Practice Question on "Signals and Systems"
Topic: Signal Energy and Power
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}&=\lim_{T\rightarrow \infty}\int_{-T}^T |\sin(2 \pi t)|^2 dt \\ &=\lim_{T\rightarrow \infty}\int_{-T}^T \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}&=\lim_{T\rightarrow \infty}\int_{-T}^T \frac{1-\cos(4 \pi t)}{2} dt \\ &=\lim_{T\rightarrow \infty}\int_{-T}^T \frac{1}{2} dt - \int_{-T}^T \frac{\cos(4\pi t)}{2} dt \end{align} $
So $ E_{\infty} = \infty $. So $ E_{\infty} = \infty $.
$ \begin{align} P_{\infty}&=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T |e^{(2jt)}|^2 dt \quad \\ &= \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 - {1 \over {2T}} (-T) \quad \\ & = \lim_{T\rightarrow \infty} {1 \over {2}} + {1 \over {2}} \quad \\ &= 1 \end{align} $
So $ P_{\infty} = 1 $.
