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<math> | <math> | ||
\begin{align} | \begin{align} | ||
| − | P_{\infty}&=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T |\sin(2\ | + | P_{\infty}&=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T |\sin(2\pi t)|^2 dt \quad \\ |
| + | |||
| + | But <math class="inline">\cos(2x) = \cos^2(x)-\sin^2(x)=1-2\sin^2(x). </math> | ||
| + | |||
| + | and therefore <math class="inline">\sin^2x = \frac{1-\cos(2x)}{2}</math>. | ||
| + | |||
&= \lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T 1 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 \Big| ^T _{-T} \quad \\ | ||
Revision as of 19:00, 20 January 2018
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?
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
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 \\ But <math class="inline">\cos(2x) = \cos^2(x)-\sin^2(x)=1-2\sin^2(x). $
and therefore $ \sin^2x = \frac{1-\cos(2x)}{2} $.
&= \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} </math>
So $ P_{\infty} = 1 $.
