That is impossible according to the definition of unit step function.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
I find this question in my signals & systems course examThat is impossible according to the definition of unit step function.
Sorry, I think my first answer is wrong.I find this question in my signals & systems course exam
WoW thanks for your help Mr. anhnhaSorry, 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.
yes thanks Mr. WbahnAs 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.
How about starting your own thread instead of hijacking someone else's.helow.. good day to u guys,.. can u help me to give a circuit for the clock using 74192 counter only..
by Jake Hertz
by Jake Hertz
by Jake Hertz