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DZ201PLC控制的恒压供水系统,毕业设计
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长春工业大学毕业设计（论文） - 1 - 附 录 集成单元积分的最优近似值函数 1.前言：积分混合电路在无线电频率系统范围中寻找应用对象。这包括单独的边频带相应位鉴别器、平衡放大器、衰减器、相位转换器和可靠的几种平衡混频器。在为微波频率中，他们经常用分布电路实现。但是，在较底频率中，他们将被集成。这篇论文认为那些展现精确积分特性的电路，假设用一个集成单元实现。 在这论文中，一个源于实现方案执行近似值函数在文献 3、 4中可以不再抑制这篇论文寻址，这还没有回答问题：给出一个操作运算的频率带和振幅平衡规则。什么是最优的解决方法，象它将被电路的最小阶数完 成。假设他将被右无源电路模式中最终实现，当然尽管他将可能用有效单元实现函数。 2.问题的陈述：一般的混合电路如图 1 所示。假设电路是无损的，如果所有的节点是完美的匹配，那么，理论指示他必须是一个方向的配合者。电源附属于一个节点，将被在两个较远的节点分开，剩余的节点被绝缘。在这篇论文中，假设激发的主要节点是节点 1，功率在节点 2和 4中被分开，接待内 3被绝缘，它已经被记载积分连接器的移动函数可以在 W项中用表达式描述： )(1 1)( 22wFwS i 3.在 F（ W）是 W的奇数函数。一个相似的表达式可以因为 *￥被分开，用他的倒数 F（ W）。 F（ W）是一个奇函数，他因为跟随￥或￥在 0频率时是整体。假设这应用于接点 4 ，因此 F（ 0） =0。这样接点变成“通过”节点，接点变成“耦合”节点，在普通的用法中它要求对于讨论的目的来说，转移函数结果在近似值等于功率因数除以指定的频率范围。因此， F（ W）近似于乘以这个范围。一个 F（ W）的理想特征在图之中显示在￥和￥分别是较底和较高频率通频带极限。 I0I012 34混合积分I0I0图一混合积分类 nts长春工业大学毕业设计（论文） - 2 - 1 W1 Wu 图二 F（ W）的理想特 设计者的任务是求出这理想指数的近似值。它不能用有限值函数产生。研究的目的也是求出最优的解决方案：考虑到方程描述的函数如下： ) . .1)(1() . .1)(1()(22212222212dWWWWWWWWAwWFwdNn 假设分子的等级和字母的第一项不同，第二项的更高求出函数的 阶数。此外，自然频率的值建议在特性因式分解中，他们是￥，￥，￥，￥等等， 考虑到此刻 W 从 0 上升的曲线，在靠近 0 点，在分子中因式的第一阶将计算出引起 F（ W）近似线性的上升。然而，随着 W 上升加深分母中的每一个三次项和分子将引起函数脉动。它可以看作一半脉动数等于函数的阶数。已经选择了 F（ W）的阶数，任务是选择自然频率的数值和常数乘以 A，象最后的脉动函数之间和形成关于整体的几何极限的切线。 nts长春工业大学毕业设计（论文） - 3 - 一个典型的解决方案如图 3 所示。对于第五阶的函数带有一个宽频无线电的100，在一个 0.110 标准化的范围，较底和较高的脉动极 限分别是被指定的 1/M或 M倍。对剩余的常规保留在这 篇论文中 图 3第五阶的合理函数 4、 F（ W）的本质的最优验证： 让 )( )()( wQ wPwF 是 n阶分母函数是根据第三部分中的方法所得。现在让￥作为第二函数，这符合以至更优于￥的近似值。整体超过指定的频率范围，也适应与 F（ W）作为 W的奇函数相同的约束条件。在通频带 F（ W）在较底与较高极限之间脉动几次， G（ W）必然在相同的通频带与 F（ W）相交至少几次。当来年感个函数是奇数的 ，那必须有几个更进一步的交叉在负数通频带，在那两个函数的脉动与第一个相交 2n+1次。换句话说，表达式： )()()()()()()()()()()()(wSwQwRwQwSwpwSwRwQwpwGwF 必须有至少 2n+1个零点。当 P（ W）或 Q（ W）是 n阶的，他跟随 P（ W）或 Q（ W）必须是至少 n+1阶的。而切，因此 G（ W）是比 F（ W）更高阶的。他所跟随 F（ W）必须是最优的和唯一的对于给定的阶数和通频带。 5、 F（ W）的几何对称性：考虑到几何学上的频率限制被应用于整体，因此￥现在考虑函数 F（ W）在第三阶段向几派生的情况，也考虑函数 F（ 1/W），这也是 n阶的。而 且在 M和 1/M极限间震荡几次，首先达到 M，当 W从 0开始上升，几为阶数，考虑函数 1/F（ 1/W）从 0 开始上升到 1/M。当 W从 0开始上升到￥它也在两个极限间震荡与原始函数 F（ W）相同的次数。他的特性与 F（ W）严密的相同，他的阶数当然也是n而且这个函数已在第四部分中证明是唯一的，它服从于 F(w)=1/F(1/w) 或 F(w) )1( wF n为奇数（ 6） 现在考虑在 n 为偶数时的情形。这时， F（ 1/W）的性质与 F（ W）完全相同。而它是相同阶数的。这个函数也是唯一的。我们 可以立刻写出 F（ W） = F（ 1/W）， n 为偶数（ 7）由于表达式 6 和 7 已经被导出，我们可以定义函数 ￥ n 阶的有理奇数函数 ，超过频率范围nb WW 1 在 1/M和 M 之间是相等的脉动。 nts长春工业大学毕业设计（论文） - 4 - 6、分析 F-m=QmQP：因为短暂的缘故，写出 F=F)1,( un wwF,P=P(w)， P=P（ W）和 Q=Q（ W）在表达式 3 对 W 求积分，得到2QQPPQF ，表达式 8 的分子的检查将揭示有最大的 2（ n-1）个零点。然而从发展 F 的程序来看，同样将 n-1 转折点。在复数通频带，因此，可以推断 F上午所有零点都是实数和简单的。总共是 2（ n-1）个。 现在考虑方程 F-m=QmQP（ 9）表达式 9 的零点可以在 F=m 处确定，从程序应用于生产 F来看，如果 n是奇数，当 W= nW时，约一个单一的零点，伴随更多的一对零点在通频带的下面，当表达式 9的分子 n阶的，不会有零 点，更深的的考虑方程 mQ QmPmF 1（ 10） F+m=QmQP（ 11） F+mQ QmQm 1（ 12）表达式 1012都有几个零点，表达式 10 在通频带上面有零点，除了一个单一的零点，所有的双零点低频边缘，而且 n 为偶数。、另外的单一零点在上带的边缘。