Examples of using Unit step function in English and their translations into Arabic
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Colloquial
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Ecclesiastic
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Computer
And the first one is the unit step function.
We have this unit step function sitting right here.
I started this video talking about the unit step function.
We have the unit step function sitting right there.
So how could Iconstruct this function right here using my unit step function?
Once you hit c, the unit step function becomes 1.
And it doesn't happen exactly like this,but it can be approximated by the unit step function.
That's our unit step function, and we want it to jump to 2.
And so given that,in the last video I showed you that if we have to deal with the unit step function, so if I said.
Until we get to c, the unit step function is zero when it's less than c.
But as soon as we hit t is equal to pi, that pi is the c in this example,as soon as we hit that, the unit step function becomes 1.
So what if I multiply the unit step function times this thing?
Well, this unit step function, when we pass pi, is only going to be equal to 1, but we want this thing to be equal to zero.
In fact, at this point, this unit step function, it has no use anymore.
Remember, this definite integral is reallyjust the area under this curve of this whole function, of the unit step function times all of this stuff.
So our function in this case is the unit step function, u sub c of t times f of t minus c dt.
The unit step function is zero until it gets to 2 times t, or of t, so, then it becomes 1 after t is equal to 2, times our function shifted by 2.
This whole thing, by definition, this unit step function is zero until we get to c.
So if we have this unit step function, this thing is going to zero out this entire integral before we get to c.
We could just takethe integral from t is equal to c to t is equal to infinity of e to the minus st, the unit step function, uc of t times f of t minus c dt.
Well, we could just add another unit step function here, something that would have been zero all along, all the way up until this point.
So let's say that I have the second derivative of my function y plus 4 times my function y is equal to sine of t minus the unit step function 0 up until 2 pi of t times sine of t minus 2 pi.
So this is going tobe minus 1/3 times the unit step function, where c is 2 pi of t times-- instead of sine of t-- sine of t minus 2pi.
And it's defined as 1--that's why we call it the unit step function-- when t is greater than or equal to c.
So what if I had written it as-- so my unit step function's at zero initially, so what if Imake it 2 minus a unit step function that starts at pi?
We showed that the Laplace transform of the unit step function t, and it goes to 1 at some value c times some function that's shifted by c to the right.
And then the other thing I said is that the unit step function, it's going to be 1 over this entire range of potential t-values, so we can just kind of ignore it.
And so the answer to my problem is going to be theunit step function-- I just throw the 2 out there-- the 2 times the unit step function times my f of t shifted by c.
Plus this very last term,which is 1/6 times the unit step function 2 pi of t, the unit step function that steps up at 2 pi times sine of-- and we have to be careful here.
Or we could just rewrite this as 2 times the unit step function starting at 2, where that's when it becomes non-zero of t times f of t minus 2. f of t minus 2 is this with t being replaced by t minus 2.