Homework 2 Solutions

Question 1

a) $ E_\infty = \lim_{T \rightarrow \infty} \int_{-T}^{T} \left|e^{-t}u(t)\right|^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} \left|e^{-t}u(t)\right|^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) $ E_\infty = \lim_{T \rightarrow \infty} \int_{-T}^{T} \left|e^{jt}u(t)\right|^2dt = \lim_{T \rightarrow \infty} \int_{0}^{T} dt = \lim_{T \rightarrow \infty} T = \infty $

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

Since the signal has infinite energy, then we expect that it has average power that is greater than zero.

c) $ E_\infty = \lim_{N \rightarrow \infty} \sum_{n=-N}^{N} \left|\frac{1}{3}u[n]\right|^2 = \lim_{N \rightarrow \infty} \sum_{n=0}^{N} \frac{1}{9} = \lim_{N \rightarrow \infty} \frac{1}{9}(N+1) = \infty $

$ P_\infty = \lim_{N \rightarrow \infty} \frac{1}{2N+1}\sum_{n=-N}^{N} \left|\frac{1}{3}u[n]\right|^2 = \lim_{N \rightarrow \infty} \frac{1}{2N+1} \sum_{n=0}^{N} \frac{1}{9} = \lim_{N \rightarrow \infty} \frac{1}{9} \cdot \frac{N+1}{2N+1} = \frac{1}{9} \cdot \frac{1}{2} = \frac{1}{18} $

Question 2

a) $ x[n+N] = e^{j\frac{3}{5}\pi(n+N-1/2)} = e^{j\frac{3}{5}\pi N} \cdot e^{j\frac{3}{5}\pi(n-1/2)} $

For $ x[n+N] $ to be equal to $ x[n] $, $ e^{j\frac{3}{5}\pi N} $ should be equal to one.

This implies that $ 3\pi N/5 = 2\pi K $, where $ k $ is an integer, or $ N=10k/3 $. Now, the smallest integer N that is not zero is 10. Then the fundamental period of this signal is 10.

b)