì)與 F（W）完全相同。而它是相同階數(shù)的。這個(gè)函數(shù)也是唯一的。我們可以立刻寫(xiě)出 F（W）= F（1/W） ，n 為偶數(shù)（7）由于表達(dá)式 6和 7 已經(jīng)被導(dǎo)出，我們可以定義函數(shù) ￥￥￥n 階的有理奇數(shù)函數(shù)，超過(guò)頻率范圍 在 1/M和 M之間是相等的脈動(dòng)。nbW~1長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 4 -6、分析 F-m= ：因?yàn)槎虝旱木壒?，?xiě)出 F=F ,P=P(w)，QmP? )1,(unwFP=P（W）和 Q=Q（W）在表達(dá)式 3對(duì) W求積分，得到 ，表達(dá)式 8的2QP????分子的檢查將揭示有最大的 2（n-1）個(gè)零點(diǎn)。然而從發(fā)展 F 的程序來(lái)看，同樣將n-1轉(zhuǎn)折點(diǎn)。在復(fù)數(shù)通頻帶，因此，可以推斷 F上午所有零點(diǎn)都是實(shí)數(shù)和簡(jiǎn)單的?？偣彩?2（n-1）個(gè)。現(xiàn)在考慮方程 F-m= （9）表達(dá)式 9的零點(diǎn)可以在 F=m處確定，從程序QmP?應(yīng)用于生產(chǎn) F來(lái)看，如果 n是奇數(shù)，當(dāng) W= 時(shí)，約一個(gè)單一的零點(diǎn)，伴隨更多nW的一對(duì)零點(diǎn)在通頻帶的下面，當(dāng)表達(dá)式 9的分子 n階的，不會(huì)有零點(diǎn)，更深的的考慮方程 （10）F+m= （11）F+ （12）表達(dá)式mQP??1QmP?mQ??110~12都有幾個(gè)零點(diǎn)，表達(dá)式 10 在通頻帶上面有零點(diǎn)，除了一個(gè)單一的零點(diǎn)，所有的雙零點(diǎn)低頻邊緣，而且 n為偶數(shù)。 、另外的單一零點(diǎn)在上帶的邊緣。表達(dá)時(shí) 11和12有和表達(dá)式 9、10 相反符號(hào)的零點(diǎn)，現(xiàn)在考慮把表達(dá)式 8，可以看到表達(dá)式 8所有單一的零點(diǎn)出現(xiàn)在相乘的函數(shù)的雙零點(diǎn)，但是后者另外的單一的零點(diǎn)在 四個(gè)通頻帶的邊緣，再比較表達(dá)式 8的分母和相乘的函數(shù)，可以看到后者有表達(dá)式 8重復(fù)的極和不再多的包含，因此它可能這樣寫(xiě)： ?? ))()(1)((122 UWmWCFFF??????C是必須的常數(shù)作為參數(shù)僅僅被定義每一邊應(yīng)該相同的極和零點(diǎn)表達(dá)式 13是一個(gè)一階的常有可分離度量的積微分方程重新整理給出： )1)(()1(( 2222 FMcdwduu ???表達(dá)式 14是一個(gè)橢圓的微分方程它可以用 Iacobi橢圓函數(shù)求解。最后，它是方便的定義一個(gè)變量，它的關(guān)于對(duì)每邊的積分常數(shù)，橢圓函數(shù)的系數(shù)是（16）當(dāng)變量 Z被武斷的選取，在這階段我們也準(zhǔn)許武斷的賦值 C最方便21UWK?的值是 0，這樣 Z= Z的軌跡可以在 Z平面上繪制當(dāng) W從 0 到無(wú)11)(CwSnuu??窮大，這圖 4中顯示，這個(gè)軌跡方便的分為 3段，第一段沿著實(shí)軸符合 W的數(shù)值從0到波段的較底邊緣，第二斷，平行于虛軸，符合 W的數(shù)值從較底到較高頻帶邊緣長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 5 -中間點(diǎn)精確的符合 W=1，最后，第三段，平行于實(shí)軸，符合 W值，從較高頻帶到無(wú)窮大，注意，在圖一中 K1和 K1分別是完整的橢圓積分的系數(shù)和互補(bǔ)的系數(shù)。現(xiàn)在把我們的注意力轉(zhuǎn)到表達(dá)式 14的右手邊，它可以被積分字同的方式下得 Z= ncmFSnc??)(1（18）這里 是積分常數(shù)，橢圓函數(shù)的系數(shù)是 （19）在 W和 Z 之間21mKn?已經(jīng)定義一個(gè)積分常數(shù)，的值不是武斷的，它從表達(dá)式 17中調(diào)換 Z和 F可得 0?nC或 （在 是完整的橢圓的積分的 處）在平行四邊形周期內(nèi)，解決方案nKn n的所有的合適點(diǎn)，對(duì)于簡(jiǎn)明性，解決方案 是被選擇的，所以 0?C（20）那是很有趣的，注意表達(dá)式 14是幾乎同樣的對(duì)于微分方程出)(1mFScZ??現(xiàn)在導(dǎo)出底通橢圓濾波器期間，明確的，那些解決方案，對(duì)于目前的目的是不要求的，不同的等級(jí)被要求的，其中一個(gè)在圖 5中舉列說(shuō)明對(duì)于 n=5的事件。他們通過(guò)在虛數(shù)方向周期的壓縮，W 曲線中被發(fā)展，在 Z平面上的軌跡是同用在圖中的相同，軌跡終止在對(duì)于任何奇數(shù) F的無(wú)窮大處，對(duì)于偶數(shù) 階段，軌跡應(yīng)該終止在 F 的 0處。8、結(jié)論它已經(jīng)顯示一個(gè)有理函數(shù)規(guī)定對(duì)于給定階數(shù)的積分混合電路最優(yōu)基數(shù)，這個(gè)函數(shù)有幾何數(shù)量規(guī)定解析的簡(jiǎn)化，對(duì)于最優(yōu)有理數(shù)的一個(gè)分析解決方案，已經(jīng)被導(dǎo)出，使他可能推算特性在早期的設(shè)計(jì)中，這個(gè)函數(shù)的竟區(qū)的就億斤微億方案已經(jīng)被介紹，對(duì)于每一個(gè)寬頻電路來(lái)說(shuō)它們自然的表明它們的電容，在實(shí)型頻率變化條件中已經(jīng)求出解決方案，這些可以用于在復(fù)數(shù)變量 的條件下，求出轉(zhuǎn)換函數(shù)，對(duì)于電路綜合是必要的。長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 6 -英文附錄Optimum approximation functions for lumped element quadrature hybridsD.P.Andrews and C.S.AitchisonAbstract:The theory of a class of rational chebychev functions that determine the response of lumped element quadrature hybrids of suitable dedign. Analytical solutions are presented, allowing the designer to determine the order of function required to meet a given divider specification and to predict the performance over frequency.1. IntroductionQuadrature hybrids find applications in a of radio frequency systems. These include: single side-band phase discriminators[1];balanced amplifiers[2], attenuators and phase shifters: and certain kinds of balanced mixer. At microwave frequencies. they are usually implemented using distributed. This paper considers those circuits that exhibit exact quadrature performance and assumes a lumped element implementation.A realisation scheme to implement the approximation functions derived in this paper is available in the literature[3.4].Although some of the theory underlying the designs is given in those References. the