I wanted to practice Impulse balancing and tried using it to solve this IAR system in one go as opposed to using its LTI attributes. The exercise wants me to find the impulse response of the system: $$y’’(t) +4y’(t)+3y(t) =x’(t) +2x(t)$$ $$h’’(t)+4h’(t)+3h(t) = \delta’(t)+2\delta(t)$$ I just need to get starting conditions. so we can integrate everything from $0^- \to 0^+ $. $$0+4h(0^+)\underbrace{-4h(0^-)}_{0: IAR} +0 = 0 +2 \implies h(0^+) = 0.5 $$ (I think the derivative of delta’s integral is 0 because of how the derivative is defined for generalized functions if we take $\phi(t)\equiv 1$)
If we integrate our original system from \$ -\infty \to t\$ we get: $$h’(t) + 4h(t) + 3\int_{-\infty}^t h(\tau) d\tau = \delta(t) + 2u(t)$$ which we can then integrate from $ 0^- \to 0^+$ and get: $$(h(0^+) -0)+0+0=1+0 \implies h(0^+) = 1 $$ I have 2 conflicting inital conditions :(. we know the solution needs to be of the form $$h(t>0) = C_1e^{-3t}+C_2e^{-t}$$ And from the initial conditions we get a system. Solving the systems leads to different coefficients than the official solution which uses the alternative method. The solution is $C_1=C_2=0.5$ Can you please tell me where I went wrong?


1 Answer 1


The mistake lies in the fact that you assumed $h'(0^+)=0$, which is not the case because $h''(t)$ contains a Dirac delta impulse. The result of the first integration should be


Your second equation


is correct. Combining $(1)$ and $(2)$ gives $h'(0^+)=-2$.

For the constants $C_1$ and $C_2$ that means

$$C_1+C_2=1\quad\textrm{and}\quad -3C_1-C_2=-2$$

which yields the correct solution $C_1=C_2=\frac12$.

Also take a look at this related question and its answer.


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