# 电子工程代写|数字信号处理代写Digital Signal Processing代考|EE123

## 电子工程代写|数字信号处理代写Digital Signal Processing代考|An Illustration of the Flexibility of the New Method

The examples discussed previously show that our method requires more number of circuit elements than Mitra’s method. Except the resistance spread in Example 1, the spreads are also seen to be higher in the networks designed by the new method. However, this is not true in general. The examples were worked out with the particular set of values $a=b=m=q=1$ and $n=p=0$. In what follows, we show that by taking advantage of the flexibility in choosing these six parameters, we are able to reduce the spreads to much lower values than the ones obtained in the circuits of Fig. $10.2 \mathrm{a}$, b. Further, it will be shown that by choosing the six parameters properly, we may also reduce the number of circuit elements.

Consider Example 1 again. Combining Eq. (10.19) with Eqs. (10.12), (10.13) and $(10.14)$ and letting $a=b=1$, we get $$\left.\begin{array}{l} \left.Y_{A}=(1 / 3)[(2+m+n) s+(3.2+2 m+1.2 n)]+(3 p+2.1 q) s /(s+1)\right] \ Y_{B}=(1 / 3)[(1+m+n) s+(0.4+2 m+1.2 n)+(3 p+2.1 q+3.9) s /(s+1)] \ Y_{E}=(1 / 3)[(m+n) s+(2 m+1.2 n)+(5.1+3 p+2.1 q) s /(s+1)] \end{array}\right}$$
We see from Eq. (10.25) that conditions in Eq. (10.15) are not necessary in this case. The complete network will be of the form shown in Fig. 10.2a with
$$\left.\begin{array}{l} R_{7}=R_{8}=R_{1}=R(\text { arbitrary) } \ C_{1}=(2+m+n) / 3, C_{2}=P+0.7 q, C_{3}=(m+n+1) / 3 \ C_{4}=p+0.7 q+1.3, C_{5}=(m+n) / 3 \text { and } C_{6}=1.7+p+0.7 q ; \ R_{1}=3 /(3.2+2 m+1.2 n), R_{2}=1 /(p+0.7 q), \ R_{3}=3 /(0.4+2 m+1.2 n), R_{4}=1 /(p+0.7 q+1.3), \ R_{5}=3 /(2 m+1.2 n) \text { and } R_{6}=1 /(1.7+p+0.7 q) \end{array}\right}$$
If we choose $m=n=p=10$ and $q=0$, we get
$$\left.\begin{array}{l} R_{7}=R_{8}=R_{9}=R \text { (arbitrary) } \ C_{1}=7.33, C_{2}=10, C_{3}=7, C_{4}=11.3, C_{5}=6.67 \text { and } C_{6}=11.7 \ R_{1}-1 / 11.73, R_{2}=1 / 10, R_{3}=1 / 10.8, R_{4}=1 / 11.3 \ R_{5}=1 / 10.67, \text { and } R_{6}=1 / 11.7 \end{array}\right}$$

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

Example 1 Consider the synthesis of a notch filter characterized by
$$T(s)=N(s) / D(s)=\left(s^{2}+2.0\right) /\left(s^{2}+0.1 s+1.2\right)$$
Choosing $Q(s)=(s+1)$ and making partial fraction expansions of $N(s) /[\operatorname{sQ}(s)]$ and $D(s) /[s Q(s)]$, we identify
$$Y_{1}=s+2, Y_{2}=3 s /(s+1), Y_{3}=s+1.2 \text { and } Y_{4}=2.1 s /(s+1)$$
Applying Eq. (10.17), we get
\begin{aligned} &Y_{A}=s+1.733+0.7 s /(s+1), Y_{C}=2 s / 3+0.8+2 s /(s+1) \text { and } \ &Y_{E}=s / 3+2 / 3+2.4 s /(s+1) \end{aligned}
The synthesis of these one ports is elementary and the complete network is shown in Fig. 10.2a.

Choosing the same $Q(s)$ and applying Mitra’s method, we get the network shown in Fig. 10.2b.
Example 2 Consider the all-pass function
$$T(s)=N(s) / D(s)=\left(-s^{3}+2 s^{2}-2 s+1\right) /\left(s^{3}+2 s^{3}+2 s+1\right)$$
Let
$$Q(s)=(s+1)(s+2)$$
Then by the usual procedure, we identify
$$\left.\begin{array}{l} Y_{1}=\frac{1}{2}+(21 / 2) s /(s+2), Y_{2}=s+6 s /(s+1), \ Y_{3}=s+\frac{1}{2} \text { and } Y_{4}=(3 / 2) s /(s+2) \end{array}\right}$$

## 电子工程代写|数字信号处理代写Digital Signal Processing代考|An Illustration of the Flexibility of the New Method

Veft. $\backslash$ begin ${$ array $}\left{{} R_{-}{7}=R_{-}{8}=R_{-}{1}=R^{\prime} \backslash\right.$ text ${$ 任意) $} \backslash C_{-}{1}=(2+m+n) / 3, C_{-}{2}=P+0.7 q, C_{-}{3}=(m+n+1) / 3 \backslash C_{-}{4}=p+0 .$

Veft.\begin } { \text { array } { { } R _ { – } { 7 } = R _ { – } { 8 } = R _ { – } { 9 } = R \backslash \text { text } { ( \text { 任意) } } \backslash C _ { – } { 1 } = 7 . 3 3 , C _ { – } { 2 } = 1 0 , C _ { – } { 3 } = 7 , C _ { – } { 4 } = 1 1 . 3 , C _ { – } { 5 } = 6 . 6 7 \backslash \text { text } { \text { 和 } } \text { C }

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

$$T(s)=N(s) / D(s)=\left(s^{2}+2.0\right) /\left(s^{2}+0.1 s+1.2\right)$$

$$Y_{1}=s+2, Y_{2}=3 s /(s+1), Y_{3}=s+1.2 \text { and } Y_{4}=2.1 s /(s+1)$$

$$Y_{A}=s+1.733+0.7 s /(s+1), Y_{C}=2 s / 3+0.8+2 s /(s+1) \text { and } \quad Y_{E}=s / 3+2 / 3+2.4 s /(s+1)$$

$$T(s)=N(s) / D(s)=\left(-s^{3}+2 s^{2}-2 s+1\right) /\left(s^{3}+2 s^{3}+2 s+1\right)$$

$$Q(s)=(s+1)(s+2)$$

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