Integration limits for unit step

ColdStart
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hello, if i have an integral like: Int( [-u(-t) + u(-t-1)] * a * dt, -Inf, +Inf)... where a is some constant.. what would be the correct initegration limits?

the most important here is to determine correctly the integration limits.. if i solve the inequalities:
-t > 0 => t < 0
-t-1 > 0 => t < -1

now get a lil bit confused... it goes from -Inf to -1? i think I am missing something?

thanks
 
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Yes, you are missing something.
Try writing your functions as [ u( -(t+1)) - u(-t) ]*a [1]

Draw a little sketch of the unit step function part. It should be clear that before t =-1 [1] is zero and after t=0 the function is zero and in between the function is 1. After this, choose your integration limits appropriately. :-)
 
Ok, let me c if i got it correct, rewriting:

u(-(t+1)) - u(-t), I am attaching picture.. the first part of this expression is shown with black, the second part which is u(-t) is shown in red..

now its clear that integral limits are from 0 to t+1?

But.. however tho.. my original problem has exp(-j*w*t) inside the inegral... i put 'a' as constant just for simplification... but now, since my limit is t+1.. it turns out that i have to change variable? but that looks weird stil... because i don't think i was supposed to arrive to this..
 

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Your picture in not correct.
u(-(t+1)) = 1 when t<= -1
u(-t) = 1 whe t<=0

When you substract u(-t) from u(-(t+1)) you should get zero everywhere except when -1<= t <=0 .
 
now i got it... thanks!
 
just one more thing regarding this problem... now that we know my integration limits are froom -1 to 0... should i place '-' sign inside the integral before constant 'a'? because the unit step is negative? or not?

i.e. should my integral be Int(-1,0 of: -a * dt) ?
 
Why do you think it is negative?
 
ok because of this:
i attached pic.
 

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Good!
My work here is done; you are all set. :-)
 
  • #10
lol what u mean? i stil didnt get something? i'd be happy if u point out what else am i missing? :)

stil have question... should i have Minus sign in front of 'a' in the integral then?
 
  • #11
Nothing more. :-)

Your integral should be int of (-1 to 0) -a dt.

I guess I should be more clear. I wanted you to be sure of yourself and not doubt your own correct deductions.
 
  • #12
thank you very much :)
 
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