Unit-step function

Discussion in 'Homework Help' started by mo2015mo, Oct 27, 2013.

  1. mo2015mo

    Thread Starter Member

    May 9, 2013
    157
    1
    Hi my friends,,


    If we have a rect function as rect(t-2/2) we can expres it like u(t-1)-u(t-3) ,, How we can exprees it as ONLY ONE unit step function :confused:
     
  2. anhnha

    Active Member

    Apr 19, 2012
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    48
    That is impossible according to the definition of unit step function.
     
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  3. mo2015mo

    Thread Starter Member

    May 9, 2013
    157
    1
    I find this question in my signals & systems course exam :confused:
     
  4. anhnha

    Active Member

    Apr 19, 2012
    774
    48
    Sorry, I think my first answer is wrong.
    One of the possible answers is u(t-1) - u(t-3) = u ( - t^2 + 4t - 3).

    u(t-1) - u(t-3) = u(f(t))

    u(t-1) - u(t-3) = 1 if t is in [1 ; 3] and 0 otherwise.
    Therefore, we have to have:

    f(t)> 0 as t is in [1 ; 3] and f(t)< 0 as t not in [1 ; 3]

    => I think there are many functions that satisfy the conditions. Here is one of them.
    f(t) = -(t-1)(t-3) = -t^2 + 4t -3.
     
    Last edited: Oct 27, 2013
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  5. WBahn

    Moderator

    Mar 31, 2012
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    4,800
    As anhnha has already figured out, the key is that the unit step function is a generic function that has a value of 1 if its argument is positive and zero if its argument is negative (and, usually, 0.5 if the argument is zero).

    We tend to think of it in terms of if time is greater than some value and that leads us to think that once it "fires" that it is a 1 from that point on. But that's not the definition. We just need use a function as the argument that happens to be positive at the points in time we want the function to be 1.
     
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  6. mo2015mo

    Thread Starter Member

    May 9, 2013
    157
    1
    WoW thanks for your help Mr. anhnha :)
     
  7. mo2015mo

    Thread Starter Member

    May 9, 2013
    157
    1
    yes thanks Mr. Wbahn :)
     
  8. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    How about starting your own thread instead of hijacking someone else's.

    I'll report your post to the moderators. They will then probably split off your post (and this one) into it's own thread and you will be good to go. So don't go start a new thread at this point. Once it is split, you will need to give us a better idea of just what you are trying to accomplish.
     
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