東京大学情報理工学系研究科電子情報学専攻 2018年度専門第5問

Author

Description

Answer the following questions about discrete signal processing. Here, \(T\) is the sampling interval.

(1) Show the definition of the \(Z\)-transform \(X(z)\) for the discrete signal series \(x_n(n = 0,1,2,\cdots)\), which is defined for \(n \ge 0\). Here, \(z\) is a complex variable.

(2) Derive the transfer function \(H(s)\) in the \(s\)-domain(the Laplace transform domain) of the circuit in Fig.1.

(3) The relationship between the Laplace transform and the \(Z\)-transform is described as \(z = e^{sT}\). Derive the following approximation.

\[ s \simeq \frac{2}{T}\frac{1 - z^{-1}}{1 + z^{-1}}. \]

You can use the following equation if necessary.

\[ e^{x} \simeq 1 + x. \]

(4) Convert \(H(s)\) to the transfer function \(H(z)\) in the \(z\)-domain by using the approximation derived in (3). Here, we assume \(T = 1\).

(5) Show a schematic of a discrete signal circuit that corresponds to \(H(z)\) in (4).

(6) By taking the same procedure, show a schematic of a discrete signal circuit for the circuit shown in Fig.2.

Kai

(1)

\[ X(z) = \sum_{n = 0}^{\infty}x(n)z^{-n} \]

(2)

\[ H(s) = \frac{V_{out}(s)}{V_{in}(s)} = \frac{\frac{1}{sC}}{R + \frac{1}{sC}} = \frac{1}{1 + s} \]

(3)

\[ z^{-1} = e^{sT} = \frac{e^{-\frac{1}{2}sT}}{e^{\frac{1}{2}sT}} = \frac{1 - \frac{1}{2}sT}{1 + \frac{1}{2}sT} \]

\[ \begin{aligned} z^{-1}(1 + \frac{1}{2}sT) &= 1 - \frac{1}{2}sT \\ z^{-1} &= 1 - \frac{1}{2}sT(1 + z^{-1}) \\ 1 - z^{-1} &= \frac{1}{2}sT(1 + z^{-1}) \\ s &= \frac{2}{T} \cdot \frac{1 - z^{-1}}{1 + z^{-1}} \end{aligned} \]

(4)

\[ \begin{aligned} H(s) &= \frac{1}{1 + s} = \frac{1}{1 + 2 \cdot \frac{1 - z^{-1}}{1 + z^{-1}}} = \frac{1 + z^{-1}}{3 - z^{-1}} \\ &= \frac{\frac{1}{3}}{1 - \frac{1}{3}z^{-1}} + \frac{\frac{1}{3}z^{-1}}{1 - \frac{1}{3}z^{-1}} \end{aligned} \]