What is the power of unit step signal with magnitude A

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Homework Help Overview

The discussion revolves around finding the power of a unit step signal represented as A u(t), where A is a constant. Participants are exploring the integration process involved in calculating the power of the signal and addressing discrepancies between their calculations and a reference answer.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the integration limits for the power calculation and the implications of the unit step function's definition on these limits. There is an exploration of the integration process and the resulting power expression.

Discussion Status

The discussion is active, with participants clarifying the definition of the unit step function and its effect on the integration limits. Some guidance has been provided regarding the area under the graph of the unit step function, and a participant expresses understanding after re-evaluating the limits of integration.

Contextual Notes

There is a mention of a discrepancy between the calculated power and a book answer, prompting questions about the integration limits and the nature of the unit step function. The variable T is described as a random time value, and the integration limits are adjusted based on the properties of u(t).

jaus tail
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Homework Statement


Find power of signal A u(t)
A is constant like 5, 6 some value.[/B]

Homework Equations


Power is integration over -T/2 to +T/2 of [x(t)]2 dt
The whole divide by T

The Attempt at a Solution


Power is integration over -T/2 to +T/2 of [A u(t)]2dt[/B]
The whole divide by T

So 1/T integral A2 dt...integral limits are -T/2 to +T/2
This comes as 1/T A2 integral dt

integral dt is t. limits are -T/2 to +T/2
so integral dt = [T/2 - ( -T/2) ] which becomes T
so Poewr is 1/T * A2 * T
which gives A2

But book answer is A2 /2
Where did the divide by 2 come from?
 
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u(t)=1 when t>0 only.
integral is area under the graph - sketch the graph and see...
 
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What are ##T## and ##u(t)## in your problem?
 
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u(t) is unit function
u(t) =1 for t > 0 or t =0
u(t) = 0 otherwise.

T is just a random time value. So -T to +T is symmetric about y axis.

Ok now i get it. Since u(t) = 0 for negative time. the integral limit changes from (-T/2 to +T/2) to ( 0 to +T/2).
Got it. Thanks.
 
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