# 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

2. ### anhnha Active Member

Apr 19, 2012
776
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

4. ### anhnha Active Member

Apr 19, 2012
776
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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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

Mar 31, 2012
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