general solution has not been published.This paper addresses the. as yet unanswered. question: Given a frequency band of operation and amplitude balance specification. what is the optimum solution such that it will be achieved with the minimum order of circuit? It is assumed that the final realisation will be in the form of a passive circuit. Although it will of course ,be possible to implement the functions using active elements.2 Statement of the problemA generic hybrid is shown in Fig.1.The circuit is assumed to be lossless. if all ports are perfectly matched .then theory dictates that it must be a directional coupler[5].Power incident to one port will be divided 長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 7 -between two further ports ,with the remaining port being isoland. In this paper,it is assumed that the primary port of excitation is port 1,with power divided between ports 2 and 4,and port 3 being isolated. It has been noted[6] that the transfer function of a quadrature coupler can be described in terms of ω by the expression (1))(1)(22wFSi???Where F(ω) is an odd function ofω. A similar expression can be derived for s21,by replacing F(ω) with its reciprocal. F(ω) being an odd function. it therefore follows that either s21 or s41 is =unity at zero frequency . it is assumed that this applies to port 4. and .hence F(0)=0. Thus port 4 becomes the through port, and port 2 becomes the coupled port .in normal parlance .it is required .for the purpose of this discussion, that the tranfer functions should result in an approximately equal power division over the specified frequency range. An idealised characteristic for F(ω) is as shown in fig.2. whereωl and ωu are the lower and upper frequency pass-band limits. respectively.I01234混合積分 I0長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 8 -Fig.1 Generic quadrature hybridF(ω)1ωl ωu ωFig.2 Idealised characteristic for F(ω)The task for the designer is to determine an approximation to this ideal characteristic. it not being possible to generate it with a finite function. it is the purpose of this investigation also to determine solutions that are optimum in the sense that the geometric deviation from unity is minimum over the specified frequency range for a given order of F(ω).by which means the through and coupled outputs will simultancously achieve the same specification.3. Proposed solutionConsider the function described in eqn.2,as follows:(2)).1)((.)( 2222dWAwWFdNn??It is supposed that the degress of the .numerator and denominater differ by one .and the higher of the two determines the order of the function. 長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 9 -Furthermore, the values of the natural frequencies suggested in the factors of the function are orderedωd1, ωn1, ωd2, ωn2,etc.Consider now the plot of F(ω) as w increases from zero.Near to zero ,the first-order factor in the numberator will dominate,causing F(ω) to increase approximately linearly. However,as w increases further,each quadratic term in the denominator and numerator will cause the function to ripple.it can be seen that the number of half ripples is equal to the order of the function.Having chosen the natural frequencies and the constant multiplier A. such that the final function ripples between and forms a tangent to geometric limits about unity.A typical solution is shown in Fig.3, for a fifth-order function,with a bandwidth ratio of 100. in a normalised range of 0.1-10. The lower and upper ripple limits are designated as l/m and m , respectively, this convention being maintianed for the remainder of this paper.Fig.3 Fifth-order rational function4. Proof of optimum nature of F(ω)Let (3))(wQPF?