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Find the z transform of x[n]=(1/2)ⁿ[n]+1/3ᵘ[n]

User Eduard
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Final answer:

The z transform of x[n]=(1/2)ⁿᵢ[n] + (1/3)ᵢ[n] is X(z) = 1/(1 - (1/2)z⁻¹) + 1/(1 - (1/3)z⁻¹), with the region of convergence |z| > 1/2.

Step-by-step explanation:

The student has asked for the z transform of the sequence x[n]=(1/2)ⁿᵢ[n] + (1/3)ᵢ[n]. To find the transform, we compute the z-transform of each term separately and then add them together.

The z-transform of (1/2)ⁿᵢ[n] is ∑ (1/2)ⁿz⁻ⁿ from n=0 to ∞, which equals to 1/(1 - (1/2)z⁻¹), provided that |z| > |1/2|. The z-transform of (1/3)ᵢ[n] is ∑ (1/3)ⁿz⁻ⁿ from n=0 to ∞, which equals to 1/(1 - (1/3)z⁻¹), provided that |z| > |1/3|.

Combining these results, the overall z-transform of x[n] is:

X(z) = 1/(1 - (1/2)z⁻¹) + 1/(1 - (1/3)z⁻¹), with the region of convergence being |z| > max(|1/2|, |1/3|), i.e. |z| > 1/2.