# 电子工程代写|数字信号处理代写Digital Signal Processing代考|EE615

## 电子工程代写|数字信号处理代写Digital Signal Processing代考|Active RC Filters Using a Single

In order to be able to express both numerator and denominator of Eq. (10.1) as the difference between two RC admittances, Mitra assumed [3].
$$Y_{A}+Y_{C}+Y_{E}=Y_{B}+Y_{D}+Y_{F}$$
The synthesis of a specified function
$$T(s)=N(s) / D(s)$$
starts by choosing a polynomial $Q(s)$ having distinct negative real roots and of degree satisfying
$$\operatorname{deg} Q(s)+1 \geq \text { higher of [deg } N(s), \operatorname{deg} D(s)]$$
The next step is to make partial fraction expansions of $[N(s), D(s), D(s)-$ $N(s)] /[s Q(s)]$; from these, the various admittances are identified as
$$\left.\begin{array}{l} N(s) / Q(s)=Y_{A}-Y_{B} \ D(s) / Q(s)=Y_{F}-Y_{E} \ {[D(s)-N(s)] / Q(s)=Y_{C}-Y_{D}} \end{array}\right}$$
It can be shown that such identification of $\mathrm{RC}$ one ports satisfies the starting assumption given by Eq. (10.2).

We present here an alternative procedure for synthesizing any real rational $T(s)$ by the circuit of Fig. $10.1$ without the need for assuming Eq. (10.2). The major advantage of the new procedure is shown to be a greater flexibility in the choice of element values, as compared to Mitra’s method.

## 电子工程代写|数字信号处理代写Digital Signal Processing代考|The New Method

We assume the upper tee of admittances in Fig. $10.1$ to obey the relations
$$Y_{D} / Y_{B}=a \text { and } \quad Y_{F} / Y_{B}=b$$
where $a$ and $b$ are arbitrary constants. In principle, $Y_{B}$ may be an arbitrary RC admittance but, for practical convenience, we may choose $Y_{B}$ to be a suitable conductance. Combining Eqs. (10.1) and (10.6), we get
$$T(s)=\frac{(a+b) Y_{A}-\left(Y_{C}+Y_{E}\right)}{b\left(Y_{A}+Y_{C}\right)-(1+a) Y_{E}}$$
Let the specified $T(s)$ be of the form of Eqs. (10.3) and let $Q(s)$ be chosen as in Mitra’s method. Then we can write $$N(s) / Q(s)=Y_{1}-Y_{2} \text { and } D(s) / Q(s)=Y_{3}-Y_{4}$$
where $Y_{1}, \ldots, Y_{4}$ are realizable RC admittances. From Eqs. (10.7) and (10.8), we get the equations
$$(a+b) Y_{A}-Y_{C}-Y_{E}=Y_{1}-Y_{2}$$
and
$$Y_{A}+Y_{C}-[(1+a) / b] Y_{E}=Y_{3} / b-Y_{4} / b$$
Adding Eqs. (10.9) and (10.10) gives
$$Y_{A}-Y_{E} / b=[1 /(1+a+b)]\left[\left(Y_{3} / b+Y_{1}\right)-\left(Y_{4} / b+Y_{2}\right)\right]$$
Let
$$Y_{A}=[1 /(1+a+b)]\left[(1+m) Y_{1}+(1+n) Y_{3} / b+p Y_{2}+q Y_{4} / b\right]$$
Then $Y_{E}$ is given by
$$Y_{E}=[b /(1+a+b)]\left[m Y_{1}+n Y_{3} / b+(1+p) Y_{2}+(1+q) Y_{4} / b\right]$$

## 电子工程代写|数字信号处理代写Digital Signal Processing代考|Active RC Filters Using a Single

$$Y_{A}+Y_{C}+Y_{E}=Y_{B}+Y_{D}+Y_{F}$$

$$T(s)=N(s) / D(s)$$

$$\operatorname{deg} Q(s)+1 \geq \text { higher of }[\operatorname{deg} N(s), \operatorname{deg} D(s)]$$

Veft. $\backslash$ beginfarray}{}} $N(s) / Q(s)=Y_{-}{A}-Y_{-}{B} \backslash D(s) / Q(s)=Y_{{}{F}-Y_{{}{E} \backslash\left{[D(s)-N(s)] / Q(s)=Y_{-}{C}-Y_{-}{D}\right} \backslash$ end ${$ array $} \backslash$ right $}$

## 电子工程代写|数字信号处理代写Digital Signal Processing代考|The New Method

$$Y_{D} / Y_{B}=a \text { and } \quad Y_{F} / Y_{B}=b$$

$$T(s)=\frac{(a+b) Y_{A}-\left(Y_{C}+Y_{E}\right)}{b\left(Y_{A}+Y_{C}\right)-(1+a) Y_{E}}$$

$$N(s) / Q(s)=Y_{1}-Y_{2} \text { and } D(s) / Q(s)=Y_{3}-Y_{4}$$

$$(a+b) Y_{A}-Y_{C}-Y_{E}=Y_{1}-Y_{2}$$

$$Y_{A}+Y_{C}-[(1+a) / b] Y_{E}=Y_{3} / b-Y_{4} / b$$

$$Y_{A}-Y_{E} / b=[1 /(1+a+b)]\left[\left(Y_{3} / b+Y_{1}\right)-\left(Y_{4} / b+Y_{2}\right)\right]$$

$$Y_{A}=[1 /(1+a+b)]\left[(1+m) Y_{1}+(1+n) Y_{3} / b+p Y_{2}+q Y_{4} / b\right]$$

$$Y_{E}=[b /(1+a+b)]\left[m Y_{1}+n Y_{3} / b+(1+p) Y_{2}+(1+q) Y_{4} / b\right]$$

myassignments-help数学代考价格说明

1、客户需提供物理代考的网址，相关账户，以及课程名称，Textbook等相关资料~客服会根据作业数量和持续时间给您定价~使收费透明，让您清楚的知道您的钱花在什么地方。

2、数学代写一般每篇报价约为600—1000rmb，费用根据持续时间、周作业量、成绩要求有所浮动(持续时间越长约便宜、周作业量越多约贵、成绩要求越高越贵)，报价后价格觉得合适，可以先付一周的款，我们帮你试做，满意后再继续，遇到Fail全额退款。

3、myassignments-help公司所有MATH作业代写服务支持付半款，全款，周付款，周付款一方面方便大家查阅自己的分数，一方面也方便大家资金周转，注意:每周固定周一时先预付下周的定金，不付定金不予继续做。物理代写一次性付清打9.5折。

Math作业代写、数学代写常见问题

myassignments-help擅长领域包含但不是全部: