Question

P3. An ideal band-pass filter is defined by the transfer function $H(\omega) = \begin{cases} 1 & 2\pi900 \le |\omega| \le 2\pi1100 \\ 0 & \text{elsewhere} \end{cases}$ a. Plot $H(\omega)$ versus $\omega$. b. Find its output $y(t)$ and its transform $Y(\omega)$ if its input is i. $x(t) = 2\cos(2\pi500t) + 2\sin(2\pi950t)$. ii. $x(t) = 2\cos(2\pi550t)\sin(2\pi450t)$. iii. $x(t) = \frac{\sin(2\pi50t)}{\pi t}$ iv. $x(t) = \frac{\sin(2\pi50t)}{\pi t}\sin(2\pi1000t)$ v. $x(t) = \sum_{n=-\infty}^{\infty} \text{rect}(2000t - 4n)$ vi. $x(t) = \delta(t)$ c. What does this filter do?

          P3. An ideal band-pass filter is defined by the transfer function
$H(\omega) = \begin{cases} 1 & 2\pi900 \le |\omega| \le 2\pi1100 \\ 0 & \text{elsewhere} \end{cases}$
a. Plot $H(\omega)$ versus $\omega$.
b. Find its output $y(t)$ and its transform $Y(\omega)$ if its input is
i. $x(t) = 2\cos(2\pi500t) + 2\sin(2\pi950t)$.
ii. $x(t) = 2\cos(2\pi550t)\sin(2\pi450t)$.
iii. $x(t) = \frac{\sin(2\pi50t)}{\pi t}$
iv. $x(t) = \frac{\sin(2\pi50t)}{\pi t}\sin(2\pi1000t)$
v. $x(t) = \sum_{n=-\infty}^{\infty} \text{rect}(2000t - 4n)$
vi. $x(t) = \delta(t)$
c. What does this filter do?

P3. An ideal band-pass filter is defined by the transfer function
H(ω) = 1 2π900 ≤ |ω| ≤ 2π1100
0 elsewhere
a. Plot H(ω) versus ω.
b. Find its output y(t) and its transform Y(ω) if its input is
i. x(t) = 2cos(2π500t) + 2sin(2π950t).
ii. x(t) = 2cos(2π550t)sin(2π450t).
iii. x(t) = (sin(2π50t))/(π t)
iv. x(t) = (sin(2π50t))/(π t)sin(2π1000t)
v. x(t) = ∑n=-∞^∞rect(2000t - 4n)
vi. x(t) = δ(t)
c. What does this filter do?

Added by Justin B.

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