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

Revision as of 10:20, 22 January 2018

Question

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

$ x(t)= e^{-2\pi jt}  $

Answer 1=

$ \begin{align} E_{\infty}&=\int_{-\infty}^\infty |e^{-2\pi jt}|^2 dt \\ &=\int_{-\infty}^\infty e^{-2\pi jt} * e^{2\pi jt} dt \\ &=\int_{-\infty}^\infty (1) dt \\ &=\infty \end{align} $

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

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