Dirac Delta Function — Question 6

Question 6 – Solution

Strategy. Post-impulse samples determine a decay rate and an effective amplitude. The area determines the impulse strength independently of its delay.

Step 1: Solve the parameterized problem. The zero datum gives $$Y=\frac{Je^{-as}}{s+\lambda},\qquad \boxed{y=JH_a(t)e^{-\lambda(t-a)}.}$$ The response has jump $J$ at $a$, solves $y'+\lambda y=0$ afterward, and has area $J/\lambda$. Its exact real transform domain is $s>-\lambda$.

Step 2: Identify the point-sample information. Both observations occur after the impulse. Taking their ratio gives $e^{-\lambda}=1/2$, so $\lambda=\ln 2$. The first observation then gives $J2^{a-2}=4$, or $$\boxed{\lambda=\ln 2,\qquad J=16\,2^{-a},\qquad 0<a<2.}$$ This is a full one-parameter family, not a unique impulse time. The strength ranges over $4<J<16$.

Step 3: Use the area and prove admissibility. The area condition now fixes $J/\ln 2=8$. Thus $$\boxed{J=8\ln 2,\qquad a=\frac{\ln(2/\ln 2)}{\ln 2},\qquad \lambda=\ln 2.}$$ Since $1/2<\ln 2<1$, we have $1<2/\ln 2<4$. Taking logarithms to base $2$ shows $0<a<2$. The area fixes $J$ uniquely, and $J=16\,2^{-a}$ is strictly decreasing in $a$, so this recovered triple is unique in the stated family.

Step 4: Verify and interpret. The recovered delay satisfies $2^a=2/\ln 2$. Therefore $$y(2)=(8\ln 2)\frac{2/\ln 2}{4}=4,\qquad y(3)=\tfrac 12y(2)=2,\qquad \int_0^\infty y\,dt=\frac{8\ln 2}{\ln 2}=8.$$ For every member of the two-sample family, all later values are $y(t)=16\,2^{-t}$ for $t\ge 2$. Thus another exact sample after $2$ would either agree with every member or reject every member; it could not separate strength from delay. The total area includes the unsampled interval after the unknown impulse and supplies the missing information.