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I have this differential equation that models a causal LTI system: $$ \ddot{v}(t) - \dot{v}(t) - 2v(t) = \ddot{u}(t) + 2\dot{u}(t) + u(t) $$

I was asked to find the impulse response both by using Laplace transform and by solving the ODE.

The first method is quite simple: set initial conditions to $0$, then apply Laplace transform to the left and right side; this brings to: $$ \begin{align} s^2V(s) - sV(s) - 2V(s) & = s^2U(s) + 2sU(s) + U(s)\\ (s^2 - s - 2)V(s)& = (s^2 + 2s + 1)U(s) \end{align} $$ Knowing that $\mathcal{L}[\delta_0](s) = 1 = U(s)$, we get $$ V(s) = \frac{s^2 + 2s + 1}{s^2 - s - 2} $$ Rewriting as partial fractions: $$ V(s) = 1 + \frac{3}{s - 2} $$ Applying the inverse Laplace transform we get the result: $$ v_{\delta_0}(t) = \delta_0(t) + 3e^{2t}\delta_{-1}(t) = h(t) $$

Trying to solve it the other way leads to a crossroads; more details in a moment. First, we need the characteristic polynomial: $$ P(s) = s^2 - s - 2 $$ Solving for $P(s) = 0$ gives: $$ s_1 = -1\quad s_2 = 2 $$ Therefore, we know $h(t)$ is of the form: $$ h(t) = d_0\delta_0(t) + d_1e^{-t}\delta_{-1}(t) + d_2e^{2t}\delta_{-1}(t) $$ since the ODE output's degree matches the input's. We now have to differentiate $h(t)$ twice and replace it into the ODE itself. The first derivative results: $$ \dot{h}(t) = d_0\dot{\delta_0}(t) + d_1(-e^{-t}\delta_{-1}(t) + e^{-t}\delta_0(t)) + d_2(2e^{2t}\delta_{-1}(t) + e^{2t}\delta_0(t)) $$ knowing that $\dot{\delta}_{-1}(t) = \delta_0(t)$. Now I had two alternatives: the first, to replace $e^{-t}\delta_0(t)$ and similars with $\delta_0(t)$, since $\delta_0(t)$ is non-zero only at $t=0$, and $e^{0} = 1$; the second, to leave it as it is and keep differentiating. I write the two paths:

  1. Not replacing: $$ \begin{align} \ddot{h}(t) & = d_0\ddot{\delta}_0(t) -d_1(-e^{-t}\delta_{-1}(t) + e^{-t}\delta_0(t)) + d_1(-e^{-t}\delta_0(t) + e^{-t}\dot{\delta}_0(t))\;+ \\ &\phantom{=}\;\; d_2(4e^{2t}\delta_{-1}(t) + 2e^{2t}\delta_0(t)) + d_2(2e^{2t}\delta_0(t) + e^{2t}\dot{\delta}_0(t)) \end{align} $$ Combining the $h(t)$s into the ODE: $$ \begin{align} d_0\ddot{\delta}_0(t) + d_1e^{-t}\delta_{-1}(t) -d_1e^{-t}\delta_0(t) - d_1e^{-t}\delta_0(t) + d_1e^{-t}\dot{\delta}_0(t)\; & + \\ 4d_2e^{2t}\delta_{-1}(t) + 2d_2e^{2t}\delta_0(t) + 2d_2e^{2t}\delta_0(t)+ d_2e^{2t}\dot{\delta}_0(t)\; & +\\ - d_0\dot{\delta_0}(t) + d_1e^{-t}\delta_{-1}(t) -d_1e^{-t}\delta_0(t) - 2d_2e^{2t}\delta_{-1}(t) -d_2e^{2t}\delta_0(t)\; & +\\ -2d_0\delta_0(t) -2d_1e^{-t}\delta_{-1}(t) -2d_2e^{2t}\delta_{-1}(t) & = \ddot{\delta}_0(t) + 2\dot{\delta}_0(t) + \delta_0(t) \end{align} $$ Many terms cancel out; we only need to solve this system of linear equations (considering $t = 0\to e^0=1$ if necessary): $$ \begin{cases} d_0 = 1\\ d_1 + d_2 - d_0 = 2\\ 3d_2 - 3d_1 - 2d_0 = 1 \end{cases} $$ which has $(d_0, d_1, d_2) = (2, 1, 1)$ as solution. In conclusion: $$ h(t) = 1\cdot\delta_0(t) + 2\cdot e^{2t}\delta_{-1}(t) + 1\cdot e^{-t}\delta_{-1}(t) = \underline{\delta_0(t) + 2e^{2t}\delta_{-1}(t) + e^{-t}\delta_{-1}(t)} $$ Hmm. Not quite like with the Laplace transform.

