東京大学 情報理工学系研究科 電子情報学専攻 2017年8月実施 専門 第5問 Author Josuke
Description Answer the following questions about discrete signal processing. Here, T T T
(1) Show the definition of the Z Z Z X ( z ) X(z) X ( z ) x n ( n = 0 , 1 , 2 , ⋯ ) x_n(n = 0,1,2,\cdots) x n ( n = 0 , 1 , 2 , ⋯ ) n ≥ 0 n \ge 0 n ≥ 0 z z z
(2) Derive the transfer function H ( s ) H(s) H ( s ) s s s
(3) The relationship between the Laplace transform and the Z Z Z z = e s T z = e^{sT} z = e s T
s ≃ 2 T 1 − z − 1 1 + z − 1 . s \simeq \frac{2}{T}\frac{1 - z^{-1}}{1 + z^{-1}}. s ≃ T 2 1 + z − 1 1 − z − 1 . You can use the following equation if necessary.
e x ≃ 1 + x . e^{x} \simeq 1 + x. e x ≃ 1 + x . (4) Convert H ( s ) H(s) H ( s ) H ( z ) H(z) H ( z ) z z z T = 1 T = 1 T = 1
(5) Show a schematic of a discrete signal circuit that corresponds to H ( z ) H(z) H ( z )
(6) By taking the same procedure, show a schematic of a discrete signal circuit for the circuit shown in Fig.2.
题目描述
回答下列离散信号处理问题，其中 T T T
(1) 对定义于 n ≥ 0 n\ge0 n ≥ 0 x n x_n x n n = 0 , 1 , 2 , … n=0,1,2,\ldots n = 0 , 1 , 2 , … Z Z Z X ( z ) X(z) X ( z ) z z z
(2) 求图 1 电路在 s s s H ( s ) H(s) H ( s )
(3) 拉普拉斯变换与 Z Z Z z = e s T z=e^{sT} z = e s T
s ≃ 2 T 1 − z − 1 1 + z − 1 . s\simeq\frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}}. s ≃ T 2 1 + z − 1 1 − z − 1 . 必要时可使用 e x ≃ 1 + x e^x\simeq1+x e x ≃ 1 + x
(4) 使用 (3) 的近似，把 H ( s ) H(s) H ( s ) z z z H ( z ) H(z) H ( z ) T = 1 T=1 T = 1
(5) 画出与 (4) 的 H ( z ) H(z) H ( z )
(6) 按相同过程，画出与图 2 电路对应的离散信号处理电路。
Kai (1)
X ( z ) = ∑ n = 0 ∞ x ( n ) z − n X(z) = \sum_{n = 0}^{\infty}x(n)z^{-n} X ( z ) = n = 0 ∑ ∞ x ( n ) z − n (2)
H ( s ) = V o u t ( s ) V i n ( s ) = 1 s C R + 1 s C = 1 1 + s H(s) = \frac{V_{out}(s)}{V_{in}(s)} = \frac{\frac{1}{sC}}{R + \frac{1}{sC}} = \frac{1}{1 + s} H ( s ) = V in ( s ) V o u t ( s ) = R + s C 1 s C 1 = 1 + s 1 (3)
z − 1 = e s T = e − 1 2 s T e 1 2 s T = 1 − 1 2 s T 1 + 1 2 s T 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} z − 1 = e s T = e 2 1 s T e − 2 1 s T = 1 + 2 1 s T 1 − 2 1 s T z − 1 ( 1 + 1 2 s T ) = 1 − 1 2 s T z − 1 = 1 − 1 2 s T ( 1 + z − 1 ) 1 − z − 1 = 1 2 s T ( 1 + z − 1 ) s = 2 T ⋅ 1 − z − 1 1 + z − 1 \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} z − 1 ( 1 + 2 1 s T ) z − 1 1 − z − 1 s = 1 − 2 1 s T = 1 − 2 1 s T ( 1 + z − 1 ) = 2 1 s T ( 1 + z − 1 ) = T 2 ⋅ 1 + z − 1 1 − z − 1 (4)
H ( s ) = 1 1 + s = 1 1 + 2 ⋅ 1 − z − 1 1 + z − 1 = 1 + z − 1 3 − z − 1 = 1 3 1 − 1 3 z − 1 + 1 3 z − 1 1 − 1 3 z − 1 \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} H ( s ) = 1 + s 1 = 1 + 2 ⋅ 1 + z − 1 1 − z − 1 1 = 3 − z − 1 1 + z − 1 = 1 − 3 1 z − 1 3 1 + 1 − 3 1 z − 1 3 1 z − 1 (5)
H ( z ) = Y ( z ) X ( z ) = 1 + z − 1 3 − z − 1 H(z) = \frac{Y(z)}{X(z)} = \frac{1 + z^{-1}}{3 - z^{-1}} H ( z ) = X ( z ) Y ( z ) = 3 − z − 1 1 + z − 1 ( 3 − z − 1 ) Y ( z ) = ( 1 + z − 1 ) X ( z ) (3 - z^{-1})Y(z) = (1 + z^{-1})X(z) ( 3 − z − 1 ) Y ( z ) = ( 1 + z − 1 ) X ( z ) Y ( z ) = 1 3 [ X ( z ) + z − 1 ( X ( z ) + Y ( z ) ) ] Y(z) = \frac{1}{3}[X(z) + z^{-1}(X(z) + Y(z))] Y ( z ) = 3 1 [ X ( z ) + z − 1 ( X ( z ) + Y ( z ))]
(6)
H ( s ) = s 1 + s H(s) = \frac{s}{1 + s} H ( s ) = 1 + s s H ( z ) = 2 ⋅ 1 − z − 1 1 + z − 1 1 + 2 ⋅ 1 − z − 1 1 + z − 1 = 2 − 2 z − 1 3 − z − 1 H(z) = \frac{2 \cdot \frac{1 - z^{-1}}{1 + z^{-1}}}{1 + 2 \cdot \frac{1 - z^{-1}}{1 + z^{-1}}} = \frac{2 - 2z^{-1}}{3 - z^{-1}} H ( z ) = 1 + 2 ⋅ 1 + z − 1 1 − z − 1 2 ⋅ 1 + z − 1 1 − z − 1 = 3 − z − 1 2 − 2 z − 1 Y ( z ) X ( z ) = 2 − 2 z − 1 3 − z − 1 \frac{Y(z)}{X(z)} = \frac{2 - 2z^{-1}}{3 - z^{-1}} X ( z ) Y ( z ) = 3 − z − 1 2 − 2 z − 1 Y ( z ) = 1 3 [ 2 X ( z ) + z − 1 ( Y ( z ) − 2 X ( z ) ) ] Y(z) = \frac{1}{3}[2X(z) + z^{-1}(Y(z) - 2X(z))] Y ( z ) = 3 1 [ 2 X ( z ) + z − 1 ( Y ( z ) − 2 X ( z ))]