表达时 11 和 12 有和表达式 9、 10相反符号的零点，现在考虑把表达式 8，可以看到表达式 8所有单一的零点出现在相乘的函数的双零点，但是后者另外的单一的 零点在 四个通频带的边缘，再比较表达式 8的分母和相乘的函数，可以看到后者有表达式 8重复的极和不再多的包含，因此它可能这样写： )()(1)(1( )()(1)(1(1 222UU WWWWmWmWCmFmFmFmFCF C 是必须的常数作为参数仅仅被定义每一边应该相同的极和零点表达式 13 是一个一阶的常有可分离度量的积微分方程重新整理给出： )1)(1()1)(1( 2222222FMmFc d Fwwwdwuu表达式 14 是一个椭圆的微分方程它可以用 Iacobi 椭圆函数求解。最后，它是方便的定义一个变量，它的关于对每边的积分常数，椭圆函数的系数是211UWK （ 16）当变量 Z 被武断的选取，在这阶段我们也准许武断的赋值 C 最方便的值是 0，这样Z=11 )(1 CwwSnWuu Z的轨迹可以在 Z平面上绘制当 W从 0 到无穷大，这图 4中显示，这个轨迹方便的分为 3段，第一段沿着实轴符合 W的数值从 0到波段的较底边缘，第二断，平行于虚轴，符合 W的数值从较底到较高频带边缘中间点精确的符合 W=1，nts长春工业大学毕业设计（论文） - 5 - 最后，第三段，平行于实轴，符合 W 值，从较高频带到无穷大，注意，在图一中 K1和 K1 分别是完整的椭圆积分的系数和互补的系数。现在把我们的注意力转到表达式14的右手 边，它可以被积分字同的方式下得 Z= ncmFSnmc )(1 （ 18）这里 是积分常数，椭圆函数的系数是 21mKn （ 19）在 W和 Z 之间已经定义一个积分常数，的值不是武断的，它从表达式 17中调换 Z和 F 可得 0nC或 nK（在nK是完整的椭圆的积分的nK处 ）在平行四边形周期内，解决方案的所有的合适点，对于简明性，解决方案 0nC是被选择的，所以 )(1 mFSnmcZ （ 20）那是很有趣的，注意表达式 14 是几乎同样的对于微分方程出现在导出底通椭圆滤波器期间，明确的，那些解决方案，对于目前的目的是不要求的，不同的等级被要求的，其中一个在图 5中举列说明对于 n=5的事件。他们通过在虚数方向周期的压缩， W曲线中被发展，在 Z 平面上的轨迹是同用在图中的相同，轨迹终止在对于任何奇数 F的无穷大处，对于偶数 阶段 ，轨迹应该终止在 F 的 0处。 8、结论 它已经显示一个有理函数规定对于给定阶数的积分混合电路最优基数，这个函数有几何数量规定解析的简化，对于最优有理数的一个分析解决方案，已经被导出，使他可能推算特性在早期的设计中，这个函数的竟区的就亿斤微亿方案已经被介绍，对于每一个宽频电路来说它们自然的表明它们的电容，在实型频率变化条件中已经求出解决方案，这些可以用于在复数变量 的条件下，求出转换函数，对于电路综合是必要的。 nts长春工业大学毕业设计（论文） - 6 - 英文附录 Optimum approximation functions for lumped element quadrature hybrids D.P.Andrews and C.S.Aitchison Abstract: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. Introduction Quadrature hybrids find applications in a of radio frequency systems. These include: single side-band phase discriminators1;balanced amplifiers2, 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 literature3.4.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 problem A 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 coupler5.Power incident to one port will be divided 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 nts长春工业大学毕业设计（论文） - 7 - between ports 2 and 4,and port 3 being isolated. It has been noted6 that the transfer function of a quadrature coupler can be described in terms of by the expression )(1 1)( 22wFwS i (1) Where F() is an odd function of . A similar expression can be derived for s21,by replacing F() with its reciprocal. F() 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() is as shown in fig.2. where l and u are the lower and upper frequency pass-band limits. respectively. I0I012 34混合积分I0I0Fig.1 Generic quadrature hybrid F() nts长春工业大学毕业设计（论文） - 8 - 1 l u Fig.2 Idealised characteristic for F() 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().by which means the through and coupled outputs will simultancously achieve the same specification. 