Todai Entrance Exam: Subject 2018 – Problem 5

(1) See Unilateral Z-transform on Wikipedia.

(2) Under superpower of Transfer Function of a Circuit:

    \[ H(s) = \frac{V_{\text{output}}}{V_{\text{input}}} = \frac{\frac{1}{Cs}}{R + \frac{1}{Cs}} = \frac{1}{RCs + 1} = \frac{1}{s + 1} \]

(3) As hinted by Wikipedia:

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


    \[ \Rightarrow s(T + zT) \simeq  2z - 2 \]


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

Q.E.D

(4)

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

(5) Thanks to Wikipedia (corresponding official source: Z Transform of Difference Equations and Direct-Form I, also Shifting Theorem of Z Transform):

Let y be the output signal, x be the input signal. From (4), we can derive that:

    \[ y(n) = \frac{1}{3}x(n) + \frac{1}{3}x(n-1) + \frac{1}{3}y(n-1) \]

(6)

    \[ H_{2}(s) = \frac{R}{R + \frac{1}{Cs}} = \frac{1}{1+ \frac{1}{RCs}} = \frac{1}{1+ \frac{1}{s}} \]

    \[ H_{2}(z) = \frac{\frac{2}{3} - \frac{2}{3}z^{-1}}{1 - \frac{1}{3}z^{-1}} \]

Let y be the output signal, x be the input signal. We get:

    \[ y(n) = \frac{2}{3}x(n) - \frac{2}{3}x(n-1) + \frac{1}{3}y(n-1) \]


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