Consider a dynamical system as bellow that is subjected to an impulse input:

$\sum_{n=0}^N {a_n x^{(n)}}= f_0 \delta(t-t_0)$

the initial values are non-resting (all non-zero):

$x(t_0^-)=x_0\hspace{0.2cm} ;\hspace{0.2cm} x'(t_0^-)=x_1\hspace{0.2cm} ;\hspace{0.2cm} ... \hspace{0.2cm} ;\hspace{0.2cm} x^{(n-1)}(t_0^-)=x_{(n-1)} $.

How can I write the correct initial values for $t=t_0^+$? This is ambiguous since integrating the differential equation above, yields:

$ \sum_{n=1}^N a_n \big[x^{(n-1)} (t_0^+)-x^{(n-1)} (t_0^-)\big] = \sum_{n=1}^N a_n c_{n-1} = f_0$

where $c_i$s are the jumps in initial conditions. In literature, it is said that only the highest order will undergo a discontinuity, yet it's not clear for me why.

  • Welcome to Engineering Stack Exchange. This seems like an assumption mde by the book. If you assume a discontinuity in the highest term, then differentiate, all the differentials will have impulses - i.e. no visible discontinuity. Therefore, this is an easy system to analyze as a first order. Other solutions exist, but this is the easiest to model. – Mark Jun 11 at 22:19
  • How can a finite order continuous model describe a delay? – Arash Jun 21 at 4:45
up vote 0 down vote accepted

I asked the question here and I found the answer myself. This is really a simple one really. Here is my answer for this question:

[...] I revisited this problem with another point of view which gives the answer right away. Since the $x$ is at least $N-1$ times differentiable, it's derivatives (up to order $N-2$) and itself must be continuous. So no shift until the last derivative. Using the my question's notation,

$c_n =0\hspace{1cm} \forall n<N-1$

$c_{N-1} =f_0/a_N$

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