長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 10 -Be a function of order n determined by the mathod in Section 3.Now let G(ω)=R(ω)/S(ω) (4)Be a second function that meets or betters the approximation of F(ω) to unity over the specified frequency range and also satisfies the same constraints as F(ω) in being an odd function ofω. As F(ω) ripples between the lower and upper limits n times in the same pass-band. As both function are odd, there must be n further crossings in the negative pass-band. Where both functions ripple about-1.In addition. The two function must cross at zero .Hence , the second function must cross the first 2n+1 times. In other words, there must be at least 2n+1 zeros of the expression:(5))()()()( wSQRpwSRQpGwF??????As either P(ω) or Q(ω) is of order n. It follows that either R(ω) or S(ω) must be at least of order n+1, and ,hence.G(ω) is of higher order than F(ω). It therfore follows that F(ω) must be optimum and unique for a given order and pass-band.5 Geometric symmetry of F(ω)Consider the frequency limits to be geometrically placed about unity, so thatωl=1/ωu. Now consider a function F(ω) of order n derived as in section 3. Consider also the function F(1/ω). This is also of order n, 長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 11 -and oscillates between the m and 1/m limits n times, reaching m first as ω inceases from zero to 1/ωl. It also oscillates between the limits the same number of times as the original function F(ω). Its behaviour is exactly the same as F(ω). Its order is. Of course, still n, and , as this function has been proved in section 4 to be unique.it follows that F(ω)=1/ F(1/ω), or F(ω)* F(1/ω)=1 n odd (6)Consider now the case where n is even. This time .the behaviour of F(1/ω) is exactly the same as F(ω),and .as they are of the same order and this function is unique. We can write immediatelyF（ω）= F（1/ω） n even (7)with wqns.6 and 7 having been derived ,we define the function Fn(ω,ωu) to be the rational odd function of ω of order n ,which is equi-ripple between 1/m over the frequency range 1/ωl-ωu6. Analytic solution of Fn(ω,ωu)For the sake of brevity write F=Fn(ω,ωu).P=P(ω) and Q=Q(ω). In eqn.3. differentiating eqn.3 with respect to ω gives(8)2QPF????inspection of the numerator of eqn.8 will reveal there to be a maximum of 2(n-1)zeros. However, from the procedure to develop F, there will be n-1 turning points in the pass-band .as F is odd, there will also be n-1 長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 12 -turning points in the negative pass-band as well. Hence ,it can be concluded that all the zeros of F are real and simple and they are 2(n-1) in number.Consider now the equationF-m= (9)QmP?zeros of eqn.9 can be identified where F=m. from the procedure used to generate F,there is a single zero when ω=ωu if n is odd, with further double zeros below that in the pass-bend . counting the double zeros twice, there are n zeros in the pass-band .as the numberator of eqn.9 is of order n.there are no more.Consider further the equations(10)mQPF??1F+m= (11)?F+ (12)mQ?1eqns. 10-12 all have n zeros. With eqn.10 having zeros in the upper pass-band ,all double except for a single zero at the lower band edge, and another single zero at the upper band edge if n is even,eqns.11 and 12 have zeros of opposite sign to eqns.9 and 10, respectively.長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 13 -Consider now the effect of multiplying together eqns.9-12.comparing the resulting function with eqn.8 appear as double zeros in the multiplied function. But the latter has ,in addition. Simple zeros at the four pass-band edges.comparing also the denominator of eqn.8 and the multiplied function. It is seen that the latter has all the poles of eqn.8 repeated and contains no more. It is therefore possible to write(13)?? ))()(1)((122 UWmWCFFF??????C is a constant necessary as the argument only determined that each side should have the same poles and zeros. Eqn .13 is a first-order differential equation with separable variables. Rearranging it gives (14))1)(()1(( 2222 FMmcdwduu ???eqn.14 is an elliptic differential equation and can be solved by the use of the jacobi elliptic functions[7].To this end ,it is convenient to define a paremeter that is equal to the integral of each side, which we shall call z. Integrating the left-hand side first gives.