Unit-step function

Thread Starter

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 :confused:
 

anhnha

Joined Apr 19, 2012
880
I find this question in my signals & systems course exam :confused:
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:

WBahn

Joined Mar 31, 2012
25,756
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.
 

Thread Starter

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 :)
 

Thread Starter

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
25,756
helow.. good day to u guys,.. can u help me to give a circuit for the clock using 74192 counter only..
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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