Unit step function

Steps build any piecewise function. A pulse equal to 1 between \(a\) and \(b\) is \(u_a(t)-u_b(t)\), and a function that changes from \(f_1\) to \(f_2\) at \(t=c\) is \(f_1(t)+\bigl(f_2(t)-f_1(t)\bigr)u_c(t)\). The value at \(t=c\) is a convention: the Laplace integral does not see a single point.

Its transform comes from an elementary integral: \(\mathcal L\{u_c\}=\int_c^\infty e^{-st}dt=\dfrac{e^{-cs}}{s}\). More generally, the second shifting theorem says \(\mathcal L\{u_c(t)f(t-c)\}=e^{-cs}F(s)\): delaying the signal by \(c\) multiplies its transform by \(e^{-cs}\). Read backwards, every factor \(e^{-cs}\) in an \(F(s)\) announces a delay when inverting.

Guiding example: for \(y'+y=u_1(t)\) with \(y(0)=0\) we get \(Y(s)=\dfrac{e^{-s}}{s(s+1)}\). Without the exponential the inverse would be \(1-e^{-t}\); with it, the response is \(u_1(t)\bigl(1-e^{-(t-1)}\bigr)\): nothing happens until \(t=1\), and then the solution starts as if time began there.

The typical mistake is applying the theorem when the argument is not shifted: in \(t\,u_2(t)\) you must write \(t=(t-2)+2\) before transforming. Mind the sign too: \(e^{-cs}\) delays, while \(F(s-c)\) (the first theorem) multiplies by \(e^{ct}\).