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# Lecture 33 ## ECE 301 ## Dr. Brown ### Question 1 $x(t) = u(t+1) - u(t-2)$ $h(t) = u(t)$ What is $y(t) = x(t) * h(t)$ ### Answer $y(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau) d\tau = \int_{-\infty}^{\infty} [u(\tau+1) - u(\tau-2)]u(t-\tau) d\tau$ **Case 1:** $t < -1$ No overlap $\implies...

# Lecture 33 ## ECE 301 ## Dr. Brown ### Question 1 $x(t) = u(t+1) - u(t-2)$ $h(t) = u(t)$ What is $y(t) = x(t) * h(t)$ ### Answer $y(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau) d\tau = \int_{-\infty}^{\infty} [u(\tau+1) - u(\tau-2)]u(t-\tau) d\tau$ **Case 1:** $t < -1$ No overlap $\implies y(t) = 0$ **Case 2:** $-1 < t < 2$ $\int_{-1}^{t} 1 d\tau = t - (-1) = t+1$ **Case 3:** $t > 2$ $\int_{-1}^{2} 1 d\tau = 2 - (-1) = 3$ ### Answer: $ y(t) = \begin{cases} 0, & t < -1 \\ t+1, & -1 < t < 2 \\ 3, & t>2 \end{cases} $ ### Question 2 $x[n] = u[n] - u[n-4]$ $h[n] = (\frac{1}{2})^n u[n]$ $y[n] = x[n] * h[n]$ ### Answer: $y[n] = \sum_{k=-\infty}^{\infty} x[k]h[n-k] = \sum_{k=-\infty}^{\infty} \{u[k] - u[k-4]\}(\frac{1}{2})^{n-k}u[n-k]$ **Case 1:** $n < 0$ No overlap $\implies y[n] = 0$ **Case 2:** $0 \leq n \leq 3$ $y[n] = \sum_{k=0}^{n} (\frac{1}{2})^{n-k} = (\frac{1}{2})^n \sum_{k=0}^{n} 2^k = (\frac{1}{2})^n \frac{1-2^{n+1}}{1-2} = 2(\frac{1}{2})^n[2^{n+1} - 1] = 2 - (\frac{1}{2})^n$ **Case 3:** $n \geq 4$ $y[n] = \sum_{k=0}^{3} (\frac{1}{2})^{n-k} = (\frac{1}{2})^n \sum_{k=0}^{3} 2^k = (\frac{1}{2})^n \frac{1-2^{4}}{1-2} = (\frac{1}{2})^n [2^4 - 1] = 15(\frac{1}{2})^n$ ### Answer: $ y[n] = \begin{cases} 0, & n < 0 \\ 2-(\frac{1}{2})^n, & 0 \leq n \leq 3 \\ 15(\frac{1}{2})^n, & n \geq 4 \end{cases} $