# 电子工程代写|数字信号处理代写Digital Signal Processing代考|ELEC3104

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

The basic circuit from which the various second-order realizations will be derived is shown in Fig. 8.1. The transfer function of this circuit can be expressed as
$$T(s)=\frac{V_{0}}{V_{i}}=\frac{Y_{A}\left(Y_{D}+Y_{F}\right)-Y_{B}\left(Y_{C}+Y_{E}\right)}{Y_{F}\left(Y_{A}+Y_{C}\right)-Y_{E}\left(Y_{B}+Y_{D}\right)}$$
In our realizations $Y_{D}=0$; therefore, the transfer function reduces to
$$T(s)=\frac{Y_{A} Y_{F}-Y_{B}\left(Y_{C}+Y_{E}\right)}{Y_{F}\left(Y_{A}+Y_{C}\right)-Y_{E} Y_{B}}$$
By choosing the time constants of series RC impedances equal, removing two series RC impedances or by removing one series RC impedance and setting the time constants of remaining two series RC impedances equal, the transfer function $T(s)$ becomes a biquadratic one
$$T(S)=\frac{a_{2} s^{2}+a_{1} s+a_{0}}{b_{2} s^{2}+b_{1} s+b_{0}}$$
Next, by setting the appropriate numerator coefficient(s) to zero, a particular filter realization, with elements less than those hitherto used, can be obtained.

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

The transfer function of $n$ units of cascaded second-order bandpass filters can be expressed as
$$T(s)=\frac{s^{n}}{\left(s-s_{1}\right)\left(s-s_{1}^{}\right)\left(s-s_{2}\right)\left(s-s_{2}^{}\right) \ldots\left(s-s_{n}\right)\left(s-s_{n}^{*}\right)}$$
where asterisk denotes conjugate, and $s_{i}=\sigma_{i}+j \omega_{i}, i=1,2, \ldots n$.
As an example, the locations of poles and zeroes of a high $Q$ bandpass filter with $n=4$ are shown in Fig. $9.1$ together with a phasor representation of Eq. (9.1) for $s=j \omega$. Note that for a high $Q,\left|\sigma_{i}\right| \ll\left|\omega_{i}\right|$, i.e. the poles are located close to the imaginary axis.

Let the frequencies under consideration be relatively large and the associated bandwidth cover a narrow range. Under such an assumption, the effect of single or multiple zeroes at the origin and the poles in the lower half $s$ plane on the transfer function are insignificant relative to the effects of the poles in the upper half-plane in the proximity of the narrow range of frequency under consideration. Thus, the transfer function of the bandpass filter given by Eq. (9.1) can be approximated for $s=j \omega$ as
$$T(s) \cong \frac{k_{0}}{\left(j \omega-s_{1}\right)\left(j \omega-s_{2}\right) \ldots\left(j \omega-s_{n}\right)}$$
where $$k_{0}=\frac{\left(j \omega_{0}\right)^{n}}{\left(j \omega_{0}-s_{1}^{}\right)\left(j \omega_{0}-s_{2}^{}\right) \ldots\left(j \omega_{0}-s_{n}^{*}\right)}$$
and $\omega_{0}$ is the algebraic mean of the upper and lower cut-off frequencies $\omega_{p}$ and $2 \omega_{0}-\omega_{p}$, respectively (defining $\omega_{0}$ in this fashion provides symmetrical response specifications on either side of $\omega_{0}$ ).

Let us make the response maximally flat at $\omega=\omega_{0}$ for Butterworth transfer function. To achieve this, we change the frequency variable from $s$ to $p$ such that
$$p=\frac{\sigma+j \bar{\omega}}{\omega_{p}-\omega_{0}}=\Sigma+j \Omega$$
where
$$\bar{\omega}-\omega-\omega_{0}$$

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

$$T(s)=\frac{V_{0}}{V_{i}}=\frac{Y_{A}\left(Y_{D}+Y_{F}\right)-Y_{B}\left(Y_{C}+Y_{E}\right)}{Y_{F}\left(Y_{A}+Y_{C}\right)-Y_{E}\left(Y_{B}+Y_{D}\right)}$$

$$T(s)=\frac{Y_{A} Y_{F}-Y_{B}\left(Y_{C}+Y_{E}\right)}{Y_{F}\left(Y_{A}+Y_{C}\right)-Y_{E} Y_{B}}$$

$$T(S)=\frac{a_{2} s^{2}+a_{1} s+a_{0}}{b_{2} s^{2}+b_{1} s+b_{0}}$$

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

$$T(s)=\frac{s^{n}}{\left(s-s_{1}\right)\left(s-s_{1}\right)\left(s-s_{2}\right)\left(s-s_{2}\right) \ldots\left(s-s_{n}\right)\left(s-s_{n}^{}\right)}$$ 其中星号表示共轭，并且 $s_{i}=\sigma_{i}+j \omega_{i}, i=1,2, \ldots n$. 例如，高的极点和零点的位置 $Q$ 带通滤波器 $n=4$ 如图所示。9.1连同等式的相量表示。(9.1) 对于 $s=j \omega$. 请注意，对于高 $Q,\left|\sigma_{i}\right| \ll\left|\omega_{i}\right|$ ，即极点靠近虚轴。 让所考虑的频率相对较大并且相关的带宽覆盖较窄的范围。在这样的假设下，原点的单个或多个零点和下 半部分的极点的影响 $s$ 相对于在所考虑的安频率范围附近的上半平面中的极点的影响，传递函数上的平面是 微不足道的。因此，由等式给出的带通滤波器的传递函数。(9.1) 可以近似为 $s=j \omega$ 作为 $$T(s) \cong \frac{k_{0}}{\left(j \omega-s_{1}\right)\left(j \omega-s_{2}\right) \ldots\left(j \omega-s_{n}\right)}$$ 在哪里 $$k_{0}=\frac{\left(j \omega_{0}\right)^{n}}{\left(j \omega_{0}-s_{1}\right)\left(j \omega_{0}-s_{2}\right) \ldots\left(j \omega_{0}-s_{n}^{}\right)}$$

$$p=\frac{\sigma+j \bar{\omega}}{\omega_{p}-\omega_{0}}=\Sigma+j \Omega$$

$$\bar{\omega}-\omega-\omega_{0}$$

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