Examples of using The unit step function in English and their translations into Hebrew
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Colloquial
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Ecclesiastic
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Computer
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Programming
We have the unit step function sitting right there.
I started this video talking about the unit step function.
Until we get to c, the unit step function is zero when it's less than c.
And we will have to shift it and multiply it by the 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.
So what if I-- my new function, I call it the unit step function up until c of t times f of t minus 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.
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.
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.
Let's say we wanted to figure out the Laplace transform of the unit step function that starts up at pi of t.
We were taking the Laplace transform of the unit step function that goes up to c, and then it's 0 up to c, and it's 1 after that, of t times some shifted function f of t minus c.
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.
We could just take the integral from t is equal to c to t is equalto infinity of e to the minus st, the unit step function, uc of t times f of t minus c dt.
So this is going to be 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.
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.
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.
So the Laplace transform of the unit step function that goes up to c times some function shifted by c is equal to e to the minus cs times the Laplace transform of just the original function times the Laplace transform of f of t.
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.
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.
I showed you that if you have the Laplace transform of the unit step function of t times some f of t shifted by some value of c, then that this is equal to e to the minus cs times F of s.
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.
And so the answer to myproblem is going to be the unit step function-- I just throw the 2 out there-- the 2 times the unit step function times my f of t shifted by c.
So we can now show that the Laplace transform of the unit step function times some function t minus c is equal to this function right here, e to the minus sc, where this c is the same as this c right here.
We have this unit step function sitting right here.
In fact, at this point, this unit step function, it has no use anymore.
This whole thing, by definition, this unit step function is zero until we get to c.
That's our unit step function, and we want it to jump to 2. This would just jump to 1 by itself.
So if we have this unit step function, this thing is going to zero out this entire integral before we get to c.