Homework 2 Solutions

Question 1

a) $ E_\infty = \lim_{T \rightarrow \infty} \int_{-T}^{T} |e^{-t}u(t)|^2dt = \lim_{T \rightarrow \infty} \int_{0}^{T} e^{-2t}dt = \lim_{T \rightarrow \infty} -\frac{1}{2}\left[e^{-2T}-e^0\right]=\frac{1}{2} $
$ P_\infty = \lim_{T \rightarrow \infty} \frac{1}{2T} \int_{-T}^{T} |e^{-t}u(t)|^2dt = \lim_{T \rightarrow \infty} \frac{1}{2T} \int_{0}^{T} e^{-2t}dt = \lim_{T \rightarrow \infty} -\frac{1}{4T}\left[e^{-2T}-e^0\right] = \lim_{T \rightarrow \infty} \frac{1-e^{-2T}}{4T}=0 $

Since the signal has finite energy, then we expect that it has zero average power.
b)