(New page: = Homework 2 Solutions = == Question 1 == a) <math class = "inline"> E_\infty = \lim_{T \rightarrow \infty} \int_{-T}^{T} |e^{-t}u(t)|^2dt = \lim_{T \rightarrow \infty} \int_{0}^{T} e^{-...) |
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== Question 1 == | == Question 1 == | ||
| − | a) <math class = "inline"> | + | a) |
| − | E_\infty = \lim_{T \rightarrow \infty} \int_{-T}^{T} |e^{-t}u(t)|^2dt | + | <math class = "inline"> |
| + | 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} \int_{0}^{T} e^{-2t}dt | ||
= \lim_{T \rightarrow \infty} -\frac{1}{2}\left[e^{-2T}-e^0\right]=\frac{1}{2} | = \lim_{T \rightarrow \infty} -\frac{1}{2}\left[e^{-2T}-e^0\right]=\frac{1}{2} | ||
| − | </math><br> | + | </math><br><br> |
| − | <math class = "inline"> | + | <math class = "inline"> |
| − | P_\infty = \lim_{T \rightarrow \infty} \frac{1}{2T} \int_{-T}^{T} |e^{-t}u(t)|^2dt | + | 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}{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}{4T}\left[e^{-2T}-e^0\right] | ||
= \lim_{T \rightarrow \infty} \frac{1-e^{-2T}}{4T}=0 | = \lim_{T \rightarrow \infty} \frac{1-e^{-2T}}{4T}=0 | ||
| − | </math> | + | </math><br><br> |
| + | |||
| + | Since the signal has '''finite energy''', then we expect that it has '''zero average power'''.<br><br> | ||
| + | b) | ||
| + | <math class = "inline"> | ||
| + | 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 | ||
| + | </math><br><br> | ||
| + | <math class = "inline"> | ||
| + | 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} | ||
| + | </math><br><br> | ||
| + | Since the signal has '''infinite energy''', then we expect that it has '''average power that is greater than zero'''.<br><br> | ||
| + | c) | ||
| + | <math class = "inline"> | ||
| + | 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 | ||
| + | </math><br><br> | ||
| + | <math class = "inline"> | ||
| + | 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} | ||
| + | </math><br><br> | ||
| + | |||
| + | == Question 2 == | ||
| + | a) | ||
| + | <math class = "inline"> | ||
| + | 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)} | ||
| + | </math><br><br> | ||
| + | For <math>x[n+N]</math> to be equal to <math>x[n]</math>, <math class="inline">e^{j\frac{3}{5}\pi N}</math> should be equal to one.<br><br> | ||
| + | This implies that <math class="inline">3\pi N/5 = 2\pi K</math>, where <math>k</math> is an integer, or <math> N=10k/3</math>. Now, the smallest integer N that is not zero is 10. Then the fundamental period of this signal is 10.<br><br> | ||
| − | |||
b) | b) | ||
Revision as of 06:18, 2 February 2011
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)
