Laplace Transform of time function

Question

solution of the question

question

Hi guys, i am trying to understand the solution of this laplace transform of the time function however i do not understand why two ramp function is used together with a negative a unit ramp, instead of using one ramp function , followed by a unit function and a negative unit ramp

In the solution photo, i have drawn what i thought would be the shape of the graph, if i had used the recommended solution (but i know this is not correct), can someone please advise me how should i derive the correct shape of the from the recommended graph? and also if it is ok to solve it using the 2nd method i mentioned

Answer 1

I feel like your reason of not understanding stems from you forgetting that functions don't just stop at a certain value. The unit step function lets you start from a certain value and have only zeroes before it. But on the other side the function will continue to exist(with probably not zero values), unless another restriction is made. Ultimately the end result is the same. I apologize if my explanation is too bloated in advance.

Try to solve the problem in parts. First from $0$ until $T$ you have the simple ramp function(analogous how $x$ works in the $x-y$ plane)

Then you have from $T$ to $2T$, here your step function still exists and we don't need that but a simple constant function, But to achieve it we need a similar function that we orignially had. We have $\color{red}{t\cdot1(t)}$ and we want it to go back to horizontal, so we need to subtract the ramp function with the same steepness (add the negative of it). But! We need to do this at $T$ not $0$, so we also multiply it by the unit function and say that the unit function is zero until $T$, then it starts. So we subtract $\color{blue}{(t-T)\cdot1(t-T)}$ from our original function, so let $f_a(t)=\color{red}{f_1(t)}-\color{blue}{f_2(t)}$ is $\color{red}{t\cdot1(t)}-\color{blue}{(t-T)\cdot1(t-T)}$

Now your function looks like this:

I recommend you play around with such stuff in Desmos (you can define the step function as $\frac{sgn(x)+1}{2}$) or some other visalising program so that you can get a feel for how these functions really play together.

Now the next step would be to have your constant function(the right side $f_a$), drop a $T$ amount in height and keep going like that until infinity on the time domain. $\color{green}{-T\cdot1(t-2T)}$ allows you to do exactly this, it takes your original/previous function ($f_a(t)$ in our case) and make it drop by $T$ and this drop(the unit step function multiplied by $-1$) should start at $2T$.

Make note that this function is not a nullifier or whatever feeling it gives you. It just transforms the orignial function, for example if you have $e^t$ then it drops the $e^t$ by $T$ at point $2T$

So your final function is $f_{final}(t)=\color{red}{t\cdot1(t)}-\color{blue}{(t-T)\cdot1(t-T)}+\color{green}{-T\cdot1(t-2T)}$

As a final step you only have take the Laplace Transform of it.

I should also highlight the fact that it doesn't matter whether you say:

• $\color{red}{f_1(t)}-\color{blue}{f_2(t)} = \color{red}{t\cdot1(t)}-\color{blue}{(t-T)\cdot1(t-T)}$
  where $f_2(t)=(t-T)\cdot1(t-T)$

or

• $\color{red}{f_1(t)}+\color{blue}{f_2(t)} = \color{red}{t\cdot1(t)}+\color{blue}{-(t-T)\cdot1(t-T)}$
  where $f_2(t)=-(t-T)\cdot1(t-T)$

They are algebraically equivalent, the difference is which one your mind prefers more. Now try to solve the second problem, i'll be sure to answer in the comments.