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# 电子代写|数字信号处理代写Digital Signal Processing代考|Analog Filter Desi

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## 电子代写|数字信号处理代写Digital Signal Processing代考|.2 The Butterworth Fil

The $2 N$ th roots of -1 are
$$\eta_k=\left(e^{j \pi} e^{j 2 \pi k}\right)^{\frac{1}{2 N}}=e^{j \frac{1+2 k}{2 N} \pi} ; \quad k=-N, \ldots, N-1 .$$
Let
$$\lambda_k=j \eta_k=e^{j \frac{\pi}{2}} \eta_k=e^{j \frac{N+1+2 k}{2 N} \pi} .$$
We are interested in those $\lambda_k$ for which $\operatorname{Re} \lambda_k<0$. This requires
$$\frac{1}{2}<\frac{N+1+2 k}{2 N}<\frac{3}{2}$$
or
$$0 \leq k \leq N-1$$
Define the degree $N$ Butterworth polynomial
$$B_N(s)=s^N+a_{N-1} s^{N-1}+\ldots+a_0$$
to be the polynomial with roots $\lambda_0, \ldots, \lambda_{N-1}$. The cases $N=4,5$ are shown in Figure 9.1:

\begin{tabular}{|c|c|}
\hline$N$ & $B_N(s)$ \
\hline 1 & $s+1$ \
\hline 2 & $s^2+1.41 s+1$ \
\hline 3 & $s^3+2.00 s^2+2.00 s+1$ \
\hline 4 & $s^4+2.61 s^3+3.41 s^2+2.61 s+1$ \
\hline 5 & $s^5+3.24 s^4+5.24 s^3+5.24 s^2+3.24 s+1$ \
\hline
\end{tabular}
Note that
$$\lambda_{k-N}=-\lambda_k ; \quad k=0, \ldots, N-1,$$
so the roots of $B_N(-s)$ are just those $\lambda_k$ that lie in $R H P$ (i.e. $\left.k=-N, \ldots,-1\right)$. It follows that the roots of $B_N(s) B_N(-s)$ are the complete set of $\lambda_k$. In other words,
$$B_N(s) B_N(-s)=(-j s)^{2 N}+1=(-1)^N s^{2 N}+1$$
The Nth order Butterworth LPF is the (causal and stable) CT system with transfer function
$$H_N(s)=\frac{1}{B_N(s)}$$
The magnitude frequency response of the filter (squared) is
\begin{aligned} \left|H_N(j \omega)\right|^2 & =H_N(j \omega) H_N^*(j \omega) \ & =H_N(j \omega) H_N(-j \omega) \ & =\frac{1}{B_N(j \omega) B_N(-j \omega)} \ & =\frac{1}{(-1)^N(j \omega)^{2 N}+1} \ & =\frac{1}{\omega^{2 N}+1}, \end{aligned}

## 电子代写|数字信号处理代写Digital Signal Processing代考|The Chebyshev F

Consider the (Type I) Chebyshev polynomials
\begin{aligned} & T_0(\omega)=1, \ & T_1(\omega)=\omega, \ & T_{N+1}(\omega)=2 \omega T_N(\omega)-T_{N-1}(\omega) . \ & \end{aligned}
Associated with these are the functions
$$\Gamma_N(s)=1+\varepsilon^2 T_N^2\left(\frac{s}{j}\right),$$
where $\varepsilon>0$ is a design parameter. The $2 N$ roots $\lambda_k$ of $\Gamma_N$ are symmetrically distributed around an ellipse with major axis equal to the imaginary axis:

Let
$$C_N(s)=2^{N-1} \varepsilon s^N+a_{N-1} s^{N-1}+\ldots+a_0$$
be the polynomial whose roots are those of $\Gamma_N(s)$ that lie in $L H P$. Then
$$\Gamma_N(s)=C_N(s) C_N(-s)$$
The transfer function of the Nth-order Chebyshev $L P F$ is
$$H_N(s)=\frac{1}{C_N(s)}$$
The frequency response is
\begin{aligned} \left|H_N(j \omega)\right|^2 & =\frac{1}{C_N(j \omega) C_N(-j \omega)} \ & =\frac{1}{\Gamma_N(j \omega)} \ & =\frac{1}{1+\varepsilon^2 T_N^2(\omega)} \ \left|H_N(j \omega)\right| & =\frac{1}{\sqrt{1+\varepsilon^2 T_N^2(\omega)}} \end{aligned}