(15)11)(CwSnWuu??which c1 is a constant of integration. And the modulus for the elliptic function is 長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 14 -(16)21UWK?as the parameter z has been chosen arbitrarily. We are at this stage, at liberty to assign c1 arbitrarily also. The most convenient value is zero, thusZ= (17))(1wSnWuu?the locus of z can be plotted on the z place as ω goes from zero to infinity ,and this is shown in Fig.4. the locus divides conveniently into three segments. The first segment along the real axis corresponds to values of ω from zero to the lower band edge. The second segment ,parallel to the imaginary axis ,corresponds to values of ω from the lower to upper band edge. With the point corresponding to ω=1 exactly mid-way. Finally ,the third segment .parallel to the real axis, corresponds to values of ω from the upper band edge to infinity ,note that ,in Fig .4 k1 and k1` are the complete elliptic integrals of the modulus and complementary modulus .respectively.turning our attention now to the right-band side of eqn.14 this can be integrated in a similar fashion to the left-hand side to give Z= (18)ncmFSnc??)(1where cn is a constant of integration. And the modulus for the elliptic function is 長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 15 -(19)21mKn?having defined a constant of integration between ω and z. the value of cn is not arbitrary . it is observed from eqn.17 that, when ω=0,z=0. also , from the derivation of F, when ω=0,F=0. substituting for z and F in eqn.18 gives either cn =0 or kn (where kn is the complete elliptic integral of kn ) . within a points. For simplicity , the solution cn=0 is chosen, and so (20))(1mFSncZ??it si interesting to note that eqn.14 is almost identical to the differential equation that arises during the derivation of low-pass elliptic filters. Clearly .those solutions are not required for the present purposes, a different class being required, one of which is illustrated in Fig.5 for the case of n=5. they are developed from the plot of ω by compression of periods in the imaginary direction, the locus in the z-plane is the same as that used in Fig.4. the locus terminates in a value of infinity for F , as it would for any odd order. For even order, the locus would terminate in a value of zero for F.ω=x ω= uwω=1m=0 ω=1/ uw長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 16 -uwk1Fig.4 locus of ω on z-plane J5ck/m F=mj4ck/m F=1/mj3ck/m F=mj2ck/m F=1/mjck/m F=mf=0 ck/mFig.5 locus of F on z-planeA useful result can be derived from the ratio of the locus side lengths in the z-plane . reading the values off Figs.4 And 5 leads,after simplification, to the equation(21)1.kn?The left-hand side of eqn.21 is a function of m and n, and the right-hand side is a function of ωU. Given two of these quantities, the third can be determined. If n is given, then eqn.21 can be translated into an algebraic relation between k1 and kn but these relations become of high order for even modest n and so are not desirable for obtaining solutions. When n=2, the relation between the moduli becomes12k??(22)(eqn.22 is known as Gauss’s transformation in the theory of elliptic functions.)