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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 电子工程代写|数字信号处理代写Digital Signal Processing代考|Triple-Frequency Impedance Matching

To bring variety into experience, let us use the transformation
$$S=k_{\infty} s+k_{1} s /\left(s^{2}+q_{1}\right)$$
Applied to D1, this gives the circuit shown in Fig. 3.3. By following the procedure given in Sect. 4 and illustrated in the previous Section, the two cubic equations to be compared are obtained as follows:
$$\omega^{3}-\left(1 / k_{\infty}\right) \omega^{2}-\left[q_{1}+\left(k_{1} / k_{\infty}\right)\right] \omega+\left(q_{1} / k_{\infty}\right)=0$$

And
\begin{aligned} &\left(\omega-\omega_{1}\right)\left(\omega+\omega_{2}\right)\left(\omega-\omega_{3}\right)=\omega^{3}-\left(\omega_{1}-\omega_{2}+\omega_{3}\right) \ &\text { or, } \quad \omega^{2}-\left(\omega_{1} \omega_{2}+\omega_{2} \omega_{3}-\omega_{1} \omega_{3}\right) \omega+\omega_{1} \omega_{2} \omega_{3}=0 \end{aligned}
Comparison gives
\begin{aligned} k_{\infty} &=\omega_{1}-\omega_{2}+\omega_{3}, q_{1}=\omega_{1} \omega_{2} \omega_{3} k_{\infty}, \text { and } \ k_{1} &=k_{\infty}\left(\omega_{1} \omega_{2}+\omega_{2} \omega_{3}-\omega_{1} \omega_{3}-q_{1}\right) \end{aligned}
The element values of the circuit in Fig. $3.3$ can now be calculated.

## 电子工程代写|数字信号处理代写Digital Signal Processing代考|Impedance Matching at Four Frequencies

In this as well as the next Section, for brevity, we shall skip the routine steps and the circuit diagrams, and shall only give the design equations. We choose the transformation
$$S=\left[k_{1} s /\left(s^{2}+q_{1}\right)\right]+\left[k_{2} s /\left(s_{m}^{2}+q_{2}\right)\right]$$
The roots of the quartic equation obtained from Eq. (3.17) should be $\omega_{1},-\omega_{2} \omega_{3}$ and $-\omega_{4}$. The equations for finding $k_{i}^{\prime} s$ and $q_{i}^{\prime} s$ are
$$\begin{gathered} k_{1}+k_{2}=-\omega_{1}+\omega_{2}-\omega_{3}+\omega_{4}=A, \text { say } \ q_{1}+q_{2}=\omega_{1} \omega_{2}-\omega_{1} \omega_{3}+\omega_{1} \omega_{4}+\omega_{2} \omega_{3}-\omega_{2} \omega_{4}+\omega_{3} \omega_{4}=B, \text { say } \ k_{1} q_{2}+k_{2} q_{1}=\omega_{2} \omega_{3} \omega_{4}-\omega_{1} \omega_{3} \omega_{4}+\omega_{1} \omega_{2} \omega_{4}-\omega_{1} \omega_{2} \omega_{3}=C, \text { say } \ q_{1} q_{2}=\omega_{1} \omega_{2} \omega_{3} \omega_{4}=D, \text { say } \end{gathered}$$

$q_{1}$ and $q_{2}$ can be found from Eqs. (3.19) and (3.21) by solving a quadratic equation. This gives two solutions, one being the reciprocal of the other. Either of them can be chosen for $q_{1}$; then the other is $q_{2}$. Substituting these values in Eq. (3.20) and combining with Eq. (3.18), one can find $k_{1}$ and $k_{2}$. The final results are
$$\left.q_{1}=(1 / 2)\left[B+\sqrt{B^{2}-4 D}\right)\right], q_{2}=D / q_{1}, k_{1}=\left(A q_{1}-C\right) /\left(q_{1}-q_{2}\right) \text {, and } k_{2}=A-k_{1}$$
It is easily shown that interchanging the values of $q_{1}$ and $q_{2}$ has the effect of interchanging the values of $k_{1}$ and $k_{2}$, so that the circuit does not change.

# 数字信号处理代考

## 电子工程代写|数字信号处理代写Digital Signal Processing代考|Triple-Frequency Impedance Matching

$$S=k_{\infty} s+k_{1} s /\left(s^{2}+q_{1}\right)$$

$$\omega^{3}-\left(1 / k_{\infty}\right) \omega^{2}-\left[q_{1}+\left(k_{1} / k_{\infty}\right)\right] \omega+\left(q_{1} / k_{\infty}\right)=0$$
$$\left(\omega-\omega_{1}\right)\left(\omega+\omega_{2}\right)\left(\omega-\omega_{3}\right)=\omega^{3}-\left(\omega_{1}-\omega_{2}+\omega_{3}\right) \quad \text { or, } \quad \omega^{2}-\left(\omega_{1} \omega_{2}+\omega_{2} \omega_{3}-\omega_{1} \omega_{3}\right) \omega+\omega_{1} \omega_{2} \omega_{3}=0$$

$$k_{\infty}=\omega_{1}-\omega_{2}+\omega_{3}, q_{1}=\omega_{1} \omega_{2} \omega_{3} k_{\infty}, \text { and } k_{1} \quad=k_{\infty}\left(\omega_{1} \omega_{2}+\omega_{2} \omega_{3}-\omega_{1} \omega_{3}-q_{1}\right)$$

## 电子工程代写|数字信号处理代写Digital Signal Processing代考|Impedance Matching at Four Frequencies

$$S=\left[k_{1} s /\left(s^{2}+q_{1}\right)\right]+\left[k_{2} s /\left(s_{m}^{2}+q_{2}\right)\right]$$

$k_{1}+k_{2}=-\omega_{1}+\omega_{2}-\omega_{3}+\omega_{4}=A$, say $q_{1}+q_{2}=\omega_{1} \omega_{2}-\omega_{1} \omega_{3}+\omega_{1} \omega_{4}+\omega_{2} \omega_{3}-\omega_{2} \omega_{4}+\omega_{3} \omega_{4}=B$, say $k_{1} q_{2}+k_{2} q_{1}=\omega_{2} \omega_{3} \omega_{4}-\omega_{1} \omega_{3} \omega_{4}+\omega_{1} \omega_{2} \omega$
$q_{1}$ 和 $q_{2}$ 可以从方程式中找到。(3.19) 和 (3.21) 通过求解二次方程。这给出了两种解决方案，一种是另一种的倒数。可以选择其中任何一个 $q_{1}$; 然后另一个是 $q_{2}$. 将这 些值代入方程式。 (3.20) 并结合方程式。(3.18)，可以找到 $k_{1}$ 和 $k_{2}$. 最终结果是
$$\left.q_{1}=(1 / 2)\left[B+\sqrt{B^{2}-4 D}\right)\right], q_{2}=D / q_{1}, k_{1}=\left(A q_{1}-C\right) /\left(q_{1}-q_{2}\right), \text { and } k_{2}=A-k_{1}$$

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