What is the power of unit step signal with magnitude A

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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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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