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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

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Author​

Josuke

Description​

Answer the following questions about discrete signal processing. Here, TT is the sampling interval.

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

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

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

s≃2T1−z−11+z−1.s \simeq \frac{2}{T}\frac{1 - z^{-1}}{1 + z^{-1}}.

You can use the following equation if necessary.

ex≃1+x.e^{x} \simeq 1 + x.

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

(5) Show a schematic of a discrete signal circuit that corresponds to H(z)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)=∑n=0∞x(n)z−nX(z) = \sum_{n = 0}^{\infty}x(n)z^{-n}

(2)​

H(s)=Vout(s)Vin(s)=1sCR+1sC=11+sH(s) = \frac{V_{out}(s)}{V_{in}(s)} = \frac{\frac{1}{sC}}{R + \frac{1}{sC}} = \frac{1}{1 + s}

(3)​

z−1=esT=e−12sTe12sT=1−12sT1+12sTz^{-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(1+12sT)=1−12sTz−1=1−12sT(1+z−1)1−z−1=12sT(1+z−1)s=2T⋅1−z−11+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}

(4)​

H(s)=11+s=11+2⋅1−z−11+z−1=1+z−13−z−1=131−13z−1+13z−11−13z−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}

(5)​

H(z)=Y(z)X(z)=1+z−13−z−1H(z) = \frac{Y(z)}{X(z)} = \frac{1 + z^{-1}}{3 - z^{-1}}
(3−z−1)Y(z)=(1+z−1)X(z)(3 - z^{-1})Y(z) = (1 + z^{-1})X(z)
Y(z)=13[X(z)+z−1(X(z)+Y(z))]Y(z) = \frac{1}{3}[X(z) + z^{-1}(X(z) + Y(z))]

(6)​

H(s)=s1+sH(s) = \frac{s}{1 + s}
H(z)=2⋅1−z−11+z−11+2⋅1−z−11+z−1=2−2z−13−z−1H(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}}
Y(z)X(z)=2−2z−13−z−1\frac{Y(z)}{X(z)} = \frac{2 - 2z^{-1}}{3 - z^{-1}}
Y(z)=13[2X(z)+z−1(Y(z)−2X(z))]Y(z) = \frac{1}{3}[2X(z) + z^{-1}(Y(z) - 2X(z))]
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