3. Proposed solution Consider the function described in eqn.2,as follows: ) . .1)(1() . .1)(1()(22212222212dWWWWWWWWAwWFwdNn (2) 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. Furthermore, the values of the natural frequencies suggested in the factors of the function are ordered d1, n1, d2, n2,etc. nts长春工业大学毕业设计（论文） - 9 - Consider now the plot of F() as w increases from zero.Near to zero ,the first-order factor in the numberator will dominate,causing F() 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 function 4. Proof of optimum nature of F( ) Let )( )()( wQ wPwF (3) Be a function of order n determined by the mathod in Section 3.Now let G( )=R( )/S( ) (4) nts长春工业大学毕业设计（论文） - 10 - Be a second function that meets or betters the approximation of F( ) to unity over the specified frequency range and also satisfies the same constraints as F( ) in being an odd function of . As F( ) 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: )()()()()()()()()()()()(wSwQwRwQwSwpwSwRwQwpwGwF (5) As either P( ) or Q( ) is of order n. It follows that either R( ) or S( ) must be at least of order n+1, and ,hence.G( ) is of higher order than F( ). It therfore follows that F( ) must be optimum and unique for a given order and pass-band. 5 Geometric symmetry of F( ) Consider the frequency limits to be geometrically placed about unity, so that l=1/ u. Now consider a function F( ) of order n derived as in section 3. Consider also the function F(1/ ). This is also of order n, and oscillates between the m and 1/m limits n times, reaching m first as inceases from zero to 1/l. It also oscillates between the limits the same number of times as the original function F( ). Its behaviour is exactly the same as F( ). nts长春工业大学毕业设计（论文） - 11 - Its order is. Of course, still n, and , as this function has been proved in section 4 to be unique.it follows that F( )=1/ F(1/ ), or F( )* F(1/ )=1 n odd (6) Consider now the case where n is even. This time .the behaviour of F(1/ ) is exactly the same as F( ),and .as they are of the same order and this function is unique. We can write immediately F（ ） = F（ 1/ ） n even (7) with wqns.6 and 7 having been derived ,we define the function Fn(,u) to be the rational odd function of of order n ,which is equi -ripple between 1/m over the frequency range 1/l -u 6. Analytic solution of Fn(,u) For the sake of brevity write F=Fn(,u).P=P() and Q=Q(). In eqn.3. differentiating eqn.3 with respect to gives 2QQPPQF (8) 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 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 equation nts长春工业大学毕业设计（论文） - 12 - F-m=QmQP(9) zeros of eqn.9 can be identified where F=m. from the procedure used to generate F,there is a single zero when =u 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 mQ QmPmF 1(10) F+m=QmQP(11) F+mQ QmQm 1(12) eqns. 