Discrete Time Fourier Transform Questions and Answers

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.

Answer: B

The signal is the unit impulse \(x(n) = \delta(n)\): one at \(n = 0\) and zero everywhere else. The discrete time Fourier transform is

\[ \begin{aligned} X(e^{j\omega}) \\ &= \sum_{n=-\infty}^{\infty} x(n)e^{-j\omega n} \\ &= \delta(0)\,e^{0} = 1 \end{aligned} \]

Only the term \(n = 0\) survives, so the transform is 1 for all frequencies: the impulse contains all frequencies equally. The values 0, 2 and 3 would belong to other signals (for example, a zero signal or multiples of the impulse). The answer is 1, option B.