IVP’s With Step Functions — Question 8

Question 8 – Solution

Strategy. The zero measurement locates the delay. Once the input is on, equally spaced samples obey a simple exponential approach to the same level.

Step 1: Solve the parameterized IVP. For $s>0$, $$Y=\frac{Ae^{-as}}{s(s+2)},\qquad \boxed{y(t)=\frac A2H_a(t)(1-e^{-2(t-a)}).}$$ The response is zero through $a$, positive for $t>a$, and continuous at $a$. Its derivative jumps by $A$, as the equation requires.

Step 2: Locate and recover the switch. Since $A>0$, the first two measurements imply $1\le a<2$. Put $r=e^{-2}$ and $C=A/2$. On each full unit interval after activation, $$y(t+1)=r\,y(t)+C(1-r).$$ Thus $2=r+C(1-r)$, and $$\boxed{A=\frac{2(2-r)}{1-r}}.$$ Using $y(2)=1$ gives $e^{-2(2-a)}=1-1/C=1/(2-r)$, hence $$\boxed{a=2-\tfrac 12\ln(2-e^{-2}).}$$

Step 3: Verify admissibility and uniqueness. Because $1<2-r<2<e^2$, one has $0<\ln(2-r)<2$, so $1<a<2$. Also $C=(2-r)/(1-r)>0$. The recovered values give $y(1)=0$ and $$y(2)=C\left(1-\frac 1{2-r}\right)=1,\qquad y(3)=r+C(1-r)=2.$$ The sample recurrence fixes $C$ uniquely. Then the exponential expression for $y(2)$ is strictly monotone in $a$, fixing the delay uniquely. Direct substitution into the response verifies the IVP on both sides of the switch.

Step 4: Test the extra datum and interpret. The same recurrence predicts $$\boxed{y(4)=2r+C(1-r)=2+e^{-2}<3.}$$ The fourth datum is inconsistent with the stated model. The limiting level is $\boxed{C=(2-e^{-2})/(1-e^{-2})}$, approached from below after activation. Its positive constant tail gives exact real transform domain $s>0$. Uniqueness here concerns the specified single positive step with known system coefficient; finitely many samples do not identify an arbitrary forcing history.