Revision as of 10:22, 22 January 2018 by Chen2156 (Talk | contribs)

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


Practice Question on "Signals and Systems"


More Practice Problems


Topic: Signal Energy and Power


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}} (2T) \quad \\ &= \lim_{T\rightarrow \infty} (1) \quad \\ &= 1 \quad \\ \end{align} $

So $ P_{\infty} = 1 $.



Answer 2


Back to ECE301 Spring 2018 Prof. Boutin

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett