# Laplace Transform of Heaviside Function

#### _N3WTON_

1. Homework Statement
Determine the Laplace transform of the given function:
$f(t) = sin(t)$ for $0 <= t < \pi$ and $f(t) = 0$ for $\pi <= t$

2. Homework Equations

3. The Attempt at a Solution
Ok, I've been having some trouble figuring out how I should write the above branched function (sorry for the format, I'm not sure how to write branched function in latex). This is what I have the function as right now:
$f(t) = (1-\sin t)u_{\pi}(t)$. I can visualize what the graph of the above function should look like, but I sometimes have trouble actually writing it. I was hoping somebody could confirm whether or not this is correct, once I have the function figured out I suspect the rest of the problem probably won't be too difficult.

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#### Zondrina

Homework Helper
The question is merely asking you to compute $L\{f(t)\}$ for $0 \leq t \leq \pi$.

The reason being, the other part of the integral converges to zero. That is:

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

#### Ray Vickson

Science Advisor
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1. Homework Statement
Determine the Laplace transform of the given function:
$f(t) = sin(t)$ for $0 <= t < \pi$ and $f(t) = 0$ for $\pi <= t$

2. Homework Equations

3. The Attempt at a Solution
Ok, I've been having some trouble figuring out how I should write the above branched function (sorry for the format, I'm not sure how to write branched function in latex). This is what I have the function as right now:
$f(t) = (1-\sin t)u_{\pi}(t)$. I can visualize what the graph of the above function should look like, but I sometimes have trouble actually writing it. I was hoping somebody could confirm whether or not this is correct, once I have the function figured out I suspect the rest of the problem probably won't be too difficult.
By trying to exploit fancy methods involving the Heaviside function, etc., you are only going to make a lot of unnecessary work for yourself. It is much more direct to just use the basic definition of the Laplace transform and apply it to your function.

Anyway, in answer to your question: NO, the formula above is incorrect. You can see that for yourself, just by asking yourself what you get for $0 < t < \pi$ and for $t > \pi$.

#### _N3WTON_

By trying to exploit fancy methods involving the Heaviside function, etc., you are only going to make a lot of unnecessary work for yourself. It is much more direct to just use the basic definition of the Laplace transform and apply it to your function.
.
I understand, but my issue is I am having trouble seeing what the function is, therefore I do not know what I am taking the Laplace transform of. The reason I arrived at my first solution was because the branched function $f(t) = 1$ for t<c and $f(t) = 0$ for t>=c is $1 - u_{c}(t)$ yes? I tried doing something similar with the above problem. I was thinking perhaps $f(t) = sin(t) - u_{\pi}(t)$?

#### Ray Vickson

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I understand, but my issue is I am having trouble seeing what the function is, therefore I do not know what I am taking the Laplace transform of. The reason I arrived at my first solution was because the branched function $f(t) = 1$ for t<c and $f(t) = 0$ for t>=c is $1 - u_{c}(t)$ yes? I tried doing something similar with the above problem. I was thinking perhaps $f(t) = sin(t) - u_{\pi}(t)$?
You said $f(t) = \sin(t)$ if $0 \leq t < \pi$, and $f(t) = 0$ if $t \geq \pi$. I honestly cannot see what possible problem you can have interpreting and plotting that. You are told exactly what is $f(t)$ for $t \in [0,\pi)$ and you are told exactly what is $f(t)$ for $t \geq \pi$.

Of course, when taking the LT we ignore the part $t < 0$; alternatively, you could take $f(t) = 0$ for $t < 0$.

#### _N3WTON_

You said $f(t) = \sin(t)$ if $0 \leq t < \pi$, and $f(t) = 0$ if $t \geq \pi$. I honestly cannot see what possible problem you can have interpreting and plotting that. You are told exactly what is $f(t)$ for $t \in [0,\pi)$ and you are told exactly what is $f(t)$ for $t \geq \pi$.

Of course, when taking the LT we ignore the part $t < 0$; alternatively, you could take $f(t) = 0$ for $t < 0$.
I'm not having a problem interpreting/plotting the function, what I am having a problem with is writing it using the unit step function...

#### _N3WTON_

Edit: I see now how stupid my questions have been, sorry :/ lol

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#### _N3WTON_

Anyhow, here is the solution I have so far:
$\mathcal{L}(f(t))= \int_{0}^{\infty} e^{-st}f(t)\,dt$
$= \int_{0}^{\pi} e^{-st}\sin t\,dt + \int_{\pi}^{\infty} e^{-st}(0)\,dt$
$\frac{e^{-st}}{s^{2}+1}[-s\sin t + \cos t]\bigr|_{0}^{\pi}$
$(\frac{e^{-st}}{s^{2}+1}[-s\sin \pi + \cos \pi]) - (\frac{1}{s^{2}+1}[-s \sin 0 + \cos 0])$
$= \frac{e^{-\pi s}+1}{s^{2}+1}$

Last edited:

#### Ray Vickson

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Anyhow, here is the solution I have so far:
$\mathcal{L}(f(t))= \int_{0}^{\infty} e^{-st}f(t)\,dt$
$= \int_{0}^{\pi} e^{-st}\sin t\,dt + \int_{\pi}^{\infty} e^{-st}(0)\,dt$
$\frac{e^{-st}}{s^{2}+1}[-s\sin t + \cos t]\bigr|_{0}^{\pi}$
$(\frac{e^{-st}}{s^{2}+1}[-s\sin \pi + \cos \pi]) - (\frac{1}{s^{2}+1}[-s \sin 0 + \cos 0])$
$= \frac{e^{-\pi s}+1}{s^{2}+1}$
I have not checked all the details, but it looks OK at a glance.

#### _N3WTON_

I have not checked all the details, but it looks OK at a glance.
great, thanks for the help sorry if I was being difficult earlier, I think I was getting hung up on a topic that didn't really even apply to this specific problem...

#### vela

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I understand, but my issue is I am having trouble seeing what the function is, therefore I do not know what I am taking the Laplace transform of. The reason I arrived at my first solution was because the branched function $f(t) = 1$ for t<c and $f(t) = 0$ for t>=c is $1 - u_{c}(t)$ yes? I tried doing something similar with the above problem. I was thinking perhaps $f(t) = sin(t) - u_{\pi}(t)$?
If you want to do it this way, you want to multiply $\sin t$ by $[1-u_\pi(t)]$. The factor $[1-u_\pi(t)]$ picks off only the first half-cycle of $\sin t$.

#### _N3WTON_

If you want to do it this way, you want to multiply $\sin t$ by $[1-u_\pi(t)]$. The factor $[1-u_\pi(t)]$ picks off only the first half-cycle of $\sin t$.
Thank you, I originally wanted to do it that way just because I wanted some practice writing branched functions in terms of the unit step function..

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