Step Functions — Question 5

Question 5 – Solution

Strategy. Subtracting a delayed copy restarts that copy’s clock. It generally does not switch off the earlier copy.

Step 1: Invert the common shape. Since $s+3=(s+1)+2$, the rational factor is the transform of $$g(v)=e^{-v}(\cos 2v+\sin 2v),\qquad v\ge 0.$$ Thus one inverse with the stipulated endpoint convention is $$\boxed{f(t)=u_2(t)g(t-2)-u_5(t)g(t-5).}$$

Step 2: Inspect the clocks and jumps. The piecewise values are zero before two, $g(t-2)$ on $[2,5)$, and $g(t-2)-g(t-5)$ on $[5,\infty)$. As $g(0)=1$, the jumps are $\boxed{+1\text{ at }2,\ -1\text{ at }5}$. In particular $f(2)=1$ and $f(5)=g(3)-1$, while the left limit at five is $g(3)$.

Step 3: Distinguish subtraction from shutoff. At $t=5$, $|g(3)|\le\sqrt 2e^{-3}<1$, so $g(3)-1\ne 0$. In fact the final piece is a nonzero exponentially decaying sinusoid, not zero. A genuinely truncated signal is instead $$h(t)=[u_2(t)-u_5(t)]g(t-2).$$ Its shutoff term has the same global argument $t-2$. To transform that term at time five, its local shape must be $g(v+3)$, not $g(v)$. The graph shows the actual continuing tail of $f$.

Step 4: Verify the transform and domain. On $t\ge 5$, put $v=t-5$. The tail is $$e^{-v}\{[e^{-3}(\cos 6+\sin 6)-1]\cos 2v +[e^{-3}(\cos 6-\sin 6)-1]\sin 2v\}.$$ The cosine coefficient is nonzero by the preceding bound. Hence absolute convergence holds for $s>-1$; at $s=-1$ a nonzero undamped sinusoid has no convergent primitive, and below it fixed-sign lobe intervals violate the Cauchy criterion. The exact domain is $\boxed{s>-1}$. On that domain each delayed copy converges and transforms to $e^{-2s}G(s)$ or $e^{-5s}G(s)$ by substitution, proving the given expression. The truly truncated $h$ would instead converge for every real $s$.