Question

Find the z-transform of the number sequence generated by sampling the time function e(t) = 2t every T seconds,

beginning at t = 0. Can you express this transform in closed form?

Answer 1

The z-transform of the number sequence generated by sampling the time function e(t) = 2t every T seconds, starting at t = 0, is E(z) = 2T / (1 - 2Tz^(-1)), which is valid inside the region of convergence where |2Tz^(-1)| < 1.

Step-by-step explanation:

The question is asking to find the z-transform of a number sequence that results from sampling a time function e(t) = 2t every T seconds, starting from t = 0. The z-transform of a discrete sequence x[n] is defined as:

X(z) = Σ (x[n] * z^(-n))

For this specific function, the sequence is x[n] = 2nT where n is an integer representing the sample number. If we sample the time function e(t) = 2t every T seconds, the sampled sequence can be represented as e[n] = 2nT, because t = nT for n = 0, 1, 2, .... The z-transform of this sequence would then be:

E(z) = Σ (2nT * z^(-n))

This is a geometric series in terms of z^(-1), and its closed form can be derived using the formula for the sum of a geometric series provided that |2Tz^(-1)| < 1. The closed form of the z-transform for the given number sequence can then be expressed as:

E(z) = 2T / (1 - 2Tz^(-1))

This expression is only valid inside the region of convergence (ROC) where |2Tz^(-1)| < 1.