# Homework Help: Laplace transform of unit step function

1. Mar 17, 2012

### NewtonianAlch

1. The problem statement, all variables and given/known data
Could someone please explain this to me? I have read several notes on it, but do not really follow the reasoning:

3. The attempt at a solution

When t = 0, -1/s*e^-st = -1/s because e^0 = 1.

When t goes to infinity is the part I do not fully understand.

Why did they take the absolute value? I'm guessing e^-jst goes to 1 because it's bounded between the - 1 and + 1 range but how did they determine a value of one when cos and sin will always be oscillating?

2. Mar 17, 2012

### Staff: Mentor

My take on it would be that it's not the absolute value but the magnitude of the vector (or phasor) quantity e-st that is being inspected. The variable s can be taken to have real and imaginary components: Rs + j Is. So that

$e^{-st} = e^{-(Rs + j Is)t} = e^{-j Is t} e^{-Rs t}$

The sin and cosine bits that you mentioned are encapsulated in the part with the imaginary exponent: That's the part of the function that is oscillatory, but its magnitude is under control because it's always 1. The negative real exponent makes for a damping term; as t → ∞ its magnitude goes to zero.

3. Mar 17, 2012

### NewtonianAlch

Edit: Re-doing

4. Mar 17, 2012

### NewtonianAlch

OK, I just realised I was doing integration like a moron.

I think I have understood it now better; so basically when t goes to infinity, it renders the expression as 0 because the "damping" term is basically approaching zero due to exponential of an increasingly large negative number which means that whole expression is multiplied by the zero...and for the case when t = 0, it's -$\frac{1}{s}$

And, 0 - (-$\frac{1}{s}$) = $\frac{1}{s}$

Is this right?

One last question. In regards to the oscillation with cos and sin, how do we know that magnitude is 1? I mean I don't understand since it's always going to be changing. It's not always going to be 1 or -1 it will be some value in-between most of the time

5. Mar 17, 2012

### Staff: Mentor

Yes, that looks good.
$$e^{-j\theta} = cos(\theta) - j\;sin(\theta)$$
and its magnitude is....

6. Mar 17, 2012

### NewtonianAlch

Mm, I'm not sure!

I guess it's 1*cos θ - 1*j*sin θ

Which means the magnitude is sqrt((+1)^2 + (-1)^2)) = 1

Is that correct?

7. Mar 17, 2012

### Staff: Mentor

$$\left| cos(\theta) - j\;sin(\theta)\right| = \sqrt{cos(\theta)^2 + sin(\theta)^2} = 1$$