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# 电子代写|数字信号处理代写Digital Signal Processing代考|EE4015 The Zero-Order Hold

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## 电子代写|数字信号处理代写Digital Signal Processing代考|The Zero-Order Hold

Define the Zero-Order Hold $(\mathrm{ZOH})$ to be the hybrid system that takes $x[n]$ into the CT signal $x(t)$ defined by
$$x(t)=x[n], \quad n T \leq t<(n+1) T .$$

In other words, each input value is held throughout the subsequent sampling interval.

Consider the unit rectangular pulse
$$w(t)=\left{\begin{array}{ll} 1, & 0 \leq t \leq 1 \ 0, & \text { else } \end{array} .\right.$$
The $\mathrm{ZOH}$ processes signals according to
$$x_0(t)=\sum_{n=-\infty}^{\infty} x[n] w\left(\frac{t-n T}{T}\right)=w\left(\frac{t}{T}\right) * x[n],$$
so the $\mathrm{ZOH}$ is an LTI hybrid system with impulse response
$$h_0(t)=w\left(\frac{t}{T}\right)=\left{\begin{array}{ll} 1, & 0 \leq t<T \ 0, & \text { else } \end{array} .\right.$$
The Fourier transform of $w(t)$ is
\begin{aligned} W(j \omega) & =\int_{-\infty}^{\infty} w(t) e^{-j \omega t} d t \ & =\int_0^1 e^{-j \omega t} d t \ & =\frac{1-e^{-j \omega}}{j \omega} \ & =\frac{e^{j \frac{\omega}{2}}-e^{-j \frac{\omega}{2}}}{j \omega} e^{-j \frac{\omega}{2}} \ & =\frac{2 \sin \frac{\omega}{2}}{\omega} e^{-j \frac{\omega}{2}} \ & =\left(\operatorname{sinc} \frac{\omega}{2 \pi}\right) e^{-j \frac{\omega}{2}}, \end{aligned}
so the $\mathrm{ZOH}$ has transfer function
$$H_0(j \omega)=\mathcal{F}\left{w\left(\frac{t}{T}\right)\right}=T W(j \omega T)=T\left(\operatorname{sinc} \frac{\omega}{\omega_s}\right) e^{-j \frac{\omega T}{2}}$$

## 电子代写|数字信号处理代写Digital Signal Processing代考|A/D and D/A Convert

The actual electronic device that performs sampling is called an analog-to-digital (A/D) converter. An A/D converter actually consists of two parts: First, the CT input is sampled. Then the resulting value (a voltage) is quantized – i.e. approximated by the nearest value taken from a given finite set. Quantization is depicted in Figure 4.11:

Here we assume the input voltage stays within $\pm 5 \mathrm{~V}$ and that the output can achieve $2^m$ possible values. The quantized signal is then represented in binary form in an $m$-bit computer register. Quantization contributes a certain degree of distortion to a signal, which can be made smaller by increasing $m$. In practice, the value of $m$ depends on the application. In control systems, $m=12$ is common, while in high-fidelity audio systems, $m=16$ is minimum. The distortion caused by the nonlinear nature of Figure $4.11$ is called quantization noise. An analytic treatment of quantization noise is possible, but it is mathematically difficult and requires the study of random processes. This is beyond our scope, so we will henceforth assume that the number of bits $m$ is sufficiently large to ensure that the effects of quantization are negligible. In other words, we will approximate the A/D converter as an ideal sampler, depicted in Figure $3.1$.

A D/A converter performs signal reconstruction. The $m$-bit binary value is converted back to a voltage with $2^m$ possible values and then passed through a ZOH. A smoothing filter may then be applied to the output. The only idealization required here is that $m$ be very large, so the ZOH can accept any DT signal $x[n]$.

## 电子代写|数字信号处理代写Digital Signal Processing代考|The Zero-Order Hold

$$x(t)=x[n], \quad n T \leq t<(n+1) T .$$

\$\$
$$w(t)=\backslash \text { left }{$$
$$1, \quad 0 \leq t \leq 10, \quad \text { else }$$

\正确的。
The $\$ \mathrm{ZOH} \$$processessignalsaccordingto$$
\begin{aligned}
& \mathrm{x}{-} 0(\mathrm{t})=\backslash \text { sum }{\mathrm{n}=-\backslash \text { infty }} \wedge{\backslash \text { infty }} \mathrm{x}[\mathrm{n}] \mathrm{w} \backslash \operatorname{left}(\backslash \text { frac }{\mathrm{tn} \mathrm{T}}{\mathrm{T}} \backslash \text { right })=\mathrm{w} \backslash \operatorname{left}(\backslash \text { frac }{\mathrm{t}}{\mathrm{T}} \backslash \text { 右 }) * \mathrm{x}[\mathrm{n}], \ & \text { sothe } \$\mathrm{ZOH} \$ \text { isanLTIhybridsystemwithimpulseresponse } \ & \mathrm{h}{-} 0(\mathrm{t})=\mathrm{w} \backslash \operatorname{left}(\backslash \text { frac }{\mathrm{t}}{\mathrm{T}} \backslash \text { right })=\backslash \text { left }{ \
& 1, \quad 0 \leq t<T 0, \quad \text { else }
\end{aligned}
$$|正确的。$$
W(j \omega)=\int_{-\infty}^{\infty} w(t) e^{-j \omega t} d t \quad=\int_0^1 e^{-j \omega t} d t=\frac{1-e^{-j \omega}}{j \omega} \quad=\frac{e^{j \frac{\omega}{2}}-e^{-j \frac{\omega}{2}}}{j \omega} e^{-j \frac{\omega}{2}}=\frac{2 \sin \frac{\omega}{2}}{\omega} e^{-j \frac{\omega}{2}} \quad=\left(\operatorname{sinc} \frac{\omega}{2 \pi}\right) e^{-j \frac{\omega}{2}}
$$sothe \ \mathrm{ZOH} \$$ hastrans fer function
$H_{-} O(j$ \omega $)=\backslash$ mathcal ${\mathrm{F}} \backslash$ left ${w \backslash$ left $(\backslash$ frac ${t}{T} \backslash$ right $) \backslash$ right $}=T W(j \backslash$ lomega
$T)=T \backslash$ left $(\backslash$ operatorname ${$ sinc $} \backslash$ frac ${\backslash$ lomega $}{$ \omega_s $} \backslash$ right $)$ e ${{-j \backslash$ frac ${\backslash$ lomega $T}{2}}$
$\$ \

## 电子代写|数字信号处理代写Digital Signal Processing代考|A/D and D/A Convert

$\mathrm{AD} / \mathrm{A}$ 转换器执行信号重构。这 $m$ 位二进制值转换回电压 $2^m$ 可能的值，然后通过 $\mathrm{ZOH}$ 。然后可以将平滑滤波器应用于输出。这里 唯一需要的理想化是 $m$ 非常大，所以 ZOH 可以接受任何 DT 信号 $x[n]$.

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