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. Consider now the effect of multiplying together paring 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 paring 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 nts长春工业大学毕业设计（论文） - 13 - no more. It is therefore possible to write )()(1)(1()()(1)(1(1222UU WWWWmWmWCmFmFmFmFCF (13) 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 )1)(1()1)(1( 2222222FMmFc d Fwwwdwuu(14) eqn.14 is an elliptic differential equation and can be solved by the use of the jacobi elliptic functions7.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. 11 )(1 CwwSnWuu (15) which c1 is a constant of integration. And the modulus for the elliptic function is 211UWK (16) 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, thus Z= )(11 wwSnW uu(17) nts长春工业大学毕业设计（论文） - 14 - the locus of z can be plotted on the z place as 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 from zero to the lower band edge. The second segment ,parallel to the imaginary axis ,corresponds to values of from the lower to upper band edge. With the point corresponding to =1 exactly mid -way. Finally ,the third segment .parallel to the real axis, corresponds to values of 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= ncmFSnmc )(1 (18) where cn is a constant of integration. And the modulus for the elliptic function is 21mKn (19) having defined a constant of integration between and z. the value of cn is not arbitrary . it is observed from eqn.17 that, when =0,z=0. also , from the derivation of F, when =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 nts长春工业大学毕业设计（论文） - 15 - )(1 mFSnmcZ (20) 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 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. =x =uw=1 m=0 =1/uwuwk 11 Fig.4 locus of on z -plane J5ck/m F=m j4ck/m F=1/m j3ck/m F=m j2ck/m F=1/m nts长春工业大学毕业设计（论文） - 16 - jck/m F=m f=0 ck/m Fig.5 locus of F on z-plane A 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 1111.kkkknnn (21) The left-hand side of eqn.21 is a function of m and n, and the right-hand side is a function of U. 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 becomes 112 12 kkk (22) (eqn.22 is known as Gausss transformation in the theory of elliptic functions.) Substituting for k1 and k2 rsing eqns. 16 and 19, gives .after expressing in terms of m. )1(21 uuwwm n=2 (23) 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 by mdB10log20)( (24) Fig.6 shows a plot of Kn/Kn against . To complete the solution to eqn.21, it is necessary to determine values for K1/K1. Fortunately, for many applications, an approximation is adequate7. When the modulus is “small”, nts长春工业大学毕业设计（论文） - 17 - that is less than 0.2, then kkk e 4log21(25) 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. U/ L) 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 Bkkn en4lo g2 11 (26) 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 attr
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