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$$H(s) = \frac{R C s}{R C s + 1}$$ If $R C s$ is much lower than 1 (i.e., $R C s \ll 1$, in Math), then you can make an approximation: $$H(s) \simeq \frac{R C s}{1}$$ Basically, the effect of the $R C s$ in the denominator becomes insignificant. So, just pick a radian frequency well below $1 / R C$ and compute the magnitude of $H(s)$ at that point -- mark ...

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Another way to see how the forward Euler method approximates a continuous-time system is by considering the "ideal" mapping of the $s$-plane to the $z$-plane (why?): $$z=e^{sT}\tag{1}$$ For frequencies that are much smaller than the sampling frequency (i.e., $|s|T\ll 1$) we can approximate $e^{sT}$ by its first order Taylor series: $$z\approx 1+sT\tag{2}$$...

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To the extent you can factor the transfer function into individual integrator sections of the general form $\frac{1}{s}$ you can make this substitution, which is an approximation of the Matched-$z$ Transform where you substitute every $s$ for $s=\frac{\ln(z)}{T}$. (map from $s$ to $z$ using $z =e^{sT}$). This results in first order forms given by $H_\mathrm{... 1 You can't implement the transfer function H(z) directly, you need to convert it to a difference equation. However, the process is trivial, so once you understand it you'll see the connection between the diagram and transfer function better. First, we need to unroll the summation. For example, we get this with m=2, for a second-order equation:$H(z) = \...

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It would help to apply some specificity to the general equation: $H(z) = \frac{b_0 + b_1z^{-1}+b_2z^{-2}}{1+a_1z^{-1}+a_2z^{-2}}$ Rearrange into a difference equation, skipping the steps for brevity: $y[n] = b_0x[n] +b_1x[n-1]+b_2x[n-1]-a_1y[n-1]-a_2y[n-2]$ Does the difference equation form show now that the Direct Form structure is derived purely by ...

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Assuming that $$H(z) = A\frac{\prod_k (1-c_kz^{-1})}{\prod_l (1-d_l z^{-1})}, \: \: R_H$$ you can perform Partial Fraction Expansion (PFE) to quickly get your impulse response $h[n]$ (what you probably call anti-transformation) using Z-transform properties and tables of Z-transform pairs. If your transfer function is not rational, such as H(z) = \mathrm{...

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First of all - I myself am not a pro in Control theory, but a mathematician - so I write what I think might be what your prof means. Part where i am quite certain: Not to be polynomial means, that there exists NO polynomial which describes the transfer function. E.g. let $H$ be your transfer function, then there exists no $n\in\mathbb{N}$ and $p\in P^n[X]$ ...

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See the image (found in this article). Inject a signal into the loop somewhere (I show it below as going between the controller and the plant). Then measure the input to the plant and the output from the plant. Note that the image is a bit confusing -- it assumes that the normal command to the loop is 0 or some constant. It would be better to actually ...

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