The Chebyshev filter has a steeper cutoff than the Butterworth filter for any value of $N$ and is, therefore, a more efficient design. However, it suffers from “ripple” in the passband. This effect can be made arbitrarily small by choosing $\varepsilon$ small, since the maximum deviation satisfies
$$\lim _{\varepsilon \rightarrow 0}\left(1-\frac{1}{\sqrt{1+\varepsilon^2}}\right)=0$$
The trade-off here is that $N$ must be increased as $\varepsilon$ is decreased in order to maintain the filter bandwidth near $\omega_B=1$. For the Chebyshev filter, taking $N \rightarrow \infty$ and $\varepsilon \rightarrow 0$ makes $\left|H_N(j \omega)\right|$ tend to the ideal LPF. However, $\angle H_N(j \omega) \rightarrow-\infty$ for each $\omega>0$.

## 电子代写|数字信号处理代写Digital Signal Processing代考|The Butterworth Fil

$$\eta_k=\left(e^{j \pi} e^{j 2 \pi k}\right)^{\frac{1}{2 N}}=e^{j \frac{12 k}{2 N} \pi} ; \quad k=-N, \ldots, N-1 .$$

$$\lambda_k=j \eta_k=e^{j \frac{\pi}{2}} \eta_k=e^{j \frac{N ! 1+2}{2 N} \pi}$$

$$\frac{1}{2}<\frac{N+1+2 k}{2 N}<\frac{3}{2}$$

$$0 \leq k \leq N-1$$

$$B_N(s)=s^N+a_{N-1} s^{N-1}+\ldots+a_0$$

$$\lambda_{k-N}=-\lambda_k ; \quad k=0, \ldots, N-1,$$

$$B_N(s) B_N(-s)=(-j s)^{2 N}+1=(-1)^N s^{2 N}+1$$
$N$ 阶 Butterworth LPF 是具有传道函数的（因果稳定的） $C T$ 系统
$$H_N(s)=\frac{1}{B_N(s)}$$

$$\left|H_N(j \omega)\right|^2=H_N(j \omega) H_N^*(j \omega) \quad=H_N(j \omega) H_N(-j \omega)=\frac{1}{B_N(j \omega) B_N(-j \omega)} \quad=\frac{1}{(-1)^N(j \omega)^{2 N}+1}=\frac{1}{\omega^{2 N}+1}$$

## 电子代写|数字信号处理代写Digital Signal Processing代考|The Chebyshev F

$$T_0(\omega)=1, \quad T_1(\omega)=\omega, T_{N+1}(\omega)=2 \omega T_N(\omega)-T_{N-1}(\omega)$$

$$\Gamma_N(s)=1+\varepsilon^2 T_N^2\left(\frac{s}{j}\right)$$

$$C_N(s)=2^{N-1} \varepsilon s^N+a_{N-1} s^{N-1}+\ldots+a_0$$

$$\Gamma_N(s)=C_N(s) C_N(-s)$$
$N$ 阶切比雪夫的传達函数 $L P F$ 是
$$H_N(s)=\frac{1}{C_N(s)}$$

$$\left|H_N(j \omega)\right|^2=\frac{1}{C_N(j \omega) C_N(-j \omega)} \quad=\frac{1}{\Gamma_N(j \omega)}=\frac{1}{1+\varepsilon^2 T_N^2(\omega)}\left|H_N(j \omega)\right| \quad=\frac{1}{\sqrt{1+\varepsilon^2 T_N^2(\omega)}}$$

$$\lim _{\varepsilon \rightarrow 0}\left(1-\frac{1}{\sqrt{1+\varepsilon^2}}\right)=0$$

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