Difference between revisions of "P infinity 2j" - Rhea

Revision as of 10:29, 21 April 2015

Question

Compute the power of the signal $ x(t)= 2j $

Share your answers below (optional)

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!

Answer 1=

$ P_{\infty}=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T | x(t) |^2 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T 4 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} 4t \Big| ^T _{-T} = \lim_{T\rightarrow \infty} {1 \over {2T}} 8T = 4 $

• looks good!

Answer 2

$ P_{\infty}=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-\infty}^\infty| x(t) |^2 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-\infty}^\infty 4 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} \infty= \frac{\infty}{\infty}=1 $

• You made a limit manipulation error.

Answer 3=

$ P_{\infty}=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T (2j)^2 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T -4 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} -4t \Big| ^T _{-T} = \lim_{T\rightarrow \infty} {1 \over {2T}} -8T = -4 $

• Power cannot be negative, so your answer cannot be correct.

Answer 4=

$ |x|^2=4 \quad 4 \frac{\infty}{\infty}=4 $

• Sorry but I don't understand your explanation.

Back to Signal Power