京都大学 情報学研究科 知能情報学専攻 2021年7月実施 専門科目 S-5

标签：

• Kyoto-University
• Z-Transform
• Discrete-Fourier-Transform

Author

祭音Myyura

Description

大学公表の原題 Let $n \in \mathbb{Z}$ be a discrete-time index. The unit impulse signal $\delta[n]$ and the unit step signal $u[n]$ are defined as follows:

$$\delta[n] = \begin{cases} 0 &(n \neq 0) \\ 1 &(n=0) \end{cases}$$

$$u[n] = \begin{cases} 0 &(n < 0) \\ 1 &(n \geq 0) \end{cases}$$

Q.1 Compute the $z$-transform $X(z)$ of a discrete-time signal $x[n]$ given below.

• (1) $x[n] = 3\delta [n] - 2\delta [n-2] + 5\delta [n-4]$
• (2) $x[n] = nu[n]$

Q.2 Judge the stability of a system whose transfer function $H(z)$ is given below and draw the correponding circuit. In addition, compute the inverse $z$-transform $h[n]$.

• (1) $H(z) = 1 + 2z^{-1} + 3z^{-2}$
• (2) $H(z) = \frac{1 + 2z^{-1}}{2 - z^{-1}}$

Q.3 Compute the discrete-time Fourier transform $F(\omega)$ of a discrete-time signal $x[n]$ given below, where $\omega$ represents a normalized angular frequency.

• (1) $x[n] = 3\delta [n] - 2\delta [n-2] + 5\delta [n-4]$
• (2) $x[n] = u[n] - u[n-6]$

题目描述

对离散时间 $n\in\mathbb Z$，单位冲激和单位阶跃定义为

$$\delta[n]=\begin{cases}1,&n=0\\0,&n\ne0,\end{cases} \qquad u[n]=\begin{cases}0,&n<0\\1,&n\ge0.\end{cases}$$

1. 求下列信号的 $z$ 变换 $X(z)$：
   1. $x[n]=3\delta[n]-2\delta[n-2]+5\delta[n-4]$；
   2. $x[n]=nu[n]$。
2. 对下列传递函数，判断系统稳定性、画出对应电路，并求逆 $z$ 变换 $h[n]$：
   1. $H(z)=1+2z^{-1}+3z^{-2}$；
   2. $H(z)=\dfrac{1+2z^{-1}}{2-z^{-1}}$。
3. 求下列信号的离散时间 Fourier 变换 $F(\omega)$，其中 $\omega$ 为归一化角频率：
   1. $x[n]=3\delta[n]-2\delta[n-2]+5\delta[n-4]$；
   2. $x[n]=u[n]-u[n-6]$。

Kai

Q.1

The bilateral $z$-transform is used.

(1)

$$X(z) = 3-2z^{-2}+5z^{-4},\qquad |z|>0$$

(2)

Note that

$$\sum_{n=-\infty}^{\infty}u[n]z^{-n} =\sum_{n=0}^{\infty}z^{-n} = \frac{1}{1-z^{-1}}$$

hence

$$\sum_{n=0}^{\infty}(-n)z^{-n-1} = \frac{-z^{-2}}{(1-z^{-1})^{2}}$$

$$\sum_{n=0}^{\infty}nz^{-n} = X(z) = \frac{z^{-1}}{(1-z^{-1})^{2}}$$

The series in (2) converges for $|z|>1$.

Q.2

For the circuit realizations below, assume a causal system initially at rest.

(1)

The finite impulse response is absolutely summable, so the system is BIBO stable.

The corresponding circuit diagram is as follows:

and

$$h[n] = \delta[n]+2\delta[n-1]+3\delta[n-2]$$

(2)

For the causal realization, the region of convergence is $|z|>1/2$, containing the unit circle, so the system is BIBO stable.

The corresponding circuit diagram is as follows:

Note that

$$\sum_{n=-\infty}^{\infty}a^{n}u[n]z^{-n} = \sum_{n=-\infty}^{\infty}u[n](az^{-1})^{n} = \sum_{n=0}^{\infty}(az^{-1})^{n} =\frac{1}{1-az^{-1}}$$

hence

$$H(z) = -2+\frac{5}{2-z^{-1}} = -2+\frac52\frac{1}{1-2^{-1}\cdot z^{-1}}$$

$$\Rightarrow h[n] = -2\delta[n]+5\cdot 2^{-n-1}u[n]$$

Without the causal assumption, the same rational expression also permits the region $|z|<1/2$, whose inverse is $h[n]=-2\delta[n]-(5/2)(1/2)^n u[-n-1]$. This left-sided response is not absolutely summable and is unstable.

Q.3

Note that

$$\begin{align} \sum_{n=-\infty}^{\infty}\delta[n-a]e^{-i\omega n} = e^{-i\omega a} \tag{*} \end{align}$$

and

$$\begin{align} \sum_{n=-\infty}^{\infty}(u[n]-u[n-a])e^{-i\omega n} &= \sum_{k=0}^{a-1}e^{-i\omega k}= \frac{1-e^{-i\omega a}}{1-e^{-i\omega}} \nonumber \\ &= \frac{e^{-i\omega a/2}\left(e^{i\omega a/2}-e^{-i\omega a/2}\right)}{e^{-i\omega/2}\left(e^{i\omega/2}-e^{-i\omega/2}\right)} \nonumber \\ &= e^{-i\omega(a-1)/2}\frac{\sin a\omega/2}{\sin \omega/2} \tag{**} \end{align}$$

(1)

By (*) we have

$$F(\omega) = 3 - 2e^{-i2\omega} + 5e^{-i4\omega}$$

(2)

By (**) we have

$$F(\omega) = e^{-i5\omega/2}\frac{\sin 3\omega}{\sin \omega/2}$$

At $\omega=2\pi k$ ($k\in\mathbb Z$), the value is $F(\omega)=6$, obtained directly from the six-sample sum or by continuity.