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 ∈ Z n \in \mathbb{Z} n ∈ Z δ [ n ] \delta[n] δ [ n ] u [ n ] u[n] u [ n ]
δ [ n ] = { 0 ( n ≠ 0 ) 1 ( n = 0 ) \delta[n] = \begin{cases}
0 &(n \neq 0) \\
1 &(n=0)
\end{cases} δ [ n ] = { 0 1 ( n = 0 ) ( n = 0 ) u [ n ] = { 0 ( n < 0 ) 1 ( n ≥ 0 ) u[n] = \begin{cases}
0 &(n < 0) \\
1 &(n \geq 0)
\end{cases} u [ n ] = { 0 1 ( n < 0 ) ( n ≥ 0 ) Q.1 Compute the z z z X ( z ) X(z) X ( z ) x [ n ] x[n] x [ n ]
(1) x [ n ] = 3 δ [ n ] − 2 δ [ n − 2 ] + 5 δ [ n − 4 ] x[n] = 3\delta [n] - 2\delta [n-2] + 5\delta [n-4] x [ n ] = 3 δ [ n ] − 2 δ [ n − 2 ] + 5 δ [ n − 4 ]
(2) x [ n ] = n u [ n ] x[n] = nu[n] x [ n ] = n u [ n ]
Q.2 Judge the stability of a system whose transfer function H ( z ) H(z) H ( z ) z z z h [ n ] h[n] h [ n ]
(1) H ( z ) = 1 + 2 z − 1 + 3 z − 2 H(z) = 1 + 2z^{-1} + 3z^{-2} H ( z ) = 1 + 2 z − 1 + 3 z − 2
(2) H ( z ) = 1 + 2 z − 1 2 − z − 1 H(z) = \frac{1 + 2z^{-1}}{2 - z^{-1}} H ( z ) = 2 − z − 1 1 + 2 z − 1
Q.3 Compute the discrete-time Fourier transform F ( ω ) F(\omega) F ( ω ) x [ n ] x[n] x [ n ] ω \omega ω
(1) x [ n ] = 3 δ [ n ] − 2 δ [ n − 2 ] + 5 δ [ n − 4 ] x[n] = 3\delta [n] - 2\delta [n-2] + 5\delta [n-4] x [ n ] = 3 δ [ n ] − 2 δ [ n − 2 ] + 5 δ [ n − 4 ]
(2) x [ n ] = u [ n ] − u [ n − 6 ] x[n] = u[n] - u[n-6] x [ n ] = u [ n ] − u [ n − 6 ]
Kai Q.1
(1)
X ( z ) = 3 − 2 z − 2 + 5 z − 4 X(z) = 3-2z^{-2}+5z^{-4} X ( z ) = 3 − 2 z − 2 + 5 z − 4 (2)
Note that
∑ n = − ∞ ∞ u [ n ] z − n = ∑ n = 0 ∞ z − n = 1 1 − z − 1 \sum_{n=-\infty}^{\infty}u[n]z^{-n} =\sum_{n=0}^{\infty}z^{-n} = \frac{1}{1-z^{-1}} n = − ∞ ∑ ∞ u [ n ] z − n = n = 0 ∑ ∞ z − n = 1 − z − 1 1 hence
∑ n = 0 ∞ ( − n ) z − n − 1 = − z − 2 ( 1 − z − 1 ) 2 \sum_{n=0}^{\infty}(-n)z^{-n-1} = \frac{-z^{-2}}{(1-z^{-1})^{2}} n = 0 ∑ ∞ ( − n ) z − n − 1 = ( 1 − z − 1 ) 2 − z − 2 ∑ n = 0 ∞ n z − n − 1 = X ( z ) = z − 1 ( 1 − z − 1 ) 2 \sum_{n=0}^{\infty}nz^{-n-1} = X(z) = \frac{z^{-1}}{(1-z^{-1})^{2}} n = 0 ∑ ∞ n z − n − 1 = X ( z ) = ( 1 − z − 1 ) 2 z − 1 Q.2
(1)
