Dirac Delta Function — Question 10

Question 10 – Solution

Strategy. Each impulse adds $I$ to the state, while the gap between impulses multiplies it by an exponential. The cycle map determines the limiting profile.

Step 1: Sum the transforms and invert. For $s>0$, the impulse series converges geometrically: $$(s+\kappa)Y=I\sum_{n=1}^\infty e^{-nTs} =\frac{Ie^{-Ts}}{1-e^{-Ts}},\qquad \boxed{Y=\frac{Ie^{-Ts}}{(s+\kappa)(1-e^{-Ts})}.}$$ Thus $$y(t)=I\sum_{n=1}^\infty H_{nT}(t)e^{-\kappa(t-nT)}.$$ Only finitely many terms are active at each time. The formula gives zero before $T$, jumps of $I$, and $y'+\kappa y=0$ between impulses.

Step 2: Solve the post-impulse recurrence. Put $q=e^{-\kappa T}\in(0,1)$. Decay followed by a jump gives $$x_n=qx_{n-1}+I,\qquad \boxed{x_n=\frac{I(1-q^n)}{1-q},\quad n\ge 0.}$$ The steady post-impulse level is $P=I/(1-q)$.

Step 3: Construct the profile and compare. For $t=nT+\tau$, $0\le\tau<T$, define $y_{\mathrm{per}}(t)=P e^{-\kappa\tau}$. Then $y(t)=x_n e^{-\kappa\tau}$, including $n=0$, and $$\boxed{y_{\mathrm{per}}(t)-y(t)=P e^{-\kappa t}.}$$ The profile’s peak $P$ is attained just after each impulse. Its pre-impulse trough is the left limit $qP$; under the right-hand convention this infimum is not attained within a cycle. Its mean is $$\frac 1T\int_0^T P e^{-\kappa\tau}\,d\tau=\boxed{\frac I{\kappa T}}.$$ The jump $P-qP=I$ is preserved in the limit, not smoothed away.

Step 4: Design a feasible spacing and interpret. With $I=\kappa=1$, the peak constraint gives $1/(1-e^{-T})\le 2$, equivalent to $T\ge\ln 2$. The mean constraint gives $1/T\ge 1$, equivalent to $T\le 1$. Therefore $$\boxed{\ln 2\le T\le 1.}$$ There is no single long-time limit: samples at $nT$ tend to $P$, while samples at $nT+T/2$ tend to the different value $Pe^{-\kappa T/2}$. Both responses have exact real transform domain $s>0$ because their persistent positive cycles prevent convergence at or below zero. The plot uses $I=\kappa=T=1$ and marks the jump endpoints explicitly.