長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 17 -Substituting for k1 and k2 rsing eqns. 16 and 19, gives .after expressing in terms of m.n=2 (23))(2uwm??Frequently, in practical applications, the amplitude inbalance in the pass-band is given as a specification. From eqn.1 and the discussion that followed, it can easily be shown that the relationship between amplitude imbalance and m is given byΔΔ (24)dB10log2)(?Fig.6 shows a plot of Kn’/Kn against Δ. To complete the solution to eqn.21, it is necessary to determine values for K1’/K1.Fortunately, for many applications, an approximation is adequate[7]. When the modulus is “small”, that is less than 0.2, then(25)kke4log21??The approximation is accurate to one part in 300 when k=0.2, and will improve as k→0. From eqn.16, the approximation is applicable to bandwidth ratios(i.e.ωU/ωL ) of 5/1 or greater. Substituting the bandwidth ratio B for 1/k1 in eqn.25 and then for K1’/K1 in eqn.21 and solving for n gives (26)ken4log21????????Equ.26 shows that the bandwidth increases exponentially with n, in contrast to many other quadrature coupler generating functions that only give an arithmetical improvement. It is this property that makes these kinds of approximation attractive for wide-band circuits.1.00.80.60.40 0.2 .04 0.6 0.8 1.0長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 18 -Determination of Fn(ωi, ωU)The function having been defined as in eqn.2, it is necessary to determine solutions for the quantities A and the natural frequencies, so that the loss to each output can be evaluated at any frequency.Consider first the natural frequencies, given by the values of ω where F equals zero or infinity. Some of these can be seen in the plot of Fig.5, where they occur at regular intervals along the imaginary axis in the z-plane. To encompass all the zeros and poles, it is necessary to consider a fundamental region in the z-plane for ω , that is to say, one that will map onto the entire ω- plane without repetition. A convenient region for this purpose is a rectangle bounded in the z-plane by the lines: x= -k1/ωU , x=k1/ω, y= -k1’/ωU, y= k1’/ωU . If this region is placed onto Fig.5, then the zero and pole positions will continue in the negative imaginary direction, repeating alternately with equalspacing.Consider first the zeros: inspection of Fig.5 reveals them to occur at values of z given by n odd )1(2).(12????nrmKjCznn even ( 27)).(rcomparing the z-plane plots of Figs.4 and 5. it can be seen that (28)mKnCwu???1applying substitutions to eqn.27 using eqns.15 and 28. a typical zero frequency can be determined thus (29)12(1knrjsur??from the theory of elliptic function. sn(jy,k)=j.sn(y,k`)/cn(y,k`)=j.sc(y.k`). and, hence ,the zeros are imaginary. As eqn.2 indicates. In addition for each zero generated by a positive value of r in eqn.27. there will be an equal and opposite zero for the negative value of r. that the zeros are in opposite pairs can also be inferred from eqn.2. there is now sufficient information to 長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 19 -determine all the natural frequencies in eqn.2. giving n odd)2(11knKrscwunr??)1(2.??n even (30)).(rfollowing the same procedure for the poles leads to the natural frequencies given by r=1… )12(knKrscwudr ???nod)1(2?(31)evr.?having determined all the zeros and poles of Fn(ω.ωu). it only remains to determine the value of the constant multiplier a in eqn.2. to define the function fully .where n is odd . it can be seen from eqn.6 that the zero and pole frequencies are reciprocal. And that Fn(1.ωu)=1. this is sufficient to define the function .which is given as n odd (32)????????122.nrdruNwwFwhere n is even . the solution is more involved. However, substituting z=ckn/m=jckn`/2m into eqn. 20 gives F=1.and .in eqn.17, gives( 33)???????11,2kndwuthere is now sufficient information to define Fn( ω,ωu) for n even . 長(zhǎng)春工業(yè)大學(xué)畢業(yè)設(shè)計(jì)（論文）- 20 -which is given as n even(34)?21212.).( nrdrnrdun wwF??? ??????????????the solution of eqn.2 has thus been derived.The results are best illustrated by an example.