東京大学 情報理工学系研究科 電子情報学専攻 2022年8月実施 専門 第5問

标签：

• Tokyo-University
• Z-Transform-of-Geometric-Sequence
• Discrete-Time-System-Transfer-and-Frequency-Response

Author

Josuke

Description

Answer the following questions about discrete signal processing.

(1) Show the definition of the $\mathcal{Z}$-transform $X(z)$ of a discrete signal sequence $x_n(n=0,1,2,\dots)$ defined for $n \ge 0$ where $z$ is a complex number.

(2) By using the definition, find the $\mathcal{Z}$-transform of a geometric sequence $x_n = p^n$ where $p$ is a real number.

(3) We consider the zero state response of the discrete time system shown in the figure below ,where $a,b$ and $c$ are real numbers. Give $X(z)$ and $Y(z)$ by using the $\mathcal{Z}$-tranform $Q(z)$ of the system's internal state $q_n$. Here, $X(z)$ and $Y(z)$ are the $\mathcal{Z}$-tranforms of the system's input $x_n$ and output $y_n$, respectively.

(4) Give the transfer function $H(z)$ of the system.

(5) Next, we find the system's frequency response by using the transfer function $H(z)$. Consider the input sequnence $x_n = e^{jn\omega T}$ sampled at the interval $T$ from a complex exponential function $x(t) = e^{j\omega t}(t \ge 0)$. Derive $Y(z)$ and then find the output sequence $y_n$.

题目描述

回答下列离散信号处理问题。

(1) 对定义于 $n\ge0$ 的离散信号序列 $x_n$（$n=0,1,2,\ldots$），写出其 $\mathcal Z$ 变换 $X(z)$ 的定义，其中 $z$ 为复数。

(2) 利用定义求实数 $p$ 所确定的几何序列 $x_n=p^n$ 的 $\mathcal Z$ 变换。

(3) 考虑上图所示离散时间系统的零状态响应，其中 $a,b,c$ 为实数。设系统内部状态 $q_n$ 的 $\mathcal Z$ 变换为 $Q(z)$，用 $Q(z)$ 表示系统输入 $x_n$、输出 $y_n$ 的 $\mathcal Z$ 变换 $X(z),Y(z)$。

(4) 求该系统的传递函数 $H(z)$。

(5) 利用 $H(z)$ 求系统的频率响应。连续复指数

$$x(t)=e^{j\omega t}\qquad(t\ge0)$$

以间隔 $T$ 采样后得到输入序列 $x_n=e^{jn\omega T}$。推导 $Y(z)$，进而求输出序列 $y_n$。

Kai

(1)

$$X(z) = \sum_{n=0}^{+\infty}x(n)z^{-n}$$

(2)

$$X(z) = \sum_{n=0}^{+\infty}p^nz^{-n} = \sum_{n=0}^{+\infty}(pz^{-1})^n = \frac{1}{1 - pz^{-1}}$$

(3)

$$\begin{aligned} X(z) &+ az^{-1}Q(z) = Q(z) \\ X(z) &= (1 - az^{-1})Q(z) \\ Y(z) &= bQ(z) + cz^{-1}Q(z) \\ &= (b + cz^{-1})Q(z) \end{aligned}$$

(4)

$$H(z) = \frac{Y(z)}{X(z)} = \frac{b + cz^{-1}}{1 - az^{-1}} = \frac{b}{1 - az^{-1}} + \frac{cz^{-1}}{1 - az^{-1}}$$

(5)

$$\begin{aligned} X(z) &= \frac{1}{1 - e^{j\omega T}z^{-1}} \qquad h[n] = ba^n + ca^{n-1} \\ Y(z) &= \frac{b + cz^{-1}}{(1 - az^{-1})(1 - e^{j\omega T}z^{-1})} \\ y[n] &= \sum_{k=0}^{+\infty}e^{j\omega kT}(ba^{n-k} + ca^{n-k-1}) \\ &= ba^{n} \sum_{k=0}^{+\infty}(e^{j\omega T}a^{-1})^k + ca^{n-1} \sum_{k=0}^{+\infty}(e^{j\omega T}a^{-1})^k \\ &= \frac{ba^n}{1 - a^{-1}e^{j\omega T}} + \frac{ca^{n-1}}{1 - a^{-1}e^{j\omega T}} \\ &=\frac{(ba + c)a^{n-1}}{1 - a^{-1}e^{j\omega T}} \end{aligned}$$