Topic: Signals and Systems
The DTFT of a sequence $x[n]$ is given by $X(e^{j\omega})$. Since $X(e^{j\omega})$ is period function of $\omega$, it can be expressed classical Fourier series as, $X(e^{j\omega})=\sum_{n=-\infty}^{\infty}C_{n}e^{jn\omega_{0}\omega}$ where $\omega_{0}$ is a fundamental frequency. Which of the following statement is correct?
- A. $\omega_{0}=\pi, C_{n}=-x[n]$
- B. $\omega_{0}=\pi, C_{n}=x[-n]$
- C. $\omega_{0}=1, C_{n}=x[-n]$
- D. $\omega_{0}=1, C_{n}=-x[-n]$
Answer: C. $\omega_{0}=1, C_{n}=x[-n]$
Explanation: $X(e^{j\omega})=\sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}$
Given $X(e^{j\omega})=\sum_{n=-\infty}^{\infty}C_{n}e^{jn\omega_{0}\omega}$
Replacing n by (-n) in the first equation:
$X(e^{j\omega})=\sum_{n=-\infty}^{\infty}x[-n]e^{j\omega n}$
Comparing it with the given series, we get $C_{n}=x[-n]$ and $\omega_{0}=1$.