# Unit-step function

#### mo2015mo

Joined May 9, 2013
157
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

#### anhnha

Joined Apr 19, 2012
885
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
That is impossible according to the definition of unit step function.

#### mo2015mo

Joined May 9, 2013
157
That is impossible according to the definition of unit step function.
I find this question in my signals & systems course exam

#### anhnha

Joined Apr 19, 2012
885
I find this question in my signals & systems course exam
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.

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#### WBahn

Joined Mar 31, 2012
26,398
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.

#### mo2015mo

Joined May 9, 2013
157
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.
WoW thanks for your help Mr. anhnha

#### mo2015mo

Joined May 9, 2013
157
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.
yes thanks Mr. Wbahn

#### WBahn

Joined Mar 31, 2012
26,398
helow.. good day to u guys,.. can u help me to give a circuit for the clock using 74192 counter only..