(A)

We first change the form of y[n] that we are given by separating the exponentials and converting the complex exponential using Euler's method.

\begin{align} y[n] &= e^{2 + j4.72135n} = e^{2}e^{j4.72135n}\\ &= e^{2}[cos(4.72135n) + jsin(4.72135n)]\\ \end{align}


Next, we use the equation for energy in discrete time. Finding the magnitude of y[n], we see that cos^2 + sin^2 = 1 Therefore the energy for infinite interval is the infinite sum of $e^{4}$ which = $\infty$

\begin{align} \\ E\infty &= \Sigma_{n=-\infty}^{\infty}|y[n]|^2\\ &= \Sigma_{n=-\infty}^{\infty}\sqrt{(e^{2})^2[cos^2(4.72135n) + jsin^2(4.72135n)]}^2\\ &= \Sigma_{n=-\infty}^{\infty}\sqrt{e^4}^2 = \Sigma_{n=-\infty}^{\infty} e^{4} = \infty \end{align}


(B)

We start with part B by noticing that the integral of the delta function is a step function. So the energy over an infinite interval is just the integral of the step function $u(t + 2) - u(t - 2)$

\begin{align} y(t) &= \int_{-\infty}^{t} \delta(t + 2) - \delta(t - 2) dt = u(t + 2) - u(t - 2) \\ so \\ E\infty &= \int_{-\infty}^{\infty} |[u(t + 2) - u(t - 2)]|^2 dt\\ &= \int_{-2}^{2} |1|^2 dt \\ &= 2 - (-2) = 4 \end{align}


## Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras. 