Laplace Transform: Answers for cos(4t-1)u(t-1) & cos4(t-1)u(t-1)

In summary, the answer for the Laplace transform of the given questions is e^-s/(s^2 + 4^2). The first question requires the use of the second shift theorem in Laplace transform and the value of 'd' must be the same. For the second question, the identity for cos(A-B) must be used. The answer for the second question is not certain. In order to receive help, gsan must first show their effort.
  • #1
gsan
22
0
what's the answer for laplace transform of the questions below?

1. cos(4t-1)u(t-1)
2. cos4(t-1)u(t-1)

I want to compare the difference between these two answer. Thanks!
 
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  • #2
gsan, you've been told before that we can not help you until you first show us your effort.
 
  • #3
Gokul43201 said:
gsan, you've been told before that we can not help you until you first show us your effort.

e^-s/(s^2 + 4^2) is the answer for i. or ii. ?
 
  • #4
Your answer is the answer for question i. This is because when we refer to the second shift theorem in laplace transform, we can see that the value of 'd' must be the same. In this question, d = 1. Then from there, we can do the laplace transform. For the question ii, I myself not very sure.
Please correct me if I'm wrong. Thank you!
 
  • #5
For part ii you need to use the idenity for cos(A-B)

cos(a - b) = cos(a)cos(b) + sin(a)sin(b)
 

1. What is the Laplace Transform?

The Laplace Transform is a mathematical operation that is used to convert a function of time into a function of complex frequency. It is often used in engineering and physics to solve differential equations and analyze systems.

2. How is the Laplace Transform applied to cos(4t-1)u(t-1) & cos4(t-1)u(t-1)?

The Laplace Transform is applied to these functions by first writing them in terms of complex exponential functions, then using the properties of the Laplace Transform to find the transformed function. In this case, the transformed functions would be expressed as a ratio of polynomials in the complex variable, s.

3. What is the purpose of using the unit step function, u(t-1), in these functions?

The unit step function, u(t-1), represents a function that is equal to 1 for values of t greater than or equal to 1, and 0 otherwise. In these particular functions, it is used to indicate that the functions are only defined for values of t greater than or equal to 1.

4. Can the Laplace Transform be used to solve any type of function?

No, the Laplace Transform can only be used to solve functions that are defined for all real values of t and have a finite number of discontinuities and maxima. Additionally, the function must decay or approach zero as t approaches infinity.

5. Are there any limitations or drawbacks to using the Laplace Transform?

One limitation of the Laplace Transform is that it cannot be applied to functions that have an infinite number of discontinuities or do not decay as t approaches infinity. Additionally, the Laplace Transform can be a complex process and may not always yield a closed-form solution, making it difficult to interpret the results.

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