京都大学 情報学研究科 知能情報学専攻 2021年8月実施 専門科目 S-5

Author

祭音Myyura

Description

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) \(x[n] = u[n] - u[n-6]\)

Kai

Q.1

(1)

\[ X(z) = 3-2z^{-2}+5z^{-4} \]

(2)

Note that

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

hence

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

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

Q.2

(1)

The system is stable since the pole of the transfer function is at \(z = 0\), which lies within the unit circle.

The corresponding circuit diagram is as follows:

and

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

(2)

The system is stable since the pole of the transfer function is at \(z = \frac{1}{2}\), which lies within the unit circle.

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

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

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

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

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