Question

I have a Laplace Transform problem and a solution (Heaviside shifting on the t-axis), but I can't understand or see some of the steps that are taking place in between. Posted below in a moment.

Answer 1

\[L\left\{(t-1)U(t-1)\right\}\]

Answer 2

(I've been reading and reading from my book, but it seems terrible IMO at explaining heaviside or showing the mechanics of it, and I can't figure out how to use it, really; my idea of what to do would be to do the following):

Since\[L\left\{f(t-a)U(t-a)\right\}=e^{-as}F(s),\]

Answer 3

\[L\left\{(t-1)U(t-1)\right\}=e^{-s}\]
That's all-but the book has that all divided by s^2.\[\frac{e^{-s}}{s^2}\]

Answer 4

I can't figure out for the life of me where that s^2 is coming from.

Answer 5

Wait a minute.

Answer 6

Well, its the function "t" being shifted. So all the s^2 part is is L[t} = 1/s^2

Answer 7

I don't understand anything additionally now from your statement. I don't, in a formulaic manner, see how the answer was achieved.

Answer 8

Sorry bout that then. Well, formula wise:
\[L[f(t-a)U(t-a)]\] is saying that you have some function, could be e^t, sint, t^2, anything, shifted along the t-axis by the same amount the unit step function is being shifted. All you're doing is now recognizing what main function is being shifted along with the unit step function. If we ignore the shift by a, we have:
\[tU(t)\]
So F(s) is going to come from the laplace transform of the function t. If I were to give a different example, say \[L \left\{ \sin(t-2)U(t-2) \right\}\] then this would be a shift of the function sint along with the unit step function. So the F(s) part of the formula would come from the laplace transform of sint, and we would have:
\[L \left\{ \sin(t-2)U(t-2) \right\} = \frac{ e^{-2s} }{ s^{2}+1 }\]

Answer 9

Another (perhaps clearer?) way of expressing the formula:
\[\mathcal{L}\left\{f(t-c)u(t-c)\right\}=e^{-cs}\mathcal{L}\left\{f(t)\right\}\]
In your question, you have \(c=1\) and \(f(t)=t\).

I find that deriving some of the formulas for myself give a better intuition of them:
\[\begin{align*}\mathcal{L}\{f(t-c)u(t-c)\}&=\int_0^{\infty} f(t-c)u(t-c)e^{-st}~dt\\\\
&=\int_c^\infty f(t-c)e^{-st}~dt\\
\text{set }p=t-c,~dp=dt\\
&=\int_{c-c}^\infty f(p)e^{-s(p+c)}~dp\\\\
&=e^{-sc}\int_0^\infty f(p)e^{-sp}~dp\\\\
&=e^{-sc}\mathcal{L}\{f(p)\}\\\\
&=e^{-sc}\mathcal{L}\{f(t)\}
\end{align*}\]
where in the last step, \(p\) and \(t\) can be interchanged because they're dummy variables at this point.

Answer 10

I'm going to be asking some elementary questions, but @SithsAndGiggles , I do not understand why in the second expression from the top, the unit step function disappears; I'm guessing it has to do with the change of the lower limit of the integral.

Answer 11

well just for your knowledge
\[u(t)=heavisude~function\]
and graphically
now if u introduce some delay then\[u(t-a)=delayed~heaviside\]
graphically
now if u multiply a "t" with a heaviside function then it will become a "ramp"\[t.u(t)=r(t)=ramp\\(t-a).u(t-a)=delayed~ramp\]
and graphically --