Question

A finite impulse response (FIR) filter has only two non-zero samples in its impulse response \(h[n]\), namely \(h[0]=h[1]=1\). The Discrete Time Fourier Transform (DTFT) of \(h[n]\) equals \(H(e^{j\omega})\), as a function of the normalized angular frequency \(\omega\). For the range \(|\omega|\leq \pi\), \(|H(e^{j\omega})|\) is equal to __________________.

• A. \(2|\cos(\omega)|\)
• B. \(2|\sin(\omega)|\)
• C. \(2|\cos(\omega/2)|\)
• D. \(2|\sin(\omega/2)|\)

Hint:

- For FIR filters, the DTFT often simplifies to a trigonometric form. - Always use Euler's formula \(e^{j\theta}+e^{-j\theta}=2\cos\theta\) to simplify. - Phase terms like \(e^{-j\omega/2}\) have unit magnitude, so they disappear in \(|H(e^{j\omega})|\).

Answer 1

The Correct Option is C

Step 1: The DTFT of \(h[n]\) is defined as \[ H(e^{j\omega})=\sum_{n=-\infty}^{\infty} h[n] e^{-j\omega n}. \] Step 2: Since \(h[0]=1\) and \(h[1]=1\), and all other samples are zero, we have \[ H(e^{j\omega}) = 1 + e^{-j\omega}. \] Step 3: Factorize the expression: \[ H(e^{j\omega}) = e^{-j\omega/2}(e^{j\omega/2}+e^{-j\omega/2})= e^{-j\omega/2}(2\cos(\omega/2)). \] Step 4: The magnitude is \[ |H(e^{j\omega})|=|e^{-j\omega/2}|\cdot |2\cos(\omega/2)| = 2|\cos(\omega/2)|. \] Thus, the answer is \(\boxed{2|\cos(\omega/2)|}\).