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The $X(z)$ calculated is correct but there are some mistakes in the steps and ROC is wrong. $X[z]$ is wrong representation of $\mathcal Z$-Transform since it is not a discrete sequence but rather a continuous function of $z$, where $z$ is a complex variable in argand plane ($z$-plane). Limit of $\sum$ changes from $n=0$ to $\infty$ to $n=1$ to $\infty$. Why?...

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For a term $\frac{1}{1-az^{-1}}$ it can either be $a^nu[n]$ with Region of Convergence(ROC) $|z| \gt |a|$ OR $-a^nu[-n-1]$ with ROC $|z| \lt |a|$. Since your input and output are both stable, your $h[n]$ is also stable. If you see its original form of $\frac{1-2z^{-1}}{1-\frac{3}{4}z^{-1}}$, it has pole at $z=3/4$, it's ROC is $|z| \gt 3/4$ so that the ...

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$H(z) = \frac{Y(z)}{X(z)} = \frac{\frac{-\frac{3}{2}z^{-1}}{(1-\frac{1}{2}z^{-1})(1-\frac{3}{4}z^{-1})}}{\frac{-\frac{3}{2}z^{-1}}{(1-\frac{1}{2}z^{-1})(1-2z^{-1})}} = \frac{1-2z^{-1}}{1-\frac{3}{4}z^{-1}}$ $H(z)= \frac{1}{1-\frac{3}{4}z^{-1}}-\frac{2z^{-1}}{1-\frac{3}{4}z^{-1}}$ $h[n] = (\frac{3}{4})^nu[n] - 2*(\frac{3}{4})^{n-1}u[n-1]$ Since $z^{-1}$(...

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One way to think about this: Everything that can be represented as a rational function in the Z-domain can also be represented as a linear difference equation in the time domain. As such at can be interpreted (and implemented) as a filter. It may be non-causal or unstable, but it's still a filter. You can certainly represent things in the z-Domain that are ...

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because even though it has more poles than zeros WRONG. The Z-transform transfer function will always have equal number of poles and zeros. Your poles are at $z = -5/4$ and $z = 1/4$. Zeros are at $z = -1/2$ and $z = \infty$. Since ROC will always be concentric circle region without including poles, there are 3 possible ROC for the given transfer function. $... 0 Only the transfer function in z-transform never gives the stability of the system because multiple systems with the same z transform can have different ROC. So you need additional information about the signal like is it causal, non-causal, stable etc. You can have multiple ROCs here. 0 That is a correct way to do it, but leaves you with (depending on what you're trying to do) an excess of delay blocks. This also works: 2 It is a bit to wrap your head around. Mathematically, if you denote the z transform as$X(z) = \mathcal{Z}\left \lbrace x_k\right \rbrace$, then when you take the transform of$x_k$after it's been delayed by one sample you get$\mathcal{Z}\left \lbrace x_{k-1}\right \rbrace = z^{-1}X(z)$. Except in strange corner cases* this always works, and you don't ... 5 It's natural consequence of applying a transform to a convolution relation. The output$y(t)$of an (continuous-time) LTI system is described by a convolution integral : $$y(t) = h(t)\star x(t) = \int_{-\infty}^{\infty} x(\tau) h(t-\tau) d\tau$$ And when you apply a Fourier transform on this relation, it turns out to be a multiplication in the transform ... 1 The Z-transform, defined as $$X(z) = \sum_{n=-\infty}^{+\infty}x[n]z^{-n}$$ expresses a discrete time signal as a sum of complex exponential signals$z^{-n}$, with non unitary amplitudes, that is $$z = re^{j\omega}, \: \: r \in \Re_+$$ Just to compare with, the Discrete Time Fourier Transform (DTFT) expresses a discrete time signal as a sum of complex ... 2 The partial fraction form helps in calculating the z-transform inverse since we can get the inverse of each term in partial fraction by inspection and hence get the inverse of the whole transfer function. The factored form can directly give poles and zeros by equating the numerator and denominator to zero. 2 The system with impulse response given by$h[n] = \cos(\pi\sqrt{n})u[n]$is BIBO-unstable because the sum$\sum_{n=-\infty}^\infty |h[n]]$diverges instead of being convergent as is needed for BIBO-stability. Note that for all positive integers$k$,$h[k^2]=\cos(\pi k)$has value$\pm 1$and so $$\sum_{n=-\infty}^\infty |h[n]] = \sum_{n=0}^\infty |\cos((\pi\... 1 The LTI system defined by the impulse response$$h[n] = \cos(\pi \sqrt{n} ) u[n] $$is unstable, as the absolute sum of the impulse response does not converge and diverges to infinity instead, i.e.;$$ \sum_{n=-\infty}^{\infty} |h[n]| = \sum_{n=0}^{\infty} |\cos(\pi \sqrt{n}) u[n]| \longrightarrow \infty$\$

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