# Solving Laplace Transform of sin[3t] ( h[t - pi/2] - h[t] )

• splitendz
In summary, Daniel is trying to find the laplace transform of:sin[3t] ( h[t - pi/2] - h[t] ) He realizes that the function is only valid for 0 <= t <= pi/2, so he needs to use the laplace transform definition. He also asks for help with inserting mathematical symbols into his post.
splitendz
Hi. I'm trying to find the Laplace transform of:

sin[3t] ( h[t - pi/2] - h[t] )

I realize that this function is only valid for 0 <= t <= pi/2.

Since the limit is not from zero to infinity (0 to pi/2 instead) how can I use the laplace tranform definition to solve this?

Also, does anyone know how to insert mathematical symbols in your post?

Thanks :)

Yes,use the LaTex code.Type the code using [ tex ] [ /tex ] (without the spacings) and the compiler will do the rest.

That "h" is Heaviside's distribution,right...?

Daniel.

Yes, the h is the heaviside function. The correct answer to the problem is -1/(s^2 + 9) * (3 + se^( (pi * s) / 2 ). I can get -3/(s^2 + 9) but not sure how they get (3 + se^( (pi * s) / 2 ) for the next term.

Do you know where I can get a list of LaTex commands?

Are you trying to compute

$$\mathcal{L}\left\{ \sin (3t) [u_{t - \pi}(t) - u_0(t)] \right\}$$?

The question that I'm asking is:

$$\mathcal{L}\left\{ \sin (3t) [ h(t - \pi/2) - h(t) ] \right\}$$

Heaviside function definition is:

$$h(t - a)=\left\{\begin{array}{cc}0,&\mbox{ if } t< a\\1, & \mbox{ if } t\geq a\end{array}\right.$$

I hope this makes it clearer to understand...

Isn't that $\mathcal{L}\left(-\sin 3t\right)$ with $t\in\left[0,\frac{\pi}{2}\left)$...?

Daniel.

Yes, it is but it's the next step that confuses me.

Laplace transform is defined as

$$\int_{0}^{\infty} e^{-st} f(t) dt$$

Since we have a different limit how do we transform the function $\mathcal{L}\left(-\sin 3t\right)$ for $t\in\left[0,\frac{\pi}{2}\left)$? I've only been learning laplace for 2 days so I'm a bit vauge, sorry if I'm asking a boring question :)

I don't see any problem

$$f(t)=\left\{\begin{array}{c}0 \ \mbox{if} \ t<0\\ -\sin 3t \ \mbox{if} \ t\in\left[0,\frac{\pi}{2}\right) \\ 0 \ \mbox{if} \ t>\frac{\pi}{2} \end{array}\right$$

Daniel.

I understand that part. It's the actual integral that I need to evaluate that I'm not sure about.

$$-\int_{0}^{\pi / 2} e^{-st} \sin 3t dt$$ Can i just do that or must the integration limits be between zero and inifinity to transform the function?

The branch function is defined on R and splitting the integral on R into 3 integrals on each interval,you'll run into that integral.

Daniel.

splitendz said:
Laplace transform is defined as

$$\int_{0}^{\infty} e^{-st}f(t)dt$$

If you really want to integrate from 0 to inf, I think you could rewrite

$$\sin(3t)u(t-\frac{\pi}{2})=-\cos(3(t-\frac{\pi}{2})u(t-\frac{\pi}{2})$$

Then use the time shifting property which says that

$$\mathcal{L}(f(t-t_0))=e^{-st_0}\mathcal{L}(f(t))$$

Now you just have to transform sin(3t) and cos(3t).

You could probably also use the differentiation property and time scaling so you would only have to actually transform sin(t).

I'm afraid it doesn't work,unless that "3" was missing from the argument of cosine.

Daniel.

Oops, my mistake.

Why is that though? It doesn't seem like messing with the time scaling of the step function should matter provided they both turn on at the same time. Isn't u(5t) the same as u(t). Actually, does u(5t) even make any sense?

Yes,i guess you're right.The function should be the same,as i depicted in post #9.

Daniel.

dextercioby said:
When pronounced incorrectly, his name is similar to that of a certain throat lozenge.

## 1. What is the Laplace Transform of sin[3t]?

The Laplace Transform of sin[3t] is (3)/(s^2 + 9).

## 2. What is the function h[t - pi/2] - h[t]?

The function h[t - pi/2] - h[t] is a step function that is equal to 1 for t > pi/2 and equal to 0 for t < pi/2.

## 3. How do you solve the Laplace Transform of sin[3t] ( h[t - pi/2] - h[t] )?

To solve the Laplace Transform of sin[3t] ( h[t - pi/2] - h[t] ), you can use the properties of the Laplace Transform and break it down into smaller parts. First, find the Laplace Transform of sin[3t] which is (3)/(s^2 + 9). Then, use the time-shifting property to shift the function by pi/2 to get the Laplace Transform of sin[3(t - pi/2)] which is (3)/(s^2 + 9) e^(-3s(pi/2)). Finally, subtract the Laplace Transform of sin[3t] from the shifted function to get the final solution of (3)/(s^2 + 9) (1 - e^(-3s(pi/2))).

## 4. What is the significance of using a step function in this problem?

The step function represents a change in the function at a specific point, in this case at t = pi/2. This allows us to break the function into two parts and solve it using the Laplace Transform properties.

## 5. Are there any real-world applications of solving Laplace Transform of sin[3t] ( h[t - pi/2] - h[t] )?

Yes, the Laplace Transform is commonly used in engineering and physics to solve differential equations and analyze systems. In this case, the function could represent a physical system with a change in behavior at t = pi/2, and solving the Laplace Transform can help understand the system's behavior and response.

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