Laplace transform Heaviside step function

In summary, the Laplace transform of the given piecewise function is ((2/s)-(1/s^2))e^(-2s) + ((2/s^2)-(1/s))e^(-3s) + ((-1/s^2)e^(-4s)) + ((1/s^2)e^(-5s)).
  • #1
Pietair
59
0

Homework Statement


What is the Laplace transform of f(t) = 0 for 0 < t < 2 and f(t) = (4-t) for 2 < t < 3 and f(t) = 1 for 3 < t < 4 and f(t) = (5-t) for 4 < t < 5 and f(t) = 0 for t > 5?

Homework Equations


7410747ec7563eab51f608f2c80a9497.png


The Attempt at a Solution


f(t) = H(t-2)(4-t) - H(t-3)(4-t) + H(t-3) - H(t-4) + H(t-4)(5-t) - H(t-5)(5-t)
f(t) = H(t-2)(4-t) + H(t-3)(2t-7) - H(t-4)(t-4) + H(t-5)(t-5)

How can I rewrite H(t-2)(4-t) + H(t-3)(2t-7) to a form suitable for Laplace transform?
 
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  • #2
For [itex]H(t-2)(4-t)[/itex], you'd like to see a [itex]t-2[/itex] in the second factor right? So just put one there, and balance it out elsewhere in the expression. Like this:

[itex]H(t-2)(4-(t-2)-2)[/itex]

See how I added and subtracted 2? This simplifies to:

[itex]H(t-2)(2-(t-2))[/itex]

Do the same for [itex]H(t-3)(2t-7)[/itex].
 
  • #3
Thank you for your reply!

Now I am getting:
[itex]
H(t-2)(2-(t-2)) + H(t-3)(2(t-3)-1) - H(t-4)(t-4) + H(t-5)(t-5)
[/itex]

And the Laplace transform:
[itex]
((2/s)-(1/s^2))e^(-2s) + ((2/s^2)-(1/s))e^(-3s) + ((-1/s^2)e^(-4s)) + ((1/s^2)e^(-5s))
[/itex]

Is this correct?
 
  • #4
Looks good to me.
 
  • #5
Thanks a lot!
 

1. What is the Laplace transform of the Heaviside step function?

The Laplace transform of the Heaviside step function, also known as the unit step function, is equal to 1/s, where s is the complex variable in the Laplace transform integral.

2. Why is the Laplace transform of the Heaviside step function useful in engineering and physics?

The Laplace transform of the Heaviside step function is useful because it allows us to transform a function from the time domain to the frequency domain. This can be used to analyze and solve differential equations in engineering and physics.

3. Can the Laplace transform of the Heaviside step function be used to solve initial value problems?

Yes, the Laplace transform of the Heaviside step function can be used to solve initial value problems. It can be used to find the solution of a differential equation with given initial conditions.

4. How does the Laplace transform of the Heaviside step function relate to the Dirac delta function?

The Laplace transform of the Heaviside step function is closely related to the Dirac delta function. In fact, the derivative of the Heaviside step function is equal to the Dirac delta function. This relationship is important in solving differential equations using Laplace transforms.

5. Is there an inverse Laplace transform for the Heaviside step function?

Yes, there is an inverse Laplace transform for the Heaviside step function. It is given by the integral of F(s)/s, where F(s) is the Laplace transform of the function. This can be used to find the original function in the time domain.

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