Step Functions — Question 2

Question 2 – Solution

Strategy. The factor $e^{-as}$ shifts the time origin of the entire accompanying inverse, not just its domain of activity.

Step 1: Derive the delayed-shape rule. For parameters where the integral converges, substitute $v=t-a$: $$\int_a^\infty e^{-st}g(t-a)\,dt =e^{-as}\int_0^\infty e^{-sv}g(v)\,dv.$$ Hence $\boxed{\mathcal L\{u_a(t)g(t-a)\}=e^{-as}G(s)}$. The effective shape is a function of elapsed time since the switch.

Step 2: Rewrite the unshifted polynomial. For $t\ge 2$, write $t=v+2$. Then $p$ has local shape $(v+2)^2=v^2+4v+4$, whereas $q$ has local shape $v^2$. Therefore $$\boxed{P(s)=e^{-2s}(\frac 2{s^3}+\frac 4{s^2}+\frac 4s),\qquad Q(s)=\frac{2e^{-2s}}{s^3},\quad s>0.}$$ Direct substitution in each defining integral verifies these expressions.

Step 3: Identify the omitted local terms. The difference is $$p-q=[4(t-2)+4]u_2(t),\qquad P-Q=e^{-2s}(4/s^2+4/s).$$ At the switch, $p(2)=4$ while $q(2)=0$, equal to their respective right limits. The switched-on signal jumps from zero to four; the delayed square starts continuously at zero with zero right slope. The student’s expression describes only the latter.

Step 4: Check domains and onset scales. Both signals have nonnegative quadratic tails, so they converge absolutely for $s>0$ and diverge for $s\le 0$. Their exact domain is $\boxed{s>0}$. Removing the delay factor gives $$e^{2s}P(s)\sim 4/s,\qquad e^{2s}Q(s)=2/s^3\quad(s\to\infty).$$ The constant local onset of $p$ produces the $1/s$ term; the quadratic local onset of $q$ produces $2/s^3$. A common switch time does not imply a common transformed shape.