The system is stable since the pole of the transfer function is at z = 0 z = 0 z = 0
The corresponding circuit diagram is as follows:
and
h [ n ] = δ [ n ] + 2 δ [ n − 1 ] + 3 δ [ n − 2 ] h[n] = \delta[n]+2\delta[n-1]+3\delta[n-2] h [ n ] = δ [ n ] + 2 δ [ n − 1 ] + 3 δ [ n − 2 ] (2)
The system is stable since the pole of the transfer function is at z = 1 2 z = \frac{1}{2} z = 2 1
The corresponding circuit diagram is as follows:
Note that
∑ n = − ∞ ∞ a n u [ n ] z − n = ∑ n = − ∞ ∞ u [ n ] ( a z − 1 ) n = ∑ n = 0 ∞ ( a z − 1 ) n = 1 1 − a z − 1 \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}} n = − ∞ ∑ ∞ a n u [ n ] z − n = n = − ∞ ∑ ∞ u [ n ] ( a z − 1 ) n = n = 0 ∑ ∞ ( a z − 1 ) n = 1 − a z − 1 1 hence
H ( z ) = − 2 + 5 2 − z − 1 = − 2 + 2 ⋅ 5 1 1 − 2 − 1 ⋅ z − 1 H(z) = -2+\frac{5}{2-z^{-1}} = -2+2\cdot 5\frac{1}{1-2^{-1}\cdot z^{-1}} H ( z ) = − 2 + 2 − z − 1 5 = − 2 + 2 ⋅ 5 1 − 2 − 1 ⋅ z − 1 1 ⇒ h [ n ] = − 2 δ [ n ] + 5 ⋅ 2 − n + 1 u [ n ] \Rightarrow h[n] = -2\delta[n]+5\cdot 2^{-n+1}u[n] ⇒ h [ n ] = − 2 δ [ n ] + 5 ⋅ 2 − n + 1 u [ n ] Q.3
Note that
∑ n = − ∞ ∞ δ [ n − a ] e − i ω n = e − i ω a \begin{align}
\sum_{n=-\infty}^{\infty}\delta[n-a]e^{-i\omega n} = e^{-i\omega a} \tag{*}
\end{align} n = − ∞ ∑ ∞ δ [ n − a ] e − iωn = e − iωa ( * ) and
∑ n = − ∞ ∞ ( u [ n ] − u [ n − a ] ) e − i ω n = ∑ k = 0 a − 1 e − i ω k = 1 − e − i ω a 1 − e − i ω = e − i ω a / 2 ( e i ω a / 2 − e − i ω a / 2 ) e − i ω / 2 ( e i ω / 2 − e − i ω / 2 ) = e − i ω ( a − 1 ) / 2 sin a ω / 2 sin ω / 2 \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} n = − ∞ ∑ ∞ ( u [ n ] − u [ n − a ]) e − iωn = k = 0 ∑ a − 1 e − iωk = 1 − e − iω 1 − e − iωa = e − iω /2 ( e iω /2 − e − iω /2 ) e − iωa /2 ( e iωa /2 − e − iωa /2 ) = e − iω ( a − 1 ) /2 sin ω /2 sin aω /2 ( ** ) (1)
By (*) we have
F ( ω ) = 3 − 2 e − i 2 ω + 5 e − i 4 ω F(\omega) = 3 - 2e^{-i2\omega} + 5e^{-i4\omega} F ( ω ) = 3 − 2 e − i 2 ω + 5 e − i 4 ω (2)
By (**) we have
F ( ω ) = e − i 5 ω / 2 sin 3 ω sin ω / 2 F(\omega) = e^{-i5\omega/2}\frac{\sin 3\omega}{\sin \omega/2} F ( ω ) = e − i 5 ω /2 sin ω /2 sin 3 ω 
