Topic: Signals and Systems
Let $X(e^{j\omega})=\sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}$ and $x[n]=\frac{1}{2\pi}\int_{-\pi}^{\pi}X(e^{j\omega})e^{j\omega n}d\omega$. If $X(e^{j\omega})=\frac{1}{(1-0.2e^{-j\omega})(1-0.1e^{-j\omega})}$, what is $x[n]$ in terms of unit discrete step function $u(n)$?
- A. $2(0.2)^nu(n)-(0.1)^nu(n)$
- B. $2(0.1)^nu(n)-(0.2)^nu(n)$
- C. $(0.2)^nu(n)-(0.1)^nu(n)$
- D. $(0.1)^nu(n)-(0.2)^nu(n)$
Answer: A. $2(0.2)^nu(n)-(0.1)^nu(n)$
Explanation: Applying partial fraction expansion to $X(e^{j\omega})$ with respect to $e^{-j\omega}$: $X(e^{j\omega}) = \frac{A}{1 - 0.2e^{-j\omega}} + \frac{B}{1 - 0.1e^{-j\omega}}$. Evaluating for $A$: $A = \left. \frac{1}{1 - 0.1e^{-j\omega}} \right\vert{}_{e^{-j\omega} = 5} = \frac{1}{1 - 0.5} = 2$. Evaluating for $B$: $B = \left. \frac{1}{1 - 0.2e^{-j\omega}} \right\vert{}_{e^{-j\omega} = 10} = \frac{1}{1 - 2} = -1$. So, $X(e^{j\omega}) = \frac{2}{1 - 0.2e^{-j\omega}} - \frac{1}{1 - 0.1e^{-j\omega}}$. Taking the inverse Discrete-Time Fourier Transform gives $x[n] = 2(0.2)^nu(n) - (0.1)^nu(n)$.