Questão 1 - Calcule a Transformada de Fourier em Tempo Discreto para as sequências a seguir
x[n]=\operatorname{sen}[2 \pi n / 8](\cos [3 \pi n / 8])^{2};

Passo 1

O enunciado pede a Transformada de Fourier em tempo discreto (DTFT) de
x[n] = \sin\!\bigl(\tfrac{2\pi n}{8}\bigr)\,\bigl(\cos\tfrac{3\pi n}{8}\bigr)^{2}.
Vamos usar identidades trigonométricas para decompor x[n] em somas de exponenciais e então aplicar a definição
X(e^{j\Omega}) = \sum_{n=-\infty}^{\infty} x[n]\,e^{-j\Omega n}.

Passo 2

Usamos \cos^{2}\theta = \tfrac{1+\cos(2\theta)}{2}. Com \theta = \tfrac{3\pi n}{8} e \sin\bigl(\tfrac{2\pi n}{8}\bigr)=\sin\bigl(\tfrac{\pi n}{4}\bigr),
x[n] = \sin\bigl(\tfrac{\pi n}{4}\bigr)\;\frac{1+\cos(\tfrac{3\pi n}{4})}{2} = \tfrac12\,\sin\bigl(\tfrac{\pi n}{4}\bigr) \;+\;\tfrac12\,\sin\bigl(\tfrac{\pi n}{4}\bigr)\cos\bigl(\tfrac{3\pi n}{4}\bigr).

Passo 3

Para o produto usamos
\sin A\cos B = \tfrac12[\sin(A+B)+\sin(A-B)]. Com A=\tfrac{\pi n}{4}, B=\tfrac{3\pi n}{4} obtemos
\sin\bigl(\tfrac{\pi n}{4}\bigr)\cos\bigl(\tfrac{3\pi n}{4}\bigr) = \tfrac12\bigl(\sin(\pi n)+\sin(-\tfrac{\pi n}{2})\bigr) = -\tfrac12\,\sin\bigl(\tfrac{\pi n}{2}\bigr)
(sabendo que \sin(\pi n)=0 para inteiro n). Logo
x[n] = \tfrac12\,\sin\bigl(\tfrac{\pi n}{4}\bigr) \;-\;\tfrac14\,\sin\bigl(\tfrac{\pi n}{2}\bigr).

Passo 4

Expressamos as funções seno em termos de exponenciais:
\sin\bigl(\tfrac{\pi n}{4}\bigr) = \tfrac{1}{2j}\bigl(e^{j\pi n/4}-e^{-j\pi n/4}\bigr), \quad \sin\bigl(\tfrac{\pi n}{2}\bigr) = \tfrac{1}{2j}\bigl(e^{j\pi n/2}-e^{-j\pi n/2}\bigr).
Então
x[n] = \tfrac{1}{4j}\bigl(e^{j\pi n/4}-e^{-j\pi n/4}\bigr) \;-\;\tfrac{1}{8j}\bigl(e^{j\pi n/2}-e^{-j\pi n/2}\bigr).

Passo 5

Pela definição da DTFT e usando
\sum_{n}e^{-j(\Omega-\Omega_{0})n}=2\pi\sum_{k}\delta(\Omega-\Omega_{0}-2\pi k), obtemos (no domínio principal)
X(e^{j\Omega}) =2\pi\Bigl[\tfrac{1}{4j}\bigl(\delta(\Omega-\tfrac{\pi}{4})-\delta(\Omega+\tfrac{\pi}{4})\bigr) -\tfrac{1}{8j}\bigl(\delta(\Omega-\tfrac{\pi}{2})-\delta(\Omega+\tfrac{\pi}{2})\bigr)\Bigr].
Fatorando,
X(e^{j\Omega}) = \tfrac{\pi}{2j}\bigl[\delta(\Omega-\tfrac{\pi}{4})-\delta(\Omega+\tfrac{\pi}{4})\bigr] -\tfrac{\pi}{4j}\bigl[\delta(\Omega-\tfrac{\pi}{2})-\delta(\Omega+\tfrac{\pi}{2})\bigr].

Resposta

X(e^{j\Omega}) = \tfrac{\pi}{2j}\bigl[\delta(\Omega-\tfrac{\pi}{4})-\delta(\Omega+\tfrac{\pi}{4})\bigr] -\tfrac{\pi}{4j}\bigl[\delta(\Omega-\tfrac{\pi}{2})-\delta(\Omega+\tfrac{\pi}{2})\bigr]

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