Difference between revisions of "Signal power energy exercise CT ECE301S18 sin" - Rhea

Latest revision as of 10:09, 22 January 2018

Question

Compute the energy $ E_\infty $ and the power $ P_\infty $ of the following continuous-time signal

$ x(t)= \sin (2 \pi t) $

What properties of the complex magnitude can you use to check your answer?

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Answer 1=

$ \begin{align} E_{\infty}&=\int_{-\infty}^\infty |\sin(2 \pi t)|^2 dt \\ &=\int_{-\infty}^\infty \sin^2(2 \pi t) dt \end{align} $

But $ \cos(2x) = \cos^2(x)-\sin^2(x)=1-2\sin^2(x). $

and therefore $ \sin^2x = \frac{1-\cos(2x)}{2} $.

$ \begin{align} E_{\infty}&=\int_{-\infty}^\infty \frac{1-\cos(4 \pi t)}{2} dt \\ &=\int_{-\infty}^\infty \frac{1}{2} dt - \int_{-\infty}^\infty \frac{\cos(4\pi t)}{2} dt \\ &\\ &=\infty \end{align} $

So $ E_{\infty} = \infty $.

$ \begin{align} P_{\infty}&=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T |\sin(2\pi t)|^2 dt \quad \\ \text{Similar to math above, the expression can be derived towards}\\ &= \lim_{T\rightarrow \infty} {1 \over {2T}} (\int_{-T}^T \frac{1}{2} dt - \int_{-T}^T \frac{1}{2} * \cos(4\pi t) dt) \quad \\ & = \lim_{T\rightarrow \infty} {1 \over {2T}} (\frac{1}{2} t \Big| ^T _{-T} - \frac{1}{8\pi} \int_{-T}^T \cos(4\pi t) d(4\pi t)) \quad \\ & = \lim_{T\rightarrow \infty} {1 \over {2T}} ((\frac{1}{2}T - \frac{1}{2}(-T)) - \frac{1}{8\pi} (\sin(4\pi t)) \Big| ^T _{-T}) \quad \\ & = \lim_{T\rightarrow \infty} {1 \over {2T}} (T - \frac{1}{8\pi} (\sin(4\pi T) - \sin(4\pi T)) \quad \\ &= \lim_{T\rightarrow \infty} {1 \over {2T}} (T) \quad \\ &= \lim_{T\rightarrow \infty} {1 \over {2}} \quad \\ &= \frac{1}{2} \quad \\ \end{align} $

So $ P_{\infty} = \frac{1}{2} $.

Answer 2

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