Find the analytical formula of the function (picture included) So we're practicing Laplace transforms and I think I came across a rather unorthodox question. The question is:
Find the analytical formula of the function:

NOTE $h(t)$ is the Heaviside function.
a) $h(t) - h(t-1)$
b) $th(t) - th(t-1)$
c) $th(t) - th(t+1)$
d) $th(t) + th(t-1)$
So, the correct answer in my workbook is $b) th(t) - th(t-1)$
However, I don't understand how I'm supposed to read the equation by looking at the photo. If the term $h(t-1)$ denotes a translation to the right (by one), why is that term negative? Why couldn't it be $ th(t) + th(t-1)$?
Thanks for taking the time to read my question!
 A: You are not supposed to read the equation from the photo. But you are supposed to be able tell which of those four possibilities produces this function.

*

*(a) is easily dismissed. $a\left(\frac 12\right) = 1$ not $\frac 12$ as in the graph.

*(d) is also easily dismissed. $d(2) = 2h(2) + 2h(1) = 4$, not $0$.

*(c) needs just a little more inspection, but not much: $c\left(-\frac 12\right) = -\frac 12\left(h\left(-\frac 12\right) - h\left(\frac 12\right)\right) = \frac 12$, not $0$.

So the answer must be (b). But we can show that (b) does indeed produce this graph:

*

*If $t < 0$, then so is $t-1$, so both $h(t)$ and $h(t-1)$ are $0$. Therefore $b(t) = 0$.

*If $0 < t < 1$, then $t-1 < 0$, so $h(t) = 1, h(t-1) = 0$ and $b(t) = t$, which matches the photo.

*if $1 < t$, then both $t, t-1 > 0$, so $h(t) = h(t-1) = 1$, and $b(t) = t(1-1) = 0$.

A:  Let's call the signal x(.) as defined by Heaviside function h(.).
 Only (b) is correct.

(a) x(.)is a signal of a step function instead of ramp.  Hence it is incorrect.
(b) x(.) is correct. it is a ramp t starting with amplitude 0 at t=0 and ending
with amplitude 1 at t=1.
  During the interval [0 1], the signal is active, x(x) ramp up from 0 to 1.  
  At t=1, the signal becomes inactive or we say it is switched off.  
  x(t)=0 when t>=1.

(c) x(.) is a signal active during the interval [-1 0] with $negative$ ramp
starting from amplitude 0 when t=-1 and reaching amplitude -1 when t=0.
Incorrect.
(d) x(.) is a ramp starting at t=-1 and double its ramp at t = 0 and continue
$forever$.
 Compared with (b), not only it is not ending at t= 0, it ramps at doubling 
 rate 2t and continue forever.
 Incorrect.

