HW1, ECE301, Fall 2008, Prof. Boutin

Problem

Given complex signal $ f(t) = \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.

Solution 3

$ E\infty=\int_{-\infty}^\infty |x(t)|^2dt $

$ E\infty=\int_{-\infty}^\infty |\cos(t)+j\sin(t)|^2dt $

$ E\infty=\int_{-\infty}^\infty |e^{j*t}|^2dt $

$ E\infty=\int_{-\infty}^\infty |e^2*e^{j*t}|dt $

$ E\infty=e^2/j*|e^{j*t}|_{-\infty}^{\infty} $

$ E\infty=e^2/j*(\infty-0) $

$ E\infty=\infty $

Compute $ P\infty $

$ x(t)=\cos(t)+j*\sin(t) $

$ P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T |x(t)|^2dt $

$ P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T |\cos(t)+j\sin(t)|^2dt $

$ P\infty=\lim_{T \to \infty}\int_{-T}^T|e^{j*t}|^2dt $

$ P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T|e^2*e^{j*t}|dt $

$ P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T|e^2*e^{j*t}|dt $

$ P\infty=\lim_{T \to \infty}\frac{\int_{-\infty}^\infty |e^2*e^{j*t}|dt}{2*T} $

$ P\infty=\frac{d(\int_{-\infty}^\infty |e^2*e^{j*t}|dt)}{d(2*T)} $

$ P\infty=\frac{e^{2*j*t}}{2} $

$ P\infty=\infty $ This answer is problematic. You are computing a quantity which is always real (not complex) and non-negative. This is because you are integrating the norm of something, and a norm is always real and non-negative. Why do you these "j" coming out of your integrals?

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