  2. Replacing: $$ \begin{align} \dot{h}(t) & = d_0\dot{\delta_0}(t) + d_1(-e^{-t}\delta_{-1}(t) + \delta_0(t)) + d_2(2e^{2t}\delta_{-1}(t) + \delta_0(t))\\ & = d_0\dot{\delta_0}(t) - d_1e^{-t}\delta_{-1}(t) +d_1\delta_0(t) + 2d_2e^{2t}\delta_{-1}(t) + d_2\delta_0(t) \end{align} $$ Now $\ddot{h}(t)$: $$ \begin{align} \ddot{h}(t) & = d_0\ddot{\delta_0}(t) -d_1(-e^{-t}\delta_{-1}(t) + \delta_{0}(t)) + d_1\dot{\delta}_0(t) + 2d_2(2e^{2t}\delta_{-1}(t) + \delta_0(t)) + d_2\dot{\delta}_{0}(t)\\ &= d_0\ddot{\delta_0}(t) + d_1e^{-t}\delta_{-1}(t) - d_1\delta_{0}(t) + d_1\dot{\delta}_0(t) + 4d_2e^{2t}\delta_{-1}(t) + 2\delta_0(t) + d_2\dot{\delta}_{0}(t) \end{align} $$ applying the same substitutions as before. Putting all together: $$ \begin{align} d_0\ddot{\delta_0}(t) + d_1e^{-t}\delta_{-1}(t) - d_1\delta_{0}(t) + d_1\dot{\delta}_0(t) + 4d_2e^{2t}\delta_{-1}(t) + 2d_2\delta_0(t) + d_2\dot{\delta}_{0}(t)\;& + \\ -d_0\dot{\delta_0}(t) + d_1e^{-t}\delta_{-1}(t) - d_1\delta_0(t) - 2d_2e^{2t}\delta_{-1}(t) - d_2\delta_0(t)\;& + \\ -2d_0\delta_0(t) -2d_1e^{-t}\delta_{-1}(t) -2d_2e^{2t}\delta_{-1}(t) & = \ddot{\delta}_0(t) + 2\dot{\delta}_0(t) + \delta_0(t) \end{align} $$ We now need to solve this, after canceling some terms: $$ \begin{cases} d_0 = 1\\ -2d_1 + d_2 - 2d_0 = 1\\ d_1 - d_0 + d_2 = 2 \end{cases} $$ which has this solution: $d_0 = 1, d_1 = 0, d_2 = 3$. Placing the coefficients we found into the original formula: $$ h(t) = 1\cdot\delta_0(t) + 0\cdot e^{-t}\delta_{-1}(t) + 3\cdot e^{2t}\delta_{-1}(t) = \underline{\delta_0(t) + 3e^{2t}\delta_{-1}(t)} $$ which is the exact solution we found with Laplace transform.

After this brief preamble, the point is that every such exercise in my workbook has the solution obtained by the first method (without substituting terms), even though it is not the same as the one found with the Laplace transform. Sometimes both solutions are reported (also when they are clearly different, instead). So, my questions are: can these solutions be considered equivalent? Isn't it wrong to perform the substitution by removing part of the signal derivatives? Why does the first method produce a different solution?

Thanks in advance to everyone who will help me out.

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  • $\begingroup$ Paul Dawkins's notes on solving differential equations (including second order ones) are great! $\endgroup$
    – Peter K.
    Apr 18 at 19:00
  • $\begingroup$ Thanks for the resource, but I have no problem solving them! As you can see, my concern was about why not simplifying the terms leads (apparently) to a wrong solution, compared to the one obtained with Laplace transform $\endgroup$
    – Lorenzo
    Apr 18 at 19:06

1 Answer 1

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The problem with your first approach is that you assume

$$f(t)\delta'(t)\stackrel{?}{=}f(0)\delta'(t)\tag{1}$$

which is wrong.

The correct equation is

$$f(t)\delta'(t)=f(0)\delta'(t)-f'(0)\delta(t)\tag{2}$$

Equation $(2)$ is easily derived as follows:

$$\big(f(t)\delta(t)\big)' = f'(t)\delta(t)+f(t)\delta'(t) = f(0)\delta'(t)$$

from which $(2)$ follows.

Using $(2)$ instead of $(1)$ in your calculations will give you the same result as with the other method.

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  • $\begingroup$ Thanks, it works! I see everything except why you set $\big(f(t)\delta(t)\big)'$ equals to $f(0)\delta'(t)$ $\endgroup$
    – Lorenzo
    Apr 18 at 20:30
  • $\begingroup$ @Lorenzo: Because $f(t)\delta(t)=f(0)\delta(t)$. $\endgroup$
    – Matt L.
    Apr 18 at 20:53
  • $\begingroup$ Now it's clear, thank you again $\endgroup$
    – Lorenzo
    Apr 18 at 20:56
  • $\begingroup$ @Lorenzo MattL provides a nice catch of your mistake, however, you should better use a quite simpler method of splitting the DE into a cascade of IIR and FIR parts (only for LTI systems), and solving them for the two stage impulse response. That will be algebraically quite neater than solving the total equation at once. $\endgroup$
    – Fat32
    Apr 18 at 23:53
  • $\begingroup$ @Fat32 thanks for the suggestion. However, this is the way we were told to solve this kind of ODEs and requested for the exam. (P.S: Actually, my original solution was the one with the substitution in advance; the wrong one was reported in workbook's solutions... It triggered me the fact they were not the same) $\endgroup$
    – Lorenzo
    Apr 19 at 4:24

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