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You are correct. If the region of convergence of a right-sided signal (like the one you have) does not include infinity, then the signal is not causal.

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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 $H_\mathrm{... 2 Another way to see how the forward Euler method approximates a continuous-time system is by considering the "ideal" mapping of the$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: $$z\approx 1+sT\tag{2}$$... 0 Multiplication with$s$in the Laplace transform domain equals differentiation in the time domain. In the discrete-time domain we can approximate differentiation by the equation $$y[n]=\frac{x[n+1]-x[n]}{T}\tag{1}$$ where$T$is the sampling interval. In the Z-transform domain, Eq.$(1)$becomes $$Y(z)=X(z)\frac{z-1}{T}\tag{2}$$ I.e., the transfer ... 1 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 $$H(z) = \mathrm{... 1 First of all - I myself am not a pro in Control theory, but a mathematician - so I write what I think might be what your prof means. Part where i am quite certain: Not to be polynomial means, that there exists NO polynomial which describes the transfer function. E.g. let H be your transfer function, then there exists no n\in\mathbb{N} and p\in P^n[X] ... 0 You are right. In the solution, the second line which performs the long division$$ H(z) = -4 + \frac{ 5 + \frac{7}{2} z^{-1} }{1 - \frac{3}{4}z^{-1} + \frac{1}{8} z^{-1} } $$is wrong and should be corrected as:$$ H(z) = -4 + \frac{ 5 - 3 z^{-1} }{1 - \frac{3}{4}z^{-1} + \frac{1}{8} z^{-1} } .$$However, the partial fraction expansion at the following ... 2 You just wrote down the \mathcal{Z}-transform of \left(\frac12\right)^nu[n], but you need the \mathcal{Z}-transform of n\left(\frac12\right)^nu[n] (note the factor n). In order to find that \mathcal{Z}-transform you can use the differentiation property (see this table):$$\mathcal{Z}\big\{nx[n]\big\}=-z\frac{dX(z)}{dz}\tag{1}$$2 Note that the order of your numerator equals the denominator, first perform a long division to simplify, before performing partial fraction expansion:$$ H(z) = \frac{ 1 + 1/6 z^{-1}}{1 - 0.25 z^{-1}} = -\frac{2}{3} + \frac{5/3}{1 - 0.25 z^{-1}} $$(Note: long divison already simplified the expression that it does not require a further partial fraction ... 1 You need to understand that a function such as F(z)=\frac{z}{z+1} doesn't have a region of convergence (ROC). Only an infinite series has a ROC. E.g., the \mathcal{Z}-transform of a sequence x[n] can be expressed by the series$$\sum_{n=-\infty}^{\infty}x[n]z^{-n}\tag{1}$$and this expression only makes sense if the series converges for certain ... 0 A difference equation simultaneously characterises a system and enables the practical computation of its output y[n] for a given input x[n] and stated initial conditions. An LCCDE (MattL Eq(1)) is a special form of a difference equation with constant coefficients; they define LTI systems by providing their impulse response h[n] as a solution to x[n] =... 0 An Nth-order linear constant-coefficient difference equation (LCCDE) is of the form$$y[n]=\sum_{k=0}^{M}b_kx[n-k]-\sum_{k=1}^{N}a_ky[n-k]\tag{1}$$It is linear because the sequences x[n] and y[n] appear linearly in (1), and it has constant coefficients because the coefficients a_k and b_k do not depend on the index n. LCCDEs are important ... 1 You need to determine the values of the complex variable z for which the series$$\sum_{n=1}^{\infty}n^2z^{-n}\tag{1}$$converges. So you need to know a few things about infinite series. For this case a simple test is the ratio test. You take the ratio of two successive elements of the series and compute the limit:$$L=\lim_{n\to\infty}\left|\frac{(n+1)... 0 Okay it's:$F[n]=m∗\frac{d[n−2]−2d[n−1]+d[n]}{T^2}$1 A rational transfer function $$H(z)=\frac{Y(z)}{X(z)}=\frac{\sum_{n=0}^{N}b[n]z^{-n}}{1+\sum_{n=1}^{N}a[n]z^{-n}}\tag{1}$$ corresponds to the following difference equation: $$y[n]=bx[n]+bx[n-1]+\ldots +b[N]x[n-N]-\\-ay[n-1]-\ldots -a[N]y[n-N]\tag{2}$$ So the current output sample$y[n]$can be computed from the current input sample$x[n]$and$N$... 2 The$\mathcal{Z}$-transform expression corresponds to a difference equation which can be solved for$v[n]$, the velocity at sample$n$. From $$V(z)=\frac{v_0T}{1+\frac{2RT}{m}-z^{-1}}\tag{1}$$ you get $$v[n]=\frac{1}{1+\frac{2RT}{m}}\big(v[n-1]+v_0T\big)\tag{2}$$ So the delay operates on the samples$v[n]$. You just choose an initial condition (e.g.,$v[-...

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In the context of digial signal signal processing, the variable $z^{-1}$ stands for a single sample delay. And more generally, the variable $z^{-d}$ stands for a delay of $d$ samples, for $d$ positive integer. (or an advance of $d$ samples when $d<0$). Whether it's an input signal such as $x[n-1]$ or an delayed output such as $y[n-2]$, or just a filter ...

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