# 电子工程代写|电子系统工程代写Digital Systems Engineering代考|ECE385

## 电子工程代写|电子系统工程代写Digital Systems Engineering代考|Equalisation

Channel distortion or system distortion may be corrected through what is known as an equaliser filter. Essentially, we use the frequency responses of the channel and the equaliser to remove the distortion introduced. Thus, we know that the undistorted frequency response is $H_{\mathrm{ud}}(\omega)=k e^{-j \omega t_{0}}$ where $k$ and $t_{0}$ are constants. Therefore, we design an equaliser, $H_{\mathrm{eq}}(\omega)$, which satisfies the equation,
$$H(\omega) H_{\mathrm{eq}}(\omega)=k e^{-j \omega t_{0}}$$
Figure $2.42$ shows how the equaliser is used. From $\sum$ quation $2.164$,
$$H_{\mathrm{eq}}(\omega)=\frac{k e^{-j \omega \mathrm{t}{0}}}{H(\omega)}$$ In the figure, \begin{aligned} X(\omega) &=H(\omega) M(\omega) \ \text { and } \hat{M}(\omega) &=k M(\omega) e^{-j \omega f{0}} \end{aligned}
where
$$m(t) \Leftrightarrow M(\omega)$$
and
$$\hat{m}(t) \Leftrightarrow \hat{M}(\omega)$$
Example 2.33 The frequency characteristic of a channel is proportional to $K e^{-\alpha \omega}$ in the frequency range $0 \leq \omega \leq \omega_{0}$. Design an equaliser to correct this frequency response.

The equaliser characteristic must he the peciprocal of this response. Thepefore
$$H_{\mathrm{eq}}(\omega) \propto \frac{k}{H(\omega)}$$
These two frequency responses and their product are shown in Fig. 2.43.

## 电子工程代写|电子系统工程代写Digital Systems Engineering代考|Nonlinear Channel Distortion

A channel may introduce nonlinearities in the signal. Such a channel may be modelled by the equation,
\begin{aligned} y(t) &=h_{0}+h_{1} x(t)+h_{2}[x(t)]^{2}+\ldots \ &=\sum_{i=0}^{\infty} h_{i}[x(t)]^{i} \end{aligned}
where some of the $h_{i}$ may be zero.
To take a concrete example, if the input to such a channel, characterised by
$$y(t)=x(t)+0.01[x(t)]^{2}$$
and the input is $A \cos \left(\omega_{0} t\right)$, having a bandwidth of $\omega_{0}$, then
\begin{aligned} y(t) &=0.01 A^{2} \cos \left(\omega_{0} t\right)^{2}+A \cos \left(\omega_{0} t\right) \ &=0.01\left(\frac{\cos \left(2 \omega_{0} t\right) A^{2}}{2}+\frac{A^{2}}{2}\right)+\cos \left(\omega_{0} t\right) A \end{aligned}
We observe that two new (small) terms have appeared, namely
$$0.01\left(\frac{\cos \left(2 \omega_{0} t\right) A^{2}}{2}+\frac{A^{2}}{2}\right)$$
one at $\omega=0$ and the other at $\omega=2 \omega_{0}$.

## 电子工程代写|电子系统工程代写Digital Systems Engineering代考|Equalisation

$$H(\omega) H_{\text {eq }}(\omega)=k e^{-j \omega t_{0}}$$
$$H_{\mathrm{eq}}(\omega)=\frac{k e^{-j \omega t 0}}{H(\omega)}$$

$$X(\omega)=H(\omega) M(\omega) \text { and } \hat{M}(\omega) \quad=k M(\omega) e^{-j \omega f 0}$$

$$m(t) \Leftrightarrow M(\omega)$$

$$\hat{m}(t) \Leftrightarrow \hat{M}(\omega)$$

$$H_{\mathrm{eq}}(\omega) \propto \frac{k}{H(\omega)}$$

## 电子工程代写|电子系统工程代写Digital Systems Engineering代考|Nonlinear Channel Distortion

$$y(t)=h_{0}+h_{1} x(t)+h_{2}[x(t)]^{2}+\ldots \quad=\sum_{i=0}^{\infty} h_{i}[x(t)]^{i}$$

$$y(t)=x(t)+0.01[x(t)]^{2}$$

$$y(t)=0.01 A^{2} \cos \left(\omega_{0} t\right)^{2}+A \cos \left(\omega_{0} t\right) \quad=0.01\left(\frac{\cos \left(2 \omega_{0} t\right) A^{2}}{2}+\frac{A^{2}}{2}\right)+\cos \left(\omega_{0} t\right) A$$

$$0.01\left(\frac{\cos \left(2 \omega_{0} t\right) A^{2}}{2}+\frac{A^{2}}{2}\right)$$
$$\text { 一个在 } \omega=0 \text { 另一个在 } \omega=2 \omega_{0} \text {. }$$

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