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You transfer function models the behavior of a system. That may be a mechanical system, an electric system, etc. In your case you have a second order low pass filter. If you calculate the step response of your system, you will notice that the system cannot follow the step. It will either grow fast, with a high slope and then start to decrease its slope ...


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Model your deterministic input as a $2 \times 1$ matrix : $$ \begin{bmatrix} u \\ T_a \\ \end{bmatrix} $$ Then your state space equations will be: $$ \begin{bmatrix} \dot x_1 \\ \dot x_2 \\ \dot x_3 \end{bmatrix} = \begin{bmatrix} -\frac 13p & \frac 13p & 0 \\ p & -(p+ \frac{c}{10}) & \frac{c}{10} \\ 0 & \frac{c}{30} & -(\...


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This is kind of hand-wavy, but you can look at this from two different perspectives: One, you can look at $z^{-1}$ as a "back-step" operator; i.e. if $X(z) = \mathcal{Z}\lbrace x_n \rbrace$, then (with a few 'i's left undotted and 't's uncrossed) $\frac{X(z)}{z} = \mathcal{Z}\lbrace x_{n-1} \rbrace$. You can also look at $s$ as a derivative operator: if $X(...


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"integrator is unstable since it has a pole on the unit circle" -- not true. This Z-1 is done all the time in CIC resamplers. The pole is --on-- the unit circle exactly, and is therefore stable. Fun fact, if you use floating point math this falls apart and the integrator blows up bc the value is not exactly 1, its 1.000000000001 (or something larger than 1)


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