I was having a conversation with my brother, a mechanical engineer, about Digital Signals processing. I was trying to explain what I had recently done in my digital controls class, and how we would use the state space model [itex] \vec{x}(k+1) = {\bf{A_d}}\vec{x}(k) + {\bf{B_d}}u(k) [/itex] in discrete time systems in place of derivatives as in [itex] \vec{\dot{x}}(t) = {\bf{A}}\vec{x}(t) + {\bf{B}}u(t) [/itex]. I had multiple ways of explaining it to him. I could have explained that fundamentally, the discrete time solution to a differential equation is not [itex] e^{-{\sigma}Tk} [/itex] but rather [itex] {\gamma}^{kT} [/itex]. And actually this might have been a fair explanation, but I was trying to make it as intuitive as possible.(adsbygoogle = window.adsbygoogle || []).push({});

The problem was that my brother was attempting to approximate a discrete time derivative [itex] \dot{x} [/itex] as [itex] \frac{x[kT] -x[(k-1)T]}{T} [/itex]. I had tried to do this too when I had first entered a DSP courses and forgot how I had made sense of the lack of this approximation technique. At this point, I would personally use laplace and z transforms to come up with s-to-z plane transformation of [itex] s = \frac{1-z^{-1}}{T} [/itex] and come up a discrete time approximation. But explain this to someone without a background in DSP is fruitless and doesn't really even touch upon the heart of the matter. Can someone help me come up with an elegant explanation for why this approximation is not used, or rather, why a whole new solution family is used (e.g., [itex] {\gamma}^{kT} [/itex]).

PS. My brother tends to get confused by any math that may obfuscate the goal to be achieved, or what the math is being used to try to solve. This is understandable, but combining this with my limited communication skills (especially oral communication) makes it difficult to communicate complex engineering concepts. Help...

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# Discrete time Derivative/Integration mechanisms in DSP.

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