The z-transform of e–t sampled at 10 Hz will be
z/(z – 0.9)
A sampling rate of 10 Hz means the sampling period is \(T = \dfrac{1}{10} = 0.1\ \text{s}\).
The sampled sequence is \(x(n) = e^{-nT} = \left(e^{-0.1}\right)^n\), which is a geometric sequence \(a^n\) with \(a = e^{-0.1}\).
Using the standard pair \(Z\{a^n\} = \dfrac{z}{z-a}\), we get \(X(z) = \dfrac{z}{z - e^{-0.1}}\).
Now \(e^{-0.1} = 0.9048 \approx 0.9\), so \(X(z) \approx \dfrac{z}{z - 0.9}\). The pole lies at z ≈ 0.9, inside the unit circle, as expected for a decaying exponential.
Hence, the answer is z/(z – 0.9).
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.................... is a client server application that identifies each host on the internet with a unique user-friendly name.
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(ii) Interface
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Codes:
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(i) Machine language
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