# Laplace transform of unit step function

• NewtonianAlch
In summary, when t = 0, the expression becomes -\frac{1}{s} due to the "damping" term. When t goes to infinity, the expression becomes 0.
NewtonianAlch

## Homework Statement

Could someone please explain this to me? I have read several notes on it, but do not really follow the reasoning:

## 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?

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.

gneill said:
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.

Edit: Re-doing

OK, I just realized 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

NewtonianAlch said:
OK, I just realized 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?
Yes, that looks good.
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

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

gneill said:
Yes, that looks good.

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

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?

NewtonianAlch said:
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?

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

That's pretty shocking.

## 1. What is the Laplace transform of a unit step function?

The Laplace transform of a unit step function is 1/s, where s is the complex frequency variable. This means that the Laplace transform of a unit step function is a function of s that can be used to analyze the behavior of a system in the frequency domain.

## 2. How is the Laplace transform of a unit step function derived?

The Laplace transform of a unit step function is derived using the definition of the Laplace transform, which involves integrating the function with respect to time and multiplying by the exponential term e^(-st). The limits of integration are from 0 to infinity, since the unit step function is defined as 0 for t < 0 and 1 for t >= 0.

## 3. What is the significance of the Laplace transform of a unit step function in control systems?

The Laplace transform of a unit step function is a fundamental tool in control systems analysis because it allows us to analyze the behavior of a system in the frequency domain. This is useful for understanding how a system responds to different input signals and for designing controllers to achieve desired system behavior.

## 4. Can the Laplace transform of a unit step function be used to solve differential equations?

Yes, the Laplace transform of a unit step function is often used to solve differential equations. By taking the Laplace transform of both sides of a differential equation, we can convert the differential equation into an algebraic equation, which is often easier to solve. The inverse Laplace transform can then be used to find the solution in the time domain.

## 5. Are there any limitations to using the Laplace transform of a unit step function?

One limitation of using the Laplace transform of a unit step function is that it assumes that the system is linear and time-invariant. This means that the system's response is proportional to the input signal and does not change over time. In real-world systems, this may not always be the case, and other methods may be needed for analysis.

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