Calculating Laplace Transform of g(t)=x(2t-5)u(2t-5)

In summary, we are trying to determine the Laplace transform of the function g(t) using the given Laplace transform of x(t). The problem involves using time shifting and frequency scaling, and the use of the unit step function. Additionally, the Laplace transform of a function at and t-b can be found by setting up an integral and making a change of variable. The use of the unit step function is necessary in this problem because it helps to determine the Laplace transform of x(2t-5), which cannot be determined solely using the Laplace transform of x(t).
  • #1
mathrocks
106
0
Ok, this is the question:

Assume that the Laplace transform of x(t) is given as X(s)=s / (2s^(2) + 1).
Determine the Laplace transform of the following function.

g(t)=x(2t-5)u(2t-5)

How do I use the transform they have given me to solve this...I guess my major problem lies using time shifting and frequency scaling

Or if you have g(t)=t^2 sin(3t)x(t)...do you ignore x(t) since you usually ignore u(t) when it's at the end of the function?
 
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  • #2
If you know the laplace transform of f(t), you how does the laplace transform of f(at) look? Or that of f(t-b)u(t-b)?

You can answer these questions generally, or you could setup the integral for the laplace transform of g(t) and make a change of variable.
 
  • #3
Galileo said:
If you know the laplace transform of f(t), you how does the laplace transform of f(at) look? Or that of f(t-b)u(t-b)?

You can answer these questions generally, or you could setup the integral for the laplace transform of g(t) and make a change of variable.


What happens to u(2t-5) though? Do you actually take the Laplace transform of that?
 
Last edited:
  • #4
"u(2t-5)"

doesn't this refer to step functions
 
  • #5
The u(2t-5) is necessary, because in the laplace transform of x(t) all information about x(t) for t<0 is lost. Since x(2t-5) is shifted to the right wrt x(t) you can't possibly know the laplace transform of x(2t-5) in terms of that of x(t). If you know x(t)=0 for t<0 then you can find the answer. That's what the unit step function does.
x(t)u(t) is just x(t) but zero for t<0. That's why you can take the lpalce transform of x(2t-5)u(2t-5) in terms of that of x(t).

If this hasn't been covered in class yet, you can just solve the problem the goold old way. Set up the integral for L(g(t))(s) and you'll see it all works out beautifully.
 

What is the Laplace Transform of g(t)=x(2t-5)u(2t-5)?

The Laplace Transform of g(t)=x(2t-5)u(2t-5) is equal to X(s)e^(-5s), where X(s) is the Laplace Transform of x(t).

How do you calculate the Laplace Transform of g(t)=x(2t-5)u(2t-5)?

To calculate the Laplace Transform of g(t)=x(2t-5)u(2t-5), you can use the formula L{g(t)}= ∫[0,∞] g(t)e^(-st) dt, where s is the complex variable.

What does x(t) represent in the Laplace Transform of g(t)=x(2t-5)u(2t-5)?

x(t) represents the input signal in the Laplace Transform of g(t)=x(2t-5)u(2t-5). It is the function that is being transformed into the frequency domain.

What is the significance of the unit step function u(t) in the Laplace Transform of g(t)=x(2t-5)u(2t-5)?

The unit step function u(t) represents a sudden change in the input signal at t=0. In the Laplace Transform, it serves as a multiplier to indicate that the signal is only present after t=0.

Can the Laplace Transform of g(t)=x(2t-5)u(2t-5) be calculated using a table or must it be done by hand?

The Laplace Transform of g(t)=x(2t-5)u(2t-5) can be calculated using a table as long as the initial value and final value of the signal are known. However, it can also be calculated by hand using the formula and integration techniques.

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