Extracting equation from a graph and Laplace's delay property

Question

Good day to everyone, I tried to solve this problem but I'm not sure about the solution I chose.

I had this function in Input of my system:

I thought that the input was like this:

$u(t)= 2[H(t-4)-H(t-6)] + (-2t+14)[H(t-6)-H(t-7)] $

The step part between 4 and 6 plus the straight line between 6 and 7

(for the line I used the formula: $y-y_1=[\frac{(y_2-y_1)}{(x_2-x_1)}](x-x_1) $ )

After using Laplace I obtained this result:

$U(s)=2(\frac{e^{-4s}}{s^2} - \frac{e^{-6s}}{s^2}) + (\frac{-2}{s^2} + \frac{14}{s})(\frac{e^{-6s}}{s^2}-\frac{e^{-7s}}{s^2})$

At the end I have the input composites by the sum of this signals:

$U(s)= \frac{2e^{-4s}}{s^2}-\frac{2e{^-6s}}{s^2}-\frac{2e^{-6s}}{s^4}+\frac{2e^{-7s}}{s^4}+\frac{14e^{-6s}}{s^3}-\frac{14e^{-7s}}{s^3} $

Are my conjections correct? I'm not quite sure about the formulation of the equation from the graph.

Answer 1

I found all the steps and now I'm sure I'm right. I'll show you the procedure:

$2H(t-4)-2H(t-6)+(14-2t)(H(t-6)-H(t-7))$ $2H(t-4)-2H(t-6)-2tH(t-6)+2tH(t-7)+14H(t-6)-14H(t-7)$

Now I expand $2tH(t-6)$ so to obtain $2(t-6)H(t-6)+12H(t-6)$ because the term show that the ramp function $2t$ starts from the point $6$ but with height $12$, hence I can't have only $2(t-6)H(t-6)$ because it'd start from $6$ with height $0$, but the step $12H(t-6)$ makes the height at $0$ just $12$. Idem for the other ramp, but where the height at $0$ is $14$.

I can make it more clear with an addendum:
If you have $$t*H(t-3)$$ you need to manipulate it so to have: $$(t-3)*H(t-3)$$ so you can apply the transform. But, to have this, you have to add $-3*H(t-3)$ and obviously also a $+3*H(t-3)$. Finally you obtain: $$$(t-3)*H(t-3)+3*H(t-3)$$ Another example could be: $$(t-2)*H(t-3)$$ that becomes: $$(t-3)*H(t-3)+1*H(t-3)$$

$2H(t-4)-2H(t-6)-2(t-6)H(t-6)-12H(t-6)+2(t-7)H(t-7)+14H(t-7)+14H(t-6)$

So, simplifying the terms with $H(t-6)$:

$2H(t-4)-2(t-6)H(t-6)+2(t-7)H(t-7)$

At this point, I implement the Laplace's transform to obtain the result:

$\frac{2e^{(-4s)}}{s}-\frac{2e^{(-6s)}}{s^2}+\frac{2e^{(-7s)}}{s^2}$