Discussion
Find the Fourier transform of a single-sided, exponential pulse \(x(n) = a^n u(n)\).
Answer: A
The discrete time Fourier transform is the sum \(X(e^{j\omega}) = \sum x(n) e^{-j\omega n}\). For \(x(n) = a^n u(n)\) the sum runs from \(n = 0\) to infinity and is a geometric series.
\[ \begin{aligned} X(e^{j\omega}) \\ &= \sum_{n=0}^{\infty} (a e^{-j\omega})^n \\ &= \frac{1}{1 - ae^{-j\omega}},\quad |a| \lt 1 \end{aligned} \]
The series converges only when \(|a| \lt 1\). The options with a plus sign in the denominator would be the transform of \((-a)^n u(n)\), and those with \(e^{+j\omega}\) have the wrong sign in the exponent of the transform kernel. The answer is option A.
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