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+ | [[Category:ECE]] | ||
+ | [[Category:ECE301]] | ||
+ | [[Category:practice problem]] | ||
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=[[Homework_1_ECE301Fall2008mboutin|HW1]], [[ECE301]], Fall 2008, [[user:mboutin|Prof. Boutin]]= | =[[Homework_1_ECE301Fall2008mboutin|HW1]], [[ECE301]], Fall 2008, [[user:mboutin|Prof. Boutin]]= | ||
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<math>P_\infty(x(t)) = \lim_{T \to \infty} (\frac{1}{2T} \int_{-T}^T |x(t)|^2\,dt)</math> | <math>P_\infty(x(t)) = \lim_{T \to \infty} (\frac{1}{2T} \int_{-T}^T |x(t)|^2\,dt)</math> | ||
− | + | ---- | |
− | ==Solution== | + | ==Solution 1== |
*<math>|x(t)| = |\cos(t) + j \sin(t)| = \sqrt{(\cos(t))^2+(\sin(t))^2} = \sqrt{1} = 1</math> | *<math>|x(t)| = |\cos(t) + j \sin(t)| = \sqrt{(\cos(t))^2+(\sin(t))^2} = \sqrt{1} = 1</math> | ||
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<math> = \lim_{T \to \infty} (\frac{1}{2T} (2T)) = \lim_{T \to \infty} (1) = 1 </math> | <math> = \lim_{T \to \infty} (\frac{1}{2T} (2T)) = \lim_{T \to \infty} (1) = 1 </math> | ||
---- | ---- | ||
+ | ==Solution 2== | ||
+ | <math>|x(t)|=|\cos(t)+\jmath\sin(t)|=\sqrt{\cos^2(t)+\sin^2(t)}=1</math> | ||
+ | |||
+ | <math>E\infty=\int_{-\infty}^\infty |x(t)|^2\,dt</math> | ||
+ | |||
+ | <math>E\infty=\int_{-\infty}^\infty |1|^2\,dt=t|_{-\infty}^\infty</math> | ||
+ | <math>E\infty=\infty</math> | ||
+ | |||
+ | <math>P\infty=lim_{T \to \infty} \ 1/(2T)\int_{-T}^{T} |x(t)|^2\,dt</math> | ||
+ | |||
+ | <math>P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int|1|^2dt</math> | ||
+ | <math>P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*t|_{-T}^T</math> <span style="color:green"> (*) </span> | ||
+ | <math>P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*(T-(-T))</math> <span style="color:green"> (*) </span> | ||
+ | <math>P\infty=lim_{T \to \infty} \ 1</math> | ||
+ | <math>P\infty=1</math> | ||
+ | |||
+ | <span style="color:green"> *I would be careful using the start symbol to denote multiplication in this context. The start symbol is typically used for denoting convolution in electrical engineering. </span> | ||
+ | ---- | ||
+ | [[Signal_energy_CT|Back to CT signal energy page]] |
Revision as of 15:13, 25 February 2015
Contents
HW1, ECE301, Fall 2008, Prof. Boutin
Problem
Given complex signal $ f(t)=e^{jt} = \cos(t) + j \sin(t) $, find $ E_\infty $ and $ P_\infty $.
Background Knowledge
$ E_\infty(x(t)) = \int_{-\infty}^\infty |x(t)|^2\,dt $
$ P_\infty(x(t)) = \lim_{T \to \infty} (\frac{1}{2T} \int_{-T}^T |x(t)|^2\,dt) $
Solution 1
- $ |x(t)| = |\cos(t) + j \sin(t)| = \sqrt{(\cos(t))^2+(\sin(t))^2} = \sqrt{1} = 1 $
- $ E_\infty(x(t)) = \int_{-\infty}^\infty |x(t)|^2\,dt = \int_{-\infty}^\infty 1\,dt = t|_{-\infty}^{\infty} = \infty $
- $ P_\infty(x(t)) = \lim_{T \to \infty} (\frac{1}{2T} \int_{-T}^T |x(t)|^2\,dt) = \lim_{T \to \infty} (\frac{1}{2T} \int_{-T}^T 1\,dt) = \lim_{T \to \infty} (\frac{1}{2T} t|_{-T}^T) $
$ = \lim_{T \to \infty} (\frac{1}{2T} (2T)) = \lim_{T \to \infty} (1) = 1 $
Solution 2
$ |x(t)|=|\cos(t)+\jmath\sin(t)|=\sqrt{\cos^2(t)+\sin^2(t)}=1 $
$ E\infty=\int_{-\infty}^\infty |x(t)|^2\,dt $
$ E\infty=\int_{-\infty}^\infty |1|^2\,dt=t|_{-\infty}^\infty $ $ E\infty=\infty $
$ P\infty=lim_{T \to \infty} \ 1/(2T)\int_{-T}^{T} |x(t)|^2\,dt $
$ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int|1|^2dt $ $ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*t|_{-T}^T $ (*) $ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*(T-(-T)) $ (*) $ P\infty=lim_{T \to \infty} \ 1 $ $ P\infty=1 $
*I would be careful using the start symbol to denote multiplication in this context. The start symbol is typically used for denoting convolution in electrical engineering.