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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{...


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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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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 ...


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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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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 ...


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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): $$\mathcal{Z}\big\{nx[n]\big\}=-z\frac{dX(z)}{dz}\tag{1}$$


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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 ...


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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 ...


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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] =...


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An $N$th-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 ...


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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)...


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Okay it's: $F[n]=m∗\frac{d[n−2]−2d[n−1]+d[n]}{T^2}$


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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: $$y[n]=b[0]x[n]+b[1]x[n-1]+\ldots +b[N]x[n-N]-\\-a[1]y[n-1]-\ldots -a[N]y[n-N]\tag{2}$$ So the current output sample $y[n]$ can be computed from the current input sample $x[n]$ and $N$...


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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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