Laplace Transform of a signal?

In summary, the conversation discusses the steps for solving a problem involving Laplace transforms using transform tables and properties. The speaker explains the process of simplifying the equation and expanding terms, using properties of exponents and the linearity of Laplace transforms. They also offer to answer any further questions.
  • #1
Chandasouk
165
0
Can someone explain the steps of this solution? In my linear systems class, we are doing Laplace transforms using transform tables and the properties. I can usually do the problems that closely resemble the table, but when they involve heavy algebraic manipulations, I don't know what to do.

http://imageshack.us/photo/my-images/593/40703948.jpg/

Why do the +1 and -1 appear? Where did the extra exponential come from? If someone can go line for line, I would appreciate it.
 
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  • #2
[itex][/itex]well, that looks a little messy, but let's do this step by step:

start from
[tex] \mathcal{L} \{ (t-2)^2 e^{-5t} u(t-1) \} = \mathcal{L} \{ (t-1-1)^2 e^{-5t+1-1} u(t-1) \} [/tex]

basically, [itex]t-2= t-1-1[/itex] and [itex] t = t + 0= t +1 -1 [/itex], straightforward enough I hope


then
[tex] \mathcal{L} \{ (t-1-1)^2 e^{-5t+1-1} u(t-1) \} = \mathcal{L} \{ [ (t-1)^2 - 2 (t-1) +1 ] e ^{-5} e^{-5(t-1)} u(t-1) \} [/tex]

the first part: [itex] (t-1-1)^2 = (t-1)^2 - 2 (t-1) +1 [/itex] is basically the same as expanding [itex] (x-1)^2 [/itex] seeing [itex]x=t-1[/itex]

the second part: [itex] e^{-5t+1-1} = e ^{-5} e^{-5(t-1)} [/itex] is a property of exponents/logs [itex]e^a e^b = e^{a+b}[/itex]

The rest is basically expanding the terms,
things you need to know is [itex] e^{-5} [/itex] is just a constant, laplace transform is a linear operation so [itex] \mathcal{L} \{ e^{-5} f(t) \} = e^{-5} \mathcal{L} \{ f(t) \} [/itex]
and [itex] \mathcal{L} \{ f(t-1) \} = e^{-s} \mathcal{L} \{ f(t) \} [/itex]

thats basically it, feel free to ask any other questions you may have :)
 

What is the Laplace Transform of a signal?

The Laplace Transform is a mathematical tool used in engineering and science to convert a function of time into a function of complex frequency. It is commonly used to analyze and solve differential equations, particularly in the field of control systems.

Why is the Laplace Transform useful?

The Laplace Transform allows for the transformation of a time-domain function into a frequency-domain function, making it easier to analyze and solve complex differential equations. It also simplifies the process of finding solutions to differential equations involving initial conditions.

How is the Laplace Transform calculated?

The Laplace Transform is calculated by integrating the function of time multiplied by the exponential function e^-st, where s is a complex variable. This integration can be solved using techniques such as partial fractions or the inverse Laplace Transform table.

What are the properties of the Laplace Transform?

Some of the properties of the Laplace Transform include linearity, time-shifting, scaling, differentiation, integration, and convolution. These properties allow for easier manipulation and analysis of functions using the Laplace Transform.

What are the applications of the Laplace Transform?

The Laplace Transform has various applications in engineering and science, particularly in the fields of control systems, electronics, and signal processing. It is also used in solving differential equations in physics and other areas of mathematics.

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